The Hall Algebra Approach to Quantum Groups

Claus Michael Ringel

E.L.A.M. Lectures 1993

Table of contents

Introduction

0. Preliminaries

1. The definition of U_{q}(n_{+}(Δ))

2. Rings and modules, path algebras of quivers

3. The Hall algebra of a finitary ring

4. Loewy series

5. The fundamental relations

6. The twisted Hall algebra

7. The isomorphism between U_{q}(n_{+}(Δ)) and H_{∗}(kΔ⃗) for Δ⃗ a Dynkin quiver

8. The canonical basis

9. The case 𝔸_{2}

10. The case 𝔸_{3} References Introduction

Given any Dynkin diagram Δ of type 𝔸_{n}, 𝔻_{n}, 𝔼_{6}, 𝔼_{7}, 𝔼_{8}, we may endow its edges with an orientation; we obtain in this way a quiver (an oriented graph) Δ⃗, and the corresponding path algebra kΔ⃗, where k is a field. We may consider the representations of Δ⃗ over k, or, equivalently, the kΔ⃗-modules. In case k is a finite field, one may define a multiplication on the free abelian group with basis the isomorphism classes of kΔ⃗-modules by counting filtrations of modules; the ring obtained in this way is called the Hall algebra H(kΔ⃗).

We denote by ℤ[v] the polynomial ring in one variable v, and we set q = v^{2}. Also, let A = ℤ[v, v^{−1}].

The free ℤ[q]-module H(Δ⃗) with basis the isomorphism classes of kΔ⃗-modules can be endowed with a multiplication so that H(Δ⃗)/(q − |k|) ≃ H(kΔ⃗), for any finite field k of cardinality |k|, thus H(Δ⃗) may be called the generic Hall algebra. The generic Hall algebra satisfies relations which are very similar to the ones used by Jimbo and Drinfeld in order to define a q-deformation U_{q}(n_{+}(Δ)) of the Kostant ℤ-form U(n_{+}(Δ)). Here, g(Δ) = n_{−}(Δ) ⊕ h(Δ) ⊕ n_{+}(Δ) is a triangular decomposition of the complex simple Lie algebra g(Δ) of type Δ. Note that U_{q}(n_{+}(Δ)) is an A-algebra, and we can modify the multiplication of H(Δ) ⊗_{ℤ[q]} A using the Euler characteristic on the Grothendieck group K_{0}(kΔ⃗) in order to obtain the twisted Hall algebra H_{∗}(Δ⃗) with

U_{q}(n_{+}(Δ)) ≃ H_{∗}(Δ⃗)

What is the advantage of the Hall algebra approach? Assume we have identified U_{q}(n_{+}(Δ)) with H_{∗}(Δ⃗). Note that the ring U_{q}(n_{+}(Δ)) is defined by generators and relations, whereas H_{∗}(Δ⃗) is a free A-module with a prescribed basis.

The presentation of U_{q}(n_{+}(Δ)) gives us a presentation for the (twisted) Hall algebra, and this may be interpreted as follows: the Jimbo-Drinfeld relations are the universal relations for comparing the numbers of composition series of modules over algebras with a prescribed quiver.

On the other hand, in H_{∗}(Δ⃗), there is the prescribed basis given by the kΔ⃗-modules, and we obtain in this way a basis for U_{q}(n_{+}(Δ)), thus normal forms for its elements, and this makes calculations in U_{q}(n_{+}(Δ)) easier. Also, the basis elements themselves gain more importance, more flavour. Since they may be interpreted as modules, one can discuss about their module theoretical, homological or geometrical properties: whether they are indecomposable, or multiplicityfree and so on.

The basis of U_{q}(n_{+}(Δ)) obtained in this way depends on the chosen orientation of Δ, and Lusztig has proposed a base change which leads to a basis which is independent of such a choice and which he calls the canonical basis. This basis also was constructed by Kashiwara and called the crystal basis of U_{q}(n_{+}(Δ)).

Here is the list of the Dynkin diagrams 𝔸_{n}, 𝔻_{n}, 𝔼_{6}, 𝔼_{7}, 𝔼_{8}:

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0. Preliminaries

Consider the following polynomials in a variable T, where n, m ∈ ℕ_{0} and m ≤ n

F^{n}(T) := \frac{(T^{n} − 1)(T^{n−1} − 1) ⋯ (T − 1)}{(T − 1)^{n}},
G_{m}^{n}(T) := \frac{(T^{n} − 1)(T^{n−1} − 1) ⋯ (T^{n−m+1} − 1)}{(T^{m} − 1)(T^{m−1} − 1) ⋯ (T − 1)}.

Note that the degree of the polynomial F^{n}(T) is \binom{n}{2}, the degree of G_{m}^{n}(T) is m(n − m).

Let k be a finite field, denote its cardinality by q_{k} = |k|. The cardinality of the set of complete flags in k^{n} is just F^{n}(q_{k}), and for 0 ≤ m ≤ n, the number of m-dimensional subspaces of k^{n} is G_{m}^{n}(q_{k}).

Let A′ = ℚ(v) be the rational function field over ℚ in one variable v, and let us consider its subring A = ℤ[v, v^{−1}]. We denote by : A′ → A′ the field automorphism with \overline{v} = v^{−1}; it has order 2, and it sends A onto itself.

We set q = v^{2}; we will have to deal with F^{n}(q) and G_{m}^{n}(q). We define

[n] := \frac{v^{n} − v^{−n}}{v − v^{−1}} = v^{n−1} + v^{n−3} + ⋯ + v^{−n+1},

thus [0] = 0, [1] = 1, [2] = v + v^{−1}, [3] = v^{2} + 1 + v^{−2}, and so on. Let

[n]! := ∏_{m=1}^{n} [ m],

\left[{n \atop m}\right] := \frac{[n]!}{[m]![n − m]!} where 0 ≤ m ≤ n.

There are the following identities:

[n] = v^{−n+1} \frac{q^{n} − 1}{q − 1}
[n]! = v^{−\binom{n}{2}}F^{n}(q)
\left[{n \atop m}\right] = v^{−m(n−m)}G_{m}^{n}(q).

1. The definition of U_{q}(n_{+}(Δ))

Let Δ = (a_{ij})_{ij} be a symmetric (n × n)-matrix with diagonal entries a_{ii} = 2, and with off-diagonal entries 0 and −1. (Such a matrix is called a simply-laced generalized Cartan matrix.)

Note that Δ defines a graph with n vertices labelled 1, 2, …, n with edges {i, j} provided a_{ij} = −1. Often we will not need the labels of the vertices, then we will present the vertices by small dots ∘. Of particular interest will be the Dynkin diagrams 𝔸_{n}, 𝔻_{n}, 𝔼_{6}, 𝔼_{7}, 𝔼_{8}.

Given Δ, we define U′_{q}(n_{+}(Δ)) as the A′-algebra with generators E_{1}, …, E_{n} and relations

E_{i}E_{j} − E_{j}E_{i} = 0  if a_{ij} = 0,
E_{i}^{2}E_{j} − (v + v^{−1})E_{i}E_{j}E_{i} + E_{j}E_{i}^{2} = 0 if a_{ij} = −1.

We denote

E_{i}^{(m)} := \frac{1}{[ m]!}E_{i}^{m}.

Let U_{q}(n_{+}(Δ)) be the A-subalgebra of U′_{q}(n_{+}(Δ)) generated by the elements E_{i}^{(m)} with 1 ≤ i ≤ n and m ≥ 0.

We denote by : U′_{q}(n_{+}(Δ)) → U′_{q}(n_{+}(Δ)) the automorphism with \overline{v} = v^{−1} and \overline{E_{i}} = E_{i} for all i.

We denote by ℤ^{n} the free abelian group of rank n with basis e_{1}, …, e_{n}. Given an element d ∈ ℤ^{n}, say d = ∑ d_{i}e_{i}, let |d| = ∑ d_{i}.

Note that the rings U_{q}(n_{+}(Δ)) and U′_{q}(n_{+}(Δ)) are ℤ^{n}-graded, where we assign to E_{i} the degree e_{i}. Given d ∈ ℤ^{n}, we denote by U_{q}(n_{+}(Δ))_{d} the set of homogeneous elements of degree d, thus

U_{q}(n_{+}(Δ)) = ⨁_{d} U_{q}(n_{+}(Δ))_{d}.

Let Δ be of the form 𝔸_{n}, 𝔻_{n}, 𝔼_{6}, 𝔼_{7}, or 𝔼_{8}. We denote by Φ = Φ(Δ) the corresponding root system. We choose a basis e_{1}, …, e_{n} of the root system, and denote by Φ^{+} the set of positive roots (with respect to this choice). The choice of the basis yields a fixed embedding of Φ into ℤ^{n}.

