Green’s Theorem on Hall Algebras

Claus Michael Ringel

ABSTRACT. Let k be a finite field and Λ a hereditary finitary k-algebra. Let P be the set of isomorphism classes of finite Λ-modules. We define a multiplication on the ℚ-space with basis P by counting the number of submodules U of a given module V with prescribed isomorphism classes both of V/U and U. In this way we obtain the so called Hall algebra H=H(Λ,ℚ) with coefficients in ℚ. Besides H, we are also interested in the subalgebra C generated by the the subset I of all isomorphism classes of simple Λmodules; this subalgebra is called the corresponding composition algebra, since it encodes the number of composition series of all Λ-modules.

Recently, J. A. Green has introduced on H (and on C) a comultiplication δ so that it becomes nearly a bialgebra. Here, “nearly” means that δ: H→H⊗H is in general not an algebra homomorphism for the usual (componentwise) multiplication on H⊗H; instead, there is a slightly twisted multiplication on H⊗H which has to be considered. The appearance of this twist should not be considered as a disadvantage: on the contrary, it is the main ingredient for Drinfeld and Lusztig in order to define the positive part U^{+} of a quantum group for an arbitrary Cartan datum. It follows that U^{+} is isomorphic to a twisted generic composition algebra.

Green’s compatibility theorem for multiplication and comultiplication on H is expressed in a marvelous formula which deals with pairs of submodules of a given module. This formula should be considered as a detailed investigation of the 3×3 Lemma in homological algebra.

This is the written version of two lectures given at the workshop preceding ICRA VII, held at U.N.A.M., Mexico, D.F. We present the complete proof of Green’s formula and outline its consequences, following closely the presentation in the original paper [G1].

1991 Mathematics Subject Classification. Primary 05A19, 16W30, 16G10, 17B37. Secondary 05A30, 16G20, 16E10, 16W50, 18G15.

C. M. Ringel

Part I. Green’s Formula

Let Λ be a ring. Let P be the set of isomorphism classes of Λ-modules of finite cardinality. For every α ∈ P let V_{α} be a representative in α. We denote by 0 both the zero module and its isomorphism class.

We recall that Λ is said to be finitary, provided Ext^{1}(M, N) is a finite set for any pair M, N of finite Λ-modules (clearly, it is sufficient to check this for finite, simple Λ-modules M, N). Throughout these notes, we usually will assume that the ring Λ is finitary.

Also, Λ is said to be hereditary, provided Ext^{2}(M, N) = 0 for all Λ-modules M, N. It is well-known that an algebra is hereditary if and only if submodules of projective modules are projective – and this explains the name.

1. The formula

Given α ∈ P, let a_{α} be the order of the automorphism group of V_{α}.

Given α, β, λ in P, let g_{αβ}^{λ} be the number of submodules B of V_{λ} such that V_{λ}/B is isomorphic to V_{α}, whereas B is isomorphic to V_{β}.

Theorem (J. A. Green). Let k be a finite field and Λ a k-algebra which is hereditary and finitary. Let α, β, α′, β′ be elements of P. Then

a_{α}a_{β}a_{α′}a_{β′} ∑_{λ} g_{αβ}^{λ}g_{α′β′}^{λ} a_{λ}^{−1} = ∑_{ρ,σ,σ′,τ} \frac{| Ext^{1}(V_{ρ}, V_{τ})|}{| Hom(V_{ρ}, V_{τ})|} g_{ρσ}^{α}g_{ρσ′}^{α′}g_{σ′τ}^{β} g_{στ}^{β′} a_{ρ}a_{σ}a_{σ′}a_{τ} .

Green’s theorem exhibits a universal formula for pairs of submodules: We consider submodules B and B′ of a module L, where L/B ∈ α, L/B′ ∈ α′, B ∈ β and B′ ∈ β′ (these four isomorphism classes α, β, α′, β′ are given). The left hand side counts the numbers of such pairs of submodules in arbitrary modules L = V_{λ}, always weighted with a_{λ}^{−1}. The right hand side involves the four subfactors L/(B + B′), B/(B ∩ B′), B′/(B ∩ B′), and B ∩ B′; their isomorphism classes are denoted by ρ, σ′, σ and τ, respectively.

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ρ σ  σ′ σ′  σ τ  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α′ β′ β′ L B+B′ B  B′ B∩B′ 0

Green’s Theorem

We stress that the right hand side of the formula only involves small modules (namely modules which occur as submodules or factor modules of the given modules V_{α}, V_{β}, V_{α′}, V_{β′}), whereas the left hand side deals with modules V_{λ}, where dim_{k} V_{λ} = dim_{k} V_{α} + dim_{k} V_{β}.

2. Examples

We usually denote the number of elements of k by q_{k}; in this section, we just write q.

A. Consider two non-isomorphic simple modules S(1), S(2) with

End(S(1)) = End(S(2)) = k and dim_{k} Ext^{1}(S(1), S(2)) = n.

For example, we may consider the path algebra Λ of a quiver with two vertices and precisely n arrows, all of which start at the same vertex and end at the other one.

Let α = α′ be the isomorphism class of S(1) and β = β′ the isomorphism class of S(2).

On the left side, all possible isomorphism classes λ have to be considered. There is the semisimple module S(1) ⊕ S(2) with automorphism group k^{∗} × k^{∗} of order (q − 1)^{2}. In addition, there are the indecomposable modules of length 2 with top S(1) and socle S(2). They have automorphism group k^{∗} of order q −1. The number

of isomorphism classes of such modules is \frac{q^{n} − 1}{q − 1}. Altogether we have 1 + \frac{q^{n} − 1}{q − 1}

isomorphism classes λ. Always, we have g_{αβ}^{λ} = g_{α′β′}^{λ} = 1. Thus the left hand side is

a_{α}a_{β}a_{α′}a_{β′} ∑_{λ} g_{αβ}^{λ}g_{α′β′}^{λ} a_{λ}^{−1}

= (q − 1)^{4}(1 · 1 · \frac{1}{(q − 1)^{2}} + \frac{q^{n} − 1}{q − 1} · 1 · 1 · \frac{1}{q − 1})

= (q − 1)^{2}q^{n}.

The right hand side has only one non-zero summand, namely for ρ the isomorphism class of S(1) and τ the isomorphism class of S(2) (and σ = σ′ = 0). Since the order of Ext^{1}(S(1), S(2)) is q^{n}, whereas Hom(S(1), S(2)) = 0 and the order of the zero group 0 is 1, we get:

\frac{| Ext^{1}(V_{ρ}, V_{τ})|}{| Hom(V_{ρ}, V_{τ})|} g_{ρσ}^{α}g_{ρσ′}^{α′}g_{σ′τ}^{β} g_{στ}^{β′} a_{ρ}a_{σ}a_{σ′}a_{τ}
= q^{n} · 1 · 1 · 1 · 1 · (q − 1) · 1 · 1 · (q − 1).

(The number n of arrows of the quiver appears in the right hand term only in the extra factor | Ext^{1}(V_{ρ}, V_{τ})|, the remaining factors are independent of n. In the left hand term, n determines the number of summands.)

B. Consider a fixed simple module S with End(S) = k, and assume that dim_{k} Ext^{1}(S, S) = n. Let α = α′ = β = β′ be the isomorphism class of S.

On the left side, the possible isomorphism classes λ are as follows: V_{λ} may be semisimple, thus of the form S ⊕S; then its automorphism group is Gl_{2}(k), of order q(q^{2} − 1)(q − 1), and we have g_{αβ}^{λ} = g_{α′β′}^{λ} = q + 1. Or else V_{λ} is an indecomposable module of length 2 with top S and socle S. Then the automorphism group is of order q(q − 1) and g_{αβ}^{λ} = g_{α′β′}^{λ} = 1. Note that the number of isomorphism classes of

such indecomposable length 2 modules is again \frac{q^{n} − 1}{q − 1}. Thus the left hand side ist

a_{α}a_{β}a_{α′}a_{β′} ∑_{λ} g_{αβ}^{λ}g_{α′β′}^{λ} a_{λ}^{−1}

= (q − 1)^{4}((q + 1) · (q + 1) · \frac{1}{q(q^{2} − 1)(q − 1)} + \frac{q^{n} − 1}{q − 1}1 · 1 · \frac{1}{q(q − 1)})

= (q − 1)^{4} · \frac{q^{n} + q}{q(q − 1)^{2}} = (q^{n−1} + 1)(q − 1)^{2}.

The right hand side has two non-zero summands: One is given for ρ and τ the isomorphism class of S, (and σ = σ′ = 0). Here we have to take into account that the order of Ext^{1}(S, S) is q^{n}, while the order of Hom(S, S) = k is q. The second summand to be considered occurs for ρ = τ = 0 and σ = σ′ the isomorphism class of S. We get:

∑ \frac{| Ext^{1}(V_{ρ}, V_{τ})|}{| Hom(V_{ρ}, V_{τ})|} g_{ρσ}^{α}g_{ρσ′}^{α′}g_{σ′τ}^{β} g_{στ}^{β′} a_{ρ}a_{σ}a_{σ′}a_{τ}

= \frac{q^{n}}{q} (q − 1)^{2} + \frac{1}{1}(q − 1)^{2} = (q^{n−1} + 1)(q − 1)^{2}.

3. Preliminary results

We exhibit several elementary counting formulae:

Let A, B, L be Λ-modules. Let E_{AB}^{L} be the set of pairs (a, b) of Λ-homomorphisms such that

0 ⟶ B ⟶^{b} L ⟶^{a} A ⟶ 0

is an exact sequence.

If dim_{k} L ≠ dim_{k} A+dim_{k} B, then E_{AB}^{L} is empty. If dim_{k} L = dim_{k} A+dim_{k} B, then E_{AB}^{L} consists of all pairs (a, b) where a: L → A is an epimorphism, b: B → L is a monomorphism and ab = 0.

Proposition 1. Let A, B, L be Λ-modules and α, β, λ their isomorphism classes, respectively. Then

|E_{AB}^{L}| = g_{αβ}^{λ}a_{α}a_{β}.

Proof: Let U be the image of b: B → L, thus b = u\underline{b}, where \underline{b}: B → U is an isomorphism and u: U → L is the inclusion map. Similarly, write a = \underline{a}p, where p: L → L/U is the canonical projection and \underline{a}: L/U → A is an isomorphism. The set of tripels (U, \underline{a}, \underline{b}) obtained in this way consists of a submodule U of L such that U belongs to the isomorphism class β, whereas L/U belongs to the isomorphism class α, and arbitrary isomorphisms \underline{a}: L/U → A and \underline{b}: B → U. Thus the number of such tripels is precisely g_{αβ}^{λ}a_{α}a_{β}.

A Λ-module is given by a pair (V, γ), where V is a k-space and γ: Λ → End_{k}(V) is a k-algebra homomorphism.

Proposition 2. Let λ ∈ P. Let V be a k-space of dimension dim_{k} V_{λ}. The number of pairs (V, γ), where γ: Λ → End_{k}(V) is a k-algebra homomorphism such that the Λ-module (V, γ) belongs to λ, is

| Aut_{k}(V)| · a_{λ}^{−1}.

Proof: Observe that G = Aut_{k}(V) operates on the set of k-algebra homomorphisms γ as follows: g ∗ γ = gγg^{−1} for g ∈ G (that means: given r ∈ Λ, let (g ∗ γ)(r) = gγ(r)g^{−1}). The orbits under this action are just the isomorphism classes, whereas the stabilizer G_{γ} of a fixed element γ is just the automorphism group of the Λ-module (V, γ). The assertion follows from the well-known fact that the number of elements in the orbit of γ is just |G| · |G_{γ}|^{−1}.

Proposition 3. Consider the following commutative diagram of Λ-module homomorphisms, with exact rows

X′ ⟶^{x} X ⟶^{x′} X′′ ⟶ 0
d′↓  ↓^{d}  ↓_{d′′}
0 ⟶ Y′ ⟶^{y} Y ⟶^{y′} Y′′.

Then the number of possible homomomorphisms d (with all the other maps x, x′, y, y′, d′, d′′ fixed) is equal to | Hom_{Λ}(X′′, Y′)|.

Proof: Let D be the set of all d ∈ Hom_{Λ}(X, Y) which satisfy dx = yd′ and y′d = d′′x′. Fix some element d_{0} ∈ D.

Given z ∈ Hom_{Λ}(X′′, Y′), then also d_{0} + yzx′ belongs to D, since x′x = 0 and y′y = 0. Since y is injective and x′ is surjective, we see that the map Hom_{Λ}(X′′, Y′) → D, which sends z to d_{0} + yzx′ is injective. The map is also surjective: Given d ∈ D, then (d − d_{0})x = 0, thus we can factor the homomorphism d − d_{0} through the cokernel of x, thus d − d_{0} = z′x′ for some z′: X′′ → Y. Now y′z′x′ = y′(d − d_{0}) = 0, the surjectivity of x′ yields y′z′ = 0, and therefore z′ can be factored through the kernel y of y′. We obtain z: X′′ → Y′ with yz = z′. Altogether we have d − d_{0} = z′x′ = yzx′.

Let A, B, L be Λ-modules, ε an element of Ext_{Λ}^{1}(A, B), and let E^{L}(ε) be the set of pairs (a, b) of Λ-homomorphisms such that

0 ⟶ B ⟶^{b} L ⟶^{a} A ⟶ 0

is an exact sequence and belongs to the class ε.

Proposition 4. Let A, B, L be Λ-modules. Let λ be the isomorphism class of L. Let ε be an element of Ext_{Λ}^{1}(A, B). Then the cardinality of E^{L}(ε) is either zero or else

\frac{a_{λ}}{| Hom_{Λ}(A, B)|}.

Proof: Assume that there exists an exact sequence

0 ⟶ B ⟶^{b} L ⟶^{a} A ⟶ 0,

which belongs to ε, thus the pair (a, b) belongs to E^{L}(ε).

Given an automorphism h of L, let

h ∗ (a, b) = (ah^{−1}, hb).

The commutative diagram

0 ⟶ B ⟶^{b} L ⟶^{a} A ⟶ 0
∥  ↓^{h}  ∥
0 ⟶ B ⟶^{hb} L ⟶^{ah^{−1}} A ⟶ 0

shows that the pair (ah^{−1}, hb) again belongs to E^{L}(ε). In this way, the group Aut_{Λ}(L) operates on the set E^{L}(ε). By the definition of the equivalence relation which gives Ext^{1}(A, B), the operation by Aut_{Λ}(L) on E^{L}(ε) is transitive. Thus, it remains to calculate the stabilizer of an element. According to Proposition 3, the order of the stabilizer of (a, b) is equal to | Hom_{Λ}(A, B)|. This completes the proof.

Given a k-space V, an exact sequence of the form

0 → B → (V, γ) → A → 0,

with a k-algebra homomorphism γ: Λ → End_{k}(V) will be called a V -sequence. Let ε be an element of Ext_{Λ}^{1}(A, B). Let E(ε) be the set of V -sequences in ε, where V is a fixed k-space of dimension dim_{k} A + dim_{k} B.

Proposition 5. Let A, B be Λ-modules. Let ε be an element of Ext_{Λ}^{1}(A, B). Then

|E(ε)| = \frac{| Aut_{k}(V)|}{| Hom_{Λ}(A, B)|}.

Proof: Let

0 ⟶ B ⟶^{b} L ⟶^{a} A ⟶ 0

be an element of ε. Let λ be the isomorphism class of L. The number of algebra homomorphisms γ: Λ → End(V) such that (V, γ) belongs to λ is given by \frac{| Aut_{k}(V)|}{a_{λ}}, according to Proposition 2.

For any such γ the number of pairs (a, b) such that (γ, a, b) belongs to E(ε) is given by \frac{a_{λ}}{| Hom_{Λ}(A,B)|}, according to Proposition 4 (note that for any γ, there is at least one such triple, since any isomorphism h: L → (V, γ) yields a commutative diagram

0 ⟶ B ⟶^{b} L ⟶^{a} A ⟶ 0
∥  ↓^{h}  ∥
0 ⟶ B ⟶^{hb} (V, γ) ⟶^{ah^{−1}} A ⟶ 0,

and this shows that the lower exact sequence again belongs to ε.)

Altogether, we see that

|E(ε)| = \frac{| Aut_{k}(V)|}{a_{λ}} · \frac{a_{λ}}{| Hom_{Λ}(A, B)|} = \frac{| Aut_{k}(V)|}{| Hom_{Λ}(A, B)|}.

4. Factorizations of homomorphisms

Let f: Y → X be a homomorphism of Λ-modules and M the image of f. Let V be a k-space of dimension dim_{k} X + dim_{k} Y − dim_{k} M.

For L = (V, γ), let F(f; L) be the set of all pairs (c, d) where c: L → X is an epimorphism, d: Y → L a monomorphism, and cd = f. Let F(f) be the set of tripels (γ, c, d) where γ: Λ → End_{k}(V) is an algebra homomorphism and (c, d) belongs to F(f; (V, γ)).

If we denote by y: Y → M the projection and by x: M → X the inclusion map (thus f = xy), then we are looking for all possible factorizations f = cd such that the commutative diagram

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is both a pushout and a pullback diagram.

Proposition. Let f: Y → X be a homomorphism of Λ-modules. Let T be the kernel and R the cokernel of f. Assume that Ext^{2}(R, T) = 0. Then

|F(f)| = | Aut_{k}(V)|\frac{| Ext_{Λ}^{1}(R, T)|}{| Hom_{Λ}(R, T)|}.

Remark. This is the only calculation where the vanishing of Ext^{2} enters. It is required to ensure that F(f) is non-empty. For a short proof that F(f) ≠ ∅, we refer to [HR].

Proof of Proposition: We denote by y′: T → Y and x: M → X the inclusion maps, by y: Y → M and x′: X → R the canonical projections; in particular, we have f = xy. We denote by ε_{0} the equivalence class of the exact sequence

0 ⟶ M ⟶^{x} X ⟶^{x′} R ⟶ 0

in Ext_{Λ}^{1}(R, M).

The exact sequence

0 ⟶ T ⟶^{y′} Y ⟶^{y} M ⟶ 0

gives rise to a long exact sequence

0 ⟶ Hom_{Λ}(R, T) ⟶ Hom_{Λ}(R, Y) ⟶ Hom_{Λ}(R, M)
⟶ Ext_{Λ}^{1}(R, T) ⟶^{Ext_{Λ}^{1}(R,y′)} Ext_{Λ}^{1}(R, Y) ⟶^{Ext_{Λ}^{1}(R,y)} Ext_{Λ}^{1}(R, M)
⟶  0

where we use that Ext_{Λ}^{2}(R, T) = 0. We denote by E the image of Ext_{Λ}^{1}(R, y′), thus we have an exact sequence

0 → Hom_{Λ}(R, T) → Hom_{Λ}(R, Y) → Hom_{Λ}(R, M) → Ext_{Λ}^{1}(R, T) → E → 0,

and consequently

|E| = | Hom_{Λ}(R, T)|^{−1} · | Hom_{Λ}(R, Y)| · | Hom_{Λ}(R, M)|^{−1} · | Ext_{Λ}^{1}(R, T)|.

Given (γ, c, d) ∈ F(f), the following diagram with d′ = x′c is commutative:

0 ⟶ Y ⟶^{d} (V, γ) ⟶^{d′} R ⟶ 0
y↓  ↓^{c}  ∥
0 ⟶ M ⟶^{x} X ⟶^{x′} R ⟶ 0.

Also, its upper row is exact: since both c and x′ are surjective, also d′ is surjective; on the other hand, we have d′d = x′cd = x′f = x′xy = 0, thus the exactness of the upper row follows from the equality dim_{k} V = dim_{k} X + dim_{k} Y − dim_{k} M = dim_{k} Y + dim_{k} R.

We define a map ω: F(f) → Ext_{Λ}^{1}(R, Y), sending the tripel (γ, c, d) to the equivalence class ω(γ, c, d) of the exact sequence

0 ⟶ Y ⟶^{d} (V, γ) ⟶^{x′c} R ⟶ 0.

We claim that the image of the map ω is just the inverse image of ε_{0} under Ext_{Λ}^{1}(R, y). On the one hand, the diagram above shows that ε_{0} is induced from ω(γ, c, d) via the homomorphism y. Conversely, assume a V -sequence ε in Ext_{Λ}^{1}(R, Y) induces ε_{0} via the homomorphism y. Thus, there is a commutative diagram with exact rows

0 ⟶ Y ⟶^{d} (V, γ) ⟶^{d′} R ⟶ 0
y↓  ↓^{c}  ∥
0 ⟶ M ⟶^{x} X ⟶^{x′} R ⟶ 0

and ε is the equivalence class of the upper row. Then (γ, c, d) belongs to F(f) and d′ = x′c, thus ε = ω(γ, c, d).

Since the image of ω is the inverse image of ε_{0} under Ext_{Λ}^{1}(R, y), and since Ext_{Λ}^{1}(R, y) is surjective, the image of ω is a residue class E′ modulo the kernel of Ext_{Λ}^{1}(R, y). But this kernel is just E, thus the image of ω has order |E′| = |E|. For any element ε in E′, the number of V -sequences in ε is

\frac{| Aut_{k}(V)|}{| Hom_{Λ}(R, Y)|}

according to Proposition 5 of Section 3. For a given V -sequence

0 ⟶ Y ⟶^{d} (V, γ) ⟶^{d′} R ⟶ 0,

the number of possible maps c satisfying cd = xy and x′c = d′ is | Hom_{Λ}(R, M)| according to Proposition 3 of Section 3. Altogether, we see that

|F(f)| = |E|\frac{| Aut_{k}(V)|}{| Hom_{Λ}(R, Y)|}| Hom_{Λ}(R, M)|

= | Aut_{k}(V)|\frac{| Ext_{Λ}^{1}(R, T)|}{| Hom_{Λ}(R, T)|}.

