The Δ-filtered modules without self-extensions for the Auslander algebra of k[T]/⟨T^{n}⟩

Thomas Brüstle, Lutz Hille, Claus Michael Ringel, Gerhard Röhrle

ABSTRACT. It is well-known that the Auslander algebra of any representation finite algebra is quasi-hereditary. We consider the Auslander algebra A_{n} of k[T]/⟨T^{n}⟩ (here, k is a field, T a variable and n a natural number). We determine all Δ-filtered A_{n}-modules without self-extensions; they can be described purely combinatorially. We show, given any Δ-filtered module N, there is (up to isomorphism) a unique Δ-filtered module M without self-extensions which has the same dimension vector. In case k is an infinite field, N is a degeneration of this module M. In particular, we see that in this case, the set of Δ-filtered modules with a fixed dimension vector is the closure of an open orbit (thus irreducible). As observed in [HiRö], the problem of describing all Δ-filtered A_{n}-modules is the same as that of describing the conjugacy classes of elements in the unipotent radical of a parabolic subgroup P of GL(m,k) under the action of P, thus we recover Richardson’s dense orbit theorem in this instance.

1. Preliminaries.

Let k be a field. We consider k-algebras, their modules, and maps between these modules. All modules are left modules, and all maps are written on the left of the argument. Thus, if X, Y and Z are modules, the composition of homomorphisms f: X → Y and g: Y → Z is denoted by gf.

Let R_{n} = k[T]/⟨T^{n}⟩, this is a representation finite algebra with precisely n isomorphism classes of indecomposable representations M(i), 1 ≤ i ≤ n, and it is convenient to choose the indexing^{1} so that M(i) is of length n − i + 1.

Let A_{n} = End(⨁_{i} M(i)) be the Auslander-algebra of R_{n}; a description of A_{n} in terms of quivers and relations is given in Section 5 below. Our indexing of the simple A_{n}-modules E(i) corresponds to that of the indecomposables M(i). Thus the indecomposable projective A_{n}-modules P (i) embed into each other as follows

P (1) ⊃ P (2) ⊃ ⋯ ⊃ P (n − 1) ⊃ P (n).

Note that A_{n} is quasi-hereditary in a unique way, its weight set is the canonically ordered set {1, 2, …, n}. We denote by E(i) the simple factor module of P (i) and we stress that the module P (1) is just the injective envelope of E(1).

We recall some well-known properties (see Lemma 7.1 of [DlRi]): For 1 ≤ i ≤ n, the standard A_{n}-module Δ(i) is the serial module of length i with socle E(1). The class F(Δ) of all Δ-filtered (or Δ-good) modules is just the set of all modules with socle generated by E(1). In particular, F(Δ) is closed under submodules and all projective A_{n}-modules belong to F(Δ).

1 With respect to the modules themselves, it would be more natural to use the actual length as index. However, this would lead to a conflict with respect to the traditional way of using indices for the weight set of a quasi-hereditary algebra: we would have to endow a sequence of natural numbers with the ordering opposite to the natural one.

Brüstle, Hille, Ringel, Röhrle

For any Δ-filtered module M, we denote by (dim_{Δ} M)_{i} the multiplicity of Δ(i) in a Δ-filtration of M (these multiplicities are uniquely determined, since we deal with the standard modules of a quasi-hereditary algebra). The (ordered) n-tuple dim_{Δ} M with coefficients (dim_{Δ} M)_{i} is said to be the Δ-dimension vector of M. The Δ-support of M is the set of indices i such that (dim_{Δ} M)_{i} ≠ 0.

The dual statements are also of interest: For 1 ≤ i ≤ n, the costandard A_{n}- module ∇(i) is the serial module of length i with top E(1). The class F(∇) of all ∇-filtered modules is just the set of all modules with top generated by E(1). Similar to the Δ-dimension vector and the Δ-support of modules in F(Δ), there is a notion of ∇-dimension vector and ∇-support for any module in F(∇), using filtrations by costandard modules.

We denote by T (i) the unique indecomposable module^{2} which has both a Δfiltration and a ∇-filtration, one composition factor of the form E(i), and all other composition factors of the form E(j) with j < i. Note that T (n) = P (1), this module is both projective and injective.

Lemma 1. The modules in F(Δ) are just the A_{n}-modules with projective dimension at most 1.

Proof: In order to bound the projective dimension of the modules in F(Δ), it is sufficient to consider the standard modules. But Δ(n) = P (n), and for 1 ≤ i < n, a projective resolution of Δ(i) is given by

0 → P (i + 1) → P (i) → Δ(i) → 0.

This shows that the projective dimension of any module in F(Δ) is at most 1.

Conversely, assume that M is an A_{n}-module with a projective resolution 0 → P_{1} → P_{0} → M → 0. We want to show that the socle of M is the direct sum of copies of E(1). Let S be a simple submodule of M. The inclusion map S ⊆ M yields an induced exact sequence 0 → P_{1} → X → S → 0, where X is a submodule of P_{0}. If this induced sequence splits, then S is isomorphic to a submodule of X, thus of P_{0}, and therefore isomorphic to E(1). If it does not split, then Ext^{1}(S, P_{1}) ≠ 0, thus there is an indecomposable projective module P (i) with Ext^{1}(S, P (i)) ≠ 0 (and i > 1). Note that P (i) embeds into P (1) and that the quotient is just T (i − 1). If we apply Hom(S, −) to the exact sequence 0 → P (i) → P (1) → T (i − 1) → 0, we obtain an exact sequence

Hom(S, T (i − 1)) → Ext^{1}(S, P (i)) → 0,

where we use again that P (1) is injective. But the only simple module S with Hom(S, T (i − 1)) ≠ 0 is S = E(1). This completes the proof.

2. The modules Δ(I) and Δ(d).

The Auslander algebra of k[T]/⟨T^{n}⟩

In this section we define some Δ-filtered modules which fail to admit any selfextension. The indecomposable ones are labelled Δ(I), where I is a non-empty subset of {1, …, n}. We also consider suitable direct sums of such modules Δ(I): they are denoted by Δ(d), where d = (d_{1}, …, d_{n}) is an n-tuple of non-negative integers.

2 Occasionally, these modules are referred to as “tilting modules”. However, this is in conflict with the original terminology. The tilting modules we are going to consider in Section 7 are the traditional ones.

Lemma 2. (1) A module M is isomorphic to a non-zero submodule of P (1) if and only if soc M = E(1). Such a module M is indecomposable and belongs to F(Δ).

(2) The map sending a submodule M of P (1) to its Δ-support affords a bijection between the set of submodules of P (1) and the subsets of {1, 2, …, n}. In particular, P (1) has precisely 2^{n} submodules.