We will have to deal with maps α: Φ^{+} → ℕ_{0}. Given such a map α, we set

dim α := ∑_{a} α(a)a ∈ ℤ^{n}

and call it its dimension vector. We denote by u(d) for d ∈ ℤ^{n} the number of maps α: Φ^{+} → ℕ_{0} with dim α = d.

Consider the ℚ-Lie-algebra n_{+}(Δ) generated by E_{1}, …, E_{n} with relations

[E_{i}, E_{j}] = 0 if a_{ij} = 0,
[E_{i}, [E_{i}, E_{j}]] = 0 if a_{ij} = −1.

(Usually, one deals with the corresponding ℂ-algebra n_{+}(Δ) ⊗_{ℚ} ℂ; here, it will be more convenient to consider the mentioned ℚ-form.)

The universal enveloping algebra U(n_{+}(Δ)) is the ℚ-algebra generated by the elements E_{1}, …, E_{n} with relations

[E_{i}, E_{j}] = 0 if a_{ij} = 0,
[E_{i}, [E_{i}, E_{j}]] = 0 if a_{ij} = −1,

thus we see:

Proposition. We have

U(n_{+}(Δ)) = U_{q}(n_{+}(Δ)) ⊗_{A} ℚ[v, v^{−1}]/(v − 1).

Of course, n_{+}(Δ) and U(n_{+}(Δ)) both are ℤ^{n}-graded, where again we assign to E_{i} the degree e_{i}. For any non-zero homogeneous element L of n_{+}(Δ), we denote by dim L its degree. It is well-known that n_{+}(Δ) has a basis E_{a} indexed by the positive roots, such that dim E_{a} = a. As a consequence, we obtain the following consequence:

Proposition. The ℚ-dimension of U(n_{+}(Δ))_{d} is u(d).

Proof: Use the theorem of Poincaré-Birkhoff-Witt.

2. Rings and modules, path algebras of quivers

Rings and modules.

Given a ring R, the R-modules which we will consider will be finitely generated right R-modules. The category of finitely generated right R-modules will be denoted by mod R.

Let R be a ring. The direct sum of two R-modules M_{1}, M_{2} will be denoted by M_{1}⊕M_{2}, the direct sum of t copies of M will be denoted by tM. The zero module will be denoted by 0_{R} or just by 0.

We write M ≃ M′, in case the modules M, M′ are isomorphic, the isomorphism class of M will be denoted by [M]. For any module M, we denote by s(M) the number of isomorphism classes of indecomposable direct summands of M.

Let R be a finite dimensional algebra over some field k. Let n = s(R_{R}), thus n is the rank of the Grothendieck group K_{0}(R) of all finite length modules modulo split exact sequences. Given such a module M, we denote its equivalence class in K_{0}(R) by dim M. There are precisely n isomorphism classes of simple R-modules S_{1}, …, S_{n}, and the elements e_{i} = dim S_{i} form a basis of K_{0}(R). If we denote the Jordan-Hölder multiplicity of S_{i} in M by [M : S_{i}], then dim M = ∑_{i}[M : S_{i}]e_{i}.

We denote by supp M the support of M, it is the set of simple modules S with [M : S] ≠ 0. (In case the simple modules are indexed by the vertices of a quiver, we also will consider supp M as a subset of the set of vertices of this quiver).

Path algebras of quivers

A quiver Δ⃗ = (Δ⃗_{0}, Δ⃗_{1}, s, t) is given by two sets Δ⃗_{0}, Δ⃗_{1}, and two maps s, t: Δ⃗_{1} → Δ⃗_{0}. The elements of Δ⃗_{0} are called vertices, the elements of Δ⃗_{1} are called arrows; given f ∈ Δ⃗_{1}, then we say that f starts in s(f) and ends in t(f), and we write f: s(f) → t(f). An arrow f with s(f) = t(f) is called a loop, we always will assume that Δ⃗ has no loops.

We denote by kΔ⃗ the path algebra of the quiver Δ⃗ over the field k. We will not distinguish between representations of Δ⃗ over k and (right) kΔ⃗-modules. Recall that a representation M of Δ⃗ over k attaches to each vertex x of Δ⃗ a vector space M_{x} over k, and to each arrow f: s(f) → t(f) a k-linear map M_{s(f)} → M_{t(f)}. For any vertex x of Δ⃗, we can define a simple kΔ⃗-module S(x) by attaching the one-dimensional k-space k to the vertex x, the zero space to the remaining vertices, and the zero map to all arrows. We stress that the number of arrows x → y is equal to dim_{k} Ext^{1}(S(x), S(y)). In case there is precisely one arrow starting in x and ending in y, there exists up to isomorphism a unique indecomposable representation of length 2 with top S(x) and socle S(y), we denote it by

\begin{matrix}S(x) \\ S(y) \end{matrix} .

In case Δ⃗ has as vertex set the set {1, 2, …, n}, we define a corresponding (n × n)- matrix Δ = (a_{ij})_{ij} as follows: let a_{ii} = 2, for all i, and let a_{ij} be the number of arrows between i and j (take the arrows i → j as well as the arrows j → i). In case there is at most one arrow between i and j we obtain a matrix as considered in section 1, and then we will call Δ the underlying graph of Δ⃗.

Let us assume that Δ⃗ is a finite quiver with n vertices, and let Λ = kΔ⃗. We assume in addition that Δ⃗ has no cyclic paths (a cyclic path is a path of length at least 1 starting and ending in the same vertex). As a consequence, Δ⃗ is finite-dimensional, and there are precisely n simple Λ-modules, namely the modules S(x), with x a vertex. Of course, if M is a representation of Δ⃗, then dim M = ∑_{x}(dim_{k} M_{x}) dim S(x) in the Grothendieck group K_{0}(Λ).

It is easy to see that Λ is hereditary, thus we can define on K_{0}(Λ) a bilinear form via

⟨dim X, dim Y ⟩ = dim_{k} Hom(X, Y) − dim_{k} Ext^{1}(X, Y)

where X, Y are Λ-modules of finite length. The corresponding quadratic form will be denoted by χ; thus for d ∈ K_{0}(Λ), we have χ(d) = ⟨d, d⟩. Of course, we have the following formula for all i, j

a_{ij} = ⟨dim S_{i}, dim S_{j}⟩ + ⟨dim S_{j}, dim S_{i}⟩

Dynkin quivers

A quiver Δ⃗ whose underlying graph is of the form 𝔸_{n}, 𝔻_{n}, 𝔼_{6}, 𝔼_{7}, 𝔼_{8} will be called a Dynkin quiver. We recall some well-known results:

Gabriel’s Theorem. Let Δ⃗ be a Dynkin quiver. The map dim yields a bijection between the isomorphism classes of the indecomposable kΔ⃗-modules and the positive roots for Δ.

Let Δ⃗ be a Dynkin quiver, and let Λ = kΔ⃗. Given a positive root a for Δ, we denote by M(a) or M_{Λ}(a) the corresponding Λ-module; thus M(a) = M_{Λ}(a) is an indecomposable Λ-module with dim M(a) = a. Similarly, given a map α: Φ^{+} → ℕ_{0}, we denote by M(α) we denote the Λ-module

M(α) = M_{Λ}(α) = ⨁_{a} α(a)M(a).

We obtain in this way a bijection between the maps Φ^{+} → ℕ_{0} and the isomorphism classes of Λ-modules of finite length (according to the Krull-Schmidt theorem).

A finite dimensional k-algebra R is called representation directed provided there is only a finite number of (isomorphism classes of) indecomposable R-modules, say M_{1}, …, M_{m}, and they can be indexed in such a way that Hom(M_{i}, M_{j}) = 0 for i > j.

Proposition. Let Δ⃗ be a Dynkin quiver. Then kΔ⃗ is representation directed.

Let Φ^{+} = { a_{1}, …, a_{m} }, we will assume that the ordering is chosen so that

Hom(M_{Λ}(a_{i}), M_{Λ}(a_{j})) ≠ 0 implies i ≤ j.

The subcategories C, D of mod Λ are said to be linearly separated provided for modules C in add C, and D in add D with dim C = dim D, we have C = 0 = D.

Lemma. The subcategories add{M(a_{1}), …, M(a_{s−1})} and add{M(a_{s}), …, M(a_{m})} are linearly separated.

3. The Hall algebra of a finitary ring.

Given a ring R, we will be interested in the finite R-modules; here a module M will be said to be finite provided the cardinality of its underlying set is finite (not just that M is of finite length). Of course, for many rings the only finite R-module will be the zero-module, but for finite rings, in particular for finite-dimensional algebras over finite fields, all finite length modules are finite modules. A ring R will be said to be finitary provided the group Ext^{1}(S_{1}, S_{2}) is finite, for all finite simple R-modules S_{1}, S_{2}. (For a discussion of finitary rings, see [R1]).

