Representation type of ${}^{\infty}_λ\mathcal{H}_μ^1$

0.27 MB·

by  Yuriy Drozd

Representation type of ∞H1
λ µ
Yuriy Drozd and Volodymyr Mazorchuk
6
0
0
Abstract
2

n
For a semi-simple finite-dimensional complex Lie algebra g we classify the rep-
a
J resentation type of the associative algebras associated with the categories ∞H1 of
λ µ

1 Harish-Chandra bimodules for g.
1

]
T 1 The result
R
h. Let g be a simple finite-dimensional complex Lie algebra with a fixed triangular decopo-
at sition, g = n− ⊕ h ⊕ n+, let λ and µ be two dominant and integral (but not necessarily
m regular) weights, let U(g) be the universal enveloping algebra of g, and let Z(g) be the
[ center of U(g). Denote by χ and χ the central characters of the Verma modules ∆(λ)
λ µ
2 and ∆(µ) respectively. Let further ∞H1 denote the full subcategory of the category of
λ µ
v
all U(g)-bimodules, which consists of all X satisfying the following conditions (see [23,
1
0 Kapitel 6]):
3
2 (1) X is finitely generated as a bimodule;
1
4
0 (2) X is algebraic, that is X is a direct sum of finite-dimensional g-modules with respect
/ to the diagonal action g 7→ (g,σ(g)), where σ is the Chevalley involution on g;
h
t
a
(3) x(z −χ (z)) = 0 for all x ∈ X and z ∈ Z(g);
m µ
v: (4) for every x ∈ X and z ∈ Z(g) there exists k ∈ N such that (z −χ (z))kx = 0.
λ
i
X
For regular µ the category ∞H1 is equivalent to a block of the BGG category O, associated
r λ µ
a with the triangular decomposition above, see [6]. For singular µ the category ∞H1 is
λ µ
equivalent toablockoftheparabolicgeneralizationO(p,Λ)ofO,studiedin[20]. Moreover,
from [20, 28] it follows that every block of O and O(p,Λ) is equivalent to some ∞H1.
λ µ
Every ∞H1 is equivalent to the module category of a properly stratified finite-dimensional
λ µ
associative algebra. The regular blocks of ∞H1 can be used to categorify a parabolic Hecke
λ µ
module, see [25].
Let W be the Weyl group of g and ρ be the half of the sum of all positive roots of g.
Then W acts on h∗ in the usual way and we recall the following dot-action of W on h∗:
w ·ν = w(ν + ρ) −ρ. Let G ⊂ W be the stabilizer of λ with respect to the dot-action,
1
and H ⊂ W be the stabilizer of µ with respect to the dot-action. We will say that the
triple (W,G,H) is associated to ∞H1. In the present paper we classify the categories ∞H1
λ µ λ µ
according to their representation type in terms of the associated triples, thus extending the
results of [21, 10, 22]. Let (W,G,H) be the triple, associated to ∞H1, and (W,G′,H′)
λ µ
be the triple, associated to some ∞H1 . Then from [5, Theorem 5.9] and [28, Theorem 11]
λ′ µ′
it follows that ∞H1 and ∞H1 are equivalent if there exists an automorphism, ϕ, of the
λ µ λ′ µ′
Coxeter system (W,S), where S is the set of simple reflections associated to our triangular
decomposition, such that ϕ(G) = G′ and ϕ(H) = H′. By the Coxeter type of a triple,
(W,G,H), we mean the triple that consists of the Coxeter types of the corresponding
components of (W,G,H). Note that, in general, the Coxeter type of the triple does not
determine the tripleina unique way (forexample, onecancomparethe cases (1e), (2d) and
(2e) in the formulation of Theorem 1.1 below). Our main result is the following statement:
Theorem 1.1. (1) The category ∞H1 is of finite type if and only if the Coxeter type of
λ µ
the associated triple is
(a) any and W = G;
(b) (A ,A ,A ), (B ,B ,B ), (C ,C ,C ), or (G ,A ,G );
n n−1 n n n−1 n n n−1 n 2 1 2
(c) (A ,e,e);
1
(d) (A ,A ,A );
n n−1 n−1
(e) (A ,A ,A ), where A is obtained from A by taking away the first and the
n n−1 n−2 n−2 n
last roots;
(f) (B ,A ,A ) or (C ,A ,A ), and G = H (in both cases);
2 1 1 2 1 1
(g) (B ,B ,B ) or (C ,C ,C ), where n ≥ 3;
n n−1 n−1 n n−1 n−1
(h) (A ,A ,e).
2 1