Proof: The equivalence stated in (1) is due to the fact that P (1) is the injective envelope of E(1). This also implies that any non-zero submodule M of P (1) is indecomposable. Such a module M belongs to F(Δ), since all projective A_{n}-modules belong to F(Δ) and F(Δ) is closed under submodules.

Let M be a submodule of P (1). First consider the case when (dim_{Δ} M)_{n} = 0. Note that A_{n−1} is the factor algebra of A_{n} modulo the trace ideal of P (n), thus the category of all A_{n}-modules N with (dim_{Δ} N)_{n} = 0 may be identified with the category of all A_{n−1}-modules. Of course, in this way T (n − 1) is just the unique indecomposable A_{n−1}-module which is both projective and injective. By induction, the Δ-support yields a bijection between the submodules of T (n−1) and the subsets of {1, 2, …, n − 1}.

Next, let (dim_{Δ} M)_{n} ≠ 0, thus Δ(n) is a submodule of M and we may consider the submodule M/Δ(n) of T (n)/Δ(n) ≃ T (n − 1). Again, using induction, we see that the Δ-support defines a bijection between the submodules of T (n) which contain Δ(n) and the subsets I of {1, 2, …, n} with n ∈ I. This completes the proof.

Dual considerations show: (1) A module M is isomorphic to a non-zero factor module of P (1) if and only if top M = E(1). Such a module M is indecomposable and belongs to F(∇).

(2) The map sending a submodule M of P (1) to its ∇-support affords a bijection between the set of factor modules of P (1) and the subsets of {1, 2, …, n}.

Definition: If I is a subset of {1, 2, …, n}, let Δ(I) be the (unique) submodule of P (1) with Δ-support I and ∇(I) the (unique) factor module of P (1) with ∇- support I.

Remark 1. For any subset I of {1, 2, …, n} with complement I′, there is an exact sequence

0 → Δ(I) → P (1) → ∇(I′) → 0.

In order to see this, we note the following: The embedding Δ(I) → P (1) has as cokernel a module with top E(1), thus this cokernel is of the form ∇(J), for some set J. Consider dimension vectors:

∑_{i=1}^{n} dim Δ(i) = dim P (1) = dim Δ(I) + dim ∇(J)

= ∑_{i∈I} dim Δ(i) + ∑_{i∈J} dim ∇(i).

Now dim ∇(i) = dim Δ(i) for all i. Since the dimension vectors dim Δ(i) are linearly independent, it follows that J = I′.

Lemma 3. Let I be a non-empty subset and j an element of {1, 2, …, n} such that i < j for each i in I. Then Ext^{1}(Δ(I), Δ(j)) is one-dimensional.

Proof: First, we note that the cokernel C of the embedding Δ(j) → P (1) has a submodule C′ isomorphic to T (j − 1) such that C/C′ has a filtration with factors ∇(j + 1), …, ∇(n). This can be seen by looking at the submodule chain Δ(j) ⊆ T (j) ⊆ T (n) = P (1); the factor module T (j)/Δ(j) is isomorphic to T (j−1), the factor module T (n)/T (j) has a filtration with factors ∇(j + 1), …, ∇(n).

Second, observe that Hom(Δ(I), ∇(r)) = 0 for all r ∉ I, in particular for all r ≥ j, thus Hom(Δ(I), C) = Hom(Δ(I), C′). Similarly, any homomorphism Δ(I) → P (1) factors through the submodule T (j − 1), since P (1)/T (j − 1) has a filtration by the modules ∇(r) with r ≥ j. This shows that the spaces Hom(Δ(I), C) and Hom(Δ(I), P (1)) have the same dimension.

If we apply Hom(Δ(I), −) to the exact sequence 0 → Δ(i) → P (1) → C → 0, we obtain an exact sequence

0 → Hom(Δ(I), Δ(i)) → Hom(Δ(I), P (1)) → Hom(Δ(I), C)
→ Ext^{1}(Δ(I), Δ(i)) → 0.

Thus we see that Ext^{1}(Δ(I), Δ(i)) has the same dimension as Hom(Δ(I), Δ(i)), but the latter space is one-dimensional.

Proposition 1. Let M be a module in F(Δ) with Δ-support in I. Then

Ext^{1}(M, Δ(I)) = 0 = Ext^{1}(Δ(I), M).

Proof: It is sufficient to consider the case where M = Δ(i) with i ∈ I. If i = n, then Δ(n) is projective, thus Ext^{1}(Δ(n), −) = 0. For 1 ≤ i < n, take the projective resolution of Δ(i) as exhibited above and apply the functor Hom(−, Δ(I)). We obtain an exact sequence

0 → Hom(Δ(i), Δ(I)) → Hom(P (i), Δ(I)) → Hom(P (i + 1), Δ(I))
→ Ext^{1}(Δ(i), Δ(I)) → 0.

Using the Δ-filtration of Δ(I), we see that dim Hom(P (i), Δ(I)) equals the number of elements j ≥ i in I. Similarly, dim Hom(P (i + 1), Δ(I)) is equal to the number of elements j > i in I, thus one less. On the other hand, Hom(Δ(i), P (1)) is one-dimensional, and all the homomorphisms Δ(i) → P (1) map into Δ(I), thus Hom(Δ(i), Δ(I)) is also one-dimensional. This shows that Ext^{1}(Δ(i), Δ(I)) = 0.

An alternative proof of Ext^{1}(Δ(i), Δ(I)) = 0 uses Remark 1: We apply the functor Hom(Δ(i), −) to the exact sequence presented there and we obtain an exact sequence

Hom(Δ(i), ∇(I′)) → Ext^{1}(Δ(i), Δ(I)) → 0.

Since Hom(Δ(i), ∇(j)) = 0 for all i ≠ j, we have Hom(Δ(i), ∇(I′)) = 0 and therefore Ext^{1}(Δ(i), Δ(I)) = 0.

In order to see that Ext^{1}(Δ(I), Δ(i)) = 0, we write I as the disjoint union of the subsets I′ = {j ∈ I | j ≤ i} and I′′ = {j ∈ I | j > i}. There is an exact sequence

0 → Δ(I′′) → Δ(I) → Δ(I′) → 0,

thus we have an exact sequence

Ext^{1}(Δ(I′), Δ(i)) → Ext^{1}(Δ(I), Δ(i)) → Ext^{1}(Δ(I′′), Δ(i)).