We assume that R is a finitary ring. We mainly will consider path algebras of finite quivers over finite fields; of course, such a ring is finitary.

Given finite R-modules N_{1}, N_{2}, …, N_{t} and M, let F_{N_{1},...,N_{t}}^{M} be the set of filtrations

M = M_{0} ⊇ M_{1} ⊇ ⋯ ⊇ M_{t} = 0

such that M_{i−1}/M_{i} is isomorphic to N_{i}, for all 1 ≤ i ≤ t. The cardinality of F_{N_{1},...,N_{t}}^{M} will be denoted by F_{N_{1},...,N_{t}}^{M} or also by ⟨N_{1}N_{2} … N_{t} ⋄| M⟩. (These cardinalities are finite, since we assume that R is finitary.)

Let H(R) be the free abelian group with basis the set of isomorphism classes [X] of finite R-modules, with a multiplication which we denote by the diamond sign ⋄

[N_{1}] ⋄ [N_{2}] := ∑_{[M]} F_{N_{1}N_{2}}^{M}[M] = ∑_{[M]}⟨N_{1}N_{2} ⋄| M⟩[M].

Given an element x ∈ H(R), we denote its t^{th} power with respect to the diamond product by x^{⋄t}.

Proposition. H(R) is an associative ring with 1.

Proof: The associativity follows from the fact that

([N_{1}] ⋄ [N_{2}]) ⋄ [N_{3}] = ∑_{[M]} F_{N_{1}N_{2}N_{3}}^{M}[M] = [N_{1}] ⋄ ([N_{2}] ⋄ [N_{3}]),

The unit element is just [0_{R}], with 0_{R} the zero module.

In case R is a finite-dimensional algebra over some finite field, we assign to the isomorphism class [M] the degree dim M ∈ ℤ^{n}. Let H(R)_{d} be the free abelian group with basis the set of isomorphism classes [M] of finite R-modules with dim M = d.

Proposition. H(R) = ⨁_{d} H(R)_{d} is a ℤ^{n}-graded ring.

Proof: We only have to observe that for F_{N_{1}N_{2}}^{M} ≠ 0, we have dim M = dim N_{1} + dim N_{2}.

From now on, let Δ⃗ be a Dynkin quiver, let k be a field. We consider Λ = kΔ⃗. Let { 1, 2, …, n } be the vertices of Δ⃗, ordered in such a way that

Ext^{1}(S_{i}, S_{j}) ≠ 0 implies i < j.

Let Φ^{+} = { a_{1}, …, a_{m} }, and we will assume that the ordering is chosen so that

Hom(M_{Λ}(a_{i}), M_{Λ}(a_{j})) ≠ 0 implies i ≤ j.

Hall polynomials

Proposition. Let α, β, γ: Φ^{+} → ℕ_{0}. There exists a polynomial ϕ_{α,γ}^{β}(q) ∈ ℤ[q] such that for any finite field k of cardinality q_{k}

F_{M_{Λ}(α)M_{Λ}(γ)}^{M_{Λ}(β)} = ϕ_{α,γ}^{β}(q_{k}).

For a proof, see [R1], Theorem 1, p.439.

The polynomials which arise in this way are called Hall polynomials.

Let Δ⃗ be a Dynkin quiver. Let H(Δ⃗) be the free ℤ[q]-module with basis the set of maps Φ^{+} → ℕ_{0}. On H(Δ⃗), we define a multiplication by

α_{1} ⋄ α_{2} := ∑_{β} ϕ_{α_{1}α_{2}}^{β}(q) · β

Proposition. H(Δ⃗) is an associative ring with 1, it is ℤ^{n}-graded (the degree of α: Φ^{+} → ℕ_{0} being dim α), and for any finite field k of cardinality q_{k}, the map α ↦ [M_{kΔ⃗}(α)] yields an isomorphism

H(Δ⃗)/(q − q_{k}) ≃ H(kΔ⃗).

4. Loewy series.

For d = (d_{1}, …, d_{n}) ∈ ℕ_{0}^{n}, let

w_{⋄}(d) := [d_{1}S_{1}] ⋄ ⋯ ⋄ [d_{n}S_{n}].

Note that the element w_{⋄}(d) only depends on the semisimple module ⨁ d_{i}S_{i} and not on the particular chosen ordering of the vertices of Δ⃗, since [d_{i}S_{i}] ⋄ [d_{j}S_{j}] = [d_{j}S_{j}] ⋄ [d_{i}S_{i}] in case Ext^{1}(S_{i}, S_{j}) = 0 = Ext^{1}(S_{j}, S_{i}).

Remark: Recall that the vertices { 1, 2, …, n } of Δ⃗ are ordered in such a way that Ext^{1}(S_{i}, S_{j}) ≠ 0 implies that i < j. Usually, there will be several possible orderings, for example in the case of 𝔸_{3} with orientation

∘_{y}

↙  ↘
∘_{x}  ∘_{z}

we have to take 1 = y but we may take 2 = x, 3 = z or else 2 = z, 3 = x. All possible orderings are obtained from each other by a finite sequence of transpositions (i, i + 1) in case Ext^{1}(S_{i}, S_{i+1}) = 0 = Ext^{1}(S_{i+1}, S_{i}).

Lemma. We have ⟨w_{⋄}(d) ⋄| M⟩ ≠ 0 if and only if dim M = d; and, in this case, ⟨w_{⋄}(d) ⋄| M⟩ = 1.

The proof is obvious.

Given a map α: Φ^{+} → ℕ_{0}, let

w_{⋄}(α) := w_{⋄}(α(a_{1})a_{1}) ⋄ ⋯ ⋄ w_{⋄}(α(a_{m})a_{m}).

The element w_{⋄}(α) does not depend on the chosen ordering of the positive roots.

Example. Consider the case 𝔸_{2}. Thus, there is given a hereditary k-algebra Λ with two simple modules S_{1}, S_{2} such that Ext^{1}(S_{1}, S_{2}) = k. There is a unique indecomposable module of length 2, and we denote it by I. There are three positive roots a_{1}, a_{2}, a_{3}, where a_{2} = a_{1} + a_{3}. If we assume that S_{2} = M_{Λ}(a_{1}) and S_{1} = M_{Λ}(a_{3}), then the ordering is as desired. For α: Φ^{+} → ℕ_{0}, we obtain the following element

w_{⋄}(α) = w_{⋄}(α(a_{1})a_{1}) ⋄ w_{⋄}(α(a_{2})a_{2}) ⋄ w_{⋄}(α(a_{3})a_{3})
= [α(a_{1})S_{2}] ⋄ [α(a_{2})S_{1}] ⋄ [α(a_{2})S_{2}] ⋄ [α(a_{3})S_{1}],

the corresponding Λ-module is M_{Λ}(α) = α(a_{1})S_{2} ⊕ α(a_{2})I ⊕ α(a_{3})S_{1}.

Lemma. We have ⟨w_{⋄}(α) ⋄| M⟩ ≠ 0 if and only if there exists a filtration

M = M_{0} ⊇ M_{1} ⊇ ⋯ ⊇ M_{m} = 0

such that dim M_{i−1}/M_{i} = α(a_{i})a_{i}.

The proof is obvious.

The set of maps Φ^{+} → ℕ_{0} will be ordered using the opposite of the lexicographical ordering: Given α, β: Φ^{+} → ℕ_{0}, we have β < α if and only if there exists some 1 ≤ j ≤ m such that β(a_{i}) = α(a_{i}) for all i < j, whereas β(a_{j}) > α(a_{j}).

Theorem 1. Let α: Φ^{+} → ℕ_{0}. Then ⟨w_{⋄}(α) ⋄| M(α)⟩ = 1. On the other hand, given a module M with ⟨w_{⋄}(α) ⋄| M⟩ ≠ 0, then M ≃ M(β) for some β ≤ α.

Before we present the proof, we need some preliminary considerations. Given α: Φ^{+} → ℕ_{0}, let us define for 0 ≤ t ≤ m, the submodule M_{t}(α) = ⨁_{i>t} α(a_{i})M_{Λ}(a_{i}) of M(α). Thus we obtain a sequence of submodules

M(α) = M_{0}(α) ⊇ M_{1}(α) ⊇ ⋯ ⊇ M_{m}(α) = 0.

Lemma. Let U be a submodule of M′ = M_{t−1}(β) such that dim M′/U = u · a_{j} for some j. Then we have j ≥ t. If j = t, then U ⊇ M_{t} (and therefore u ≤ β(a_{t})), there is an isomorphism M′/U ≃ uM(a_{j}), and U = M_{t}(β) ⊕ U′, with U′ ≃ (β(a_{t}) − u)M(a_{t}).