(2) The category ∞H1 is tame if and only if the Coxeter type of the associated triple is
λ µ
(a) (A ,A ×A ,A ), (A ,e,A ), (B ,e,B ), (G ,e,G ), (B ,A ,B ), (C ,A ,C ), or
3 1 1 3 2 2 2 2 2 2 3 2 3 3 2 3
(D ,D ,D ) where n ≥ 4;
n n−1 n
(b) (B ,A ,A ) or (C ,A ,A ), and G 6= H (in both cases);
2 1 1 2 1 1
(c) (A ,A ,A ×A ), n > 2;
n n−1 1 n−2
(d) (A ,A ,A ), n > 2, where A is included into A and contains either the
n n−1 n−2 n−2 n−1
first or the last root of A ;
n
(e) (A ,A ,A ), n > 2, where A is not included into A ;
n n−1 n−2 n−2 n−1
(f) (A ,A ,e), (B ,A ,e), (C ,A ,e).
3 2 2 1 2 1
(3) In all other cases the category ∞H1 is wild.
λ µ
2
For regular µ Theorem 1.1 gives the classification of the representation type of the
blocks of the category O obtained in [21] (see also [10] for a different proof). Formally, we
do not use any results from [21] and [10], however, the main idea of our proof is similar to
the one of [10].
In the case H = W (i.e. µ is most singular) Theorem 1.1 reduces to the classification of
the representation type for the algebra C(W,G) of G-invariants in the coinvariant algebra
associated to W. This result was obtained in [22] and, in fact, our argument in the present
paper is based upon it.
The last important ingredient in the proof of Theorem 1.1, the latter being presented
in Section 3, is the classification of the representation type of all centralizer subalgebras in
the Auslander algebra A of k[x]/(xn). This classification is given in Section 2. Two series
n
of centralizer subalgebras, namely those considered in Lemma 2.7 and Lemma 2.8, seem
to be rather interesting and non-trivial.
The paper finishes with an extension of Theorem 1.1 to the case of a semi-simple Lie
algebra g. This is presented in Section 4, where one more interesting tame algebra arises.
We would like to finish the introduction with a remark that just recently a first step to-
wards the classification of the representation type of the blocks of Rocha-Caridi’s parabolic
analogue O of O was made in [7]. The next step would be to complete this classification
S
and then to classify the representation type of the “mixed” version of O and O(p,Λ). As
S
the results of [7] and of the present paper suggest, this might give some interesting tame
algebras in a natural way.
2 Representation type of the centralizer subalgebras
in the Auslander algebra of k[x]/(xn)
In the paper we will compose arrows of the quiver algebras from the right to the left. Let
k be an algebraically closed field. Recall that, according to [17], every finite-dimensional
associative k-algebra has either finite, tame or wild representation type. In what follows
we will call the latter statement the Tame and Wild Theorem. The algebras, which are
not of finite representation type, are said to be of infinite representation type.
Let A = (A ,A ) be a k-linear category. An A-module, M, is a functor from A to
ob mor
the category of k-vector spaces. In particular, for x ∈ A and α ∈ A we will denote by
ob mor
M(x) and M(α) the images of x and α under M respectively.
For a positive integer n > 1 let A be the algebra given by the following quiver with
n
relations:
a1 a2 an−1 a b = b a , i = 1,...,n−2,
1 jj ** 2 jj ++ ...kk ++n i i i+1 i+1
a b = 0.
b1 b2 bn−1 n−1 n−1
The algebra A is the Auslander algebra of k[x]/(xn) (see for example [13, Section 7]). For
n
X ⊂ {2,3,...,n} let e denote the direct sum of all primitive idempotents of A , which
X n
3
corrrespond to the vertexes from {1} ∪ X. Set AX = e A e . The main result of this
n X n X
section is the following:
Theorem 2.1. (i) The algebra AX has finite representation type if and only if X ⊂
n
{2,n}.
(ii) The algebra AX has tame representation type if and only if either n > 3 and X = {3},
n
{2,3}, {n−1}, {n−1,n}, or n = 4 and X = {2,3,4}.