This completes the proof.

∗ ∗ ∗

We consider now Green’s formula in detail. First, we interpret the left hand side, then the right hand side of the formula (both multiplied with a fixed factor, namely | Aut_{k}(V)|, where V is a k-space). We will see that the left hand side counts suitable “crosses”, while the right hand side counts pairs consisting of a “frame” and the factorization of a map derived from it. Here, the frames and the crosses are parts of the 3 × 3 configuration in Homological Algebra (consisting of three rows and three columns all of which are short exact sequences). It will be easy to attach to every cross a corresponding frame. The main interest lies in the fact that one can count the number of crosses which yield the same frame.

We fix elements α, β, α′ and β′ of P. We want to derive Green’s formula. We can assume that

dim_{k} V_{α} + dim_{k} V_{β} = dim_{k} V′_{α} + dim_{k} V′_{β},

since otherwise the formula is trivially satisfied: both sides are zero (since g_{αβ}^{λ} ≠ 0 implies that dim_{k} V_{α} + dim_{k} V_{β} = dim_{k} V_{λ}).

Fix Λ-modules A ∈ α, B ∈ β, A′ ∈ α′ and B′ ∈ β′. Also, fix a k-space V of dimension dim_{k} A + dim_{k} B.

5. Green’s formula: The left hand side

Let γ: Λ → End_{k}(V) be a k-algebra homomorphism and L = (V, γ). We consider the set

Q(L) = E_{AB}^{L} × E_{A′B′}^{L},

and we denote by Q the union of the sets Q(V, γ). Thus Q is the set of quintets (γ, a, b, a′, b′) where γ: Λ → End_{k}(V) is a k-algebra homomorphism, and a, b, a′, b′ are Λ-module homomorphisms such that we obtain exact sequences

\begin{matrix}0 & ⟶ & B & ⟶^{b} & (V, γ) & ⟶^{a} & A & ⟶ & 0, \\ 0 & ⟶ & B′ & ⟶^{b′} & (V, γ) & ⟶^{a′} & A′ & ⟶ & 0. \end{matrix}

The data given by the elements of Q may be visualized as forming the following

crosses: B

↓^{b}
B′ ⟶^{b′} (V, γ) ⟶^{a′} A′
↓^{a}

A Proposition.

|Q| = | Aut_{k}(V)| · a_{α}a_{β}a_{α′}a_{β′} ∑_{λ} g_{αβ}^{λ}g_{α′β′}^{λ} a_{λ}^{−1}.

Proof: Consider some L = (V, γ). According to Proposition 1 of Section 3, we know

|E_{AB}^{L} × E_{A′B′}^{L}| = g_{αβ}^{λ}a_{α}a_{β} · g_{α′β′}^{λ}a_{α′}a_{β′}.

According to Proposition 2 of Section 3, the number of choices for γ with (V, γ) ∈ λ is | Aut_{k}(V)| · a_{λ}^{−1}. Thus the number of elements of Q is equal to

∑_{λ} | Aut_{k}(V)| · a_{λ}^{−1} · g_{αβ}^{λ}a_{α}a_{β} · g_{α′β′}^{λ}a_{α′}a_{β′}.

6. Green’s formula: The right hand side

We introduce the set R of all sextets (S, T, e_{1}, e_{2}, e_{3}, e_{4}), where S is a submodule of A, where T is a submodule of B, where (e_{3}, e_{1}) belongs to E_{ST}^{B′} , and (e_{4}, e_{2}) belongs to E_{A/S,B/T}^{A′}.

The data given by elements of R may be visualized as frames of the following kind:

T ⟶^{u_{T}} B ⟶^{q_{T}} B/T
e_{1}↓  ↓^{e_{2}}
B′  A′
e_{3}↓  ↓^{e_{4}}
S ⟶_{u_{S}} A ⟶_{q_{S}} A/S

For the submodule S of A, we have denoted the inclusion map S → A by u_{S}, and the canonical projection A → A/S by q_{S}. Similarly, for the submodule T of B, we have denoted by u_{T} : T → B the inclusion and by q_{T} : B → B/T the projection.

Let B, B′ be Λ-modules and e: T → B′ a monomorphism, where T is a submodule of B. We denote by B ⊔_{e} B′ the pushout of the inclusion map u_{T} : T → B and the map e, thus

B ⊔_{e} B′ = B ⊕ B′/{(m, −e(m))| m ∈ T }.

C. M. Ringel

We denote by u_{B}: B → B ⊔_{e}B′ and u_{B′}: B′ → B ⊔_{e}B′ the canonical inclusion maps. Thus, we have the following commutative square:

T ⟶^{u_{T}} B
e↓ ↓^{u_{B}}
B′ ⟶_{u_{B′}} B ⊔_{e} B′.

We also consider the dual situation. We start with Λ-modules A and A′, and an epimorphism e′: A′ → A/S, where S is a submodule of A. We denote by A ⊓^{e′} A′ the pullback of the projection map q_{S}: A → A/S and e′, thus

A ⊓^{e′} A′ = {(m, m′)| m ∈ A, m′ ∈ A′, m + S = e′(m′)}.

We denote by p_{A}: A ⊓^{e′} A′ → A and p_{A′}: A ⊓^{e′} A′ → A′ the canonical projections.

Given Ξ = (S, T, e_{1}, e_{2}, e_{3}, e_{4}) ∈ R, we define

Y = Y (Ξ) = B ⊔_{e_{1}} B′, X = X(Ξ) = A ⊓^{e_{4}} A′.

Lemma. Given Ξ = (S, T, e_{1}, e_{2}, e_{3}, e_{4}) ∈ R, there exists a unique map f = f(Ξ): Y (Ξ) → X(Ξ) satisfying

p_{A}fu_{B} = 0,  p_{A}fu_{B′} = u_{S}e_{3},
p_{A′}fu_{B} = e_{2}q_{T} ,  p_{A′}fu_{B′} = 0.

The kernel of f is isomorphic to T and the cokernel of f is isomorphic to A/S.

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Proof: Since

0 · u_{T} = 0 = u_{S}e_{3} · e_{1},

and since Y is the pushout of u_{T} and e_{1}, there exists f_{A}: Y → A satisfying

f_{A}u_{B} = 0 and f_{A}u_{B′} = u_{S}e_{3}.

Similarly, since

e_{2}q_{T} · u_{T} = 0 = 0 · e_{1},

there exists f_{A′}: Y → A′ satisfying

f_{A′}u_{B} = e_{2}q_{T} and f_{A′}u_{B′} = 0.

Now we have

q_{S}f_{A}u_{B} = 0 = e_{4}e_{2}q_{T} = e_{4}f_{A′}u_{B},
q_{S}f_{A}u_{B′} = q_{S}u_{S}e_{3} = 0 = e_{4}f_{A′}u_{B′}.

As a consequence, since X is the pullback of q_{S} and e_{4}, there exists f: Y → X such that

p_{A}f = f_{A}, and p_{A′}f = f_{A′}.

It follows that f satisfies the required equations.

The uniqueness of f comes from the fact that [u_{B} u_{B′}]: B ⊕ B′ → Y is an

epimorphism and [ \begin{matrix}p_{A} \\ p_{A′} \end{matrix} ] : X → A ⊕ A′ is a monomorphism.

Finally, we have to determine kernel and cokernel of f. We claim that the following sequence is exact:

0 ⟶ T ⟶^{u_{B}u_{T}} Y ⟶^{f} X ⟶^{q_{S}p_{A}} A/S ⟶ 0.

First of all, observe that fu_{B}u_{T} = 0, since

p_{A}fu_{B}u_{T} = 0 · u_{T} = 0,
p_{A′}fu_{B}u_{T} = e_{2}q_{T} u_{T} = 0.

Similarly, we see that q_{S}p_{A}f = 0. On the other hand, the homomorphism

[ \begin{matrix}p_{A} \\ p_{A′} \end{matrix} ] · f · [u_{B} u_{B′}] = [ \begin{matrix}0 & u_{S}e_{3} \\ e_{2}q_{T} & 0 \end{matrix} ] = [ \begin{matrix}0 & u_{S} \\ e_{2} & 0 \end{matrix} ] · [ \begin{matrix}q_{T} & 0 \\ 0 & e_{3} \end{matrix} ]

has image isomorphic to B/T ⊕ S. Since [ \begin{matrix}p_{A} \\ p_{A′} \end{matrix} ] is a monomorphism, whereas

[u_{B} u_{B′}] is an epimorphism, also the image of f is isomorphic to B/T ⊕ S. But we have

dim_{k} Y = dim_{k} B + dim_{k} B′ − dim_{k} T
= dim_{k}(B/T ⊕ S) + dim_{k} T.

This shows that the sequence is exact at Y. A similar argument shows that the sequence is exact at X.

For any Λ-module L, let O(L) be the set of all octets (S, T, e_{1}, e_{2}, e_{3}, e_{4}, c, d), where Ξ = (S, T, e_{1}, e_{2}, e_{3}, e_{4}) belongs to R such that (c, d) belongs to F(f(Ξ); L).

Let O be the set of all tuples (γ, S, T, e_{1}, e_{2}, e_{3}, e_{4}, c, d) where γ: Λ → End_{k}(V) is a k-algebra homomorphism, and (S, T, e_{1}, e_{2}, e_{3}, e_{4}, c, d) belongs to O(V, γ).

Proposition.

|O| = | Aut_{k}(V)| · ∑_{ρ,σ,σ′,τ} \frac{| Ext^{1}(V_{ρ}, V_{τ})|}{| Hom(V_{ρ}, V_{τ})|} g_{ρσ}^{α}g_{ρσ′}^{α′}g_{σ′τ}^{β} g_{στ}^{β′} a_{ρ}a_{σ}a_{σ′}a_{τ} .

Proof: Fix submodules S of A and T of B, let τ be the isomorphism class of T, σ the isomorphism class of S, σ′ the isomorphism class of B/T and ρ the isomorphism class of A/S. Then the number of sextets of the form Ξ = (S, T, e_{1}, e_{2}, e_{3}, e_{4}) in R (with fixed S, T) is equal to

g_{ρσ′}^{α′}g_{στ}^{β′} a_{ρ}a_{σ}a_{σ′}a_{τ} ,

according to Proposition 1 of Section 3.

The previous Lemma asserts that the kernel of f(Ξ) is isomorphic to V_{τ} , while its cokernel is isomorphic to V_{ρ}. According to Section 4, the number of pairs (c, d) in F(f(Ξ)) is

| Aut_{k}(V)| · \frac{| Ext^{1}(V_{ρ}, V_{τ})|}{| Hom(V_{ρ}, V_{τ})|}.

Of course, the number of choices (S, T) with fixed ρ, σ, σ′, τ is

g_{ρσ}^{α}g_{σ′τ}^{β} .

Adding up all the possibilities, we obtain the formula as stated.

7. More about pushouts and pullbacks

Proposition. Let B, B′, L be Λ-modules. There is a bijection η between the set of pairs (b, b′) where b: B → L and b′: B′ → L are monomorphisms, and the set of triples (T, e, d) where T is a submodule of B, such that both e: T → B′ and d: B ⊔_{e} B′ → L are monomorphisms.

Here, η(b, b′) = (T, e, d), where T = b^{−1}b′(B′). The map e is defined by e(m) = (b′)^{−1}b(m) for m ∈ T, and d is defined by du_{B} = b, du_{B′} = b′.

Moreover, η^{−1} is defined by η^{−1}(T, e, d) = (du_{B}, du_{B′}).

Proof: Let b: B → L and b′: B′ → L be monomorphisms. Of course, T = b^{−1}b′(B′) is a submodule of B, and e(m) = (b′)^{−1}b(m) for m ∈ T defines a monomorphism e: T → B′. Thus, the pushout B ⊔_{e} B′ is defined. Since we have the following commutative diagram

T ⟶^{u_{T}} B
e↓  ↓^{b}
B′ ⟶_{b′} L,

there exists a unique d: B ⊔_{e} B′ → L such that du_{B} = b, du_{B′} = b′.

Consider the following diagram:

T  u_{T}  ^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.} B B′  b′  ^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.} L e^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}  b^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.} Y  ^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.} u_{B}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.} u_{B′} ^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}  ^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.} d^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}

where Y = B ⊔_{e} B′.

In order to see that d is a monomorphism, note that the map [ \begin{matrix}u_{T} \\ −e \end{matrix} ] : T →

B ⊕ B′ is the kernel of [b b′]: B ⊕ B′ → L, and the map [u_{B} u′_{B}]: B ⊕ B → B ⊔_{e} B′

is the cokernel of [ \begin{matrix}u_{T} \\ −e \end{matrix} ].

The converse recipe is easier: given a triple (T, e, d), where T is a submodule of B, and both maps e: T → B′ and d: B ⊔_{e} B′ → L are monomorphisms, we form the pair η′(T, e, d) = (du_{B}, du_{B′}). Since the maps d, u_{B}, u_{B′} are monomorphisms, also du_{B} and du_{B′} are monomorphisms.

We claim that η is bijective with inverse η′. Of course, we trivially have η′η(b, b′) = (b, b′). Consider a triple (T, e, d) and let b = du_{B}, b′ = du_{B′}. One easily checks that T = b^{−1}b′(B′), and that e(m) = (b′)^{−1}b(m) for m ∈ T. As a consequence, η(b, b′) = (T, e, d). This completes the proof.

We present a dual version of the previous Proposition:

Dual Proposition. Let A, A′, L be Λ-modules. There is a bijection η between the set of pairs (a, a′) where a: L → A and a′: L → A′ are epimorphisms, and the set of triples (S, e′, c), where S is a submodule of A and e′: A′ → A/S and

c: L → A ⊓^{e′} A′ are epimorphisms.

Here, η(a, a′) = (S, e′, c), where S = ab′(B′), the map e′ being defined by

e′(m′) = a(a′)^{−1}(m′) + S for m′ ∈ A′, and c the unique homomorphism L → A ⊓^{e′} A′ satisfying p_{A}c = a, p_{A′}c = a′.

Moreover, η^{−1} is defined by η^{−1}(S, e′, c) = (p_{A}c, p_{A′}c).

Note that ab′(B′) = a(Ker a′).

Consider the following diagram:

L  a′  ^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.} A′ A  q^{S}  ^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.} A/S a^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}  e′^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.} X^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}p_{A}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}  ^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.} p_{A′}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.} c ^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}

where X = A⊓^{e′}A′. Of course, the composition p = q_{S}p_{A}(= e′p_{A′}) is an epimorphism

p: A ⊓^{e′} A′ → A/S and its kernel is just S ⊕ Ker(e′).

8. Proof of Green’s formula

Proposition. There exists a bijection

η: Q(L) ⟶ O(L).

Proof: Given (a, b) ∈ E_{AB}^{L} and (a′, b′) ∈ E_{A′B′}^{L}, let η(b, b′) = (T, e_{1}, d) and η(a, a′) = (S, e_{4}, c), as defined above. Thus, we have diagrams which are pushouts and pullbacks

\begin{matrix}T & ⟶^{u_{T}} & B \\ ^{e_{1}}↓ & & ↓^{u_{B}} \\ B′ & ⟶_{u_{B′}} & Y \end{matrix}  \begin{matrix}X & ⟶^{p_{A′}} & A′ \\ ^{p_{A}}↓ & & ↓^{e_{4}} \\ A & ⟶_{q_{S}} & A/S \end{matrix}

and maps d: Y → L, c: L → X such that

du_{B} = b,  p_{A}c = a,
du_{B′} = b′, p_{A′}c = a′.

We denote by q_{T} : B → B/T the canonical projection, and by u_{S}: S → A the inclusion.

Note that u_{T} is the kernel of a′b, thus there exists a monomorphism e_{2} : B/T → A′ such that

a′b = e_{2}q_{T} .

Similarly, q_{S} is the cokernel of ab′, thus there exists an epimorphism e_{3} : B′ → S such that

ab′ = u_{S}e_{3},

We have

e_{4}e_{2} = 0 and e_{3}e_{1} = 0

(for example, e_{4}e_{2}q_{T} = e_{4}a′b = q_{S}ab = 0). Actually, both sequences

\begin{matrix}0 & ⟶ & B/T & ⟶^{e_{2}} & A′ & ⟶^{e_{4}} & A/S & ⟶ & 0, \\ 0 & ⟶ & T & ⟶^{e_{1}} & B′ & ⟶^{e_{3}} & S & ⟶ & 0 \end{matrix}

are exact: since e_{1}, e_{2} are monomorphisms, e_{3}, e_{4} are epimorphisms and e_{4}e_{1} = 0 = e_{3}e_{2}, we know that

dim_{k} A′ ≥ dim_{k} B/T + dim_{k} A/S and
dim_{k} B′ ≥ dim_{k} T + dim_{k} S.

Adding up these two inequalities, we obtain

dim_{k} L = dim_{k} A′ + dim_{k} B′
≥ dim_{k} B/T + dim_{k} A/S + dim_{k} T + dim_{k} S
= dim_{k} A + dim_{k} B = dim_{k} L.

In this way, we see that

dim_{k} A′ = dim_{k} B/T + dim_{k} A/S, and
dim_{k} B′ = dim_{k} T + dim_{k} S.

Altogether we see that (e_{3}, e_{1}) belongs to E_{ST}^{B′} , and (e_{4}, e_{2}) to E_{A/S,B/T}^{A′}, thus Ξ = (S, T, e_{1}, e_{2}, e_{3}, e_{4}) belongs to R.

Remark. Here, we use the following result (see [Mi], Proposition I.16.5.): The diagram

0  0  0
↓  ↓  ↓
0 ⟶ T ⟶^{u_{T}} B ⟶^{q_{T}} B/T ⟶ 0
e_{1}↓  ↓^{b}  ↓^{e_{2}}
0 ⟶ B′ ⟶^{b′} L ⟶^{a′} A′ ⟶ 0
e_{3}↓  ↓^{a}  ↓^{e_{4}}
0 ⟶ S ⟶_{u_{S}} A ⟶_{q_{S}} A/S ⟶ 0
↓  ↓  ↓
0  0  0

is commutative with exact rows and exact columns if and only if the upper left square is a pullback, the lower right square is a pushout, q_{T} is an epimorphism, e_{2} is a monomorphism with e_{2}q_{T} = a′b, while e_{3} is an epimorphism, and u_{S} is a monomorphism with u_{S}e_{3} = ab′.

Insert into this diagram also

Y = B ⊔_{e_{1}} B′ and X = A ⊓^{e_{4}} A′

and the corresponding maps u_{B}, u′_{B}, d and p_{A}, p′_{A}, c :

T  u_{T}  ^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.} B  q_{T}  ^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.} B/T B′  b′  ^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.} L  a′  ^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.} A′ S  u^{S}  ^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.} A  q^{S}  ^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.} A/S e_{1}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}  b^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}  e_{2}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.} e_{3}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}  a^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}  e_{4}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.} Y  ^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.} u_{B}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.} u_{B′} ^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}  ^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.} d^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}  X^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}p_{A}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}  ^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.} p_{A′}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.} c ^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}

We have

p_{A}cdu_{B} = ab = 0  p_{A}cdu_{B′} = ab′ = u_{S}e_{3}
p_{A′}cdu_{B} = a′b = e_{2}q_{T}  p_{A′}cdu_{B′} = a′b′ = 0.

Thus cd = f(Ξ). It follows that the octet η(a, b, a′, b′) = (S, T, e_{1}, e_{2}, e_{3}, e_{4}, c, d) belongs to O(L).

Conversely, given an octet (S, T, e_{1}, e_{2}, e_{3}, e_{4}, c, d) in O(L), form η^{−1}(T, e_{1}, d) = (b, b′) and η^{−1}(S, e_{4}, c) = (a, a′). We claim that

ab = 0, and a′b′ = 0.

Namely, let f = f(S, T, e_{1}, e_{2}, e_{3}, e_{4}). Then

ab = p_{A}cdu_{B} = p_{A}fu_{B} = 0 and a′b′ = p_{A′}cdu_{B′} = p_{A′}fu_{B′} = 0.

This shows that (a, b) belongs to E_{AB}^{L} and that (a′, b′) belongs to E_{A′B′}^{L}. Let η′(S, T, e_{1}, e_{2}, e_{3}, e_{4}, c, d) = (a, b, a′, b′). As we have seen, this element belongs to Q(L).

By construction, we have η′η(a, b, a′, b′) = (a, b, a′, b′) and it is easy to see that also ηη′(S, T, e_{1}, e_{2}, e_{3}, e_{4}, c, d) = (S, T, e_{1}, e_{2}, e_{3}, e_{4}, c, d). This completes the proof.

Proof of Green’s Theorem: According to Section 8, we have |Q| = |O|, and by Sections 5 and 6, this yields the formula, with both sides multiplied by | Aut_{k}(V)|.

Remark. Also the following converse of Green’s theorem is true: In case Green’s formula is satisfied for all finite Λ-modules, then the category of finite Λmodules is hereditary.