Clearly, Ext^{1}(Δ(I′′), Δ(i)) = 0, since Ext^{1}(Δ(j), Δ(i)) = 0 for i ≤ j. It remains to be seen that Ext^{1}(Δ(I′), Δ(i)) = 0. Let J = I′ \ {i}, and consider the exact sequence

0 → Δ(i) → Δ(I′) → Δ(J) → 0;

we denote its equivalence class in Ext^{1}(Δ(J), Δ(i)) by ϵ. If we apply Hom(−, Δ(i)) to this sequence, we obtain an exact sequence

Hom(Δ(i), Δ(i)) ⟶^{δ} Ext^{1}(Δ(J), Δ(i)) ⟶ Ext^{1}(Δ(I′), Δ(i)) ⟶ 0,

since Ext^{1}(Δ(i), Δ(i)) = 0. According to Lemma 3, we know that the vector space Ext^{1}(Δ(J), Δ(i)) is one-dimensional. The connecting homomorphism δ sends the identity morphism of Δ(i) to ϵ and this is a non-zero element; it follows that δ is surjective. This shows that Ext^{1}(Δ(I′), Δ(i)) = 0.

Corollary. Let I, J be subsets of {1, 2, …, n} such that I ⊆ J or I ⊇ J. Then

Ext^{1}(Δ(I), Δ(J)) = 0.

Definition: For any d ∈ ℕ_{0}^{n}, we define Δ(d) inductively as follows: Let I be the support of d and let d_{I} be the characteristic function on I. Then Δ(I) is defined and by induction on the Δ-dimension also Δ(d−d_{I}). Let Δ(d) = Δ(I)⊕Δ(d−d_{I}). We obtain in this way a Δ-filtered module Δ(d) with Δ-dimension vector d such that

Ext^{1}(Δ(d), Δ(d)) = 0.

In a similar way, we may define the modules ∇(d).

It is convenient to use the following visualization for Δ(d). Let d = (d_{1}, …, d_{n}) and m = ∑_{i} d_{i}. We start with an arrangement of m vertices in n columns, such that the j-th column contains precisely d_{j} vertices. More precisely, using the convention for labelling matrix entries, there is a vertex with row index i and column index j precisely when 1 ≤ i ≤ d_{j}. For example, for n = 4 and d = (3, 1, 5, 1), we work with the following arrangement:

•  •  •  • •  • •  • • •

Next, we connect these vertices by horizontal edges, whenever possible; in this manner, the rows give rise to the connected components of a graph:

•  •  •  • •  • •  • • •

Note that the rows represent the various summands Δ(I) of Δ(d); a row with vertices in the columns i_{1}, …, i_{s} corresponds to a direct summand of the form Δ({i_{1}, …, i_{s}}). The graph exhibits nicely the direct decomposition of Δ(d) into indecomposables: In our example d = (3, 1, 5, 1), we see that

Δ((3, 1, 5, 1)) = Δ({1, 2, 3, 4}) ⊕ Δ({1, 3}) ⊕ Δ({1, 3}) ⊕ Δ({3}) ⊕ Δ({3}).

(In case some of the d_{i}’s are zero, it may be necessary to add the labelling of the columns, in order to be able to recover d from the diagram.)

3. The Δ-filtered modules without self-extensions

The aim of this section is the following characterization of Δ-filtered modules without self-extensions.

Theorem 1. For a Δ-filtered module M with Δ-dimension vector d, the following conditions are equivalent:

(i) Ext^{1}(M, M) = 0. (ii) M is isomorphic to Δ(d).

The implication (ii) ⟹ (i) is a direct consequence of the last Corollary. The proof of the implication (i) ⟹ (ii) will be done in four steps. We will

work with classes M(J, s) of Δ-filtered modules, where J is a subset of {1, 2, …, n} of cardinality t, and s is an integer with 1 ≤ s ≤ t. The class M(J, s) is defined as follows: If J = {j_{1} < j_{2} < ⋯ < j_{t}}, then M(J, s) is the class of all Δfiltered A_{n}-modules M with Δ-support J which have a factor module of the form Δ({j_{1}, j_{2}, …, j_{s}}).

(a) If M belongs to M(J, s), then either s = t or else there is an extension 0 → M → N → M → 0 such that N belongs to M(J, s + 1).

Proof: Let us consider a module M in M(J, s), where s < t. Denote by φ an epimorphism M → Δ(I), say with kernel M′.

Since s + 1 belongs to the Δ-support of M, there are submodules M′′ ⊂ M′ of M such that M′/M′′ = Δ(s + 1) and such that M/M′ is Δ-filtered with Δsupport in I. Let ε: M′ → Δ(s + 1) be an epimorphism with kernel M′′. Finally, let ι : M′ → M be the inclusion map.

Let I′ = I ∪ {j_{s+1}}. Since j_{s+1} > i for all i ∈ I, there is an exact sequence 0 → Δ(s + 1) → Δ(I′) → Δ(I) → 0. This sequence is used as the upper row of the following commutative diagram with exact rows:

0 ⟶ Δ(s + 1) ⟶ Δ(I′) ⟶ Δ(I) ⟶ 0
∥  ↑^{φ′}  ↑^{φ}
0 ⟶ Δ(s + 1) ⟶ X ⟶ M ⟶ 0
ε↑  ↑_{ε′}  ∥
0 ⟶ M′ ⟶ Y ⟶ M ⟶ 0
ι↓  ↓_{ι′}  ∥
0 ⟶ M ⟶ N ⟶ M ⟶ 0

The second row is induced from the first one by φ. In order to obtain the third row, we note that with ε also Ext^{1}(M, ε): Ext^{1}(M, M′) → Ext^{1}(M, Δ(s + 1) is surjective, since M has projective dimension at most 1. Finally, the last row is induced from the third one, using ι.

Since N is an extension of M by itself, the Δ-support of N is the same as the Δ-support of M, thus equal to J. We may assume that ι′ is again an inclusion map, thus we have obtained a module N with a submodule Y and a surjection φ′ε′: Y → Δ(I′), let N′ be its kernel. We see that N/N′ has the submodule Y/N′ isomorphic to Δ(I′) and the corresponding factor module is (N/N′)/(Y/N′) ≃ N/Y ≃ M/M′. But M/M′ is Δ-filtered with Δ-support in I, thus N/N′ is just the direct sum of Δ(I′) and M/M′, according to Proposition 1. This shows that Δ(I′) is a factor module of N. It follows that N belongs to M(J, s + 1).

(b) If M is any Δ-filtered module with Δ-support J, and if Ext^{1}(M, M) = 0, then M^{2^{s−1}} belongs to M(J, s), for each 1 ≤ s ≤ t.

Proof by induction on s. For s = 1, we only have to note that any Δ-filtered module M with Δ-support J has a factor module Δ(j_{1}), where j_{1} is the minimal element of J. Now assume we know that M^{2^{s−1}} belongs to M(J, s) and that s < t. By (a), there is an exact sequence 0 → M^{2^{s−1}} → N → M^{2^{s−1}} → 0 such that N belongs to M(J, s + 1). Since Ext^{1}(M, M) = 0, this sequence splits, thus N ≃ M^{2^{s}}.

(c) If M is any Δ-filtered module with Δ-support J, and if Ext^{1}(M, M) = 0, then Δ(J) is a direct summand of M.