Proof. We can assume u > 0.

Let us first assume that M′/U ≃ u · M(a_{j}). First of all, we show that j ≥ t. For j < t, we have Hom(M(a_{i}), M(a_{j}) = 0 for all i ≥ t, thus Hom(M′, M(a_{j})) = 0, whereas there is given a non-zero map M′ → M′/U ≃ u · M(a_{j}). Now assume j = t. Using the same argument, we see that the composition of the inclusion map M_{t}(β) → M′ and the projection map M′ → M′/U has to be zero, since Hom(M(a_{i}), M(a_{t})) = 0 for i > t. This shows that U ⊇ M_{t}, and consequently u ≤ β(a_{t}). The canonical projection β(a_{t})M(a_{t}) ≃ M′/M_{t}(β) → M′/U splits, thus U/M_{t}(β) ≃ (β(a_{t})−u)M(a_{t}). But then also the projection U → U/M_{t}(β) splits (since Ext^{1}(M(a_{t}), M(a_{i})) = 0 for all i > t). This shows the existence of a direct complement U′ in U to M_{t}(β), and we have U′ ≃ U/M_{t}(β) ≃ (β(a_{t})−u)M(a_{t}).

In general, we can write M′/U = M(γ) for some γ: Φ^{+} → ℕ_{0}. Choose s minimal with γ(a_{s}) > 0. Let U ⊆ V ⊂ M_{t−1}(β) such that M_{t−1}(β)/V = M_{s}(γ). Then M_{t−1}(β)/V ≃ M(γ)/M_{s}(γ) ≃ γ(a_{s})M(a_{s}), and we can apply the previous considerations. We see that s ≥ t, and if s = t, then γ(a_{t}) ≤ β(a_{t}). By assumption, u·a_{j} = dim M′/U = ∑_{i} γ(a_{i})a_{i} = ∑_{i≥s} γ(a_{i})a_{i}, with non-negative coefficients u and γ(a_{i}). We cannot have j < s, since {M(a_{1}), …, M(a_{s−1})} and {M(a_{s}), …, M(a_{m})} are linearly separated. This shows that j ≥ s ≥ t. Now assume j = t, thus we have j = s = t. We must have γ(a_{t}) ≤ u, and thus we can write (u − γ(a_{t})) · a_{t} = ∑_{i>t} γ(a_{i})a_{i} with non-negative coefficients (u − γ(a_{t})) and γ(a_{i}). Now, we use that {M(a_{1}), …, M(a_{t})} and {M(a_{t+1}), …, M(a_{m})} are linearly separated in order to conclude that u − γ(a_{t}) = 0 and γ(a_{i}) = 0 for all i > t. This shows that M′/U ≃ γ(a_{t})M(a_{t}) and therefore our first considerations do apply.

Proof of Theorem 1: First of all, the sequence

M(α) = M_{0}(α) ⊇ M_{1}(α) ⊇ ⋯ ⊇ M_{m}(α) = 0

shows that ⟨w_{⋄}(α) ⋄| M(α)⟩ ≠ 0.

Let as assume that ⟨w_{⋄}(α) ⋄| M(β)⟩ ≠ 0 for some α, β: Φ^{+} → ℕ_{0}. Thus, we know that there exists a sequence

M(β) = M_{0} ⊇ M_{1} ⊇ ⋯ ⊇ M_{m} = 0

such that dim M_{i−1}/M_{i} = α(a_{i})a_{i}. Let β(a_{i}) = α(a_{i}) for all i < j. By induction, we claim that M_{i} = M_{i}(β) for i < j. Assume, we know that M_{i−1} = M_{i−1}(β). According to Lemma, the only submodule M_{i}(= U) of M_{i−1}(β) with dim M_{i−1}(β)/M_{i} = β(a_{i}) is M_{i} = M_{i}(β). In particular, we have M_{j−1} = M_{j−1}(β). Again, using Lemma, we see that we must have α(a_{j}) ≤ β(a_{j}), this shows that α ≥ β. Also, we see that for α = β, we have M_{i} = M_{i}(β) for all i, thus ⟨w_{⋄}(α) ⋄| M(α)⟩ = 1. This completes the proof.

5. The fundamental relations

Lemma. Let S_{i}, S_{j} be simple R-modules with

Ext^{1}(S_{i}, S_{j}) = 0, Ext^{1}(S_{j}, S_{i}) = 0.

Then we have

[S_{i}] ⋄ [S_{j}] = [S_{j}] ⋄ [S_{i}].

The proof is obvious.

Lemma. Let k be a finite field of cardinality q_{k}. Let R be a k-algebra. Let S_{i}, S_{j} be simple R-modules such that

Ext^{1}(S_{i}, S_{i}) = 0, Ext^{1}(S_{j}, S_{j}) = 0, Ext^{1}(S_{i}, S_{j}) = k, Ext^{1}(S_{j}, S_{i}) = 0.

Then

[S_{i}]^{⋄2} ⋄ [S_{j}] − (q_{k} + 1)[S_{i}] ⋄ [S_{j}] ⋄ [S_{i}] + q_{k}[S_{j}] ⋄ [S_{i}]^{⋄2} = 0,
[S_{i}] ⋄ [S_{j}]^{⋄2} − (q_{k} + 1)[S_{i}] ⋄ [S_{j}] ⋄ [S_{i}] + q_{k}[S_{j}]^{⋄2} ⋄ [S_{i}] = 0.

Proof: Since Ext^{1}(S_{i}, S_{j}) = k, there exists an indecomposable module M of length 2 with top S_{i} and socle S_{j}. Taking into account the assumptions Ext^{1}(S_{i}, S_{i}) = 0 = Ext^{1}(S_{j}, S_{i}), we see that there are just two isomorphism classes of modules of length three with 2 composition factors of the form S_{i} and one of the form S_{j}, namely X = M ⊕ S_{i} and Y = 2S_{i} ⊕ S_{j}. It is easy to check that

[S_{i}]^{⋄2} ⋄ [S_{j}] = (q_{k} + 1)[X]+(q_{k} + 1)[Y],
[S_{i}] ⋄ [S_{j}] ⋄ [S_{i}] =  [X]+(q_{k} + 1)[Y],
[S_{j}] ⋄ [S_{i}]^{⋄2} =  (q_{k} + 1)[Y].

This yields the first equality. Similarly, there are the two isomorphism classes of modules with 2 composition factors of the form S_{j} and one of the form S_{i}, namely X′ = M ⊕ S_{j} and Y′ = S_{i} ⊕ 2S_{j}. It is easy to check that

[S_{i}] ⋄ [S_{j}]^{⋄2} = (q_{k} + 1)[X′]+(q_{k} + 1)[Y′],
[S_{i}] ⋄ [S_{j}] ⋄ [S_{i}] =  [X′]+(q_{k} + 1)[Y′],
[S_{j}]^{⋄2} ⋄ [S_{i}] =  (q_{k} + 1)[Y′].

This yields the second equality.

As an immediate consequence we obtain:

Proposition. The elements [S_{i}] of H(Δ⃗) satisfy the following relations: Let i < j. If there is no arrow from i to j, then

[S_{i}] ⋄ [S_{j}] − [S_{j}] ⋄ [S_{i}] = 0,

if there is an arrow i → j, then

[S_{i}]^{⋄2} ⋄ [S_{j}] − (q + 1)[S_{i}] ⋄ [S_{j}] ⋄ [S_{i}] + q[S_{j}] ⋄ [S_{i}]^{⋄2} = 0,
[S_{i}] ⋄ [S_{j}]^{⋄2} − (q + 1)[S_{i}] ⋄ [S_{j}] ⋄ [S_{i}] + q[S_{j}]^{⋄2} ⋄ [S_{i}] = 0.

More general, if we start with simple modules S_{i}, S_{j} satisfying

Ext^{1}(S_{i}, S_{i}) = 0, Ext^{1}(S_{j}, S_{j}) = 0,
Ext^{1}(S_{i}, S_{j}) = k^{t}, Ext^{1}(S_{j}, S_{i}) = 0.

for some t, then we obtain relations which are similar to the Jimbo-Drinfeld relations which are used to define quantum groups for arbitrary symmetrizable generalized Cartan matrices. See [R3].

6. The twisted Hall algebra

In the ring A = ℤ[v, v^{−1}], we denote the element v^{2} by q. In this way, we have fixed an embedding of ℤ[q] into A. Consider the A-module

H_{∗}(Δ⃗) = H(Δ⃗) ⊗_{ℤ[q]} A.