9. Green’s formula rewritten

Remember that Λ is a k-algebra, where k is a finite field, say with q_{k} elements. We denote by K_{0}(Λ) the Grothendieck group of all finite Λ-modules modulo exact sequences. For any finite Λ-module M, we denote by dim M the corresponding element of K_{0}(Λ). Let I ⊆ P be the set of isomorphism classes of simple Λ-modules. For i ∈ I, a representative of the isomorphism class will usually be denoted by S(i), instead of V_{i}.

The Jordan-Hölder Theorem asserts that K_{0}(Λ) is the free abelian group on the set I of isomorphism classes of simple Λ-modules, or better, on the set of dimension vectors dim S(i), and we may write

dim M = ∑_{i∈I}[M : S(i)] dim S(i),

for any finite Λ-module M. Here, [M : S(i)] is the Jordan-Hölder multiplicity of S(i) in M. It is the number of factors isomorphic to S(i) in any fixed composition series of M.

Given Λ-modules M, N, let

⟨M, N⟩ = dim_{k} Hom(M, N) − dim_{k} Ext^{1}(M, N).

Since Λ is hereditary, ⟨M, N⟩ depends only on dim M, dim N. Thus, we can write

⟨dim M, dim N⟩ = ⟨M, N⟩,

and in this way, we obtain a bilinear form on K_{0}(Λ) with values in ℤ. Also, if we consider elements α, β ∈ P, then we may write

⟨α, β⟩ = ⟨V_{α}, V_{β}⟩.

Of course, since |k| = q_{k}, we have

q_{k}^{⟨M,N⟩} = \frac{| Hom(M, N)|}{| Ext^{1}(M, N)|}.

In particular, the extra factor | Ext^{1}(V_{ρ}, V_{τ})|·| Hom(V_{ρ}, V_{τ})|^{−1} appearing in Green’s formula may be written in the form

\frac{| Ext^{1}(V_{ρ}, V_{τ})|}{| Hom(V_{ρ}, V_{τ})|} = q_{k}^{−⟨V_{ρ},V_{τ} ⟩}.

Now, let v_{k} be a square root of q_{k}. The symmetrization of ⟨−, −⟩ will be denoted by •, thus

x•y = ⟨x, y⟩ + ⟨y, x⟩

for x, y ∈ K_{0}(Λ). Let

\overline{g}_{αβ}^{λ} = v_{k}^{⟨α,β⟩}g_{αβ}^{λ}.

Green’s formula may be rewritten as follows:

Theorem* (J. A. Green). Let k be a finite field and Λ a k-algebra which is hereditary and finitary. Let α, β, α′, β′ be elements of P. Then

a_{α}a_{β}a_{α′}a_{β′} ∑_{λ} \overline{g}_{αβ}^{λ}\overline{g}_{α′β′}^{λ} a_{λ}^{−1} = ∑_{ρ,σ,σ′,τ} v_{k}^{σ•σ′} \overline{g}_{ρσ}^{α}\overline{g}_{ρσ′}^{α′} \overline{g}_{σ′τ}^{β} \overline{g}_{στ}^{β′} a_{ρ}a_{σ}a_{σ′}a_{τ} .

Proof: The left hand term differs from the left hand term of Green’s first formula just by v_{k}^{⟨α,β⟩+⟨α′,β′⟩}.

We consider now the right hand side:

∑_{ρ,σ,σ′,τ}v_{k}^{σ•σ′} \overline{g}_{ρσ}^{α}\overline{g}_{ρσ′}^{α′} \overline{g}_{σ′τ}^{β} \overline{g}_{στ}^{β′} a_{ρ}a_{σ}a_{σ′}a_{τ}
= ∑_{ρ,σ,σ′,τ}v_{k}^{σ•σ′+e}g_{ρσ}^{α}g_{ρσ′}^{α′}g_{σ′τ}^{β} g_{στ}^{β′} a_{ρ}a_{σ}a_{σ′}a_{τ} ,

where

e = ⟨ρ, σ⟩ + ⟨ρ, σ′⟩ + ⟨σ′, τ⟩ + ⟨σ, τ⟩.

Consider a non-zero summand. It follows from g_{ρσ}^{α}g_{ρσ′}^{α′}g_{σ′τ}^{β} g_{στ}^{β′} ≠ 0 that we have

dim V_{α} = dim V_{ρ} + dim V_{σ},
dim V_{α′} = dim V_{ρ} + dim V_{σ′},
dim V_{β} = dim V_{σ′} + dim V_{τ} ,
dim V_{β′} = dim V_{σ} + dim V_{τ} ,

and therefore

⟨α, β⟩ = ⟨ρ, σ′⟩ + ⟨ρ, τ⟩ + ⟨σ, σ′⟩ + ⟨σ, τ⟩,
⟨α′, β′⟩ = ⟨ρ, σ⟩ + ⟨ρ, τ⟩ + ⟨σ′, σ⟩ + ⟨σ′, τ⟩.

Thus,

σ•σ′ + e = ⟨σ, σ′⟩ + ⟨σ′, σ⟩ + e
= ⟨α, β⟩ + ⟨α′, β′⟩ − 2⟨ρ, τ⟩.

C. M. Ringel

Altogether, we see that the right hand term is of the form

∑_{ρ,σ,σ′,τ} v_{k}^{σ•σ′} \overline{g}_{ρσ}^{α}\overline{g}_{ρσ′}^{α′} \overline{g}_{σ′τ}^{β} \overline{g}_{στ}^{β′} a_{ρ}a_{σ}a_{σ′}a_{τ}

= v_{k}^{⟨α,β⟩+⟨α′,β′⟩} ∑_{ρ,σ,σ′,τ} v_{k}^{−2⟨ρ,τ⟩}g_{ρσ}^{α}g_{ρσ′}^{α′}g_{σ′τ}^{β} g_{στ}^{β′} a_{ρ}a_{σ}a_{σ′}a_{τ} .

This is just the right hand term of Green’s first formula, multiplied by v_{k}^{⟨α,β⟩+⟨α′,β′⟩}. Altogether, we see that the second formula is a direct consequence of the first one.

Part II. Hall algebras

Let Λ be a finitary ring. Below, we recall that we may use the numbers g_{αβ}^{λ} as structure constants for a ℤ-algebra with basis P; this algebra is called the Hall algebra H(Λ) of Λ with coefficients in ℤ.

More generally, fix some commutative ring K (with unit element 1_{K}). We may consider also the Hall algebra H(Λ, K) of Λ with coefficients in K; it is the free K-module KP with a basis indexed by P, and again with structure constants g_{αβ}^{λ}.

On the other hand, it also will be of interest to consider a suitable coalgebra structure on KP. Let us assume that Λ is a k-algebra, where k is a finite field of order q_{k}. We will see below that the numbers

h_{λ}^{αβ} = g_{αβ}^{λ}a_{α}a_{β}a_{λ}^{−1},

will belong to ℤ[q_{k}^{−1}], thus in case q_{k} · 1_{K} is invertible in K, we may use these elements h_{λ}^{αβ} as structure constants for a comultiplication.

In case Λ is a hereditary k-algebra, Green’s formula expresses a compatibility relation between the described multiplication and comultiplication: we obtain what we call a (K, v, χ)-bialgebra.

1. Preliminaries: Twisting the multiplication of a graded algebra.

Let K be a commutative ring. If I is a set, we denote by KI the free Kmodule with basis I, or better, with a basis indexed by I. In particular, ℤI is the free abelian group with basis I. The elements of ℤI can be written in the form x = (x_{i})_{i}, with integers x_{i}, so that x_{i} = 0 for almost all i ∈ I. We write x ≥ 0 provided x_{i} ≥ 0 for all i ∈ I. Also, we write x > 0, provided we have both x ≥ 0 and x ≠ 0.

We are going to consider ℤI-graded algebras (and later also ℤI-graded coalgebras). We start with a ℤI-graded K-module

A = ⨁_{x∈ℤI} A_{x}

(thus, this is a direct decomposition of K-modules); this will be the underlying set. We say that A is positive provided that A_{x} ≠ 0 only for x ≥ 0, and that A_{0} = K.

For A = (A, μ) to be a ℤI-graded K-algebra, we require that μ is an associative multiplication on A with a unit element, and that μ respects the grading (thus μ maps A_{x} ⊗A_{y} into A_{x+y}, for x, y ∈ ℤI; in particular, the unit element has to belong to A_{0}.) We usually will just write ab instead of μ(a, b).

Let A be a ℤI-graded K-algebra. Given a bilinear form ϕ on ℤI and an invertible element v ∈ K, we are going to define a twisted multiplication ∗ on A, and we denote the new algebra by A_{[v,ϕ]}.

In order to form A_{[v,ϕ]}, we will not change the underlying K-module nor the grading, thus

A_{[v,ϕ]} = ⨁_{x∈ℤI} A_{x}.

On this K-module, we introduce a new multiplication ∗ as follows: For a ∈ A_{x}, b ∈ A_{y}, where x, y ∈ ℤI, we define

a ∗ b = v^{ϕ(x,y)} ab;

(instead of ϕ(x, y) we also will write ϕ(a, b), where, as before, a ∈ A_{x}, and b ∈ A_{y} are homogeneous elements of A). The new multiplication ∗ again is associative, as one easily checks (using the bilinearity of ϕ); in fact, one calculates for homogeneous elements a, b, c both a ∗ (b ∗ c) and (a ∗ b) ∗ c and shows in this way:

a ∗ b ∗ c = v^{ϕ(a,b)+ϕ(a,c)+ϕ(b,c)} abc,

more generally: if a_{1}, a_{2}, …, a_{n} are homogeneous elements of A, then

a_{1} ∗ a_{2} ∗ ⋯ ∗ a_{n} = v^{∑_{i<j} ϕ(a_{i},a_{j})} a_{1}a_{2} ⋯ a_{n}.

Also, the unit element 1 of A (which belongs to A_{0}) will be a unit element for the new multiplication ∗ as well, since ϕ(0, x) = 0 = ϕ(x, 0) for all x ∈ ℤI.

Lemma. Let f: A → B be a homomorphism of ℤI-graded K-algebras. If we consider f as a map A_{[v,ϕ]} → B_{[v,ϕ]}, then this is also a homomorphism of ℤI-graded K-algebras.

Proof. Let a, a′ ∈ A be homogeneous. Then ϕ(a, a′) = ϕ(f(a), f(a′)), thus we have

f(a ∗ a′) = f(v^{ϕ(a,a′)}aa′) = v^{ϕ(a,a′)}f(aa′) = v^{ϕ(a,a′)}f(a)f(a′) = f(a) ∗ f(a′).

2. The (K, v, χ)-bialgebras.

Let K be a commutative ring. We are going to consider besides ℤI-graded K-algebras also ℤI-graded K-coalgebras.

For a ℤI-graded K-coalgebra A = (A, δ), we again assume that there is given a ℤI-graded K-module

A = ⨁_{x∈ℤI} A_{x}

and we require that

δ: A → A ⊗_{K} A

is a coassociative comultiplication on A with a counit, and that δ respects the grading (thus δ(A_{z}) ⊆ ⨁_{x+y=z} A_{x} ⊗_{K} A_{y}). We often will use the socalled Sweedler notation: we write δ(a) = ∑ a_{1} ⊗ a_{2}, instead of δ(a) = ∑_{i} a_{i1} ⊗ a_{i2}. We usually will assume that A_{0} = K so that the counit will just be the projection onto A_{0}.

In order to describe a possible interrelation between a multiplication and a comultiplication, we assume that there are given two bilinear forms χ′, χ′′ on ℤI with values in ℤ and in addition an invertible element v ∈ K. Let A be a ℤI-graded algebra.

Given the pair χ = (χ′, χ′′) of bilinear forms on ℤI and v ∈ K, we consider a corresponding map (ℤI)^{4} → ℤ, which we also denote by χ and which is defined as follows:

χ(x_{1}, x_{2}, x_{3}, x_{4}) = χ′(x_{1}, x_{4}) + χ′′(x_{2}, x_{3}),

we say that χ is given by the pair (χ′, χ′′). Note that this map χ is a bilinear form on (ℤI)^{2}. (A characterization of the maps (ℤI)^{4} → ℤ which are given by pairs (χ′, χ′′) will be presented at the beginning of Part IV.)

Of course, we may consider A⊗A as a (ℤI)^{2}-graded algebra, where for x, y ∈ ℤI, we have (A ⊗ A)_{(x,y)} = A_{x} ⊗ A_{y}. Thus, given a pair χ = (χ′, χ′′) of bilinear forms on ℤI, we may use the corresponding bilinear form χ on (ℤI)^{2} in order to twist the multiplication of A ⊗ A and we obtain in this way the algebra (A ⊗ A)_{[v,χ′,χ′′]} = (A ⊗ A)_{[v,χ]}. To repeat: the algebra (A ⊗ A)_{[v,χ′,χ′′]} is given by A ⊗ A together with the multiplication ∗ defined by

(a_{1} ⊗ a_{2}) ∗ (b_{1} ⊗ b_{2}) = v^{χ′(a_{1},b_{2})+χ′′(a_{2},b_{1})} a_{1}b_{1} ⊗ a_{2}b_{2}.

By definition, a (K, v, χ)-bialgebra is of the form A = (A, μ, δ), where A is a K-module with a direct decomposition A = ⨁_{x∈ℤI} A_{x}, such that (A, μ) is a ℤIgraded algebra, (A, δ) is a ℤI-graded coalgebra and such that on the one hand, the counit ϵ satisfies ϵ(1) = 1, and, on the other hand,

δ: A → (A ⊗ A)_{[v,χ]}

is an algebra homomorphism. The (K, v, χ)-bialgebra A is said to be positive provided A is a positive ℤI-graded K-module. (In dealing with (K, v, (χ′, χ′′))- bialgebras, we will delete the inner brackets and speak of (K, v, χ′, χ′′)-bialgebras.)

Lemma 1. Let A be a positive (K, v, χ)-bialgebra. Then, for a ∈ A_{i} with i ∈ I, we have δ(a) = a ⊗ 1 + 1 ⊗ a.

Proof. Let i ∈ I. We have (A ⊗ A)_{i} = A_{i} ⊗ A_{0} + A_{0} ⊗ A_{i} = A_{i} ⊗ K + K ⊗ A_{i}. Now, given a ∈ A_{i}, the element δ(a) belongs to (A⊗A)_{i}, thus δ(a) = a′ ⊗1+1⊗a′′, where a′, a′′ are suitable elements of A_{i}. The fact that the projection A → A_{0} is the counit implies that a′ = a = a′′.

Lemma 2. Let A, B be positive (K, v, χ)-bialgebras. Assume that A is generated as a K-algebra by ⨁_{i∈I} A_{i}. Then any homomorphism f: A → B of ℤI-graded algebras is also a homomorphism of coalgebras.

Proof: Let f: A → B be a homomorphism of ℤI-graded algebras. Then f ⊗ f: A ⊗ A → B ⊗ B is a homomorphism of ℤI-graded algebras, and therefore also f ⊗ f: (A ⊗ A)_{[v,χ]} → (B ⊗ B)_{[v,χ]} is a homomorphism of ℤI-graded algebras. Consider the following diagram of homomorphisms of ℤI-graded algebras and algebra homomorphisms:

A ⟶^{δ} (A ⊗ A)_{[v,χ]}
f↓  ↓^{f⊗f}
B ⟶_{δ} (B ⊗ B)_{[v,χ]}

In order to show that it commutes, we only have to consider a generating set of A. Let a ∈ A_{i} with i ∈ I. Then we know that δ(a) = a ⊗ 1 + 1 ⊗ a. Also, f(a) ∈ B_{i}, and therefore δ(f(a)) = f(a) ⊗ 1 + 1 ⊗ f(a). Also, (f ⊗ f)δ(a) = f(a) ⊗ 1 + 1 ⊗ f(a) thus (f ⊗ f)δ(a) = δf(a), and therefore (f ⊗ f)δ = δf.

3. The Hall algebra H(Λ, K).

Let Λ be a ring. Recall that P denotes the set of isomorphism classes of Λ-modules of finite cardinality and let I be the set of all isomorphism classes of finite simple Λ-modules, it is a subset of P. Let us consider the free K-module H(Λ, K) = KP with basis {u_{α}| α ∈ P}.

In order to be able to define a multiplication on H(Λ, K), we want to assume that Λ is finitary, and we use the numbers g_{αβ}^{λ} as structure constants: thus, for α, β ∈ P, let

u_{α}u_{β} = ∑_{λ} g_{αβ}^{λ} u_{λ};

note that this is a finite sum, since we assume that Λ is finitary.

We also want to see that H(Λ, K) is a ℤI-graded algebra. For x ∈ ℤI, let P(x) be the set of isomorphism classes α such that dim V_{α} = x. Of course, P(x) is empty unless x ≥ 0 and P(0) just consists of the isomorphism class of the zero module.

Let H(Λ, K)_{x} = KP(x), the free K-module with basis elements u_{α}, where α ∈ P(x).

Lemma. H(Λ, K) is a ℤI-graded algebra. Its unit element is 1 = u_{0}.

Proof: The multiplication respects the grading, since g_{αβ}^{λ} ≠ 0 implies that dim V_{λ} = dim V_{α} + dim V_{β}.

The multiplication is associative: Namely, given elements α, β, γ, λ in P, let g_{αβγ}^{λ} be the number of filtrations

U′ ⊆ U ⊆ V_{λ} such that V_{λ}/U ∈ α, U/U′ ∈ β, U′ ∈ γ.

Then it is easy to see that we have

∑_{ν} g_{αν}^{λ}g_{βγ}^{ν} = g_{αβγ}^{λ} = ∑_{μ} g_{αβ}^{μ}g_{μγ}^{λ}.

Of course, the left hand term is the coefficient of u_{λ} in u_{α}(u_{β}u_{γ}), the right hand term is the coefficient of u_{λ} in (u_{α}u_{β})u_{γ}.

Finally, u_{0} is a right unit for this multiplication, since every module M has a unique submodule isomorphic to the zero module, and the corresponding factor module is isomorphic to M. Similarly, u_{0} is a left unit.

As mentioned above, we will denote the Hall algebra of Λ with coefficients in ℤ just by H(Λ).

4. The comultiplication.

Let k be a finite field of order q_{k}, let Λ be a k-algebra. We now assume that the element q_{k} · 1_{K} is invertible in K. We will write ℤ_{k} = ℤ[q_{k}^{−1}].

In order to define a comultiplication on H(Λ, K) = KP, we need the following consideration: We consider the numbers

h_{λ}^{αβ} = g_{αβ}^{λ}a_{α}a_{β}a_{λ}^{−1},

we want to use them as structure constants for a comultiplication. The definition of h_{λ}^{αβ} involves the inverse of a_{λ}. Let us show that actually it is sufficient to require (as we do) that the element q_{k} · 1_{K} is invertible in K.

Given Λ-modules A, B, L, let us denote by Ext_{Λ}^{1}(A, B)^{L} the set of equivalence classes in Ext_{Λ}^{1}(A, B) of exact sequences of the form

0 ⟶ B ⟶^{b} L ⟶^{a} A ⟶ 0.

The following observation seems to be due to Riedtmann [Ri] and Peng [P].

Proposition. Let α, β, λ ∈ P. Then

h_{λ}^{αβ} = \frac{| Ext_{Λ}^{1}(V_{α}, V_{β})^{V_{λ}}|}{| Hom_{Λ}(V_{α}, V_{β})|}.

Proof. Let A = V_{α}, B = V_{β}, L = V_{λ}. Consider the set E_{AB}^{L} of exact sequences of the form

0 ⟶ B ⟶^{b} L ⟶^{a} A ⟶ 0.

According to Propositions 1 and 4 in Section I.3, we have

g_{αβ}^{λ}a_{α}a_{β} = |E_{AB}^{L}| = ∑_{ε∈Ext_{Λ}^{1}(A,B)^{L}} |E^{L}(ε)| = | Ext_{Λ}^{1}(A, B)^{L}|\frac{a_{λ}}{| Hom_{Λ}(A, B)|}.

Corollary. The coefficients h_{λ}^{αβ} belong to ℤ_{k}.

Proof: Let n = dim_{k} Hom_{Λ}(V_{α}, V_{β}). Then

q_{k}^{n} · h_{λ}^{αβ} = | Ext_{Λ}^{1}(V_{α}, V_{β})^{V_{λ}}| ∈ ℤ.

Since we assume that q_{k}·1_{K} is invertible in K, we can define the comultiplication δ on H(Λ, K) by

δ: H(Λ, K) ⟶ H(Λ, K) ⊗_{K} H(Λ, K)

given by

δ(u_{λ}) = ∑_{α,β} h_{λ}^{αβ} u_{α} ⊗ u_{β} (= ∑_{α,β} g_{αβ}^{λ}a_{α}a_{β}a_{λ}^{−1} u_{α} ⊗ u_{β}).

Note that this always is a finite sum, since for every λ ∈ P, the number of pairs (α, β) with g_{αβ}^{λ} ≠ 0 is finite. (Thus, in order to consider the comultiplication, we do not need the assumption on Λ to be finitary.)

Let us mention the following special case: in case V_{λ} = V_{i} is a simple module, we have

δ(u_{i}) = u_{i} ⊗ 1 + 1 ⊗ u_{i}.