Proof: From (b) we know that M^{2^{t−1}} belongs to M(J, t), thus M^{2^{t−1}} has a factor module of the form Δ(J). Since J is the Δ-support of M^{2^{t−1}}, this factor module is actually a direct summand, thanks to Proposition 1. But Δ(J) is indecomposable, thus it is already a direct summand of M.

(d) Let M be a Δ-filtered module with Δ-support J and with Ext^{1}(M, M) = 0. From (c) we know that M = Δ(J) ⊕ \overline{M} for some module \overline{M}. Of course, \overline{M} is again Δ-filtered and satisfies Ext^{1}(\overline{M}, \overline{M}) = 0. By induction on the Δ-dimension, we see that \overline{M} is isomorphic to Δ(dim_{Δ} \overline{M}), and therefore, M is isomorphic to Δ(J) ⊕ Δ(dim_{Δ} \overline{M}) = Δ(dim_{Δ} M). This proves the remaining implication (i) ⟹ (ii) of Theorem 1.

Proposition 2. Let I be a subset and j an element of {1, 2, …, n} such that i < j for each i in I. Let M be Δ-filtered with Δ-support in J = I ∪ {j}. Then either Δ(J) is a direct summand of M or else any surjective map M → Δ(I) splits.

Proof: Let I = {j_{1} < j_{2} < ⋯ < j_{t−1}} and j = j_{t}, thus J = {j_{1} < j_{2} < ⋯ < j_{t}}. Let M be Δ-filtered with Δ-support in J. Consider a surjective map f: M → Δ(I). Let M′ be the largest submodule of M which is a direct sum of copies of Δ(j). Of course, the kernel U of f contains M′ and the induced map M/M′ → Δ(I) splits, according to Proposition 1. Thus, there exists a submodule U′ of M such that U + U′ = M and U ∩ U′ = M′. Note that U′/M′ is isomorphic to M/U = Δ(I). If the sequence 0 → M′ → U′ → Δ(I) → 0 splits, then we obtain a direct complement U′′ to M′ in U′, and therefore U ⊕ U′′ = M. In this case, f splits. But if the sequence does not split, then U′ is isomorphic to the direct sum of Δ(J) and copies of Δ(j), since Ext^{1}(Δ(I), Δ(j)) is one-dimensional. In particular, M contains a submodule U′′ isomorphic to Δ(J). But such a submodule is a direct summand of M by Proposition 1.

4. Homomorphisms between modules of the form Δ(I)

Lemma 4. Let I = {i_{1} > i_{2} > ⋯ > i_{s}} and J = {j_{1} > j_{2} > ⋯ > j_{t}} be subsets of {1, 2, …, n}. The following conditions are equivalent:

(i) Δ(I) ⊆ Δ(J) (as submodules of P (1)). (ii) There is a monomorphism Δ(I) → Δ(J). (iii) We have: s ≤ t and i_{r} ≤ j_{r} for 1 ≤ r ≤ s.

Proof: The implications (iii) ⟹ (i) and (i) ⟹ (ii) are obvious. Let us consider the implication (ii) ⟹ (iii). Note that Δ(I) has a submodule of the form Δ(i_{s}). Assume that there exists a monomorphism Δ(I) → Δ(J). Then Δ(i_{s}) must embed into Δ(J), but this implies i_{1} ≤ j_{1}. The map induces a monomorphism Δ(I)/Δ(i_{1}) → Δ(J)/Δ(j_{1}), this allows us to use induction.

Corollary. If I ⊆ J, then Δ(I) ⊆ Δ(J).

For our next lemma, we recall the definition of the rational completion of a module M. Let Q(M) be an injective envelope of M. The rational completion R(M) of M is the intersection of the kernels of all endomorphisms ϕ of Q(M) which vanish on M; in particular, we always have M ⊆ R(M) ⊆ Q(M). In case Q(M) is of finite length, this submodule R(M) of Q(M) is characterized as follows: it contains M, no composition factor of R(M)/M embeds into M, whereas every composition factor of the socle of Q(M)/R(M) does embed into M.

Lemma 5. Let I be a subset of {1, 2, …, n} of cardinality t ≥ 1. Then Δ(I) ⊆ P (n + 1 − t) and this is the rational completion of Δ(I).

Proof: Note that P (i) = Δ({i, i + 1, …, n}), thus P (n + 1 − t) = Δ({n − t + 1, n − t + 2, …, n}). Now I is a subset of {1, 2, …, n} of cardinality t, whereas {n − t + 1, n − t + 2, …, n} is the subset of the t largest elements of {1, 2, …, n}. According to Lemma 4 we see that Δ(I) ⊆ P (n + 1 − t).

In order to see that the inclusion Δ(I) ⊆ P (n + 1 − t) is a rational completion, we first note that the socle of Δ(I) is equal to E(1) and that P (n + 1 − t) is a submodule of the injective envelope P (1) of Δ(I). The Jordan-Hölder multiplicity of the simple module E(1) in both modules Δ(I) and P (n + 1 − t) is t, thus E(1) is not a subfactor of the factor module P (n + 1 − t)/Δ(I). On the other hand, for t < n, we know that P (1)/P (n + 1 − t) = T (n − t), thus its socle is equal to E(1), whereas P (1)/P (n + 1 − t) = 0 for t = n.

Lemma 6. Let I = {i_{1} < i_{2} < ⋯ < i_{s}} and J = {j_{1} < j_{2} < ⋯ < j_{t}} be subsets of {1, 2, …, n} such that I ⊆ J or J ⊆ I. Then dim Hom(Δ(I), Δ(J)) = min(s, t).

Proof: Let r = min(s, t). Let I^{q} = {i_{1} < i_{2} < ⋯ < i_{q}} for 1 ≤ q ≤ r. Clearly, Δ(I^{q}) is a factor module of Δ(I). We are going to show that Δ(I^{q}) is a submodule of Δ(J). First, assume that I ⊆ J. Then I^{q} ⊆ I ⊆ J shows that Δ(I^{q}) ⊆ Δ(J). Second, assume that J ⊆ I. Then J_{q} = {j_{t−q+1} < j_{t−q+2} < ⋯ < j_{t}} is a subset of J, thus Δ(J_{q}) ⊆ Δ(J). We may consider both I^{q} and J_{q} as subsets of I of the same cardinality q. Since I^{q} are the q smallest elements of I, it follows that i_{p} ≤ j_{t−q+p}, for all 1 ≤ p ≤ q, thus Δ(I^{q}) ⊆ Δ(J_{q}), according to Lemma 3. Thus, we see that Δ(I^{q}) ⊆ Δ(J_{q}) ⊆ Δ(J). Altogether, the projection Δ(I) → Δ(I^{q}) composed with the inclusions Δ(I^{q}) ⊆ Δ(J) yields a homomorphism with image Δ(I^{q}). Clearly, these homomorphisms are linearly independent. This shows that dim Hom(Δ(I), Δ(J)) ≥ min(s, t).