In H_{∗}(Δ⃗), we introduce a new multiplication ∗ by

[N_{1}] ∗ [N_{2}] := v^{dim_{k} Hom(N_{1},N_{2})−dim_{k} Ext^{1}(N_{1},N_{2})}[N_{1}] ⋄ [N_{2}]
= v^{⟨dim N_{1},dim N_{2}⟩}[N_{1}] ⋄ [N_{2}]

where N_{1}, N_{2} are Λ-modules.

The following assertion is rather obvious:

Proposition. The free A-module H_{∗}(Δ⃗) with the multiplication ∗ is an associative algebra with 1, and ℤ^{n}-graded.

We call H_{∗}(Δ⃗) (with this multiplication) the twisted Hall algebra of Δ⃗. For any element x, we denote its t^{th} power with respect to the ∗ multiplication by x^{∗(t)}.

Using induction, one shows that

[N_{1}] ∗ [N_{2}] ∗ ⋯ ∗ [N_{m}] = v^{∑_{i<j}⟨dim N_{i},dim N_{j}⟩}[N_{1}] ⋄ [N_{2}] ⋄ … ⋄ [N_{m}].

Example. Assume there is an arrow i → j. Then

[S_{i}] ∗ [S_{j}] = v^{−1}[S_{i}] ⋄ [S_{j}] = v^{−1}([ \begin{matrix}S_{i} \\ S_{j} \end{matrix} ] + [S_{i} ⊕ S_{j}]),
[S_{j}] ∗ [S_{i}] = [S_{i} ⊕ S_{j}],

thus

[ \begin{matrix}S_{i} \\ S_{j} \end{matrix} ] = v [S_{i}] ∗ [S_{j}] − [S_{i} ⊕ S_{j}] = v [S_{i}] ∗ [S_{j}] − [S_{j}] ∗ [S_{i}].

Proposition. The elements [S_{i}] of H_{∗}(Δ⃗) satisfy the following relations:

[S_{i}] ∗ [S_{j}] − [S_{j}] ∗ [S_{i}] = 0  if a_{ij} = 0,
[S_{i}]^{∗(2)} ∗ [S_{j}] − (v + v^{−1})[S_{i}] ∗ [S_{j}] ∗ [S_{i}] + [S_{j}] ∗ [S_{i}]^{∗(2)} = 0 if a_{ij} = −1.

Proof: In case a_{ij} = 0, we must have Ext^{1}(S_{i}, S_{j}) = 0 = Ext^{1}(S_{j}, S_{i}), and therefore we have ⟨dim S_{i}, dim S_{j}⟩ = 0 = ⟨dim S_{j}, dim S_{i}⟩.

Now assume a_{ij} = −1. First, consider the case when i < j, thus dim_{k} Ext^{1}(S_{i}, S_{j}) = 1, and Ext^{1}(S_{j}, S_{i}) = 0. In this case, we have

⟨dim S_{i}, dim S_{j}⟩ = −1, and ⟨dim S_{j}, dim S_{i}⟩ = 0.

Also, ⟨dim S_{i}, dim S_{i}⟩ = 1, thus

[S_{i}]^{∗(2)} ∗ [S_{j}] = v^{−1}[S_{i}]^{⋄2} ⋄ [S_{j}],
[S_{i}] ∗ [S_{j}] ∗ [S_{i}] = [S_{i}] ⋄ [S_{j}] ⋄ [S_{i}],
[S_{j}] ∗ [S_{i}]^{∗(2)} = v[S_{j}] ⋄ [S_{i}]^{⋄2},

thus

[S_{i}]^{∗(2)} ∗ [S_{j}] − (v+v^{−1})[S_{i}] ∗ [S_{j}] ∗ [S_{i}] + [S_{j}] ∗ [S_{i}]^{∗(2)}
= v^{−1}[S_{i}]^{⋄2} ⋄ [S_{j}] − (v + v^{−1})[S_{i}] ⋄ [S_{j}] ⋄ [S_{i}] + v[S_{j}] ⋄ [S_{i}]^{⋄2}
= v^{−1}([S_{i}]^{⋄2} ⋄ [S_{j}] − (q + 1)[S_{i}] ⋄ [S_{j}] ⋄ [S_{i}] + q[S_{j}] ⋄ [S_{i}]^{⋄2})
= 0.

Similarly, if j < i, so that dim_{k} Ext^{1}(S_{j}, S_{i}) = 1, and Ext^{1}(S_{i}, S_{j}) = 0, then

[S_{i}]^{∗(2)} ∗ [S_{j}] − (v+v^{−1})[S_{i}] ∗ [S_{j}] ∗ [S_{i}] + [S_{j}] ∗ [S_{i}]^{∗(2)}
= v[S_{i}]^{⋄2} ⋄ [S_{j}] − (v + v^{−1})[S_{i}] ⋄ [S_{j}] ⋄ [S_{i}] + v^{−1}[S_{j}] ⋄ [S_{i}]^{⋄2}
= v^{−1}(q[S_{i}]^{⋄2} ⋄ [S_{j}] − (q + 1)[S_{i}] ⋄ [S_{j}] ⋄ [S_{i}] + [S_{j}] ⋄ [S_{i}]^{⋄2})
= 0.

Also in general, the fundamental relations in H(Δ⃗) give rise to the Jimbo-Drinfeld relations in H_{∗}(Δ⃗), see [R6].

Divided powers. Given an indecomposable module X, let

[X]^{∗(t)} := \frac{1}{[ t]!}[X]^{∗(t)},

we claim that this is an element of H_{∗}(Λ). Namely:

[X]^{∗(t)} = v^{\binom{t}{2}}[X]^{⋄(t)}
= v^{\binom{t}{2}}F^{t}(q)[tX]
= v^{t(t−1)}[ t]![tX],

(since F^{t}(q) = v^{\binom{t}{2}}[ t]!). Thus

[X]^{(∗(t))} = \frac{1}{[ t]!}[X]^{∗(t)} = v^{t(t−1)}[tX]

.

Using divided powers, we can rewrite the fundamental relations

[S_{i}]^{∗(2)} ∗ [S_{j}] − (v + v^{−1})[S_{i}] ∗ [S_{j}] ∗ [S_{i}] + [S_{j}] ∗ [S_{i}]^{∗(2)} = 0

as follows:

[S_{i}]^{(∗2)} ∗ [S_{j}] − [S_{i}] ∗ [S_{j}] ∗ [S_{i}] + [S_{j}] ∗ [S_{i}]^{(∗2)} = 0

Recall that the vertices { 1, 2, …, n } of Λ are ordered in such a way that

Ext^{1}(S_{i}, S_{j}) ≠ 0 implies i < j.

In case M is semisimple, say M = ⨁ d_{i}S(i), we have

[M] = [d_{n}S_{n}] ⋄ ⋯ ⋄ [d_{1}S_{1}] = [d_{n}S_{n}] ∗ ⋯ ∗ [d_{1}S_{1}],

since for i > j we have Hom(d_{i}S_{i}, d_{j}S_{j}) = 0 = Ext^{1}(d_{i}S_{i}, d_{j}S_{j}). Also, recall that

[tS_{i}] = v^{−t(t−1)}[S_{i}]^{(∗t)},

thus

[M] = [d_{n}S_{n}] ∗ ⋯ ∗ [d_{1}S_{1}] = v^{− ∑ d_{i}(d_{i}−1)}[S_{n}]^{(∗d_{n})} ∗ ⋯ ∗ [S_{1}]^{(∗d_{1})}.

The words w_{∗}(d), w_{∗}(α).

Recall that { 1, 2, …, n } is the set of vertices vertices of Λ, ordered in such a way that Ext^{1}(S_{i}, S_{j}) ≠ 0 implies i < j.

Also, recall that Φ^{+} = { a_{1}, …, a_{m} } is the set of positive roots and we assume that the ordering is chosen so that

Hom(M_{Λ}(a_{i}), M_{Λ}(a_{j})) ≠ 0 implies i ≤ j.

Using the multiplication ∗, we define for d ∈ ℕ_{0}^{n} and α: Φ^{+} → ℕ_{0}

w_{∗}(d) := [S_{1}]^{∗(d_{1})} ∗ ⋯ ∗ [S_{n}]^{∗(d_{n})},
w_{∗}(α) := w_{∗}(α(a_{1})a_{1}) ∗ ⋯ ∗ w_{∗}(α(a_{m})a_{m}).

Lemma. We have

w_{∗}(α) = v^{r(α)}w_{⋄}(α), with r(α) := − dim_{k} M_{Λ}(α) + dim_{k} End(M_{Λ}(α))

.