As before, we consider H(Λ, K) as a ℤI-graded module. Of course, H(Λ, K)_{0} = K, and we denote by ϵ the projection modulo the ideal ⨁_{x>0} H(Λ, K)_{x} onto H(Λ, K)_{0}.

Lemma. H(Λ, K) is a ℤI-graded coalgebra with counit ϵ.

It is clear that δ respects the grading: namely, if h_{λ}^{αβ} ≠ 0, then we must have dim V_{λ} = dim V_{α} + dim V_{β}.

Let us show that this comultiplication is coassociative:

(1 ⊗ δ)δ(u_{λ}) = (1 ⊗ δ)(∑_{α,ν} g_{αν}^{λ} a_{α}a_{ν}a_{λ}^{−1} u_{α} ⊗ u_{ν})

= ∑_{α,ν} ∑_{β,γ} g_{αν}^{λ}a_{α}a_{ν}a_{λ}^{−1} g_{βγ}^{ν}a_{β}a_{γ}a_{ν}^{−1} u_{α} ⊗ u_{β} ⊗ u_{γ}

= ∑_{α,β,γ} g_{αβγ}^{λ}a_{α}a_{β}a_{γ}a_{λ}^{−1} u_{α} ⊗ u_{β} ⊗ u_{γ}

Green’s Theorem

and similarly

(δ ⊗ 1)δ(u_{λ}) = ∑_{α,β,γ} g_{αβγ}^{λ}a_{α}a_{β}a_{γ}a_{λ}^{−1} u_{α} ⊗ u_{β} ⊗ u_{γ}.

Also, it is easy to see that the K-linear map ϵ: H(Λ, K) → K is a counit.

5. The compatibility of multiplication and comultiplication.

Let k be a finite field of order q_{k}, let Λ be a k-algebra which is hereditary and finitary. Let K be a commutative ring with q = q_{k} · 1_{K} being invertible in K.

Let us write H = H(Λ, K). We consider also the tensor product H ⊗ H (over K) as an algebra with multiplication ∗ defined as follows:

(u_{α} ⊗ u_{β}) ∗ (u_{α′} ⊗ u_{β′}) = \frac{| Ext^{1}(V_{α}, V′_{β})|}{| Hom(V_{α}, V′_{β})|} μ(u_{α} ⊗ u_{α′}) ⊗ μ(u_{β} ⊗ u_{β′});

thus, we deal with (H ⊗ H)_{[q,−⟨−,−⟩,0]}.

Corollary 1 to Green’s formula. The map δ: H → (H ⊗ H)_{[q,−⟨−,−⟩,0]} is an algebra homomorphism.

Proof: Consider the following diagram

H ⊗ H ⟶^{μ} H
δ⊗δ↓
H ⊗ H ⊗ H ⊗ H ↓^{δ}
ζ′ ↘
H ⊗ H ⊗ H ⊗ H ⟶_{μ⊗μ} H ⊗ H

where ζ′ is defined by

ζ′(u_{α} ⊗ u_{β} ⊗ u_{α′} ⊗ u_{β′}) = \frac{| Ext^{1}(V_{α}, V′_{β})|}{| Hom(V_{α}, V′_{β})|} u_{α} ⊗ u_{α′} ⊗ u_{β} ⊗ u_{β′}.

Green’s formula expresses the fact that this diagram commutes: Indeed, consider u_{α} ⊗ u_{β} ∈ H ⊗ H. Then

δμ(u_{α} ⊗ u_{β}) = δ(∑_{λ} g_{αβ}^{λ}u_{λ})

= ∑_{λ} g_{αβ}^{λ} ∑_{α′,β′} g_{α′β′}^{λ}a_{α′}a_{β′}a_{λ}^{−1} u_{α′} ⊗ u_{β′}

= ∑_{α′,β′} (a_{α′}a_{β′} ∑_{λ} g_{αβ}^{λ}g_{α′β′}^{λ}a_{λ}^{−1}) u_{α′} ⊗ u_{β′}

whereas

(μ ⊗ μ) ζ′ (δ ⊗ δ)(u_{α} ⊗ u_{β})

= (μ ⊗ μ) ζ′( ∑_{ρ,σ,σ′,τ} g_{ρσ}^{α}g_{σ′τ}^{β} a_{ρ}a_{σ}a_{σ′}a_{τ} a_{α}^{−1}a_{β}^{−1}u_{ρ} ⊗ u_{σ} ⊗ u_{σ′} ⊗ u_{τ} )

= (μ ⊗ μ)( ∑_{ρ,σ,σ′,τ} \frac{| Ext^{1}(V_{ρ}, V_{τ})|}{| Hom(V_{ρ}, V_{τ})|} g_{ρσ}^{α}g_{σ′τ}^{β} a_{ρ}a_{σ}a_{σ′}a_{τ} a_{α}^{−1}a_{β}^{−1}u_{ρ} ⊗ u_{σ′} ⊗ u_{σ} ⊗ u_{τ} )

= ∑_{ρ,σ,σ′,τ} \frac{| Ext^{1}(V_{ρ}, V_{τ})|}{| Hom(V_{ρ}, V_{τ})|} g_{ρσ}^{α}g_{σ′τ}^{β} a_{ρ}a_{σ}a_{σ′}a_{τ} a_{α}^{−1}a_{β}^{−1}(∑_{α′} g_{ρσ′}^{α′}u_{α′}) ⊗ (∑_{β′} g_{στ}^{β′} u_{β′})

= ∑_{α′,β′} (a_{α}^{−1}a_{β}^{−1} ∑_{ρ,σ,σ′,τ} \frac{| Ext^{1}(V_{ρ}, V_{τ})|}{| Hom(V_{ρ}, V_{τ})|} g_{ρσ}^{α}g_{σ′τ}^{β} g_{ρσ′}^{α′}g_{στ}^{β′} a_{ρ}a_{σ}a_{σ′}a_{τ} ) u_{α′} ⊗ u_{β′}.

Of course, (μ ⊗ μ) ζ′ is just the multiplication map of (H ⊗ H)_{[q,−⟨−,−⟩,0]}. The commutativity of the diagram expresses the fact that

δ: H → (H ⊗ H)_{[q,−⟨−,−⟩,0]}

is an algebra homomorphism.

We may read the diagram also in another way: We may consider H ⊗ H as a coalgebra using the comultiplication

δ′ = ζ′ (δ ⊗ δ);

since we use the same twist, we will denote this coalgebra by (H ⊗ H)_{[q,−⟨−,−⟩,0]}, again. We may reformulate the commutativity of the diagram as follows:

Corollary 2. The multiplication map μ: (H⊗H)_{[q,−⟨−,−⟩,0]} → H is a coalgebra homomorphism.

In the case of the ordinary Hall algebra H(ℤ̂_{p}), where p is a prime number, the comultiplication has been considered by Geissinger and Zelevinsky (see [Z] p.116).

Theorem. Let k be a finite field of order q_{k}. Let Λ be a k-algebra which is hereditary and finitary. Let K be a commutative ring such that q = q_{k} · 1_{K} is invertible in K. Then H(Λ, K) is a (K, q, −⟨−, −⟩, 0)-bialgebra.

6. The twisted Hall algebras.

Again, we start with a k-algebra Λ, where k is a finite field. Let us assume now that K contains an invertible element v such that v^{2} = q_{k} · 1_{K}, where q_{k} = |k|.

Let H_{∗}(Λ, K) = KP. In case Λ is finitary, we endow H_{∗}(Λ, K) with the multiplication

u_{α} ∗ u_{β} = ∑_{λ} \overline{g}_{αβ}^{λ} u_{λ};

we recall that

\overline{g}_{αβ}^{λ} = v^{⟨α,β⟩}g_{αβ}^{λ}.

Similarly, let

\overline{h}_{λ}^{αβ} = v^{⟨α,β⟩}h_{λ}^{αβ} = v^{⟨α,β⟩}g_{αβ}^{λ}a_{α}a_{β}a_{λ}^{−1}.

Without any assumption on Λ, we may consider H_{∗}(Λ, K) as a coalgebra using the comultiplication

δ_{∗}(u_{λ}) = ∑_{α,β} \overline{h}_{λ}^{αβ} u_{α} ⊗ u_{β} (= ∑_{α,β} v^{⟨α,β⟩}g_{αβ}^{λ}a_{α}a_{β}a_{λ}^{−1} u_{α} ⊗ u_{β}).

Recall that • denotes the symmetrization of the bilinear form ⟨−, −⟩ on the Grothendieck group ℤI = K_{0}(Λ) of Λ.

Theorem. Let k be a finite field of order q_{k}. Let Λ be a k-algebra which is hereditary and finitary. Let K be a commutative ring which contains an invertible element v such that v^{2} = q_{k} · 1_{K}. Then H_{∗}(Λ, K) is a (K, v, 0, •)-bialgebra.

Proof: Similar to the case of H(Λ, K).

Of particular interest is the case where K is equal to ℤ_{∗k} = ℤ[v_{k}, v_{k}^{−1}], here v_{k} is a root of q_{k} = |k|. We write H_{∗}(Λ) instead of H_{∗}(Λ, ℤ_{∗k}).

7. The bilinear form on H(Λ, K) and H_{∗}(Λ, K).

We define a symmetric bilinear form on the free K-module KP as follows: For α ∈ P, let t_{α} = \frac{|V_{α}|}{a_{α}}. Let

(u_{α}, u_{β}) = \{ \begin{matrix}t_{α} & if α = β \\ 0 & otherwise. \end{matrix}

In case we deal with a ℤI-graded algebra A which is endowed with a bilinear form (−, −), we also will consider a corresponding bilinear form on A ⊗ A, again denoted by (−, −) and defined by

(a_{1} ⊗ a_{2}, b_{1} ⊗ b_{2}) = (a_{1}, b_{1})(a_{2}, b_{2})

(thus, the form (−, −) is defined componentwise, without any twist). A bilinear form (−, −) on A will be said to respect the ℤI-grading provided we have (a, b) = 0 for homogeneous elements a, b ∈ A with |a| ≠ |b|.

Proposition. For all elements a, b, c in H(Λ, K), we have

(a, bc) = (δ(a), b ⊗ c).

We can reformulate the result as follows: The comultiplication δ is left adjoint to the multiplication.

Proof: It is sufficient to consider the case a = u_{α}, b = u_{β}, c = u_{γ}. By definition, bc = ∑_{λ} g_{βγ}^{λ}u_{λ}, thus

(u_{α}, u_{β}u_{γ}) = (u_{α}, ∑_{λ} g_{βγ}^{λ}u_{λ})
= g_{βγ}^{α} (u_{α}, u_{α})
= g_{βγ}^{α} t_{α}.

On the other hand, δ(u_{α}) = ∑_{μν} g_{μν}^{α}a_{μ}a_{ν}a_{α}^{−1}u_{μ} ⊗ u_{ν}, thus

(δ(u_{α}), u_{β} ⊗ u_{γ}) = (∑_{μν} g_{μν}^{α}a_{μ}a_{ν} \frac{1}{a_{α}} u_{μ} ⊗ u_{ν}, u_{β} ⊗ u_{γ})

= g_{βγ}^{α}a_{β}a_{γ} \frac{1}{a_{α}} (u_{β} ⊗ u_{γ}, u_{β} ⊗ u_{γ})

= g_{βγ}^{α}a_{β}a_{γ} \frac{1}{a_{α}} t_{β}t_{γ}

= g_{βγ}^{α}a_{β}a_{γ} \frac{1}{a_{α}} \frac{|V_{β}|}{a_{β}} \frac{|V_{γ}|}{a_{γ}}

= g_{βγ}^{α} \frac{|V_{α}|}{a_{α}}

= g_{βγ}^{α} t_{α}.

Here, we have used that for g_{βγ}^{α} ≠ 0 we must have |V_{α}| = |V_{β}| · |V_{γ}|.

There is the analogous result for H_{∗}(Λ, K).

Proposition*. For all elements a, b, c in H_{∗}(Λ, K), we have

(a, b ∗ c) = (δ_{∗}(a), b ⊗ c).

Again, we may reformulate the result: The comultiplication δ_{∗} is left adjoint to the multiplication of H_{∗}(Λ, K).

Remark. Instead of using the coefficients t_{α} = |V_{α}|a_{α}^{−1} one may consider (as Green [G1] does) t′_{α} = a_{α}^{−1}. Of course, the corresponding calculations will become easier. However, our choice will give precisely the bilinear form used by Lusztig on U^{+} (at least for a basic k-algebra Λ), since for i ∈ I we get

t_{i} = \frac{|V_{i}|}{a_{i}} = \frac{q_{k}^{d_{i}}}{q_{k}^{d_{i}} − 1},

where d_{i} = dim_{k} End(V_{i}), and q_{k} = |k|. The coefficients t_{α} play a prominent role in Lusztig’s description of the canonical basis of U^{+} in terms of the Lusztig form on U^{+}, see [L], Theorem 14.2.3.

Part III. The composition algebras

Let Λ be a finitary k-algebra. We have defined above the Hall algebra H(Λ). By definition, the elements u_{α}, with α ∈ P form a ℤ-basis of H(Λ). Recall that I denotes the subset of P given by the isomorphism classes of the simple Λ-modules, and we may consider the subring C(Λ) generated by the elements u_{i} with i ∈ I. This subalgebra is called the composition algebra of Λ with coefficients in ℤ. We are going to study properties of this subring.

We want to give an interpretation of this subring C(Λ). Let (i_{1}, i_{2}, …, i_{n}) be a sequence of elements from I. Then we can form the product u_{i_{1}}u_{i_{2}} ⋯ u_{i_{n}} in H(Λ), thus this product is a linear combination

u_{i_{1}}u_{i_{2}} ⋯ u_{i_{n}} = ∑_{λ∈P} c^{λ}u_{λ},

and clearly, c^{λ} counts the number of composition series of V_{λ} which are of the form

V_{λ} = L_{0} ⊃ L_{1} ⊃ L_{2} ⊃ ⋯ ⊃ L_{n} = 0

with L_{j−1}/L_{j} belonging to the isomorphism class i_{j}. We see that C(Λ) encodes the numbers of possible composition series of arbitrary Λ-modules, this explains the name.

We want to derive some relations between the generators u_{i}. In order to do so, we need to recall some notation.

1. Quantum binomial coefficients.

Consider the ring ℤ[ṽ, ṽ^{−1}], with ṽ a variable. The elements of ℤ[ṽ, ṽ^{−1}] will be called Laurent polynomials in the variable ṽ. Given such a Laurent polynomial φ = φ(ṽ), and an integer d, one denotes by φ_{d} the Laurent polynomial which is obtained from φ by inserting ṽ^{d}, thus φ_{d}(ṽ) = φ(ṽ^{d}). We denote q̃ = ṽ^{2}.

In this section, we will write v instead of ṽ and q instead of q̃.

For 0 ≤ t ≤ n, let

[n] = \frac{v^{n} − v^{−n}}{v − v^{−1}} (= v^{n−1} + v^{n−3} + … v^{−n+1}),
[n]! = ∏_{t=1}^{n}[t] ,
[ \begin{matrix}n \\ t \end{matrix} ] = \frac{[n]!}{[t]![n − t]!}.

and also

| n] = \frac{q^{n} − 1}{q − 1} (= q^{n−1} + ⋯ + q + 1) = v^{n−1}[n] ,
| n]! = ∏_{t=1}^{n} | t]  = v^{\binom{n}{2}}[n]! ,
| \begin{matrix}n \\ t \end{matrix} ] = \frac{| n]!}{| t]! · | n − t]!}  = v^{t(n−t)} [ \begin{matrix}n \\ t \end{matrix} ] .

Lemma. For any element z in a ℤ[q]-algebra, we have

∏_{j=0}^{n−1} (1 + q^{j}z) = ∑_{t=0}^{n} q^{\binom{t}{2}} | \begin{matrix}n \\ t \end{matrix} ] z^{t}.

The proof is rather straight-forward, using induction. See for example [J].

Corollary 1. The elements | \begin{matrix}n \\ t \end{matrix} ] belong to ℤ[q].

Proof: The left hand side of the formula exhibited in Lemma can be written in the form ∑_{t=0}^{n} c_{t}z^{t}, where c_{t} is a polynomial in the variable q. This coefficient c_{t} is the sum of all products q^{j_{1}}q^{j_{2}} ⋯ q^{j_{t}}, where 0 ≤ j_{1} < j_{2} < ⋯ < j_{t} ≤ n−1. In particular, such a product is always divisible by q^{\binom{t}{2}}. Thus, c_{t} = q^{\binom{t}{2}}c′_{t}, where c′_{t} is

a polynomial in the variable q. The lemma asserts that c′_{t} = | \begin{matrix}n \\ t \end{matrix} ] .

Corollary 2. The elements [ \begin{matrix}n \\ t \end{matrix} ] belong to ℤ[v, v^{−1}].

Corollary 3. There is the following equality:

∑_{t=0}^{n}(−1)^{t}q^{\binom{t}{2}} | \begin{matrix}n \\ t \end{matrix} ] = 0.

Proof. Take z = −1. Then we obtain (1 + q^{0}(−1)) = 0 as a factor on the left hand side, namely as the factor with index 0, thus the product is zero.

Interpretation. Let k be a finite field with q_{k} elements. Recall that the expressions

| n] , | n]! and | \begin{matrix}n \\ t \end{matrix} ]

are polynomials in the variable q with integer coefficients, thus we may insert for q integers, for example the number q_{k}; as usual, we will write | n](c) for the evaluation

of the polynomial | n] = | n](q) at the number c. Note that | n], | n]!, | \begin{matrix}n \\ t \end{matrix} ] usually

will be considered as polynomials in the variable q (in contrast to [n], [n]!, [ \begin{matrix}n \\ t \end{matrix} ]

which are Laurent polynomials in the variable v and q = v^{2}). The convention ϕ_{d}(v) = ϕ(v^{d}) will be used only for Laurent polynomials in the variable v, thus we write [n]_{2} instead of \frac{v^{2n}−v^{−2n}}{v^{2}−v^{−2}} , but we write | n](q^{2}) instead of \frac{q^{2n}−1}{q^{2}−1}.

Let V be an n-dimensional k-space. Then | n](q_{k}) is the number of onedimensional subspaces of V, and | n]!(q_{k}) is the number of complete linear flags in V (a complete linear flag in V is a sequence of subspaces 0 = V_{0} ⊂ V_{1} ⊂ ⋯ ⊂ V_{n} = V

where V_{i} is of dimension i for all 0 ≤ i ≤ n). Finally, | \begin{matrix}n \\ m \end{matrix} ] (q_{k}) is the number of

m-dimensional subspaces of V.

2. The fundamental relations for H(Λ) and H_{∗}(Λ)

Consider elements i ≠ j in I, and let us assume that Ext^{1}(S(j), S(i)) = 0. Then we define

d_{i} = dim_{k} End(S(i)),
e(i, j) = dim_{k} Ext^{1}(S(i), S(j)),
n(i, j) = d_{i}^{−1}e(i, j) + 1,
n(j, i) = d_{j}^{−1}e(i, j) + 1.

Note that in case also the opposite assumption Ext^{1}(S(i), S(j)) = 0 is satisfied, then we will have e(i, j) = 0 = e(j, i), and therefore n(i, j) = 1 = n(j, i).

Proposition. Consider elements i ≠ j in I. We assume that Ext^{1}(S(j), S(i)) = 0. If we also have Ext^{1}(S(i), S(i)) = 0, then the following relation

∑_{t=0}^{n(i,j)}(−1)^{t}q_{k}^{d_{i}\binom{t}{2}} | \begin{matrix}n(i, j) \\ t \end{matrix} ] (q_{k}^{d_{i}}) u_{i}^{n(i,j)−t}u_{j}u_{i}^{t} = 0

is satisfied in H(Λ).

If we have Ext^{1}(S(j), S(j)) = 0, then the relation

∑_{t=0}^{n(j,i)}(−1)^{t}q_{k}^{d_{j} \binom{t}{2}} | \begin{matrix}n(j, i) \\ t \end{matrix} ] (q_{k}^{d_{j}}) u_{j}^{t}u_{i}u_{j}^{n(j,i)−t} = 0

is satisfied in H(Λ).

Proof: We indicate the proof of the second relation as presented in [R2], the proof of the first one is, of course, similar.

We fix the pair (i, j). As abbreviation, let d = d_{j} and n = n(j, i). Observe that e(i, j) = d(n − 1). First, we have to calculate the product u_{j}^{t}u_{i}u_{j}^{n−t} in H(Λ). This product has to be written as a linear combination c_{t}^{λ}u_{λ} with coefficients c_{t}^{λ} ∈ ℤ and λ ∈ P. Then we have to show that for every λ ∈ P, we have

∑_{t=0}^{n}(−1)^{t}q_{k}^{d\binom{t}{2}} | \begin{matrix}n \\ t \end{matrix} ] (q_{k}^{d}) c_{t}^{λ} = 0.

Thus, we fix some λ ∈ P. Let L be a Λ-module in the isomorphism class λ. In order that c_{t}^{λ} ≠ 0, we see that L must have a composition series with n factors of the form S(j) and one factor isomorphic to S(i). Let N be a direct summand of L of minimal length which has a composition factor S(i). Then L = N ⊕S(j)^{m} for some m. Here, we use that Ext^{1}(S(j), S(j)) = 0. Using again this condition and the fact that Ext^{1}(S(j), S(i)) = 0, we see that the radical of N is of the form S(j)^{n−m}, and N modulo its radical is isomorphic to S(i). In particular, it follows that L modulo its radical is of the form S(i) ⊕ S(j)^{m}. Finally, since dim_{k} Ext^{1}(S(i), S(j)) < dn, it follows that m ≥ 1.