Any homomorphism Δ(I) → Δ(J) can be extended uniquely to a homomorphism P (n + 1 − s) → P (n + 1 − t), since these modules are just the rational completions of Δ(I) and Δ(J), respectively. But dim Hom(P (n + 1 − s), P (n + 1 − t)) = min(s, t). This completes the proof.

Corollary. Let d = (d_{1}, …, d_{n}) ∈ ℕ^{n}. If the support of d is contained in I, then

dim Hom(Δ(I), Δ(d)) = ∑_{i∈I} d_{i} = dim Hom(Δ(d), Δ(I)).

Proposition 3. Let d = (d_{1}, …, d_{n}) ∈ ℕ^{n}. Then dim End(Δ(d)) = ∑ d_{i}^{2}.

Proof: Let I be the support of d, and d′ = d − d_{I} (where again d_{I} is the characteristic function on I). Then Δ(d) = Δ(I) ⊕ Δ(d′). By induction, we can assume that dim End(Δ(d′)) = ∑_{i∈I}(d_{i} − 1)^{2}. It follows from the Corollary that dim Hom(Δ(I), Δ(d′)) = ∑_{i∈I}(d_{i} − 1) = dim Hom(Δ(d′), Δ(I)) and that dim End(Δ(I)) = |I|. It remains to see that

∑_{i∈I} d_{i}^{2} = ∑_{i∈I}(d_{i} − 1)^{2} + 2 ∑_{i∈I}(d_{i} − 1) + |I|.

As a special case, we recover the well-known fact that the dimension of A_{n} is equal to ∑_{i=1}^{n} i^{2}, since Δ((1, 2, …, n)) is just the regular representation of A_{n}.

5. Orbit closures

Assume now that k is an infinite field. Let Q be the quiver of the algebra A_{n}; its set of vertices is Q_{0} = {1, 2, …, n}, the set Q_{1} consists of the arrows α_{a}: a → a−1, for 1 < a ≤ n and β_{a}: a → a+1, for 1 ≤ a < n. In general, given an arrow γ ∈ Q_{1}, let s(γ) be the vertex where it starts, and t(γ) the vertex where it ends. Also, there is the set ρ of relations β_{a−1}α_{a} = α_{a+1}β_{a} for 1 < a < n and β_{n−1}α_{n} = 0.

Given an n-tuple d = (d_{1}, …, d_{n}) of natural numbers, let v = (v_{1}, …, v_{n}), where v_{a} = ∑_{r≥a} d_{r}. Choose a k-space V_{a} of dimension v_{a}. We consider the vector space

R(Q; V) = ∏_{γ∈Q_{1}} Hom_{k}(V_{s(γ)}, V_{t(γ)}),

its elements are written in the form m = (m_{γ})_{γ∈Q_{1}}, where m_{γ}: V_{s(γ)} → V_{t(γ)} is a klinear map; these elements m, or, more precisely, the pairs (V, m), may be considered as representations of the path algebra of the quiver Q. Let R(Q, ρ, Δ; V) be the subset of R(Q; V) of all tuples m = (m_{γ})_{γ∈Q_{1}} subject to the requirements, first, that the relations ρ are satisfied, and second, that all the maps m_{α_{a}} are injective. Clearly, we obtain in this way representations of the algebra A_{n} (since the relations ρ are satisfied) which are Δ-filtered (since we require that the maps m_{α_{a}} are injective), and the Δ-dimension vector of any such representation is d. Also, conversely, for any Δ-filtered A_{n}-module with Δ-dimension vector d, there is an isomorphic one which belongs to R(Q, ρ, Δ; V). The group G(V) = ∏_{a∈Q_{0}} GL(V_{a}) operates on R(Q, ρ, Δ; V) via conjugation: given an element g = (g_{a})_{a} with g_{a} ∈ GL(V_{a}), then g ∗ m = (g_{t(γ)}m_{γ}g_{s(γ)}^{−1})_{γ}. Two elements in R(Q, ρ, Δ; V) belong to the same G(V)-orbit if and only if the corresponding representations are isomorphic.

Theorem 2. Let M be a Δ-filtered module with Δ-dimension vector d. Then M is in the Zariski closure of the orbit of Δ(d).

Proof: Let d = (d_{1}, …, d_{n}). Denote by O the set of elements of R(Q, ρ, Δ; V) which are isomorphic to Δ(d). We want to show that M belongs to the Zariski closure \overline{O} of O.

For 1 ≤ a ≤ n, let W_{a} be a vector space of dimension d_{a} and let V_{a} = ⨁_{r≥a} W_{r}. The canonical inclusion map V_{a} → V_{a−1} will be denoted by ι_{a}, and we consider the subset R_{ι} = R_{ι}(Q, ρ, Δ; V) of all elements (m_{γ})_{γ∈Q_{1}} of R(Q, ρ, Δ; V) such that m_{α_{a}} = ι_{a}. Also, let O_{ι} = O ∩ R_{ι}. Any Δ-filtered module with Δ-dimension vector d is isomorphic to a module belonging to R_{ι}. Thus, up to isomorphism (or, equivalently, up to the operation of the group G(V)) we may assume that both M and Δ(d) are elements of R_{ι}.

We observe the following properties: First of all, R_{ι} is a linear subspace of R(Q; V) (this follows from the fact that the relations in ρ are homogeneous and linear in the β-maps). In fact, the elements m of R_{ι} can be described as follows (here, we write β_{a} instead of m_{β_{a}}) : choose arbitrary linear maps β_{ij}: W_{j} → W_{i}, for 1 ≤ j < i ≤ n, and let β_{ij} = 0 otherwise; if we set β_{a} = (β_{ij})_{a+1≤i≤n,a≤j≤n}, then this gives an element of R_{ι}, and all the elements are obtained in this way. It follows that the k-dimension of the vector space R_{ι} is just ∑_{i>j} d_{i}d_{j}.

Let P be the subgroup of G(V) of all elements g = (g_{a})_{a} with g_{a} ∈ GL(V_{a}) such that g_{a−1}ι_{a}g_{a}^{−1} = ι_{a} for 2 ≤ a ≤ n. As in the case of R_{ι}, we can describe the elements g of P as sequences of matrices in block form, where again g = (g_{a})_{a} is uniquely determined by g_{1}, and it follows that P is an affine group of dimension ∑_{i≥j} d_{i}d_{j}.