Proof: We have for d = ∑ d_{i}e_{i}
w_{∗}(d) = [S_{1}]^{(∗d_{1})} ∗ ⋯ ∗ [S_{n}]^{(∗d_{n})}
= v^{∑ d_{i}^{2}−∑ d_{i}}[d_{1}S_{1}] ∗ ⋯ ∗ [d_{n}S_{n}]
= v^{∑ d_{i}^{2}−∑ d_{i}−∑_{i→j} d_{i}d_{j}}[d_{1}S_{1}] ⋄ ⋯ ⋄ [d_{n}S_{n}]
= v^{χ(d)−|d|}[d_{1}S_{1}] ⋄ ⋯ ⋄ [d_{n}S_{n}],
= v^{χ(d)−|d|}w_{⋄}(d)

where we have used that Hom(d_{i}S_{i}, d_{j}S_{j}) = 0 for i > j and dim_{k} Ext^{1}(d_{i}S_{i}, d_{j}S_{j}) = d_{i}d_{j} for i → j. We apply this for d = α(a_{i})a_{i}. We note that

χ(α(a_{i})a_{i}) = dim_{k} End(M_{Λ}(α(a_{i})a_{i})),

and

|α(a_{i})a_{i}| = dim_{k} M_{Λ}(α(a_{i})a_{i}),

therefore

w_{∗}(α(a_{i})a_{i}) = v^{dim_{k} End(M_{Λ}(α(a_{i})a_{i}))−dim_{k} M_{Λ}(α(a_{i})a_{i})}w_{⋄}(α(a_{i})a_{i}).

On the other hand,

w_{∗}(α) = w_{∗}(α(a_{1})a_{1}) ∗ ⋯ ∗ w_{∗}(α(a_{m})a_{m})
= v^{r′}w_{∗}(α(a_{1})a_{1}) ⋄ ⋯ ⋄ w_{∗}(α(a_{m})a_{m})

with

r′ = ∑_{i<j}⟨α(a_{i})a_{i}, α(a_{j})a_{j}⟩
= ∑_{i<j} dim_{k} Hom(α(a_{i})M_{Λ}(a_{i}), α(a_{j})M_{Λ}(a_{j}))
= dim_{k} rad End(M_{Λ}(α)),

here we use that for i < j, we have Ext^{1}(α(a_{i})M_{Λ}(a_{i}), α(a_{j})M_{Λ}(a_{j})) = 0, and that for i > j, we have Hom(α(a_{i})M_{Λ}(a_{i}), α(a_{j})M_{Λ}(a_{j})) = 0. Altogether, we see that

w_{∗}(α) = v^{dim_{k} rad End(M_{Λ}(α))}w_{∗}(α(a_{1})a_{1}) ⋄ ⋯ ⋄ w_{∗}(α(a_{m})a_{m}) = v^{r}w_{⋄}(α),

with

r = r′ + ∑_{i} dim_{k} End(M_{Λ}(α(a_{i})a_{i})) − ∑_{i} dim_{k} M_{Λ}(α(a_{i})a_{i})
= dim_{k} End(M_{Λ}(α)) − dim_{k} M_{Λ}(α).

This completes the proof.

By definition, H_{∗}(Λ) is the free A-module with basis elements the isomorphism classes [M] of the finite Λ-modules. It seems to be worthwhile to consider besides these elements [M] also their multiples

⟨M⟩ := v^{− dim_{k} M+dim_{k} End(M)}[M].

Example.

⟨ \begin{matrix}S_{i} \\ S_{j} \end{matrix} ⟩ = v^{−2+1} [ \begin{matrix}S_{i} \\ S_{j} \end{matrix} ] = v^{−1}(v [S_{i}] ∗ [S_{j}] − [S_{j}] ∗ [S_{i}]) = [S_{i}] ∗ [S_{j}] − v^{−1}[S_{j}] ∗ [S_{i}].

Theorem 1′.

w_{∗}(α) = ⟨M_{Λ}(α)⟩ + ∑_{β<α} g_{αβ}⟨M_{Λ}(β)⟩ with g_{αβ} ∈ A

Proof: This is a direct consequence of Theorem 1.

Lemma.

⟨M_{Λ}(α)⟩ = ⟨α(a_{1})M_{Λ}(a_{1})⟩ ∗ ⋯ ∗ ⟨α(a_{m})M_{Λ}(a_{m})⟩

Proof:

⟨α(a_{1})M_{Λ}(a_{1})⟩ ∗ ⋯ ∗ ⟨α(a_{m})M_{Λ}(a_{m})⟩

= v^{− ∑ |α(a_{i})a_{i}|+∑ α(a_{i})^{2}}[α(a_{1})M_{Λ}(a_{1})] ∗ ⋯ ∗ [α(a_{m})M_{Λ}(a_{m})]

= v^{− ∑ |α(a_{i})a_{i}|+∑ α(a_{i})^{2}}v^{dim_{k} rad End(M(α))}[α(a_{1})M_{Λ}(a_{1})] ⋄ ⋯ ⋄ [α(a_{m})M_{Λ}(a_{m})]

= v^{− dim_{k} M(α)+dim_{k} End(M(α))}[M_{Λ}(α)]

= ⟨M_{Λ}(α)⟩ Example. Let us consider the explicit expression for w_{∗}(d), where d ∈ ℕ_{0}^{n}.

w_{∗}(d) = ∑_{dim β=d} v^{−δ(β)}⟨M_{Λ}(β)⟩ with δ(β) := dim_{k} Ext^{1}(M_{Λ}(β), M_{Λ}(β)). Proof: We have

w_{∗}(d) = v^{χ(d)−|d|}w_{⋄}(d) = v^{χ(d)−|d|} ∑_{dim β=d}[M_{Λ}(β)],

since any module M_{Λ}(β) with dim β = d has a unique filtration of type w_{⋄}(d). But

χ(d) − |d| = dim_{k} End(M_{Λ}(β)) − dim_{k} Ext^{1}(M_{Λ}(β), M_{Λ}(β)) − |d|
= −δ(β) + r(β).

Thus,

w_{∗}(d) = v^{χ(d)−|d|} ∑_{dim β=d}[M_{Λ}(β)]

= ∑_{dim β=d} v^{−δ(β)}v^{r(β)}[M_{Λ}(β)]

= ∑_{dim β=d} v^{−δ(β)}⟨M_{Λ}(β)⟩.

More generally, given α, β: Φ^{+} → ℕ_{0}, we have to consider

δ(β; α) = dim_{k} Ext^{1}(M(β), M(β)) − dim_{k} Ext^{1}(M(α), M(α))
= dim_{k} End(M(α)) − dim_{k} End(M(β)),

of course, we have δ(β) = δ(β; 0).

7. The isomorphism between U_{q}(n_{+}(Δ)) and H_{∗}(kΔ⃗) for Δ⃗ a Dynkin quiver

Proposition. The elements [S_{i}]^{∗(t)} with 1 ≤ i ≤ n and t ≥ 1 generate H_{∗}(Δ⃗) as a A-algebra.

Proof: Let H′ be the A-algebra generated by the elements [S_{i}]^{∗(t)} with 1 ≤ i ≤ n and t ≥ 1. By induction on dim α, we show that ⟨M_{Λ}(α)⟩ belongs to H′.

If the support of α contains more than one element, then we use the formula

⟨M_{Λ}(α)⟩ = ⟨α(a_{1})M_{Λ}(a_{1})⟩ ∗ ⋯ ∗ ⟨α(a_{m})M_{Λ}(a_{m})⟩.

By induction, all the elements ⟨α(a_{i})M_{Λ}(a_{i})⟩ belong to H′, thus also ⟨M_{Λ}(α)⟩, and thefore [M_{Λ}(α)] belong to H′.

In case the support of α consists of the unique element a_{i}, let d = α(a_{i})a_{i}, thus M_{Λ}(α) = M_{Λ}(d), and we know that

w_{∗}(d) = ⟨M_{Λ}(α)⟩ + ∑_{{dim β=d \atop β≠α}} v^{−δ(β)}⟨M_{Λ}(β)⟩.

The support of any β with dim β = d and β ≠ α contains more than one element; as we have seen, this implies that the corresponding elements ⟨M_{Λ}(β)⟩ belong to H′. Since also w_{∗}(d) is in H′, we conclude that ⟨M_{Λ}(α)⟩ belongs to H′.

Of course, with ⟨M_{Λ}(α)⟩ also [M_{Λ}(α)] belongs to H′. This completes the proof.

The fundamental relations show that we may define a ring homomorphism

η: U_{q}(n_{+}(Δ)) → H_{∗}(Δ⃗)

by η(E_{i}) = [S_{i}]. The Lemma above shows that this map is surjective.