The information obtained so far is sufficient to calculate c_{t}^{λ}. Namely, we have to find the number of composition series

L = L_{0} ⊃ L_{1} ⊃ ⋯ ⊃ L_{n+1} = 0

such that the first t factors are of the form S(j), the next one is of the form S(i), and the last n − t ones are again of the form S(j). Clearly, if t > m, then c_{t}^{λ} = 0, since L modulo its radical is of the form S(i) ⊕ S(j)^{m}.

Thus, let t ≤ m. If (L_{i})_{i} is a composition series of L as required, then we must have L_{t} ⊇ N. Thus, we first choose a submodule L_{t}/N of L/N of length m − t, and then in L/L_{t} a complete flag of submodules. The module L_{t} will have a unique submodule L_{t+1} with factor module isomorphic to S(i), thus, finally, we have to choose a complete flag of submodules in L_{t+1}. The number of possible submodules of the form L_{t} is the number of (m − t)-dimensional subspaces in an m-dimensional

vectorspace over the field k′ = End(S(j)), thus equal to | \begin{matrix}m \\ m − t \end{matrix} ] (q_{k}^{d}). The number

of complete flags in L/L_{t} is | t]!(q_{k}^{d}) and the number of complete flags in L_{t+1} is | n−t]!(q_{k}^{d}). This shows that c_{t}^{λ} is the evaluation at q_{k}^{d} of the following polynomial:

| \begin{matrix}m \\ m − t \end{matrix} ] · | t]! · | n − t]! = \frac{| m]!}{| m − t]!| t]!}| t]!| n − t]! = \frac{| m]!| n − t]!}{| m − t]!}.

Thus:

∑_{t=0}^{n}(−1)^{t}q_{k}^{d\binom{t}{2}} | \begin{matrix}n \\ t \end{matrix} ] (q_{k}^{d}) c_{t}^{λ} = ∑_{t=0}^{m}(−1)^{t}q_{k}^{d\binom{t}{2}} (\frac{| n]!}{| t]!| n − t]!} · \frac{| m]!| n − t]!}{| m − t]!}) (q_{k}^{d})

= | n]!(q_{k}^{d}) ∑_{t=0}^{m}(−1)^{t}q_{k}^{d\binom{t}{2}} \frac{| m]!}{| t]!| m − t]!}(q_{k}^{d})

= | n]!(q_{k}^{d}) ∑_{t=0}^{m}(−1)^{t}q_{k}^{d\binom{t}{2}} | \begin{matrix}m \\ t \end{matrix} ] (q_{k}^{d}) = 0

This completes the proof.

∗ ∗ ∗

Proposition. Consider elements i ≠ j in I. We assume that Ext^{1}(S(j), S(i)) = 0. If we also have Ext^{1}(S(i), S(i)) = 0, then the relation

∑_{t=0}^{n(i,j)}(−1)^{t} [ \begin{matrix}n(i, j) \\ t \end{matrix} ]_{d_{i}} u_{i}^{n(i,j)−t}u_{j}u_{i}^{t} = 0

is satisfied in H_{∗}(Λ).

If we have Ext^{1}(S(j), S(j)) = 0, then the relation

∑_{t=0}^{n(j,i)}(−1)^{t} [ \begin{matrix}n(j, i) \\ t \end{matrix} ]_{d_{j}} u_{j}^{t}u_{i}u_{j}^{n(i,j)−t} = 0

is satisfied in H_{∗}(Λ).

Proof: We only have to rewrite the multiplication in H_{∗}(Λ) using the multiplication of H(Λ, ℤ_{∗k}) and the twisting factors. See [R4].

3. The composition algebra and the twisted composition algebra.

Recall that the composition algebra C(Λ) of Λ with coefficients in ℤ is the subring of H(Λ) which is generated by the elements u_{i} with i ∈ I. We note the following: The composition algebra C(Λ) is a free ℤ-module, since submodules of free ℤ-modules are free.

As an abelian group, the composition algebra C(Λ) is generated by the set of monomials u_{i_{1}}u_{i_{2}} … u_{i_{n}} with i_{1}, i_{2}, …, i_{n} ∈ I and n ≥ 0. Since the generating elements u_{i} are homogeneous elements, the composition algebra C(Λ) again is a ℤI-graded algebra, and we have

C(Λ) = ⨁_{x∈ℤI} C(Λ)_{x}.

For x ∈ ℤI, the ℤ-module C(Λ)_{x} is generated by the monomials u_{i_{1}}u_{i_{2}} … u_{i_{n}} of degree x. Of course, the monomial u_{i_{1}}u_{i_{2}} … u_{i_{n}} has degree x = (x_{i})_{i} if and only if the element i occurs precisely x_{i} times in the sequence (i_{1}, i_{2}, …, i_{n}).

Consider now a commutative ring K. We define

C(Λ, K) = C(Λ) ⊗_{ℤ} K.

Since C(Λ) is a free ℤ-module, it follows that C(Λ, K) is a free K-module. Also, C(Λ, K) is again a ℤI-graded algebra, and we have

C(Λ, K) = ⨁_{x∈ℤI} C(Λ, K)_{x}

where C(Λ, K)_{x} is generated, as a K-module, by the monomials u_{i_{1}}u_{i_{2}} … u_{i_{n}} of degree x. We call C(Λ, K) the composition algebra of Λ with coefficients in K.

Let us assume now that q_{k} · 1_{K} is invertible in K. We consider first the special case ℤ_{k} = ℤ[q_{k}^{−1}]. The inclusion map ι: C(Λ) → H(Λ) induces an injective map

ι_{k} = ι ⊗ 1: C(Λ, ℤ_{k}) ⟶ H(Λ) ⊗_{ℤ} ℤ_{k} = H(Λ, ℤ_{k}),

thus we may consider C(Λ, ℤ_{k}) as a subalgebra of H(Λ, ℤ_{k}). Also, since both C(Λ, ℤ_{k}) and H(Λ, ℤ_{k}) are free ℤ_{k}-modules, it follows that ι_{k} ⊗ ι_{k} again is injective, thus we also may consider C(Λ, ℤ_{k}) ⊗_{ℤ_{k}} C(Λ, ℤ_{k}) as subalgebra of H(Λ, ℤ_{k}) ⊗_{ℤ_{k}} H(Λ, ℤ_{k}).

As we have seen above, H(Λ, ℤ_{k}) is a coalgebra via the comultiplication δ with

δ(u_{λ}) = ∑_{α,β} g_{αβ}^{λ}a_{α}a_{β}a_{λ}^{−1} u_{α} ⊗ u_{β}.

We have δ(u_{i}) = u_{i} ⊗ 1 + 1 ⊗ u_{i}, thus δ maps C(Λ, ℤ_{k}) into the ℤ_{k}-subalgebra of (H(Λ, ℤ_{k}) ⊗ H(Λ, ℤ_{k}))_{[q_{k},−⟨−,−⟩,0]} generated by the elements u_{i} ⊗ 1 and 1 ⊗ u_{i} with i ∈ I. But the latter is just (C(Λ, ℤ_{k}) ⊗ C(Λ, ℤ_{k}))_{[q_{k},−⟨−,−⟩,0]}; thus we see that δ induces a comultiplication

C(Λ, ℤ_{k}) ⟶ (C(Λ, ℤ_{k}) ⊗ C(Λ, ℤ_{k}))_{[q_{k},−⟨−,−⟩,0]}

again denoted by δ. In this way, C(Λ, ℤ_{k}) becomes a (ℤ_{k}, q_{k}, −⟨−, −⟩, 0)-bialgebra.

For a general commutative ring K with q = q_{k} · 1_{K} invertible, we note that C(Λ, K) = C(Λ, ℤ_{k}) ⊗_{ℤ_{k}} K. Since C(Λ, ℤ_{k}) is a (ℤ_{k}, q_{k}, −⟨−, −⟩, 0)-bialgebra, it follows that C(Λ, K) is a (K, q, −⟨−, −⟩, 0)-bialgebra.

Theorem. Let k be a finite field of order q_{k}. Let Λ be a k-algebra which is hereditary and finitary. Let K be a commutative ring such that q = q_{k} · 1_{K} is invertible in K. Then C(Λ, K) is a (K, q, −⟨−, −⟩, 0)-bialgebra. As a K-module, C(Λ, K) is free.

∗ ∗ ∗

We introduce now the corresponding twisted composition algebra. We consider only commutative rings K which contain an invertible element v such that v^{2} = q_{k} · 1_{K}. In particular, we will have to deal with the commutative ring ℤ_{∗k} = ℤ[v_{k}, v_{k}^{−1}], where v_{k} denotes a root of q_{k} = |k|.

As before, Λ is a finitary k-algebra. Let C_{∗}(Λ) be the ℤ_{∗k}-subalgebra of H_{∗}(Λ) generated by the elements u_{i}, with i ∈ I, it will be said to be the twisted composition algebra of Λ (with coefficients in ℤ_{∗k}). As in the case of the composition algebra itself, we see: The twisted composition algebra C_{∗}(Λ) is a free ℤ_{∗k}-module, since also ℤ_{∗k} is a principal ideal domain.

As a ℤ_{∗k}-module, the composition algebra C_{∗}(Λ) is generated by the set of monomials u_{i_{1}}u_{i_{2}} … u_{i_{n}} with i_{1}, i_{2}, …, i_{n} ∈ I and n ≥ 0. Since the generating elements u_{i} are homogeneous elements, the composition algebra C_{∗}(Λ) again is a ℤI-graded algebra, and we have

C_{∗}(Λ) = ⨁_{x∈ℤI} C_{∗}(Λ)_{x};

of course, for x ∈ ℤI, the ℤ_{∗k}-module C_{∗}(Λ)_{x} is generated by the monomials u_{i_{1}}u_{i_{2}} … u_{i_{n}} of degree x.

Next, we want to see that C_{∗}(Λ) also is a coalgebra. Consider the inclusion map ι_{∗}: C_{∗}(Λ) → H_{∗}(Λ). Since both C_{∗}(Λ) and H_{∗}(Λ) are free ℤ_{∗k}-modules, it follows that ι_{∗} ⊗ ι_{∗} again is injective, thus we also may consider C_{∗}(Λ) ⊗_{ℤ_{∗k}} C_{∗}(Λ) as subalgebra of H_{∗}(Λ) ⊗_{ℤ_{∗k}} H_{∗}(Λ).

As we have seen above, H_{∗}(Λ) is a coalgebra via the comultiplication δ_{∗} with

δ_{∗}(u_{λ}) = ∑_{α,β} \overline{g}_{αβ}^{λ}a_{α}a_{β}a_{λ}^{−1} u_{α} ⊗ u_{β}.

We have δ_{∗}(u_{i}) = u_{i} ⊗ 1 + 1 ⊗ u_{i}, thus δ_{∗} maps C_{∗}(Λ) into the ℤ_{∗k}-subalgebra of (H_{∗}(Λ) ⊗ H_{∗}(Λ))_{[v_{k},0,•]} generated by the elements u_{i} ⊗ 1 and 1 ⊗ u_{i} with i ∈ I. But this subalgebra is just (C_{∗}(Λ) ⊗ C_{∗}(Λ))_{[v_{k},0,•]}; thus we see that δ_{∗} induces a comultiplication

C_{∗}(Λ) ⟶ (C_{∗}(Λ) ⊗ C_{∗}(Λ))_{[v_{k},0,•]}

again denoted by δ_{∗}. In this way, C_{∗}(Λ) becomes a (ℤ_{∗k}, v_{k}, 0, •)-bialgebra.

Consider now a commutative ring K which contains an element v such that v^{2} = q_{k} · 1_{K} and we consider K as a ℤ_{∗k}-algebra, with v_{k} · 1_{K} = v. We define

C_{∗}(Λ, K) = C_{∗}(Λ) ⊗_{ℤ_{∗k}} K.

Since C_{∗}(Λ) is a free ℤ_{∗k}-module, it follows that C_{∗}(Λ, K) is a free K-module. Also, C_{∗}(Λ, K) is again a ℤI-graded algebra, and we have

C_{∗}(Λ, K) = ⨁_{x∈ℤI} C_{∗}(Λ, K)_{x}

where C_{∗}(Λ, K)_{x} is generated, as a K-module, by the monomials u_{i_{1}}u_{i_{2}} … u_{i_{n}} of degree x. We call C_{∗}(Λ, K) the twisted composition algebra of Λ with coefficients in K.

Theorem*. Let k be a finite field of order q_{k}. Let Λ be a k-algebra which is hereditary and finitary. Let K be a commutative ring which contains an invertible element v such that v^{2} = q_{k} · 1_{K}. Then C_{∗}(Λ, K) is a (K, v, 0, •)-bialgebra. As a K-module, C_{∗}(Λ, K) is free.

4. Euler forms and Cartan data.

We are going to consider some special bilinear forms on a free abelian group ℤI which occur in the representation theory of algebras. For simplicity, we assume that I is a finite set. Note that a bilinear form on ℤI is uniquely determined by an arbitrary function I × I → ℤ.

An Euler form is a pair (ω, d) consisting of a bilinear form ω on ℤI with values in ℤ and a function d: I → ℕ_{1} such that the following three properties are satisfied for all i, j ∈ I:

(a) ω(i, j) is divisible by both d_{i} and d_{j};

(b) ω(i, j) ≤ 0 for i ≠ j;

(c) d_{i}^{−1}ω(i, i) ≤ 1. An Euler datum is a pair (•, d) consisting of a symmetric bilinear form • on ℤI

with values in ℤ and a function d: I → ℕ_{1} such that the following two properties are satisfied:

(a) \frac{i•i}{2d_{i}} ∈ \{1, 0, −1, −2, … \}, for all i ∈ I.

(b) \frac{i•j}{d_{i}} ∈ \{0, −1, −2, … \}, for all i ≠ j in I.

Let (ω, d) be an Euler form, let • be the symmetrization of ω, thus x•y = ω(x, y) + ω(y, x). Then (•, d) is an Euler datum.

Conversely, let (•, d) be an Euler datum. We fix some total ordering ≺ on the set I. We denote by ω = ω_{(•,≺)} the following bilinear form on ℤI

ω(i, j) = \{ \begin{matrix}i•j & if i ≺ j, \\ \frac{1}{2} i•i & if i = j, \\ 0 & if i ≻ j. \end{matrix}

Then (ω, d) is an Euler form, and its symmetrization is the given Euler datum.

The Euler form (ω, d) is said to be without short cycles, provided we have, first of all, ω(i, i) = d_{i}, and secondly, ω(i, j)ω(j, i) = 0 for i ≠ j. Of course, given an Euler form (ω, d) without short cycles, the function d is uniquely determined by ω, thus we do not have to mention d.

Remark. For a general Euler form (ω, d), the function d is not uniquely determined, for example consider the zero bilinear form ω on ℤ itself. A more interesting example is given for I = {1, 2}, by ω(1, 2) = −2 and ω(i, j) = 0 otherwise. In this case, any one of d_{1}, d_{2} may be equal to either 1 or 2.

An Euler datum (•, d) with i•i = 2d_{i} for all i is said to be a Cartan datum. Of course, dealing with a Cartan datum, the function d is uniquely determined by (•), thus we do not have to mention d. A symmetric bilinear form (•) is a Cartan datum if and only if the following two conditions are satisfied: (a′) i•i ∈ {2, 4, 6, … }, for all i ∈ I. (b′) 2\frac{i•j}{i•i} ∈ \{0, −1, −2, … \}, for all i ≠ j in I.

We have shown above:

Lemma. Given an Euler form ω without short cycles, its symmetrization will be a Cartan datum, and any Cartan datum arises in this way.

Remark. Cartan data occur as basic ingredient in the book of Lusztig [L]; there is a direct correspondence between the set of Cartan data and the set of pairs consisting of a symmetrizable generalized Cartan matrices and a corresponding symmetrization, as considered in the book of Kac [K]. For any Euler datum (•, d), the set I is used only as an index set, and we may replace it by any other set of the same cardinality. In particular, if I is a set with n elements, we may replace I by the set {1, 2, …, n}. Note that the Cartan data (•) with set I = {1, …, n} correspond bijectively to the (symmetrizable) generalized Cartan matrices with a given symmetrization. We recall that a generalized Cartan matrix (a_{ij})_{ij} of size n with a symmetrization d = (d_{1}, …, d_{n}) is given by n^{2} integers a_{ij} and n positive integers d_{i} (with 1 ≤ i, j ≤ n), such that

(1) a_{ii} = 2 for all i;

(2) a_{ij} ≤ 0 for all i ≠ j; and

(3) d_{i}a_{ij} = d_{j}a_{ji} for all i ≠ j. Given a Cartan datum (\{1, …, n\}, •), then we define d_{i} = \frac{i•i}{2}, and a_{ii} = 2, for all 1 ≤ i ≤ n, whereas, for i ≠ j, we define a_{ij} = 2\frac{i•j}{i•i} . In this way, we obtain a generalized Cartan matrix (a_{ij})_{ij} of size n with a symmetrization d = (d_{1}, …, d_{n}). Conversely, let (a_{ij})_{ij} be a generalized Cartan matrix of size n with a symmetrization d = (d_{1}, …, d_{n}). Then we define i•j = d_{i}a_{ij} for all i, j.

Proposition. Let k be a field. Let Λ be a k-algebra with only a finite set I of isomorphism classes of simple Λ-modules. For any i ∈ I, let S(i) be a module in the isomorphism class i. We assume that for i, j ∈ I, the k-spaces Hom_{Λ}(S(i), S(j)) and Ext_{Λ}^{1}(S(i), S(j)) are finite-dimensional. For i, j ∈ I, let

⟨i, j⟩ = dim_{k} Hom_{Λ}(S(i), S(j)) − dim_{k} Ext_{Λ}^{1}(S(i), S(j)),
d_{i} = dim_{k} End(S(i)).

Then E(Λ) = (⟨−, −⟩, d) is an Euler form.

The symmetrization of E(Λ) is a Cartan datum if and only if Ext_{Λ}^{1}(S(i), S(i)) = 0, for all i.

Also, E(Λ) has no short cycles, provided we have first of all Ext_{Λ}^{1}(S(i), S(i)) = 0, for all i, and secondly, if Ext_{Λ}^{1}(S(i), S(j)) ≠ 0 for some pair i ≠ j, then we have Ext_{Λ}^{1}(S(j), S(i)) = 0.

Proof: Since the modules S(i) are simple and pairwise non-isomorphic, we see that End_{Λ}(S(i)) is a division ring, for any i, and that Hom_{Λ}(S(i), S(j)) = 0, for i ≠ j.

Let e_{ij} be the k-dimension of Ext_{Λ}^{1}(S(i), S(j)). For all i, j, we may consider Ext_{Λ}^{1}(S(i), S(j)) as an End_{Λ}(S(j))-End_{Λ}(S(i))-bimodule. In this way, we see that e_{ij} is divisible both by d_{i} and by d_{j}.

We have ⟨i, i⟩ = d_{i} − e_{ii}, thus

\frac{⟨i, i⟩}{d_{i}} = 1 − \frac{e_{ii}}{d_{i}} .

Finally, it is clear that i•i = 2d_{i} if and only if Ext_{Λ}^{1}(S(i), S(i)) = 0.

Remark. Assume that the simple Λ-modules S(i), S(j) are finite-dimensional. Then clearly Hom_{Λ}(S(i), S(j)) is a finite-dimensional k-space. But one should keep in mind that there are obvious examples of (even hereditary) k-algebras Λ with only finite-dimensional simple Λ-modules, such that Ext_{Λ}^{1}(M, N) may not be finitedimensional for some pair M, N of finite-dimensional Λ-modules. For example, take

Λ = [ \begin{matrix}k & V \\ 0 & k \end{matrix} ], where V is an infinite-dimensional k-space (with k acting centrally

on V). However, the required finiteness condition will be satisfied in case we start with a k-algebra Λ which is finite-dimensional:

Corollary. Let Λ be a finite-dimensional hereditary k-algebra. Let I be the set of isomorphism classes of simple Λ-modules. Then E(Λ) is defined, and its symmetrization is a Cartan datum.

Proof. Since Λ is finite-dimensional, all simple Λ-modules are finite-dimensional, and Ext_{Λ}^{1}(M, N) is finite-dimensional for all finite-dimensional Λ-modules M, N. This shows that E(Λ) is defined. Since Λ is in addition also hereditary, one knows that Ext_{Λ}^{1}(S, S) = 0 for all simple Λ-modules S. Thus, the symmetrization of E(Λ) is a Cartan datum.

Example. Consider the algebra Λ = [ \begin{matrix}k & 0 & 0 \\ k & k & 0 \\ k & k & k \end{matrix} ] of all lower triangular (3×3)-

matrices with coefficients in k. As index set, we have I = {1, 2, 3}, and we have first of all, ⟨i, i⟩ = 1, for all i; secondly, we have ⟨1, 2⟩ = −1 = ⟨2, 3⟩; and finally ⟨i, j⟩ = 0 for the remaining pairs i, j. Of course, this is usually called the case A_{3}, with diagram

∘  ∘  ∘

In general, given a quiver Q without loops, and consider Λ = kQ, the path algebra of Q. If Q does not have oriented cycles, then Λ is finite dimensional, and the vertex set of the quiver Q may be used as index set for the isomorphism classes of the simple Λ-modules. It follows that in this case E(Λ) = (ω, d), where we have d_{i} = 1 for all i, and where −ω(i, j) is the number of arrows i → j, for i ≠ j.