Obviously, the restriction to P of the operation of G(V) on R maps R_{ι} into itself and O_{ι} is a P -orbit. The stabilizer P_{0} of Δ(d) in P is just the automorphism group of Δ(d), thus the group of units of the endomorphism ring End(Δ(d)). The dimension of this group P_{0} is the same as the that of End(Δ(d)), thus, according to Proposition 3, equal to ∑ d_{a}^{2}. The dimension of the orbit O_{ι} is the difference of dim P and dim P_{0}, thus equal to ∑_{i>j} d_{i}d_{j}. As a consequence, O_{ι} is dense in R_{ι}. This shows that M belongs to the closure of O_{ι}.

Remark 2. The subgroup P of GL(V_{1}) used in the proof is a parabolic subgroup, and R_{ι} is the Lie algebra of the unipotent radical of P. To be more precise: As we have mentioned, any element m of R_{ι} is uniquely determined by the map β_{1}, the additional maps β_{a} being just restrictions of β_{1}; similarly, any element g of P is uniquely determined by g_{1}, again the additional components g_{a} are just restrictions of g_{1}. The projection of P given by g ↦ g_{1} yields an embedding of P into GL(V_{1}), its image is the parabolic subgroup defined by the flag of subspaces V_{n} ⊆ V_{n−1} ⊆ ⋯ ⊆ V_{2} ⊆ V_{1}, and we may identify R_{ι} with the Lie algebra of the unipotent radical of P. These arguments just reproduce some of the considerations in [HiRö]. We present an explicit example at the end of Section 8.

Corollary. The algebraic varieties R(Q, ρ, Δ; V) are irreducible.

Remark 3. For any Q_{0}-graded vector space V, we denote by R(Q, ρ; V) the set of all elements in R(Q; V) which satisfy the relations ρ. Of course, the elements of R(Q, ρ; V) can be considered as representations of A_{n} with dimension vector v = (v_{1}, …, v_{n}), where v_{a} = dim V_{a}, for any a ∈ Q_{0}. As before, the orbits of the group action of G(V) on R(Q, ρ; V) via conjugation just yield the isomorphism classes. Note that v is the dimension vector of some Δ-filtered module M if and only if v_{1} ≥ v_{2} ≥ ⋯ ≥ v_{n}, and then the Δ-dimension vector is d, where d_{a} = v_{a} − v_{a+1} for 1 ≤ a < n and v_{n} = d_{n}.

By Theorem 2, the orbit of Δ(d) is dense in R(Q, ρ, Δ; V); in particular, R(Q, ρ, Δ; V) is irreducible. Since Ext^{1}(Δ(d), Δ(d)) = 0, the orbit of Δ(d) is also open in R(Q, ρ; V), and thus its closure in R(Q, ρ; V) yields an irreducible component of R(Q, ρ; V). However, in general there will be several irreducible components in R(Q, ρ; V). Namely, the dimension vectors of Δ-filtered modules are at the same time the dimension vectors of ∇-filtered modules, thus we may similarly consider the nabla-analogues ∇(d) of the modules Δ(d). Clearly, the closure of the orbit of ∇(d) will also be an irreducible component of R(Q, ρ; V), and it will be different from the closure of the orbit of Δ(d), except in case Δ(d) = ∇(d), thus in case the module Δ(d) is a direct sum of modules of the form T (a_{i}).

6. More closures

Again, we assume that the field k is infinite. We consider the sets M(J, s) defined at the beginning of Section 3.

Proposition 4. Let J be a subset of {1, 2, …, n} of cardinality t and let 1 ≤ s ≤ t. The set M(J, s − 1) is contained in the closure of the set M(J, s).

Proof: Consider a module M with Δ-dimension vector d which belongs to M(J, s − 1). We want to show that M belongs to the closure of M(J, s). Thus, assume that M does not belong to M(J, s). Let J = {j_{1} < ⋯ < j_{t}}. We consider the set I = {j_{1} < ⋯ < j_{s−1}} and have to deal with the two vertices i = j_{s−1} and j = j_{s}. Let M′ be the maximal submodule of M with a Δ-filtration with factors of the form Δ(j′), where j′ > j. According to the splitting lemma, we know that we can find submodules X, Y of M such that X + Y = M, X ∩ Y = M′, and X/M′ is isomorphic to Δ(I). Choose submodules M′ ⊂ X′ ⊆ X and M′ ⊂ Y′ ⊆ Y such that X′/M′ is isomorphic to Δ(i) and Y′/M′ is isomorphic to Δ(j).

Let v = (v_{1}, …, v_{n}) be the dimension vector of M and fix vector spaces V_{a} of dimension v_{a}. Take a realization (m_{γ})_{γ∈Q_{1}} of M in R(Q, ρ, Δ; V). We are going to work with bases B_{a} of the vector spaces V_{a} which are compatible with the action of the monomorphisms m_{α_{a}}. In order to do so, we start with a Δ-filtration of M, say

0 = M^{(0)} ⊂ M^{(1)} ⊂ ⋯ ⊂ M^{(u)} = M

and choose for M^{(i)}/M^{(i−1)} ≃ Δ(a_{i}) an element of M_{a_{i}}^{(i)} \ M_{a_{i}}^{(i−1)} and take these elements as well as their images under the compositions of the various maps m_{α_{a}}. In this way, we obtain all the bases B_{a} simultaneously.

The Δ-filtration which we want to use should be a refinement of the following chain of submodules of M

(∗)  0 ⊆ M′ ⊂ Y′ ⊂ X′ + Y′ ⊆ X + Y′ ⊆ M.

Note that the factors Y′/M′ ≃ Δ(j) and (X′ +Y′)/Y′ ≃ X′/M′ ≃ Δ(i) are already standard modules, thus do not have to be refined. Let x = x_{i} be the chosen base elements for the factor Y′/M′, thus x_{i} is an element of Y′_{i} \ M′_{i}. Let y = y_{j} be the chosen base elements for the factor (X′ + Y′)/Y′, we may assume that y_{i} is an element of X′_{i} \ M′_{i}. Inductively, we define x_{a−1} = m_{α_{a}}(x_{a}), for 1 < a ≤ i, and y_{a−1} = m_{α_{a}}(y_{a}), for 1 < a ≤ j.

Let us consider the subset R_{m}(Q, ρ, Δ; V) of all elements m′ in R(Q, ρ, Δ; V) with m′_{γ} = m_{γ} for all γ = α_{i}; note that R_{m}(Q, ρ, Δ; V) is a linear subspace, since the relations in ρ are linear, once we fix the maps α_{i}. To be more precise, we introduce an addition +_{α} on R_{m}(Q, ρ, Δ; V) by

(m′ +_{α} m′′)_{γ} = \{ \begin{matrix}m_{γ} & if γ = α_{i}, \\ m′_{γ} + m′′_{γ} & if γ = β_{i}. \end{matrix}

Similarly, there is a scalar multiplication ·_{α} on R_{m}(Q, ρ, Δ; V) defined by

(λ ·_{α} m′)_{γ} = \{ \begin{matrix}m_{γ} & if γ = α_{i}, \\ λm′_{γ} & if γ = β_{i}, \end{matrix}

where m′ belongs to R_{m}(Q, ρ, Δ; V) and λ ∈ k.