Theorem. The map η: U_{q}(n_{+}(Δ)) → H_{∗}(Δ⃗) is an isomorphism.

We have to show that η is also injective. Let A′′ = ℚ[v, v^{−1}], and U′′ = U′′_{q}(n_{+}(Δ)) the A′′-subalgebra of U′_{q}(n_{+}(Δ)) generated by the elements E_{i}^{(t)} with 1 ≤ i ≤ n and t ≥ 0.

Also, let H′′_{∗}(Δ⃗) = H_{∗}(Δ⃗) ⊗_{A} A′′. Of course, the map η extends in a unique way to a map η′′: U′′ → H′′_{∗}(Δ⃗) (thus η′′|U_{q}(n_{+}(Δ)) = η). It remains to be seen that η′′ is injective. Both U′′ and H′′_{∗}(Δ⃗) are ℤ^{n}-graded, and η′′ respects this graduation, thus, for d ∈ ℤ^{n}, there is the corresponding map η′′_{d} : U′′_{d} → H′′_{∗}(Δ⃗)_{d}, and we show that all these maps η′′_{d} are injective.

The A′′-module U′′_{d} is torsionfree (since it is a submodule of U′_{q}(n_{+}(Δ))) and finitely generated. Since A′′ is a principal ideal domain, we see that U′′_{d} is a free A′′-module. In order to calculate its rank, we consider the factor module U′′_{d}/(v − 1). As we have seen in section 1, we can identify U′′_{d}/(v − 1) with U(n_{+}(Δ))_{d}, thus it has ℚ-dimension u(d). It follows that U′′_{d} is a free A′′-module of rank u(d). On the other hand, H′′_{∗}(Δ⃗)_{d} is the free A′′-module with basis the set of maps α: Φ^{+} → ℕ_{0} satisfying dim α = d, thus it also is a free A′′-module of rank u(d). But any surjective map between free A′′-modules of equal rank has to be an isomorphism. This completes the proof.

In our further considerations, it sometimes will be useful to identify U_{q}(n_{+}(Δ)) and H_{∗}(Δ⃗) via the map η. Under this identification, the generator E_{i} corresponds to the isomorphism class [S_{i}].

8. The canonical basis

For any pair β < α of maps Φ^{+} → ℕ_{0}, Theorem 1′ gives an element g_{αβ} ∈ A. Let g_{αα} = 1, and g_{αβ} = 0 in the remaining cases. We may consider g = (g_{αβ})_{αβ} as a matrix using some total ordering of the indices; it is the base change matrix between the basis given by the elements ⟨M_{Λ}(α)⟩ and the basis given by the elements w_{∗}(α). Note that we may assume that g is a unipotent lower triangular matrix. Let \overline{g} be obtained from g by applying the automorphism , and g′ the inverse of \overline{g}. Since \overline{w_{∗}(α)} = w_{∗}(α), we see that

w_{∗}(α) = \overline{w_{∗}(α)} = ∑_{β} \overline{g_{αβ}}\overline{⟨M_{Λ}(β)⟩},

thus

\overline{⟨M_{Λ}(α)⟩} = ∑_{β} g′_{αβ}w_{∗}(β) = ∑_{β} ∑_{γ} g′_{αβ}g_{βγ}⟨M_{Λ}(γ)⟩.

Let us denote by h = g′g the matrix product, then h is again a unipotent lower triangular matrix, and \overline{h} = h^{−1}.

There exists a unique unipotent lower triangular matrix u = (u_{αβ})_{α,β} with off-diagonal entries in ℤ[v^{−1}] without constant term, such that u = \overline{u}h (see [L6], 7.10, or also [D], 1.2).

The desired basis is

C(α) := ⟨M_{Λ}(α)⟩ + ∑_{β≺α} u_{αβ}⟨M_{Λ}(β)⟩ with u_{αβ} ∈ v^{−1}ℤ[v^{−1}]

. this is called the canonical basis of H_{∗}(Δ⃗) or also of U_{q}(n_{+}(Δ)).

Note that by construction the elements of the canonical basis are invariant under the automorphism , since

\overline{C(α)} = ∑_{β} \overline{u_{αβ}}\overline{⟨M_{Λ}(β)⟩}

= ∑_{β,γ} \overline{u_{αβ}}h_{βγ}⟨M_{Λ}(γ)⟩

= ∑_{β} u_{αβ}⟨M_{Λ}(β)⟩ = C(α).

In fact, the element C(α) is characterized by the two properties

C(α) := ⟨M_{Λ}(α)⟩ + ∑_{β≺α} u_{αβ}⟨M_{Λ}(β)⟩ with u_{αβ} ∈ v^{−1}ℤ[v^{−1}],

and

\overline{C(α)} = C(α)

.

In particular, any monomial will satisfy the second property, thus in order to show that a monomial belongs to the canonical basis, we only have to verify the first property.

9. The case 𝔸_{2}

We consider the quiver

1 ⟶ 2.

There are three positive roots a_{1} = (1, 0), a_{2} = (1, 1), a_{3} = (0, 1), with corresponding indecomposable modules S_{1} = M(1, 0), M(1, 1), S_{2} = M(0, 1). (For simplicity, we sometimes will denote the isomorphism class [S_{1}] by 1, the isomorphism class [S_{2}] by 2.)

The Auslander-Reiten quiver is of the form

M(1, 1)
↗  ↘
M(0, 1)  ⋯  M(1, 0)

Let

M(c, r, s) = cM(0, 1) ⊕ rM(1, 1) ⊕ sM(1, 0),

note that M(c, r, s) has dimension vector (c + r, r + s), it is given by a linear map

M(c, r, s)_{1} = k^{s+r} ⟶ k^{r+c} = M(c, r, s)_{2}

of rank r (thus, s is the dimension of its kernel, c the dimension of its cokernel). We may visualize M(c, r, s) as follows:

1  ∘ ⋯ ∘ ∘ ⋯ ∘
↓ ⋯ ↓
2  ∘ ⋯ ∘ ∘ ⋯ ∘
_{c} _{r} _{s}

Let ϵ(c, r, s) = dim_{k} End M(c, r, s), thus
ϵ(c, r, s) = c^{2} + r^{2} + s^{2} + cr + rs,

and for 0 ≤ i ≤ r,

ϵ(c + i, r − i, s + i) − ϵ(c, r, s) = i(i + c + s).

Claim:

⟨[cS_{2}] ⋄ [(r + s)S_{1}] ⋄ [rS_{2}] ⋄| M(c + i, r − i, s + i)⟩ = G_{i}^{c+i}.

Proof: We take an r-dimensional subspace U of the (c + r)-dimensional space M(c, r, s)_{2} such that U contains a fixed (r − i)-dimensional subspace V (the image of the given map M(c + i, r − i, s + i)_{1} → M(c + i, r − i, s + i)_{2}), thus in the (c + i)-dimensional space M(c + i, r − i, s + i)_{2}/V, we choose an arbitrary i-dimensional subspace.

Similarly:

⟨[rS_{1}] ⋄ [(c + r)S_{2}] ⋄ [sS_{1}] ⋄| M(c + i, r − i, s + i)⟩ = G_{s}^{s+i}.

Proof: Here, we take an s-dimensional subspace in the (s + i)-dimensional kernel of the map M(c + i, r − i, s + i)_{1} → M(c + i, r − i, s + i)_{2}, and the number of such subspaces is G_{s}^{s+i}.

It follows that

2^{(∗c)} ∗ 1^{(∗(r+s))} ∗ 2^{(∗r)} = ∑_{i=0}^{r} v^{−i(i+c+s)}G_{i}^{c+i}⟨M(c + i, r − i, s + i)⟩

and

1^{(∗r)} ∗ 2^{(∗(c+r))} ∗ 1^{(∗s)} = ∑_{i=0}^{r} v^{−i(i+c+s)}G_{s}^{s+i}⟨M(c + i, r − i, s + i)⟩

Note that in both expressions, the coefficient of ⟨M(c, r, s)⟩ itself is 1. Consider the coefficients of the summands with index i > 0. Since G_{i}^{c+i} has degree ic, we see that for c ≤ s, the coefficient v^{−i(i+c+s)}G_{i}^{c+i} belongs to v^{−1}ℤ[v^{−1}], similarly, for c ≥ s, the coefficient v^{−i(i+c+s)}G_{s}^{s+i} belongs to v^{−1}ℤ[v^{−1}].