If the quiver Q has oriented cycles, then we have to consider instead of the path algebra kQ its completion Λ = \overline{kQ} at the ideal generated by the arrows; again, the vertex set of the quiver Q may be used as index set for the isomorphism classes of the simple Λ-modules, and E(Λ) = (ω, d) is given by d_{i} = 1 for all i, and by −ω(i, j) being the number of arrows i → j, for i ≠ j.

This shows that it is easy to construct for an arbitrary Euler form (ω, d) with d_{i} = 1 for all i ∈ I an algebra Λ with E(Λ) = (ω, d).

As we will see, all Euler forms (ω, d) can be realized in the form E(Λ) for some k-algebra Λ, at least for suitable fields k. Note that if k is algebraically closed, then we always will have d_{i} = 1 for all i, thus the only Euler forms (ω, d) which can be realized, are those with d_{i} = 1, for all i ∈ I. On the other hand, suppose there is given a field extension k ⊆ K with [K : k] = n, and consider the k-

algebra Λ = [ \begin{matrix}k & 0 & 0 \\ k & k & 0 \\ K & K & K \end{matrix} ] . As index set, we again have I = \{1, 2, 3\}, and we have

⟨i, i⟩ = 1, for i = 1, 2 and ⟨3, 3⟩ = n; secondly, we have ⟨1, 2⟩ = −1, and ⟨2, 3⟩ = n; and finally ⟨i, j⟩ = 0 for the remaining pairs i, j. Of course, for n = 2, this is called the case C_{3}, with diagram

∘  ∘  ∘^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}

We are interested in the case when k is a finite field. In this case, any Euler form can be realized in the form E(Λ), where Λ is some k-algebra:

Proposition. Let (ω, d) be an Euler form. Let k be a finite field. Then there exists a hereditary k-algebra Λ such that E(Λ) = (ω, d).

Proof. We denote by \overline{k} an algebraic closure of k. For any n ∈ ℕ_{1}, let F (n) be the subfield of \overline{k} with [F (n) : k] = n. (Of course, F (n) is the splitting field of the equation X^{q_{k}^{n}} − X, where q_{k} is the cardinality of k, thus F (n) is uniquely determined.) Note that for n = n_{1}n_{2}, we have F (n_{1}) ⊆ F (n) and [F (n) : F (n_{1})] = n_{2}.

In particular, if m = n_{1}n_{2}m′, where n_{1}, n_{2}, m′ are positive integers, then F (m) is an F (n_{1})-F (n_{2})-bimodule (using the field multiplication as scalar multiplications). For m = 0, let F (m) be the zero module. Then we have the general assertion: if m = n_{1}n_{2}m′, where n_{1}, n_{2} are positive integers and m′ is a non-negative integer, F (m) is an F (n_{1})-F (n_{2})-bimodule.

We define a k-species S = (F_{i},_{i}M_{j})_{i,j∈I} as follows: For i ≠ j, let e_{ij} = −ω(i, j); and let e_{ii} = −ω(i, i) + d_{i}. Let

F_{i} = F (d_{i}) and _{i}M_{j} = F (e_{ij}).

The tensor algebra T = T (S) is a hereditary k-algebra. For i ∈ I, we consider the corresponding simple T -module S(i) (here, S(i) is the representation of S which is given by the field F (i) at the vertex i, by zero at the remaining vertices, and by using only zero maps). In case T is infinite-dimensional, there will be additional simple T -modules, thus we have to choose for Λ a suitable localisation of T so that the modules S(i) are the only simple Λ-modules. We claim that E(Λ) = (ω, d).

Clearly, we have End(S(i)) = F (i), and we can identify

Ext^{1}(S(i), S(j)) = Hom_{k}(_{i}M_{j}, k).

This finishes the proof.

Remark. Our assumption on I to be finite, is not essential. However, in order to realize homologically bilinear forms on ℤI where I is infinite, it seems natural to consider hereditary length categories in general. Recall that a length category is an abelian category where all objects have finite length. Of course, given any ring Λ, the category of all Λ-modules of finite length is a length category.

5. The generic composition algebras.

For any Euler form (ω, d), we are going to construct a socalled generic composition algebra with coefficients in K_{0} = ℤ[q̃, q̃^{−1}], where q̃ is a variable.

Let K be a set of finite fields and assume that the order of these fields is not bounded. For any k ∈ K, there exists a k-algebra Λ_{k} such that E(Λ_{k}) = (ω, d). Note that the simple modules for these algebras are indexed by our fixed set I.

We consider the product

Π = ∏_{k∈K} H(Λ_{k}, ℤ_{k});

its elements will be written in the form a = (a^{(k)})_{k} with a^{(k)} ∈ H(Λ_{k}, ℤ_{k}); we call a^{(k)} the coefficient of a with index k. Consider the central element q̃ with q̃^{(k)} = q_{k} and its inverse q̃^{−1} (its coefficient with index (k) is q_{k}^{−1}). Note that in this way Π may be regarded as a ℤ[q̃, q̃^{−1}]-algebra. (Note that the ring homomorphism ℤ[q̃, q̃^{−1}] → Π which sends q̃ to q̃ is obviously injective; this has allowed us to denote the element (q_{k})_{k} of Π just by q̃.)

We define some additional elements in Π. For i ∈ I, let ũ_{i} = (ũ_{i}^{(k)})_{k} where ũ_{i}^{(k)} = u_{i} (the generator corresponding to the simple Λ_{k}-module with index i). We denote by C(ω, d) the subring of Π generated by the elements q̃, q̃^{−1} and ũ_{i} with i ∈ I. The ℤ[q̃, q̃^{−1}]-algebra C(ω, d) will be called a generic composition algebra of type (ω, d).

∗ ∗ ∗

In order to construct a twisted generic composition algebra, again we start with an Euler form (ω, d). Let K_{∗0} = ℤ[ṽ, ṽ^{−1}], where also ṽ is a variable. (Of course, one may ask why we distinguish the rings ℤ[ṽ, ṽ^{−1}] and ℤ[q̃, q̃^{−1}]: the reason is that we want to have q̃ = ṽ^{2}).

The twisted generic compositon algebra will be a K_{∗0}-algebra. Again, let K be a set of finite fields and assume that the order of these fields is not bounded, and for any k ∈ K, we take a k-algebra Λ_{k} such that E(Λ_{k}) = (ω, d).

We consider now the product

Π_{∗} = ∏_{k∈K} H_{∗}(Λ_{k});

(note that H_{∗}(Λ_{k}) = H_{∗}(Λ_{k}, ℤ_{∗k})). Consider this time the central element ṽ with ṽ^{(k)} = v_{k} and its inverse ṽ^{−1} (the coefficient with index (k) being v_{k}^{−1}). Note that in this way Π_{∗} may be regarded as a K_{∗0}-algebra.

As above, we define the elements ũ_{i} in Π_{∗}, for i ∈ I, thus ũ_{i} = (ũ_{i}^{(k)})_{k} where ũ_{i}^{(k)} = u_{i} is the generator corresponding to the simple Λ_{k}-module with index i. And we denote by C_{∗}(ω, d) the K_{∗0}-subalgebra of Π_{∗} generated by these elements ũ_{i} with i ∈ I. The K_{∗0}-algebra C_{∗}(ω, d) will be called a twisted generic composition algebra of type (ω, d).

Lemma. There is a K_{∗0}-algebra homomorphism

(C(ω, d) ⊗_{K_{0}} K_{∗0})_{[ṽ,ω]} ⟶ C_{∗}(ω, d)

which maps ũ_{i} ⊗ 1 to ũ_{i}.

Proof. On the one hand, consider for any k ∈ K the composition of the canonical ring homomorphisms

C(ω, d) ⟶ Π ⟶ H(Λ_{k}) ⟶ H(Λ_{k}) ⊗ ℤ_{∗k}

and also

K_{∗0} ⟶^{π_{k}} ℤ_{∗k} ⟶ H(Λ_{k}) ⊗ ℤ_{∗k},

where π_{k}(ṽ) = v_{k}. They obviously combine to a ring homomorphism

C(ω, d) ⊗_{ℤ} K_{∗0} ⟶ H(Λ_{k}) ⊗ ℤ_{∗k},

since the images of the two compositions commute. Since the images both of q̃⊗ 1 and 1 ⊗ q̃ are equal, namely q_{k} ⊗ 1 = 1 ⊗ q_{k}, we see that we can replace the tensor product sign ⊗_{ℤ} by ⊗_{K_{0}}.

According to Lemma II.1, the map

C(ω, d) ⊗_{K_{0}} K_{∗0} ⟶ H(Λ_{k}) ⊗ ℤ_{∗k}

may also be considered as an algebra homomorphism

(C(ω, d) ⊗_{K_{0}} K_{∗0})_{[ṽ,ω]} ⟶ (H(Λ_{k}) ⊗ ℤ_{∗k})_{[v_{k},ω]} = H_{∗}(Λ_{k}).

The family of these maps yields a corresponding map

(C(ω, d) ⊗_{K_{0}} K_{∗0})_{[ṽ,ω]} → ∏_{k∈K} H_{∗}(Λ_{k})

and its image is the K_{∗0}-algebra generated by the elements ũ_{i}, thus just C_{∗}(ω, d).

Part IV. Quantum Groups

The following are given: First of all, a set I and a pair χ = (χ′, χ′′) of bilinear forms on ℤI with values in ℤ. Second, let K be a commutative ring and let v ∈ K be an invertible element.

We will consider the free K-algebra F = K⟨I⟩ generated by the set I; the generator corresponding to i ∈ I will be denoted by θ_{i}. Note that F is a ℤI-graded algebra, with θ_{i} of degree i.

Depending on v and χ, we will define on F a comultiplication δ: F → F ⊗ F, so that (F, δ) is a ℤI-graded coalgebra. The definition follows Lusztig [L], and is arranged in such a way that F becomes a (K, v, χ)-bialgebra. All the results presented in this part are based on the book [L] of Lusztig.

1. Preliminaries on bilinear forms.

We are going to construct some (K, v, χ)-bialgebras. Before we do this, let us consider pairs of bilinear forms on a free abelian group. Thus, let us assume that there are given a set I and a pair χ = (χ′, χ′′) of bilinear forms on ℤI with values in ℤ.

Recall that χ: (ℤI)^{4} → ℤ is defined by

χ(x_{1}, x_{2}, x_{3}, x_{4}) = χ′(x_{1}, x_{4}) + χ′′(x_{2}, x_{3}),

and that we say that χ is given by the pair (χ′, χ′′).

In general, let χ: (ℤI)^{n} → ℤ be a (set) map. Given 1 ≤ r < s ≤ n, we denote by L(r, s) the following property: If x_{i}, y_{i}, z_{i} are integers such that x_{r} = y_{r} + z_{r}, x_{s} = y_{s} + z_{s}, and x_{i} = y_{i} = z_{i} for the remaining i, then

χ(x_{1}, …, x_{n}) = χ(y_{1}, …, y_{n}) + χ(z_{1}, …, z_{n}),

The maps χ: (ℤI)^{4} → ℤ which are given by pairs of bilinear forms on ℤI are characterized by the following conditions: L(1, 2), L(1, 3), L(2, 4), L(3, 4).

As an example, let us show that χ: (ℤI)^{4} → ℤ given by the pair (χ′, χ′′) satisfies the property L(1, 2):

χ(y_{1} + z_{1}, y_{2} + z_{2}, x_{3}, x_{4}) = χ′(y_{1} + z_{1}, x_{4}) + χ′′(y_{2} + z_{2}, x_{3})
= χ′(y_{1}, x_{4}) + χ′(z_{1}, x_{4}) + χ′′(y_{2}, x_{3}) + χ′′(z_{2}, x_{3})
= χ(y_{1}, y_{2}, x_{3}, x_{4}) + χ(z_{1}, z_{2}, x_{3}, x_{4})

Similarly, χ also satisfies the conditions L(1, 3), L(2, 4), L(3, 4). Conversely, assume that χ: (ℤI)^{4} → ℤ satisfies the properties L(1, 2), L(1, 3), L(2, 4), L(3, 4). Then L(1, 2) and L(3, 4) yield

χ(x_{1}, x_{2}, x_{3}, x_{4}) = χ(x_{1}, 0, x_{3}, x_{4}) + χ(0, x_{2}, x_{3}, x_{4})

= χ(x_{1}, 0, x_{3}, 0) + χ(x_{1}, 0, 0, x_{4}) + χ(0, x_{2}, x_{3}, 0) + χ(0, x_{2}, 0, x_{4}); using L(2, 4) we see that χ(x_{1}, 0, x_{3}, 0) = χ(x_{1}, 0, x_{3}, 0) + χ(x_{1}, 0, x_{3}, 0), and therefore χ(x_{1}, 0, x_{3}, 0) = 0; similarly, we see that L(1, 3) yields χ(0, x_{2}, 0, x_{4}) = 0. As a consequence,

χ(x_{1}, x_{2}, x_{3}, x_{4}) = χ(x_{1}, 0, 0, x_{4}) + χ(0, x_{2}, x_{3}, 0),

and clearly the two terms on the right are bilinear forms on ℤI.

The conditions L(1, 2) and L(3, 4) just mean that χ is a bilinear form on (ℤI)^{2}. Thus: If χ: (ℤI)^{4} → ℤ is given by pairs of bilinear forms on ℤI, then χ is a bilinear form on (ℤI)^{2}.

If χ is a bilinear form on (ℤI)^{2}, then χ̃ defined by

χ̃(x_{1}, x_{2}, x_{3}; y_{1}, y_{2}, y_{3}) = χ(x_{1}, x_{2}; y_{1}, y_{2}) + χ(x_{1}, x_{3}; y_{1}, y_{3}) + χ(x_{2}, x_{3}; y_{2}, y_{3})

for x_{1}, x_{2}, x_{3}, y_{1}, y_{2}, y_{3} ∈ ℤI is a bilinear form on (ℤI)^{3}. Again, the proof is straightforward.

We extend a previous convention as follows: Given any ℤI-graded algebra A, a (set) map χ: (ℤI)^{n} → ℤ, and elements a_{1}, …, a_{n} such that a_{i} is homogeneous of degree x_{i}, then we write χ(a_{1}, …, a_{n}) = χ(x_{1}, …, x_{n}), in case this is welldefined and not confusing. Sometimes it will be helpful to denote the degree of a homogeneous element a ∈ A by |a|, thus |a| ∈ ℤI and a ∈ A_{|a|} (of course, |a| is well-defined only in case a ≠ 0).

2. The free K-algebra on I as a (K, v, χ)-bialgebra.

Recall that a fixed set I is given. We denote by ⟨I⟩ the free semigroup (with 1) generated by I, the generator corresponding to i ∈ I will be denoted by θ_{i}. Thus, the elements of ⟨I⟩ are words in the letters θ_{i} (i ∈ I): there is the empty word which is denoted by 1, and there are the words θ_{i_{1}}θ_{i_{2}} ⋯ θ_{i_{n}} of length n ≥ 1, with i_{1}, i_{2}, …, i_{n} ∈ I. The multiplication in ⟨I⟩ is just the concatenation of words. The free K-algebra F = K⟨I⟩ generated by I is just the semigroup algebra of ⟨I⟩ over K; it is the free K-module with basis ⟨I⟩, and the multiplication of these base elements is the concatenation of words.

We consider F = K⟨I⟩ as a ℤI-graded algebra, with the generator θ_{i} being of degree i. We denote by ϵ: F → F_{0} = K the canonical projection.

In order to define the comultiplication δ, we consider the algebra (F ⊗ F)_{[v,χ]}; recall that it is given by the tensor product F ⊗ F with the multiplication

(a_{1} ⊗ a_{2}) ∗ (b_{1} ⊗ b_{2}) = v^{χ(a_{1},a_{2},b_{1},b_{2})}a_{1}b_{1} ⊗ a_{2}b_{2}.

As we have seen above, this is an associative multiplication, since χ is a bilinear form on (ℤI)^{2}.

Let δ be the algebra homomorphism F → (F ⊗ F)_{[v,χ]} defined by δ(θ_{i}) = θ_{i} ⊗ 1 + 1 ⊗ θ_{i}.

Lemma. With this comultiplication δ the K algebra F = K⟨I⟩ becomes a (K, v, χ)-bialgebra; its counit is ϵ.

We have to verify the following: first, the comultiplication is coassociative; and secondly, ϵ is the corresponding counit; after these verifications, it will be trivially true that we have obtained a (K, v, χ)-bialgebra.

Recall that the bilinear form χ on (ℤI)^{2} gives rise to a corresponding bilinear form χ̃ on (ℤI)^{3}. Thus, given a ℤI-graded K-algebra, we may consider the (ℤI)^{2}- graded K-algebra A ⊗ A (with componentwise multiplication (a_{1} ⊗ a_{2})(b_{1} ⊗ b_{2}) = a_{1}b_{1} ⊗ a_{2}b_{2}) and its twisted version (A ⊗ A)_{[v,χ]} (as considered in the last chapter), as well as the (ℤI)^{3}-graded algebra A ⊗ A ⊗ A (again with componentwise multiplication) and its twisted version (A ⊗ A ⊗ A)_{[v,χ̃]}. Of course, all these algebras A ⊗ A, (A ⊗ A)_{[v,χ]}, A ⊗ A ⊗ A and (A ⊗ A ⊗ A)_{[v,χ̃]} can be considered also as ℤI-graded algebras using the “total degree”: For example, for z ∈ ℤI, we let

(A ⊗ A)_{z} = ⨁_{x+y=z} A_{x} ⊗ A_{y}.

Lemma. Let A be a ℤI-graded K-algebra. Let δ: A → (A ⊗ A)_{[v,χ]} be a homomorphism of ℤI-graded algebras. Then also 1 ⊗ δ and δ ⊗ 1 are algebra homomorphisms (A ⊗ A)_{[v,χ]} → (A ⊗ A ⊗ A)_{[v,χ̃]}.

Proof: Consider elements a_{1}, a_{2}, b_{1}, b_{2} ∈ A. Let
δ(a_{2}) = ∑ a_{21} ⊗ a_{22},
δ(b_{2}) = ∑ b_{21} ⊗ b_{22},

thus

(1 ⊗ δ)(a_{1} ⊗ a_{2}) = ∑ a_{1} ⊗ a_{21} ⊗ a_{22},
(1 ⊗ δ)(b_{1} ⊗ b_{2}) = ∑ b_{1} ⊗ b_{21} ⊗ b_{22}.

The product of these two elements is

(1⊗δ)(a_{1} ⊗ a_{2}) · (1 ⊗ δ)(b_{1} ⊗ b_{2})

= ∑ v^{χ(a_{1},a_{21},b_{1},b_{21})+χ(a_{1},a_{22},b_{1},b_{22})+χ(a_{21},a_{22},b_{21},b_{22})}a_{1}b_{1} ⊗ a_{21}b_{21} ⊗ a_{22}b_{22}.

On the other hand

(1 ⊗ δ)((a_{1} ⊗ a_{2})(b_{1} ⊗ b_{2}))
= (1 ⊗ δ)v^{χ(a_{1},a_{2},b_{1},b_{2})}(a_{1}b_{1} ⊗ a_{2}b_{2})
= v^{χ(a_{1},a_{2},b_{1},b_{2})}(a_{1}b_{1} ⊗ ∑(a_{21} ⊗ a_{22})(b_{21} ⊗ b_{22}))
= ∑ v^{χ(a_{1},a_{2},b_{1},b_{2})+χ(a_{21},a_{22},b_{21},b_{22})}a_{1}b_{1} ⊗ a_{21}b_{21} ⊗ a_{22}b_{22}.

We have to show that the two exponents of v are equal. Note that the last summands of the exponents are equal, thus it remains to see that

χ(a_{1}, a_{2}, b_{1}, b_{2}) = χ(a_{1}, a_{21}, b_{1}, b_{21}) + χ(a_{1}, a_{22}, b_{1}, b_{22}),

but this is just the linearity condition L(2, 4).

This shows that 1 ⊗ δ is multiplicative; similarly, one shows that δ ⊗ 1 is multiplicative.

Proof of Lemma: The maps (δ ⊗ 1)δ, (1 ⊗ δ)δ from F to (F ⊗ F ⊗ F)_{[v,χ̃]} are algebra homomorphisms and they coincide on the generators θ_{i} (both send θ_{i} to θ_{i} ⊗1⊗1+1⊗θ_{i} ⊗1+1⊗1⊗θ_{i}), thus they are equal. This shows the coassociativity.

Next, we show that ϵ is a counit. One easily checks that ϵ ⊗ 1 and 1 ⊗ ϵ are algebra homomorphisms (F ⊗ F)_{[v,χ]} → F since F_{x} ≠ 0 only for x ≥ 0. It follows that the algebra maps (ϵ ⊗ 1)δ and (1 ⊗ ϵ)δ both are equal to the identity, since they map any θ_{i} to itself.