For example, we may consider the element m^{0} of R_{m}(Q, ρ, Δ; V) defined by m_{γ}^{0} = m_{γ} for γ = α_{i} and m_{γ}^{0} = 0 for γ = β_{i}; this is just the zero element, and it corresponds to a representation of A_{n} which has the same dimension vector as M, but is the direct sum of standard modules.

A special element of R_{m}(Q, ρ, Δ; V) which turns out to be of interest, denoted by m^{xy}, is defined as follows: First of all, m_{γ}^{xy} = m_{γ} for γ = α_{i}. Second, we require that m_{β_{i}}^{xy}(x_{i}) = y_{i+1}, for 1 ≤ a ≤ i. Finally, m_{γ}^{xy}(z) = 0 for all other elements z of the bases B_{a}. This element m^{xy} yields a representation of A_{n} which is the direct sum of one copy of Δ({i, j}) and additional standard modules.

Now we can define the desired one-parameter family: For λ ∈ k, let

m(λ) = m +_{α} (λ ·_{α} m^{xy}).

The A_{n}-module corresponding to m(λ) is denoted by M(λ).

Of course, M(0) = M. It remains to be shown that for λ ≠ 0, the module M(λ) belongs to M(J, s). Note that by construction, the chain of subspaces (∗) is still a chain of submodules, for any parameter λ, but we should add the parameter λ in brackets (−) in order to be specific:

0 ⊆ M′(λ) ⊂ Y′(λ) ⊂ (X′ + Y′)(λ) ⊆ (X + Y′)(λ) ⊆ M(λ).

Actually, M′(λ) = M′, and also M(λ)/Y′(λ) = M/Y′, these modules remain unchanged. In particular, M(λ)/Y′(λ) = M/Y′ contains (X+Y′)(λ)/Y′(λ) = (X+ Y′)/Y′ as a submodule, and this submodule is isomorphic to Δ(I). Let λ ≠ 0. In this case, the module (X + Y′)(λ)/M′(λ) is no longer the direct sum of Δ(I) and Δ(j), it is a non-split extension of Δ(j) by Δ(I). Since M(λ)/(X +Y′)(λ) = M/(X +Y′) has Δ-support in I ∪ {j}, it follows that the subfactor (X + Y′)(λ)/M′(λ) in M(λ) is isomorphic to Δ(I ∪ {j}), according to Lemma 3, and it is a direct summand of M(λ)(λ)/M′(λ) according to Proposition 1. In particular, we see that M(λ) maps onto Δ(I ∪ {j}), thus M(λ) belongs to M(J, s), as claimed.

We have shown that all the modules M(λ) with λ ≠ 0 belong to M(J, s), thus M = M(0) belongs to the closure of M(J, s).

Observe that the proof of Proposition 4 is based on explicit degenerations along affine lines. We are going to use Proposition 4 in order to present a second proof of Theorem 2. In this way, we describe an explicit method of degenerations.

This alternative proof proceeds by induction on the dimension of M, and we use a second induction on s, where M belongs to M(J, s), this time in reversed order. Thus, let J be the Δ-support of M, and assume that J has cardinality t.

Proposition 4 asserts that for 1 ≤ s ≤ t, the set M(J, s − 1) is contained in the closure of the set M(J, s). Thus, inductively, we see that M(J, s) is contained in the closure of M(J, t).

It remains to be seen that M(J, t) is contained in the closure of Δ(d). If M ≠ 0 belongs to M(J, t), then M = Δ(J) ⊕ M′ for some Δ-filtered module M′ of smaller dimension. By induction, M′ belongs to the closure of the orbit of Δ(d′), where d′ is the Δ-dimension vector of M′. Since M′ belongs to the closure of Δ(d′), it follows that M = Δ(J) ⊕ M′ belongs to the closure of the orbit of Δ(d) = Δ(J) ⊕ Δ(d′).

7. The tilting modules Δ(π)

Let S_{n} be the set of all permutations of {1, 2, …, n}. For a fixed π ∈ S_{n} let I_{π}(s) = {π^{−1}s, …, π^{−1}n}, for 1 ≤ s ≤ n and define

Δ(π) := ⨁_{s=1}^{n} Δ(I_{π}(s)).

Observe that (dim_{Δ} Δ(π))_{i} = πi for each 1 ≤ i ≤ n.

We use the following definition of a tilting module which is equivalent to the original one given in [HaRi]; this is a module of projective dimension at most 1, without self-extensions, and such that the number of isomorphism classes of indecomposable summands is equal to the number of simple A_{n}-modules. A module M is said to be basic provided it can be written as a direct sum of pairwise nonisomorphic indecomposable modules.

Theorem 3. The modules Δ(π) for π ∈ S_{n} furnish a complete list of all the isomorphism classes of basic tilting modules.

Proof: It is clear that the modules Δ(π) satisfy all the defining properties of a basic tilting module. Conversely, let M be a basic tilting module. Since M has projective dimension at most 1, we see that M belongs to F(Δ). Thus M is isomorphic to Δ(d) for some d. This means that M is isomorphic to a direct sum of modules Δ(I(s)), where I(s) are subsets of {1, 2, …, n}, and that these subsets are pairwise comparable. Since M is basic, these subsets have to be also pairwise different. Since the number of summands Δ(I(s)) has to be n, we see that the subsets I(s) have to form a complete chain of subsets of the set {1, 2, …, n}, thus M is isomorphic to a module of the form Δ(π) for some permutation π in S_{n}.

We are going to calculate the endomorphism rings of the basic tilting modules:

Theorem 4. The endomorphism ring of any basic tilting module is isomorphic to A_{n}.

Proof. Consider the module Δ(π) for some π ∈ S_{n}. Since I_{π}(s) has cardinality n − s + 1, the rational completion of Δ(I_{π}(s)) is the projective module P (s), according to Lemma 5. Thus, the canonical embedding Δ(π) ⊆ ⨁_{s=1}^{n} P (s) = P is the rational completion. This shows that the endomorphism ring E of Δ(π) embeds into the endomorphism ring the regular representation P of A_{n}, thus into the opposite of A_{n}. Since A_{n} is isomorphic to its opposite algebra, we see that E can be considered as a subalgebra of A_{n}. On the other hand, Proposition 3 applied to d = (π^{−1}1, …, π^{−1}n) shows that dim E = ∑_{i=1}^{n} i^{2} = dim A_{n}. Thus E = A_{n}.

Corollary. The endomorphism ring of a module of the form Δ(d) is equal to the endomorphism ring of a projective A_{n}-module.