Let us consider the formulae in case c = s. In this case, the right hand sides coincide, since G_{i}^{s+i} = G_{s}^{s+i}. Thus, we see:

2^{(∗s)} ∗ 1^{(∗(r+s))} ∗ 2^{(∗r)} = 1^{(∗r)} ∗ 2^{(∗(s+r))} ∗ 1^{(∗s)}

This shows the following: Proposition. The canonical basis of U_{q}(n_{+}(𝔸_{2})) consists of the following elements:

take the monomials 2^{(∗c)}∗1^{(∗(r+s))}∗2^{(∗r)} with c ≤ s and the monomials 1^{(∗r)}∗2^{(∗(c+r))}∗1^{(∗s)} with c > s.

10. The case 𝔸_{3}. Consider the following quiver

∘
↙  ↘
∘  ∘

denote the source by 2, the sinks by 1 and 3, respectively.

The indecomposable representations have the following dimension vectors

a = (100),
b = (001),
c = (111),
d = (011),
e = (110),
f = (010).

The Auslander-Reiten quiver is of the form

\begin{aligned}a & & ⋯ & & d \\ & ↘ & & ↗ & & ↘ \\ & & c & & ⋯ & & f \\ & ↗ & & ↘ & & ↗ \\ b & & ⋯ & & e \end{aligned}

Consider the dimension vector (xyz), with positive integers x, y, z. Let α: Φ → ℕ_{0} with

M(α) = M(c) ⊕ (x − 1)M(a) ⊕ (y − 1)M(f) ⊕ (z − 1)M(b).

We want to determine C(α).

Let β, β′, γ: Φ → ℕ_{0} with

M(β) = M(d) ⊕ xM(a) ⊕ (y − 1)M(f) ⊕ (z − 1)M(b),
M(β′) = M(e) ⊕ (x − 1)M(a) ⊕ (y − 1)M(f) ⊕ zM(b),
M(γ) = xM(a) ⊕ yM(f) ⊕ zM(b).

We have

ϵ(α) = x^{2} − x + y^{2} − y + z^{2} − z + 1,
ϵ(β) = x^{2} + y^{2} − y + z^{2} − z + 1,
ϵ(β′) = x^{2} − x + y^{2} − y + z^{2} + 1,
ϵ(γ) = x^{2} + y^{2} + z^{2}.

Thus, we see that

ϵ(β) − ϵ(α) = x,
ϵ(β′) − ϵ(α) = z,
ϵ(γ) − ϵ(α) = x + y + z − 1.

On the other hand,

⟨[S_{2}] ⋄ [xS_{1}] ⋄ [zS_{3}] ⋄ [(y − 1)S_{2}] ⋄| ⟨M(β)⟩⟩ = 1
⟨[S_{2}] ⋄ [xS_{1}] ⋄ [zS_{3}] ⋄ [(y − 1)S_{2}] ⋄| ⟨M(β′)⟩⟩ = 1
⟨[S_{2}] ⋄ [xS_{1}] ⋄ [zS_{3}] ⋄ [(y − 1)S_{2}] ⋄| ⟨M(γ)⟩⟩ = G_{y−1}^{y}.

It follows that 2 ∗ 1^{(∗x)} ∗ 3^{(∗z)} ∗ 2^{(∗(y−1))} = ⟨M(α)⟩ + v^{−x}⟨M(β)⟩ + v^{−z}⟨M(β′)⟩ + v^{−(x+z)}[y]⟨M(γ)⟩.

The two coefficients v^{−x}, v^{−z} belong to v^{−1}ℤ[v^{−1}]. In case x+z ≥ y, also the last coefficient v^{−(x+z)}[y] belongs to v^{−1}ℤ[v^{−1}]. Thus we see:

If x + z ≥ y, then C(α) = 2 ∗ 1^{(∗x)} ∗ 3^{(∗z)} ∗ 2^{(∗(y−1))}

In case x + z < y, we use the following equality

v^{−(x+z)}[y] = [y − x − z] + v^{−y}[x + z],

in order to see that

2 ∗ 1^{(∗x)}∗3^{(∗z)} ∗ 2^{(∗(y−1))} − [y − x − z]1^{(∗x)} ∗ 3^{(∗z)} ∗ 2^{(∗y)}
= ⟨M(α)⟩ + v^{−x}⟨M(β)⟩ + v^{−z}⟨M(β′)⟩ + v^{−y}[x + z]⟨M(γ)⟩.

Note that the last coefficient v^{−y}[x + z] belongs to v^{−1}ℤ[v^{−1}].

For x + z < y, C(α) = 2 ∗ 1^{(∗x)} ∗ 3^{(∗z)} ∗ 2^{(∗(y−1))} − [y − x − z]1^{(∗x)} ∗ 3^{(∗z)} ∗ 2^{(∗y)}

Lemma. If c ≥ a + d, c ≥ b + e, then 1^{∗(a)} ∗ 3^{∗(b)} ∗ 2^{∗(c)} ∗ 1^{∗(d)} ∗ 3^{∗(e)} belongs to the canonical basis.

Proof: Let w_{∗} = 1^{∗(a)} ∗3^{∗(b)} ∗2^{∗(c)} ∗1^{∗(d)} ∗3^{∗(e)}, and w_{⋄} = 1^{⋄(a)} ⋄3^{⋄(b)} ⋄2^{⋄(c)} ⋄1^{⋄(d)} ⋄3^{⋄(e)} Let M = M(d, c, e) be the generic module with dimension vector (d, c, e), let S = aS_{1}⊕bS_{3}. Since d ≤ a+d ≤ c, we see that Hom(M, S_{1}) = 0. Similarly, Since e ≤ b+e ≤ c, we see that Hom(M, S_{2}) = 0. Thus Hom(M, S) = 0. Let N = S ⊕ M. It follows that ⟨w_{⋄} ⋄| N⟩ = 1.

Now, consider any module N′ with ⟨w ⋄| N′⟩ ≠ 0. It follows that N′ maps surjectively to S, and, since S is projective, S is a direct summand of N′. Let i, j be maximal so that S′ = (a + i)S_{1} ⊕ (b + j)S_{3} is a direct summand of N′, say N′ = S′ ⊕ M′. Note that we have Hom(M′, S′) = 0. Let M′′ be the generic module with dimension vector equal to the dimension vector of M′. Let ϵ, ϵ′, ϵ′′ be the dimension of the endomorphism rings of N, N′, and N′′ = S′ ⊕ M′′ respectively. Then ϵ′ ≥ ϵ′′.

Note that

ϵ′′ = dim_{k} End(S′) + dim_{k} End(M′′) + dim_{k} Hom(S′, M′′)
= q(S′) + q(M′′) + ⟨S′, M′′⟩
= q(S′ ⊕ M′′) − ⟨M′′, S′⟩
= q(a + d, c, b + e) + dim Ext^{1}(M′′, S′)

where we first have used that Hom(M′′, S′) = 0, then that Ext^{1}(S′, M′′) = 0, and finally again that Hom(M′′, S′) = 0.

Let us show that dim_{k} Ext^{1}(M′′, S′) = (a + i)(c − d + i) + (b + j)(c − e + j). Note that M′′ has no direct summand of the form S_{1} or S_{3}, thus the number of indecomposable direct summands in any direct decomposition is just dim_{k} M_{2} = c, whereas the number of indecomposable direct summands with dimension vector (111) or (110) is dim_{k} M_{1} = d−i. Thus, the number of indecomposable direct summands with dimension vector (011) or (010) is c − d + i. It follows that dim_{k} Ext^{1}(M′′, S_{1}) = c − d + i. Similarly, dim_{k} Ext^{1}(M′′, S_{2}) = c − e + i.

As a consequence,

ϵ′′ = q(a + d, c, b + e) + dim_{k} Ext^{1}(M′′, S′)
= q(a + d, c, b + e) + (a + i)(c − d + i) + (b + j)(c − e + j).

In particular, we also see that

ϵ = q(a + d, c, b + e) + a(c − d) + b(c − e).

Therefore,

ϵ′ − ϵ ≥ ϵ′′ − ϵ = (a + i)(c − d + i) + (b + j)(c − e + j) − a(c − d) − b(c − e)
= i(a + c − d + i) + j(b + c − d + i) ≥ i(2a + i) + j(2b + j),

since we assume that c ≥ a + d, and c ≥ b + d. In particular, in case (i, j) ≠ (0, 0), we see that

ϵ′ − ϵ > 2(ai + bj).

On the other hand, we clearly have

⟨w ⋄| N′⟩ = G_{a}^{a+i}G_{b}^{b+j},

and this is a polynomial of degree 2(ai + bj). The coefficient of w_{∗} = 1^{∗(a)} ∗ 3^{∗(b)} ∗ 2^{∗(c)} ∗ 1^{∗(d)} ∗ 3^{∗(e)} at ⟨N′⟩ is v^{−ϵ′+ϵ}G_{a}^{a+i}G_{b}^{b+j}, thus it belongs to v^{−1}ℤ[v^{−1}]. This completes the proof.

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