3. Lusztig forms on F.

Given a graded ℤI-module A = ⨁_{x∈ℤI} A_{x}, a bilinear form (−, −) : A⊗A → K is said to respect the grading provided we have (a, b) = 0 for a ∈ A_{x}, b ∈ A_{y} and x ≠ y in ℤI.

Proposition. Assume that for any i ∈ I, there is given an element t_{i} ∈ K. There exists a unique bilinear form (−, −)_{t} on the (K, v, χ)-bialgebra F = K⟨I⟩ with the following properties:

(0) The bilinear form (−, −)_{t} respects the grading.

(1) We have (θ_{i}, θ_{i})_{t} = t_{i} for all i ∈ I,

(2) For δ(a) = ∑ a_{1} ⊗ a_{2}, we have

(a, bc)_{t} = ∑(a_{1}, b)_{t}(a_{2}, c)_{t}.

We write just:

(a, bc)_{t} = (δ(a), b ⊗ c)_{t},

where we introduce a corresponding bilinear form on F ⊗ F which works componentwise and we denote it again by (−, −)_{t}. Thus, for a_{1}, a_{2}, b_{1}, b_{2} ∈ F, we have (a_{1} ⊗ a_{2}, b_{1} ⊗ b_{2})_{t} = (a_{1}, b_{1})_{t}(a_{2}, b_{2})_{t}.

This result has been shown (at least for special pairs χ) by Lusztig [L]. We call (−, −)_{t} the Lusztig form on F = K⟨I⟩ with respect to v, χ and t.

Proof of the existence:

The algebra F = ⨁_{x∈ℤI} F_{x} is a graded algebra, and the parts F_{x} are finitely generated free k-modules. We may consider the corresponding dual k-modules F_{x}^{∗} and we define

F^{∗} = ⨁_{x∈ℤI} F_{x}^{∗},

with multiplication given by δ and with comultiplication given by the multiplication of F. Let ϕ_{i}, i ∈ I be the dual basis (inside ⨁_{i∈I} F_{i}^{∗}) for the generating set θ_{i}, i ∈ I. Given an element ϕ ∈ F_{x}^{∗}, we denote the image of a ∈ F_{x} under ϕ by (a)ϕ.

Remark. The elements ϕ_{i}, i ∈ I usually will not form a generating set of F^{∗}; in contrast, one should be interested in the subalgebra generated by the elements ϕ_{i}. The algebra F^{∗} is called the shuffle algebra for I by J.A.Green, see [G2].

Consider the algebra homomorphism

α: F → F^{∗} defined by (θ_{i})α = ϕ_{i}t_{i}.

Define

(a, b)_{t} = (a)(b)α

(this means: apply α to b, this gives a linear form (b)α, and now apply this linear form to a; in case a, b are homogeneous and have different degrees, then this is by definition 0.)

Clearly, (−, −)_{t} is a bilinear form, with the following properties: if a, b are homogeneous, and (a, b)_{t} ≠ 0, then a, b have the same degree, thus (−, −)_{t} respects the grading of F. Also, we have (θ_{i}, θ_{i})_{t} = (θ_{i})ϕ_{i}t_{i} = t_{i}, thus also condition (1) is satisfied. In addition, we have the following equality:

(a, bc)_{t} = a(bc)α = a(bα ∗ cα)

where ∗ is the multiplication in F^{∗} (let us recall how the multiplication in F^{∗} is defined: let ϕ, ψ belong to F^{∗}. Then ϕ ∗ ψ maps a ∈ F onto δ(a)(ϕ ⊗ ψ), where δ is the comultiplication of F.) Now assume that δ(a) = ∑ a_{1} ⊗ a_{2}. We can continue our calculation

a(bα ∗ cα) = ∑(a_{1})(b)α · (a_{2})(c)α = ∑(a_{1}, b)_{t}(a_{2}, c)_{t}.

This shows that also condition (2) is satisfied.

The proof of the unicity will use the following Lemma:

Lemma. Assume that A is a ℤI-graded k-algebra. Let δ: A → (A ⊗ A)_{[v,χ]} be a homomorphism of ℤI-graded algebras. Assume that there exists a bilinear form (−, −) on A with values in k such that

(a, bc) = (δ(a), b ⊗ c)

for all a, b, c ∈ A.

Let a, a′ be homogeneous elements of A and let

δ(a) = ∑ a_{1} ⊗ a_{2}, δ(a′) = ∑ a′_{1} ⊗ a′_{2}.

Then, given b, b′ ∈ A, we have

(aa′, bb′) = ∑ v^{χ(a_{1},a_{2},a′_{1},a′_{2})}(a_{1}a′_{1}, b)(a_{2}a′_{2}, b′).

Proof: Since δ preserves the ℤI-grading, the various elements a_{1}, a_{2}, a′_{1}, a′_{2} all are homogeneous, again. Since δ is multiplicative,

δ(aa′) = δ(a)δ(a′)
= ∑(a_{1} ⊗ a_{2}) · (a′_{1} ⊗ a′_{2})
= ∑ v^{χ(a_{1},a_{2},a′_{1},a′_{2})}a_{1}a′_{1} ⊗ a_{2}a′_{2},

and therefore

(aa′, bb′) = (δ(aa′), b ⊗ b′)
= ∑ v^{χ(a_{1},a_{2},a′_{1},a′_{2})}(a_{1}a′_{1} ⊗ a_{2}a′_{2}, b ⊗ b′)
= ∑ v^{χ(a_{1},a_{2},a′_{1},a′_{2})}(a_{1}a′_{1}, b)(a_{2}a′_{2}, b′).

Proof of the unicity of the form (−, −)_{t}. Let (−, −) and (−, −)′ be two bilinear forms with the properties (0), (1), (2). If a, b belong to A_{0} ⊕ ⨁_{i∈I} A_{i}, then by assumption (a, b) = (a, b)′. Consider now A_{x} where x = (x_{i})_{i} belongs to ℤI, all x_{i} ≥ 0, and ∑ x_{i} ≥ 2. We show that the forms (−, −) and (−, −)′ agree on A_{x}. Let a, a′, b, b′ be homogeneous of non-zero degree such that aa′ and bb′ belong to A_{x}. Let

δ(a) = ∑ a_{1} ⊗ a_{2}, δ(a′) = ∑ a′_{1} ⊗ a′_{2}.

The Lemma asserts that

(aa′, bb′) = ∑ v^{χ(a_{1},a_{2},a′_{1},a′_{2})}(a_{1}a′_{1}, b)(a_{2}a′_{2}, b′);

and similarly

(aa′, bb′)′ = ∑ v^{χ(a_{1},a_{2},a′_{1},a′_{2})}(a_{1}a′_{1}, b)′(a_{2}a′_{2}, b′)′.

By induction, we know that

(a_{1}a′_{1}, b) = (a_{1}a′_{1}, b)′ and
(a_{2}a′_{2}, b′) = (a_{2}a′_{2}, b′)′.

Thus we see that (aa′, bb′) = (aa′, bb′)′.

4. Scalar extension.

Let η: K → K′ be a homomorphism of commutative rings. Let v be invertible in K and η(v) = v′. Let (−, −)_{1} be the bilinear form on K⟨I⟩ for v, χ and 1 = (1)_{i}. Let (−, −)′_{t} be the bilinear form on K′⟨I⟩ for v′, χ and t = (t_{i})_{i}. If x = (x_{i})_{i} belongs to ℤI (thus x_{i} ∈ ℤ, and almost all of these coefficients are zero), then we put t^{x} = ∏_{i∈I} t_{i}^{x_{i}}.

We define η: K⟨I⟩ → K′⟨I⟩ by η(v) = v′ and η(θ_{i}) = θ_{i}.

Corollary 1. Let x ∈ ℤI. Then, for all a, b ∈ K⟨I⟩_{x}, we have

(η(a), η(b))′_{t} = η((a, b)_{1}) · t^{x}.

Proof: Let F = K⟨I⟩ and F′ = K′⟨I⟩. First of all, we note that

η ⊗ η: (F ⊗_{K} F )_{[v,χ]} → (F′ ⊗_{K′} F′)_{[v′,χ]}

is an algebra homomorphism. It follows that

δ′η = (η ⊗ η)δ,

since these two maps coincide on the generators θ_{i}.

Since F′ is the free K′-module with basis ⟨I⟩, we may define a K′-bilinear form (−, −)′′ on F′ by just specifying the values on pairs from ⟨I⟩. For x ∈ ℤI, we denote by ⟨I⟩_{x} the set of words of degree x. Recall that the monomial θ_{i_{1}}θ_{i_{2}} … θ_{i_{n}} has degree x = (x_{i})_{i} if and only if the element i occurs precisely x_{i} times in the sequence (i_{1}, i_{2}, …, i_{n}).

The set ⟨I⟩ will be considered as a K-basis of F; if θ is an element of ⟨I⟩, then the corresponding element of F′ will be denoted by η(θ). Of course, the element η(θ) is just the “same” element as θ, but, in this way, we distinguish the basis elements of F and F′.

If θ, θ′ are monomials of degree x for some x ∈ ℤI, let

(η(θ), η(θ′))′′ = η((θ, θ′)_{1}) · t^{x}

(note that the last product has only finitely many factors different from 1); for x ≠ y in ℤI and θ ∈ ⟨I⟩_{x}, θ′ ∈ ⟨I⟩_{y}, we define (η(θ), η(θ′))′′ = 0. We claim that (−, −)′′ = (−, −)′_{t}.

First of all, the bilinear form (−, −)′′ respects the grading. Second, we clearly have (η(θ_{i}), η(θ_{i}))′′ = t_{i}. Thus it remains to consider the adjunction property (a, bc)′′ = (δ′(a), b ⊗ c)′′, for a, b, c ∈ F′ (note that we will denote the comultiplication of F′ by δ′). It is sufficient to verify this equality for a, b, c being monomials. Let |a| = x, |b| = y, |c| = z. We can assume that x = y + z. Let δ(a) = ∑ a_{1} ⊗ a_{2}. Then δ′η(a) = ∑ η(a_{1}) ⊗ η(a_{2}). We see:

(δ′η(a), η(b) ⊗ η(c))′′· = ∑(η(a_{1}), η(b))′′(η(a_{2}), η(c))′′
= ∑ η((a_{1}, b)_{1})t^{y} · η((a_{2}, c)_{1})t^{z}
= η(∑(a_{1}, b)_{1}(a_{2}, c)_{1}) · t^{x}
= η((a_{1}, b)_{1}(a_{2}, c)_{1}) · t^{x}
= η((a, bc)_{1}) · t^{x}
= (η(a), η(b)η(c))′′.

Since (−, −)′′ satisfies all the defining properties for (−, −)_{t}, we conclude that these forms are equal.

There are two special cases which should be mentioned.

Corollary 2. Let x ∈ ℤI. The values on F_{x} of the Lusztig form (−, −)_{t} with respect to v, χ and t are given by

(a, b)_{t} = (a, b)_{1} · t^{x}.

Of particular interest is (as J.A.Green has pointed out) the case K_{0} = ℤ[q̃, q̃^{−1}], where q̃ is a variable; the corresponding Lusztig form on K_{0}⟨I⟩ with respect to q̃, χ and 1 will be called the generic Lusztig form for χ.

Corollary 3. Let K_{0} = ℤ[q̃, q̃^{−1}]. Let (−, −)_{1} be the generic Lusztig form on K_{0}⟨I⟩. Let η: K_{0} → K be a ring homomorphism, and let (−, −)′_{1} the Lusztig form on K⟨I⟩ with respect to η(q̃), χ and 1. Then we have for a, b ∈ K_{0}⟨I⟩

(η(a), η(b))′_{1} = (a, b)_{1}(η(q̃)).

Note that (a, b)_{1}(η(q̃)) means the following: (a, b)_{1} belongs to ℤ[q̃, q̃^{−1}], is thus a Laurent polynomial in the variable q̃ and we insert η(q̃) ∈ K into this Laurent polynomial.

5. The definition of U^{+}(K, v, χ).

As before, let K be a commutative ring, v ∈ K an invertible element. Let I be a set and χ = (χ′, χ′′) a pair of bilinear forms on ℤI.

Let (−, −) = (−, −)_{1} be the Lusztig form on F = K⟨I⟩ with respect to v, χ and t_{i} = 1 for all i ∈ I. Let I(K, v, χ) be the set of elements b ∈ F such that (a, b) = 0 for all a ∈ F.

Lemma. I(K, v, χ) is an ideal of F.

Proof. Clearly, I(K, v, χ) is a K-submodule. Let b ∈ I(K, v, χ) and c ∈ F. Then, for a ∈ F with δ(a) = ∑ a_{1} ⊗ a_{2} we have

(a, bc) = (δ(a), b ⊗ c) = ∑(a_{1}, b)(a_{2}, c) = 0,
(a, cb) = (δ(a), c ⊗ b) = ∑(a_{1}, c)(a_{2}, b) = 0.

Definition: Let

U^{+}(K, v, χ) = F/I(K, v, χ).

For special values of χ, we obtain in this way just the positive part of a corresponding “quantum group”, see Lusztig [L]. Typical cases will be considered in the last section.

Proposition 1. Let A be a positive (K, v, χ)-bialgebra and assume that A is generated by elements a_{i} ∈ A_{i}, with i ∈ I. In addition, let (−, −) be a bilinear form on A which respects the grading and such that (a, bc) = (δ(a), b ⊗ c) for all a, b, c ∈ A, and let us assume that no element t_{i} = (a_{i}, a_{i}) is a zero-divisor in K.

Then the kernel of the K-algebra homomorphism ξ: K⟨I⟩ → A with ξ(θ_{i}) = a_{i} is contained in the ideal I(K, v, χ) and thus ξ induces an algebra homomorphism \overline{ξ}: A → U^{+}(K, v, χ).

Proof: Write F = K⟨I⟩. According to section II.1, we know that ξ: F → A is also a coalgebra homomorphism. We define a bilinear form (−, −)′ on F by

(a, b)′ = (ξ(a), ξ(b)) for a, b ∈ F.

This bilinear form respects the grading of F, and we have

(a, bc)′ = (ξ(a), ξ(bc))
= (ξ(a), ξ(b)ξ(c))
= (δξ(a), ξ(b) ⊗ ξ(c))
= (ξδ(a), ξ(b) ⊗ ξ(c))
= (δ(a), b ⊗ c)′.

The unicity of the Lusztig form on F with respect to v, χ and t implies that (−, −)′ = (−, −)_{t}.

We want to show that Ker ξ ⊆ I(K, v, χ). Since ξ respects the grading, it is sufficient to consider b ∈ F_{x} ∩ Ker ξ. Let a ∈ F_{x}. Then

(a, b)_{1} · t^{x} = (a, b)_{t} = (a, b)′ = (ξ(a), ξ(b)) = 0.

This shows that b belongs to I(K, v, χ).

∗ ∗ ∗

In general, the Lusztig forms on F will not be symmetric. We show the following:

Proposition 2. If χ′′ is symmetric, then the Lusztig form (−, −)_{1} on F is also symmetric.

Proof: We assume that χ′′ is symmetric and we show that (−, −)_{1} also satisfies

(aa′, c)_{1} = (a ⊗ a′, δ(c))_{1}

for all a, a′, c ∈ A.

This equality is easily verified in case c belongs to F_{0} ⊕ ⨁_{i∈I} F_{i}, thus, it is sufficient to consider the case where c = bb′ with homogeneous elements b, b′ of non-zero degree. Write

δ(a) = ∑ a_{1} ⊗ a_{2},  δ(a′) = ∑ a′_{1} ⊗ a′_{2},
δ(b) = ∑ b_{1} ⊗ b_{2},  δ(b′) = ∑ b′_{1} ⊗ b′_{2}.

Using the Lemma and induction, we have

(aa′, bb′)_{1} = ∑ v^{χ(a_{1},a_{2},a′_{1},a′_{2})}(a_{1}a′_{1}, b)_{1}(a_{2}a′_{2}, b′)_{1}
= ∑ v^{χ(a_{1},a_{2},a′_{1},a′_{2})}(a_{1} ⊗ a′_{1}, δ(b))_{1}(a_{2} ⊗ a′_{2}, δ(b′))_{1}
= ∑ v^{χ(a_{1},a_{2},a′_{1},a′_{2})}(a_{1} ⊗ a′_{1}, b_{1} ⊗ b_{2})_{1}(a_{2} ⊗ a′_{2}, b′_{1} ⊗ b′_{2})_{1}
= ∑ v^{χ(a_{1},a_{2},a′_{1},a′_{2})}(a_{1}, b_{1})_{1}(a′_{1}, b_{2})_{1}(a_{2}, b′_{1})_{1}(a′_{2}, b′_{2})_{1}.

On the other hand, we have

δ(bb′) = δ(b)δ(b′)
= ∑(b_{1} ⊗ b_{2}) · (b′_{1} ⊗ b′_{2})
= ∑ v^{χ(b_{1},b_{2},b′_{1},b′_{2})}b_{1}b′_{1} ⊗ b_{2}b′_{2},

and therefore

(a ⊗ a′, δ(bb′))_{1} = ∑ v^{χ(b_{1},b_{2},b′_{1},b′_{2})}(a ⊗ a′, b_{1}b′_{1} ⊗ b_{2}b′_{2})_{1}
= ∑ v^{χ(b_{1},b_{2},b′_{1},b′_{2})}(a, b_{1}b′_{1})_{1}(a′, b_{2}b′_{2})_{1}
= ∑ v^{χ(b_{1},b_{2},b′_{1},b′_{2})}(δ(a), b_{1} ⊗ b′_{1})_{1}(δ(a′), b_{2} ⊗ b′_{2})_{1}
= ∑ v^{χ(b_{1},b_{2},b′_{1},b′_{2})}(a_{1} ⊗ a_{2}, b_{1} ⊗ b′_{1})_{1}(a′_{1} ⊗ a′_{2}, b_{2} ⊗ b′_{2})_{1}
= ∑ v^{χ(b_{1},b_{2},b′_{1},b′_{2})}(a_{1}, b_{1})_{1}(a_{2}, b′_{1})_{1}(a′_{1}, b_{2})_{1}(a′_{2}, b′_{2})_{1}
= ∑ v^{χ(b_{1},b_{2},b′_{1},b′_{2})}(a_{1}, b_{1})_{1}(a′_{1}, b_{2})_{1}(a_{2}, b′_{1})_{1}(a′_{2}, b′_{2})_{1}.

We claim that for non-zero (a_{1}, b_{1})_{1}(a′_{1}, b_{2})_{1}(a_{2}, b′_{1})_{1}(a′_{2}, b′_{2})_{1}, the coefficients

v^{χ(a_{1},a_{2},a′_{1},a′_{2})} and v^{χ(b_{1},b_{2},b′_{1},b′_{2})}

coincide. However (a_{1}, b_{1})_{1}(a′_{1}, b_{2})_{1}(a_{2}, b′_{1})_{1}(a′_{2}, b′_{2})_{1} ≠ 0 implies that

|a_{1}| = |b_{1}|, |a′_{1}| = |b_{2}|, |a_{2}| = |b′_{1}|, |a′_{2}| = |b′_{2}|

and by assumption

χ(a_{1}, a_{2}, a′_{1}, a′_{2}) = χ(b_{1}, b′_{1}, b_{2}, b′_{2}) = χ(b_{1}, b_{2}, b′_{1}, b′_{2}).

We see that also the form (−, −)′ defined by (a, b)′ = (b, a)_{1} satisfies the conditions (0), (1) (2) of a Lusztig form, thus (−, −)′ = (−, −)_{1}, thus we have (a, b)_{1} = (b, a)_{1} for all a, b.

Part V. The Isomorphism

Let • be a Cartan datum which is defined on ℤI. We can consider it as the symmetrization of an Euler form ω without short cycles. In Part III, we have constructed to (ω, d) generic composition algebras and twisted generic composition algebras. In Part IV, we have seen that given any pair of bilinear forms on ℤI, for example (−ω, 0) or (0, •), one may endow the corresponding free algebra K⟨I⟩ with a symmetric bilinear form whose radical is an ideal, and we have denoted the corresponding factor algebra by U^{+}. The aim of this final part is to show that under some constraints the same algebras will have been constructed in these two different ways.

Let ℚ(ṽ) be the field of rational functions in one variable ṽ. Let q̃ = ṽ^{2}. As before, we consider the following subrings of ℚ(ṽ)

K_{0} = ℤ[q̃, q̃^{−1}] and K_{∗0} = ℤ[ṽ, ṽ^{−1}].

1. Quantum Serre relations.

In Part III, the fundamental relations for H(Λ) and H_{∗}(Λ) have been exhibited. These relations only involve the generators of the corresponding composition algebras, thus, we obtain in this way relations which are valid for the generators ũ_{i} of the generic composition algebra and the twisted generic composition algebra.

Let ω be an Euler form without short cycles. (Thus, ω: ℤI × ℤI → ℤ is a bilinear form, d_{i} = ω(i, i) is positive and divides both ω(i, j) and ω(j, i) and, for i ≠ j, we have ω(i, j)ω(j, i) = 0.) We define for i ≠ j

n(i, j) = \{ \begin{matrix}−d_{i}^{−1}ω(i, j) + 1 & if ω(j, i) = 0, \\ −d_{i}^{−1}ω(j, i) + 1 & if ω(i, j) = 0, \end{matrix}

Let • be the symmetrization of ω. Then we have

n(i, j) = −d_{i}^{−1} i•j + 1,

thus n(i, j) only depends on •.