8. The description of the Richardson classes in GL(m, k)

In this section let k be an algebraically closed field and m ∈ ℕ. Let P be a parabolic subgroup of GL(m, k) with unipotent radical P_{u}. Then P has a Levi decomposition P = LP_{u}. A Levi subgroup L of P is (isomorphic to) a product of general linear groups, say GL(d_{i}, k), for 1 ≤ i ≤ n, with m = ∑ d_{i}. The ordered tuple d = (d_{1}, …, d_{n}) uniquely determines the conjugacy class of P in GL(m, k). To indicate this, we write P = P (d).

By a fundamental result of R.W. Richardson [R] a parabolic subgroup P of any reductive group G always admits an open dense orbit when operating on P_{u} as well as on its Lie algebra p_{u} = Lie P_{u}. The associated unipotent class in G is known as the Richardson class C_{P} of P in G, likewise for the adjoint action of P on p_{u}.

Following Spaltenstein [S], we give an explicit description of the Richardson class C_{P} associated to P in terms of partitions for GL(m, k). A partition μ = (1^{c_{1}}, 2^{c_{2}}, …) of m canonically defines a unipotent class C_{μ} whose elements have c_{i} Jordan blocks of size i for i = 1, 2, . . .. λ(d) = λ = (λ_{1} ≥ λ_{2} ≥ …) be the partition of m associated to the tuple d (i.e., λ_{i} = d_{τi} for some permutation τ of {1, 2, …, n}). Let λ^{∗} be the partition dual to λ (i.e., the partition whose parts are the lengths of the columns of the Young tableau of λ). Then C_{P} = C_{λ^{∗}}, where P = P (d) and λ = λ(d), see [S], II §5. In particular, the Richardson class of P only depends on the partition derived from the tuple d, but not on d itself; this means that C_{P} only depends on the conjugacy class of a Levi subgroup L of P in GL(m, k) as opposed to that of P, see [S], II 3.7. Moreover, in GL(m, k) every unipotent class is the Richardson class of some parabolic subgroup [S], III §1.1.

It has been observed in [HiRö] that the problem of describing the conjugacy classes of elements of P_{u} or p_{u} under the action of P = P (d) is the same as that of describing the isomorphism classes of all Δ-filtered A_{n}-modules with Δ-dimension d. Consequently, Theorem 2 above recovers Richardson’s theorem for GL(m, k). Moreover, with our detailed description of the modules Δ(d) in the dense G(V)- orbit on the variety R(Q, ρ, Δ; V), we may obtain an explicit representative of the open dense P -orbit on p_{u} where P = P (d). We illustrate this with an example.

Example. Let n = 4 and consider the 4-tuple d = (3, 1, 5, 1). At the end of Section 2 we have attached to d a graph which now turns out to be very useful:

•  •  •  • •  • •  • • •

As mentioned above, the rows represent the various summands Δ(I) of Δ(d); a row with vertices in the columns i_{1}, …, i_{s} corresponds to a direct summand of the form Δ({i_{1}, …, i_{s}}).

Note that Δ(d) is an A_{4}-module, thus a representation (V_{1}, …, V_{4}, α_{i}, β_{i}) of the following quiver:

1 ⟶^{β_{1}} ; ⟵_{α_{2}} 2 ⟶^{β_{2}} ; ⟵_{α_{3}} 3 ⟶^{β_{3}} ; ⟵_{α_{4}} 4

Recall that V_{1} is the direct sum W_{1} ⊕ ⋯ ⊕ W_{4} (with W_{i} the cokernel of α_{i+1}). We may consider the vertices of the diagram as basis vectors of V_{1} as follows: the vertices in the i-th column represent basis vectors of W_{i} for 1 ≤ i ≤ 4. It is of importance to observe the relevance of the edges: an edge, say v w, indicates that α_{2}β_{1} maps the basis vector v to the basis vector w. Thus, this diagram illustrates the operation of α_{2}β_{1} on each of the 5 summands. As a consequence, we can use the diagram in order to write down a matrix for α_{2}β_{1} explicitly. We label the basis vectors v_{1}, …, v_{10} column-wise, as follows:

1  4  5  10 2  6 3  7 8 9

The resulting matrix of the nilpotent endomorphism α_{2}β_{1} in Lie GL(V_{1}) = End(V_{1}) is

0 0  0 0  0 1 1 1 1  1 · · · · · · · · · · · · · · · · · · · · · · ·····^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}_{.}_{.}  ^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}_{.}_{.}

The only non-zero entries of this matrix are at the positions (i, j) with i j. Thus, it is straightforward to write down this representing matrix for any given d. (We have exhibited the matrix together with its block structure: the diagonal blocks as well as the blocks below the diagonal are zero matrices; the remaining blocks form the “p_{u}-part”, the 0’s inside the “p_{u}-part” are represented by dots, where, as before, P = P (d) and Lie P_{u} = p_{u}.)

Thanks to Theorem 2 this is a matrix representative of the Richardson class associated to the adjoint action of P on p_{u}, namely of the element α_{2}β_{1} in p_{u}. According to Remark 2 above, Δ(d) is uniquely determined by the linear map β_{1}, or also by the map α_{2}β_{1} : V_{1} → V_{1}.

As we have mentioned in the last paragraph, the partition λ = (5, 3, 1, 1) associated to d uniquely determines the Richardson class of P. However, the nilpotent element given by the Jordan blocks of the dual partition λ^{∗} of λ usually does not lie in p_{u}. This happens only when d = λ.

9. References

[DlRi] V. Dlab, C. M. Ringel: The module theoretical approach to quasi–hereditary algebras. In: Representations of Algebras and Related Topics (ed. H. Tachikawa and S. Brenner). London Math. Soc. Lecture Note Series 168. Cambridge University Press (1992), 200–224.

[HaRi] D. Happel, C. M. Ringel: Tilted algebras. Trans. Amer. Math. Soc. 274 (1982), 399–443.

[HiRö] L. Hille, G. Röhrle: A classification of parabolic subgroups of classical groups with a finite number of orbits on the unipotent radical. Preprint 97–103, SFB 343, Bielefeld (1997).

[R] R. W. Richardson: Conjugacy classes in parabolic subgroups of semisimple algebraic groups. Bull. London Math. Soc. 6 (1974), 21–24.

[S] N. Spaltenstein: Classes unipotentes et Sous-groupes de Borel. Springer Lecture Note Series in Mathematics 946 (1982).

Brüstle, Ringel, Röhrle: Fakultät für Mathematik, Universität Bielefeld,

POBox 100 131, D-33 501 Bielefeld, Germany Hille: Mathematisches Seminar, Universität Hamburg

Bundesstr. 55, D-20 146 Hamburg, Germany E-mail addresses:

bruestle@mathematik.uni-bielefeld.de
hille@math.uni-hamburg.de
ringel@mathematik.uni-bielefeld.de
roehrle@mathematik.uni-bielefeld.de