First, let us consider the case of the generic composition algebra C(ω). By definition, C(ω) may be considered as a factor algebra of the free algebra K_{0}⟨I⟩.

We consider a general commutative ring K with a fixed invertible element q ∈ K as a K_{0}-algebra, with q̃ acting on K by multiplying with q. Then also the free algebra K⟨I⟩ becomes a K_{0}-algebra and we may consider the ideal J (K, q, ω) in K⟨I⟩ generated by the elements

∑_{t=0}^{n(i,j)}(−1)^{t}q^{d_{i}\binom{t}{2}} | \begin{matrix}n(i, j) \\ t \end{matrix} ] (q^{d_{i}}) θ_{i}^{n(i,j)−t}θ_{j}θ_{i}^{t},

∑_{t=0}^{n(j,i)}(−1)^{t}q^{d_{j} \binom{t}{2}} | \begin{matrix}n(j, i) \\ t \end{matrix} ] (q^{d_{j}}) θ_{j}^{t}θ_{i}θ_{j}^{n(j,i)−t},

for all pairs i ≠ j with ω(j, i) = 0. We recall that the coefficients | \begin{matrix}n(i, j) \\ t \end{matrix} ] (q^{d_{i}})

belong to K_{0} (they even belong to ℤ[q̃]), according to III.1. The generating elements of J (K, q, ω) will be said to be quantum Serre relations. We define

Û^{+}(K, q, ω) = K⟨I⟩/J (K, q, ω).

Remark. The quantum Serre relations exhibited above are certain homogeneous elements in the free algebra K_{0}⟨I⟩, they are linear combinations of monomials of the form θ_{i}^{t′}θ_{j}θ_{i}^{t} with fixed t + t′, and the coefficients are in K_{0}, thus Laurent polynomials in the variable q̃. If we evaluate q̃ at 1, we obtain usual Serre relations as they are used in order to obtain the Serre presentation of the positive part of a semisimple complex Lie algebra or, more generally, the positive part of a Kac-Moody Lie algebra.

We see the following: The kernel of the canonical projection

K⟨I⟩ ⟶ C(ω) ⊗_{K_{0}} K

which sends θ_{i} to ũ_{i} ⊗ 1 contains the twosided ideal J (K, q, ω). Thus:

Proposition. Let ω be an Euler form without short cycles. Let K be a commutative ring and q ∈ K. Then there exists a homomorphism of K-algebras

π^{K}: Û^{+}(K, q, ω) → C(ω) ⊗_{K_{0}} K

which maps the residue class of θ_{i} onto ũ_{i} ⊗ 1.

∗ ∗ ∗

Assume now that there is given an invertible element v in K, so that we may consider K as a K_{∗0}-algebra, with ṽ operating on K by the multiplication with v.

We denote by J_{∗}(K, v, •) the twosided ideal of K⟨I⟩ generated by the elements

∑_{t=0}^{n(i,j)}(−1)^{t} [ \begin{matrix}n(i, j) \\ t \end{matrix} ]_{d_{i}} θ_{i}^{n(i,j)−t}θ_{j}θ_{i}^{t}

for all i ≠ j. Again, we call these generators quantum Serre relations and we put

Û_{∗}^{+}(K, v, •) = K⟨I⟩/J_{∗}(K, v, •).

Remark. The second kind of quantum Serre relations which have been exhibited as generators of J_{∗}(K, v, •) are homogeneous elements in the free algebra K_{∗0}⟨I⟩, thus linear combinations of monomials of the form θ_{i}^{t′}θ_{j}θ_{i}^{t} with fixed t + t′, and the coefficients are in K_{∗0}, thus Laurent polynomials in the variable ṽ. Again, the evaluation of ṽ at 1 yields usual Serre relations.

Proposition*. Let ω be an Euler form without short cycles. Let • be its symmetrization. Let K be a commutative ring and v ∈ K invertible. There exists a homomorphism of K-algebras

π_{∗}^{K}: Û_{∗}^{+}(K, v, •) → C_{∗}(ω) ⊗_{K_{∗0}} K

which maps the residue class of θ_{i} onto ũ_{i} ⊗ 1.

Again, this follows from part III: The kernel of the canonical map

K⟨I⟩ → C_{∗}(ω) ⊗_{K_{∗}0} K

which sends θ_{i} to ũ_{i} ⊗ 1 contains the twosided ideal J_{∗}(K, v, •).

∗ ∗ ∗

Finally, let us compare the two algebras Û^{+}(K, v^{2}, ω) and Û_{∗}^{+}(K, v, •).

Lemma. Let ω be an Euler form without short cycles. Let • be its symmetrization. Let K be a commutative ring and v ∈ K invertible. Then there exists an isomorphism of K-algebras

Û_{∗}^{+}(K, v, •) → Û^{+}(K, v^{2}, ω)_{[v,ω]}

which maps the residue class of θ_{i} modulo J_{∗}(K, v, •) to the residue class of the same element θ_{i} modulo J (K, v^{2}, ω).

Proof: We consider the K-algebra A = K⟨I⟩_{[v,ω]}. Since K⟨I⟩ is a free algebra, there exists an algebra homomorphism

η: K⟨I⟩ ⟶ A

with η(θ_{i}) = θ_{i}. Since ω has no short cycles, we can assume that ω(j, i) = 0, therefore n(i, j) = −d_{i}^{−1}ω(i, j) + 1, and thus ω(i, j) = −d_{i}(n(i, j) − 1). We write n = n(i, j). Let us calculate η(θ_{i}^{n−t}θ_{j}θ_{i}^{t}). We have

η(θ_{i}^{n−t}θ_{j}θ_{i}^{t}) = v^{\binom{n}{2}ω(i,i)+(n−t)ω(i,j)} θ_{i}^{n−t}θ_{j}θ_{i}^{t}
= v^{\binom{n}{2}ω(i,i)−(n−t)d_{i}(n−1)} θ_{i}^{n−t}θ_{j}θ_{i}^{t}
= v^{\binom{n}{2}ω(i,i)−nd_{i}(n−1)}v^{d_{i}t(n−1)} θ_{i}^{n−t}θ_{j}θ_{i}^{t}.

It follows that the quantum Serre relation

∑_{t=0}^{n}(−1)^{t} [ \begin{matrix}n \\ t \end{matrix} ]_{d_{i}} θ_{i}^{n−t}θ_{j}θ_{i}^{t}

is mapped unter η onto a multiple of

∑_{t=0}^{n}(−1)^{t}([ \begin{matrix}n \\ t \end{matrix} ] v^{t(n−1)})_{d_{i}}θ_{i}^{n−t}θ_{j}θ_{i}^{t} = ∑_{t=0}^{n}(−1)^{t}(| \begin{matrix}n \\ t \end{matrix} ] v^{t(t−1)})_{d_{i}}θ_{i}^{n−t}θ_{j}θ_{i}^{t}

(the scalar factor v^{\binom{n}{2}ω(i,i)−nd_{i}(n−1)} does not depend on t). It follows that η yields a K-algebra homomorphism

Û_{∗}^{+}(K, v, •) → Û^{+}(K, v^{2}, ω)_{[v,ω]}

and this has to be an isomophism, since we can use the same argument in order to construct a corresponding K-algebra homomorphism

Û^{+}(K, v^{2}, ω) → Û_{∗}^{+}(K, v, •)_{[v,−ω]}.

This completes the proof.

2. Green’s map from C(ω, d) or C_{∗}(ω, d) onto some U^{+}.

Proposition. Let (ω, d) be an Euler form on ℤI. There exists a homomorphism of K_{0}-algebras

ψ: C(ω, d) → U^{+}(K_{0}, q̃, −ω, 0)

with ũ_{i} being mapped to the residue class of θ_{i}.

Proof. Let K be a family of finite fields k with q_{k} = |k| being unbounded. For every field k, let Λ_{k} be a k-algebra with E(Λ_{k}) = (ω, d).

We recall that the generic composition algebra C(ω, d) is the K_{0}-subalgebra of ∏_{k∈K} H(Λ_{k}, ℤ_{k}) generated by the elements ũ_{i}. Thus, there exists a K_{0}-algebra homomorphism

ξ: K_{0}⟨I⟩ → ∏_{k∈K} H(Λ_{k}, ℤ_{k})

such that ξ(θ_{i}) = ũ_{i}; its image is C(ω, d).

We may compose ξ with the canonical projections onto the factors H(Λ_{k}, ℤ_{k}) and obtain corresponding maps

ξ_{k}: K_{0}⟨I⟩ → H(Λ_{k}, ℤ_{k}).

Note that we have

Ker ξ = ⋂_{k∈K} Ker ξ_{k}.

We claim that

Ker ξ ⊆ I(K_{0}, q̃, −ω, 0).

This then implies that C(ω, d) = K_{0}⟨I⟩/ Ker ξ maps onto U^{+}(K_{0}, q̃, −ω, 0), with the residue class of θ_{i} modulo Ker ξ being mapped onto the residue class of θ_{i} modulo the ideal I(K_{0}, q̃, −ω, 0).

We can factor the map ξ_{k} as follows:

K_{0}⟨I⟩ ⟶^{η_{k}} ℤ_{k}⟨I⟩ ⟶^{ξ′_{k}} H(Λ_{k}, ℤ_{k}),

where η_{k} is the scalar extension map (it sends θ_{i} to θ_{i}), and where ξ′_{k} is the ℤ_{k}- algebra homomorphism which sends θ_{i} to u_{i}. The map ξ′_{k} is the one considered in Section IV.5, thus we know that

Ker ξ′_{k} ⊆ I(ℤ_{k}, q_{k}, −ω, 0).

On the other hand, according to IV.4, for any pair of elements a, b ∈ K_{0}⟨I⟩, we know that

(η_{k}(a), η_{k}(b))_{1}^{(k)} = (a, b)_{1}(q_{k}),

where (−, −)_{1} is the generic Lusztig form on K_{0}⟨I⟩ for (−ω, 0), whereas (−, −)_{1}^{(k)} is the Lusztig form on ℤ_{k}⟨I⟩ with respect to q_{k}, (−ω, 0) and 1.

Now assume there is given b ∈ Ker ξ, thus η_{k}(b) belongs to Ker ξ′_{k} and therefore to I(ℤ_{k}, q_{k}, −ω, 0). If now a ∈ K_{0}⟨I⟩ is an arbitrary element, then the Laurent polynomial (a, b)_{1} satisfies

(a, b)_{1}(q_{k}) = (η_{k}(a), η_{k}(b))_{1}^{(k)} = 0

for all k ∈ K. Since there are infinitely many such q_{k}, it follows that (a, b)_{1} is the zero element in K_{0}. This completes the proof.

∗ ∗ ∗

Proposition*. Let (ω, d) be an Euler form on ℤI. Let (•, d) be the corresponding symmetrization. There exists a homomorphism of K_{∗0}-algebras

ψ_{∗}: C_{∗}(ω, d) → U^{+}(K_{∗0}, v, 0, •)

with ũ_{i} being mapped to the residue class of θ_{i}.

Proof. Let K be a family of finite fields k with q_{k} = |k| being unbounded. For every field k, let Λ_{k} be a k-algebra with E(Λ_{k}) = (ω, d).

The twisted generic composition algebra C_{∗}(ω, d) is the K_{∗0}-subalgebra of ∏_{k∈K} H_{∗}(Λ_{k}) generated by the elements ũ_{i}. Thus, there exists a K_{∗0}-algebra homomorphism

ξ_{∗}: K_{∗0}⟨I⟩ → ∏_{k∈K} H_{∗}(Λ_{k})

such that ξ_{∗}(θ_{i}) = ũ_{i}; its image is C_{∗}(ω, d)

We compose ξ_{∗} with the canonical projection onto the factors H_{∗}(Λ_{k}) and obtain

ξ_{∗k}: K_{∗0}⟨I⟩ → H_{∗}(Λ_{k}).

We have

Ker ξ_{∗} = ⋂_{k∈K} Ker ξ_{∗k}.

We claim that

Ker ξ_{∗} ⊆ I(K_{∗0}, ṽ, 0, •).

This then implies that C_{∗}(ω, d) = K_{∗0}⟨I⟩/ Ker ξ_{∗} maps onto U^{+}(K_{∗0}, ṽ, 0, •), with the residue class of θ_{i} modulo Ker ξ_{∗} being mapped onto the residue class of θ_{i} modulo the ideal I(K_{∗0}, ṽ, 0, •).

We can factor the map ξ_{∗k} as follows:

K_{∗0}⟨I⟩ ⟶^{η_{k}} ℤ_{∗k}⟨I⟩ ⟶^{ξ′_{∗k}} H_{∗}(Λ_{k}),

where η_{k} is the scalar extension map (it sends θ_{i} to θ_{i}), and where ξ′_{∗k} is the ℤ_{∗k}- algebra homomorphism which sends θ_{i} to u_{i}. The map ξ′_{∗k} is the one considered in section IV.5, thus we know that

Ker ξ′_{∗k} ⊆ I(ℤ_{∗k}, v_{k}, 0, •).

On the other hand, according to IV.4, for any pair of elements a, b ∈ K_{∗0}⟨I⟩, we know that

(η_{k}(a), η_{k}(b))_{1}^{(k)} = (a, b)_{1}(v_{k}),

where (−, −)_{1} is the generic Lusztig form on K_{∗0}⟨I⟩ for (0, •), whereas (−, −)_{1}^{(k)} is the Lusztig form on ℤ_{∗k}⟨I⟩ with respect to v_{k}, (0, •) and 1.

Now assume there is given b ∈ Ker ξ_{∗}, thus η_{k}(b) belongs to Ker ξ′_{∗k} and therefore to I(ℤ_{∗k}, v_{k}, 0, •). If now a ∈ K_{∗0}⟨I⟩ is an arbitrary element, then the Laurent polynomial (a, b)_{1} satisfies

(a, b)_{1}(v_{k}) = (η_{k}(a), η_{k}(b))_{1}^{(k)} = 0

for all k ∈ K. Since there are infinitely many such v_{k}, it follows that (a, b)_{1} is the zero element in K_{∗0}. This completes the proof.

3. From Û^{+} to C to U^{+}.

Let ω be an Euler form without short cycles and with symmetrization •. Recall that we obtain in this way an arbitrary Cartan datum •.

On the one hand, we have the ℚ(ṽ)-algebra homomorphism π = π^{ℚ(ṽ)}

π: Û^{+}(ℚ(ṽ), q̃, ω) → C(ω) ⊗_{K_{0}} ℚ(ṽ),

which sends the residue class of θ_{i} to ũ_{i} ⊗ 1, for any i ∈ I. On the other hand, there is given Green’s map

ψ: C(ω) → U^{+}(K_{0}, q̃, −ω, 0);

tensoring with the field of rational functions ℚ(ṽ), we obtain

ψ ⊗ 1: C(ω) ⊗_{K_{0}} ℚ(ṽ) → U^{+}(K_{0}, q̃, −ω, 0) ⊗_{K_{0}} ℚ(ṽ)

Also, we may consider the canonical map

μ: U^{+}(K_{0}, q̃, −ω, 0,) ⊗_{K_{0}} ℚ(ṽ) → U^{+}(ℚ(v), q̃, −ω, 0)

given by multiplication.

Altogether we have three surjective ℚ(ṽ)-algebra homomorphisms

Û^{+}(ℚ(ṽ), q̃, ω)
↓^{π}
C(ω) ⊗_{K_{0}} ℚ(ṽ)
↓^{ψ⊗1}
U^{+}(K_{0}, q̃, −ω, 0,) ⊗_{K_{0}} ℚ(ṽ)
↓^{μ}
U^{+}(ℚ(ṽ), q̃, −ω, 0)

and the composition sends the residue class of θ_{i} (modulo the ideal J (ℚ(ṽ), q̃, ω) generated by the quantum Serre relations) to the residue class of θ_{i} (modulo the ideal I(ℚ(ṽ), q̃, −ω, 0)).

∗ ∗ ∗

In the twisted case, we have corresponding maps, as follows. First of all, there is π_{∗} = π_{∗}^{ℚ(ṽ)}

π_{∗}: Û_{∗}^{+}(ℚ(ṽ), ṽ, •) → C_{∗}(ω) ⊗_{K_{∗0}} ℚ(ṽ),

which sends the residue class of θ_{i} to ũ_{i} ⊗ 1, for any i ∈ I. On the other hand, there is given Green’s map

ψ_{∗}: C_{∗}(ω) → U^{+}(K_{∗0}, ṽ, 0, •);

tensoring with the field of rational functions ℚ(ṽ), we obtain

ψ_{∗} ⊗ 1: C_{∗}(ω) ⊗_{K_{∗0}} ℚ(ṽ) → U^{+}(K_{∗0}, ṽ, 0, •) ⊗_{K_{∗0}} ℚ(ṽ)

Finally, we consider again the multiplication map

μ_{∗}: U^{+}(K_{∗0}, ṽ, 0, •) ⊗_{K_{∗0}} ℚ(ṽ) → U^{+}(ℚ(ṽ), ṽ, 0, •).

Altogether, we get

Û_{∗}^{+}(ℚ(ṽ), ṽ, •)
↓^{π_{∗}}
C_{∗}(ω) ⊗_{K_{∗0}} ℚ(ṽ)
↓^{ψ_{∗}⊗1}
U^{+}(K_{∗0}, ṽ, 0, •) ⊗_{K_{∗0}} ℚ(ṽ)
↓^{μ_{∗}}
U^{+}(ℚ(ṽ), ṽ, 0, •)

and the composition sends the residue class of θ_{i} (modulo the ideal J_{∗}(ℚ(ṽ), ṽ, •) generated by the quantum Serre relations) to the residue class of θ_{i} (modulo the ideal I(ℚ(ṽ), ṽ, 0, •)).

4. Lusztig’s Theorem.

We recall the following:

Theorem (Lusztig). Let • be a Cartan datum. Then the ideal I(ℚ(ṽ), ṽ, 0, •) is generated by the quantum Serre relations.

This means that

I(ℚ(ṽ), ṽ, 0, •) = J_{∗}(ℚ(ṽ), ṽ, •).

For the proof, we refer to Lusztig [L], Theorem 33.1.3. The proof is based on the representation theory of Kac-Moody Lie-algebras; it uses some fundamental results in the book of Kac [K], namely the Kac-Weyl character formula 10.4 and its corollary 10.4.

5. Green’s Theorem.

Theorem. Let ω be an Euler form without short cycles. Then the three ℚ(ṽ)- algebra maps

\begin{matrix}Û_{∗}^{+}(ℚ(ṽ), ṽ, •) & ⟶^{π_{∗}} & C_{∗}(ω) ⊗_{K_{∗0}} ℚ(ṽ) \\ C_{∗}(ω) ⊗_{K_{∗0}} ℚ(ṽ) & ⟶^{ψ_{∗}⊗1} & U^{+}(K_{∗0}, ṽ, 0, •) ⊗_{K_{∗0}} ℚ(ṽ) \\ U^{+}(K_{∗0}, ṽ, 0, •) ⊗_{K_{∗0}} ℚ(ṽ) & ⟶^{μ_{∗}} & U^{+}(ℚ(ṽ), ṽ, 0, •) \end{matrix}

are isomorphisms.

Proof: All the maps mentioned are surjective and the composition is the identity. Thus, all these maps must be bijective.

In section III.5, we have constructed a map

(C(ω, d) ⊗ K_{∗0})_{[ṽ,ω]} ⟶ C_{∗}(ω, d)

of K_{∗0}-algebras. Scalar extension yields a corresponding map

τ: (C(ω, d) ⊗_{K_{0}} ℚ(ṽ))_{[ṽ,ω]} → C_{∗}(ω, d) ⊗_{K_{∗}0} ℚ(ṽ)

Corollary. Let ω be an Euler form without short cycles. Then the following ℚ(ṽ)-algebra maps

\begin{matrix}Û^{+}(ℚ(ṽ), q̃, ω) & ⟶^{π} & C(ω) ⊗_{K_{0}} ℚ(ṽ) \\ (C(ω) ⊗_{K_{0}} ℚ(ṽ))_{[ṽ,ω]} & ⟶^{τ} & C_{∗}(ω) ⊗_{K_{∗}0} ℚ(ṽ) \end{matrix}

are isomorphisms.

Proof: According to Lemma II.1, we may consider π also as an algebra homomorphism

π: Û^{+}(ℚ(ṽ), q̃, ω)_{[ṽ,ω]} → (C(ω) ⊗_{K_{0}} ℚ(ṽ))_{[ṽ,ω]}.

In addition, consider the isomorphism of ℚ(ṽ)-algebras

η: Û_{∗}^{+}(ℚ(ṽ), ṽ, •) → Û^{+}(ℚ(ṽ), ṽ^{2}, ω)_{[ṽ,ω]}

constructed in V.1. Then, the composition τπη is just the map π_{∗} considered in Theorem, and there we have seen that π_{∗} is an isomorphism. Since all three maps τ, π, η are surjective, it follows that they also are isomorphisms. This completes the proof.

Acknowledgement

The author is endebted to J.A.Green for several interesting discussions. He also thanks T. Brüstle, H. Krause, G. Röhrle, J. Schröer, Xiao Jie and Zhang Pu for reading carefully parts of the manuscript.

Main Reference

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Fakultät für Mathematik, Universität Bielefeld, POBox 100 131, D-33 501 Bielefeld

E-mail address: ringel@mathematik.uni-bielefeld.de