(iii) The algebra AX is wild in all other cases.
n
To prove Theorem 2.1 we will need the following lemmas:
Lemma 2.2. The algebra A{m} has infinite representation type for m ∈ {3,...,n−1} and
n
n ≥ 4.
Proof. The algebra A{m} is given by the following quiver with relations:
n
ax = ya, xb = by,
a
x 991 jj ++mff y ab = ym−1, ba = xm−1, (1)
b yn−m+1 = 0,
where x = b a , y = b a , a = a ...a , b = b ...b . Modulo the square of the
1 1 m m m−1 1 1 m−1
radical A{m} gives rise to the following diagram of infinite type:
n
1 m.
1 qMqMqMqMqMqMqMqMqMqMqMqMqMm
Hence A{m} has infinite representation type as well.
n
Lemma 2.3. The algebra AX is wild for X = {3,m}, where m > 4.
n
Proof. In this case the algebra AX is given by the following quiver with relations:
n
ax = ya, xb = by,
y
sy = zs, yt = tz,
a (cid:6)(cid:6) s
x 1 jj ** 3 jj ++m z ab = y2, ba = x2,
99 ff
b t st = zm−3, ts = ym−3,
zn−m+1 = 0,
where x = b a , y = b a , z = b a , a = a a , b = b b , s = a ...a , t = b ...b .
1 1 3 3 m m 2 1 1 2 m−1 3 3 m−1
Note that z = 0 if m = n. Modulo the square of the radical AX gives rise to the following
n
diagram:
1 3 m
1 qMqMqMqMqMqMqMqMqMqMqMqMqM3 qMqMqMqMqMqMqMqMqMqMqMqMqMm(cid:31)(cid:31)(cid:31)
4
(where the dashed line disappears in the case m = n). With or without the dashed line
the diagram is not an extended Dynkin quiver and hence is wild (see [14, 12]). Hence AX
n
is wild as well.
Lemma 2.4. The algebra AX is wild for X = {2,n−1} and n ≥ 5.
n
Proof. To make the quivers in the proof below look better we set m = n−1. The algebra
AX is given by the following quiver with relations:
n
sab = xs, abt = tx,
a s
1 jj **2 jj ++ m x st = 0, ts = (ab)n−3,
ff
b t x2 = 0,
where a = a , b = b , s = a ...a , t = b ...b , x = b a . The universal covering
1 1 n−2 2 2 n−2 n−1 n−1
of AX has the wild fragment (a hereditary algebra, whose underlined quiver is not an
n
extended Dynkin diagram, see [14, 12]) indicated by the dotted arrows in the following
picture:
... ... ...
1 a //2 s //m
1 wwooooooobaooooooo// 2 s //m(cid:15)(cid:15)x
z
b zz
zz x
z
z
1 ww a //2 s zzzz //m(cid:15)(cid:15)
z
z
b t zz
zz x
z
z
11 wwwwooooooobaaooooooo//// 22 }}}}zzzzzztzzzzzzz ss ////mm(cid:15)(cid:15)(cid:15)(cid:15)x
... ... ...
Hence AX is wild as well.
n
Lemma 2.5. The algebra A{3,4} is wild.
5
Proof. The algebra A{3,4} is given by the following quiver with relations:
5
ax = tsa, xb = bts,
a s
x 1jj **3jj ** 4 ba = x2, ab = (ts)2,
99
b t (st)2 = 0,
5
where a = a a , b = b b , s = a , t = b , x = b a . The universal covering of A{3,4} has the
2 1 1 2 3 3 1 1 5
wild fragment (a hereditary algebra, whose underlined quiver is not an extended Dynkin
diagram, see [14, 12]) indicated by the dotted arrows in the following picture:
... ... ...
a s
1 // 3 // 4
x t
(cid:15)(cid:15)
a ww s
1 //3 // 4
x1(cid:15)(cid:15) xx b a //3wwooooooostooooooo// 4
... ... ...
Hence A{3,4} is wild as well.
5
Lemma 2.6. The algebra A{m} is wild for m ∈ {4,...,n−2} and n ≥ 6.
n
Proof. The algebra A{m} is given by (1). We consider its quotient B given by the additional
n
relations x3 = y3 = ab = ba = 0 (which is possible because of our restrictions on m and
n). Then the universal covering of B exists and has the following fragment,
m 1 ,
AAAAAyAAA ~~}}a}}}}}} (cid:31)(cid:31)(cid:31) ?????x???(cid:31)(cid:31)
m (cid:31) 1 m 1
AAAAAyAAA (cid:15)(cid:15)(cid:31)(cid:31)(cid:31) (cid:127)(cid:127)(cid:127)(cid:127)a(cid:127)(cid:127)(cid:127)(cid:127)(cid:127)(cid:127) =====x==(cid:30)(cid:30) (cid:127)(cid:127)(cid:127)(cid:127)(cid:127)b(cid:127)(cid:127)(cid:127)(cid:127)(cid:127) AAAAAyAAA (cid:127)(cid:127)(cid:127)(cid:127)a(cid:127)(cid:127)(cid:127)(cid:127)(cid:127)(cid:127) =====x==(cid:30)(cid:30)
m 1 m 1
which is wild by [31]. This implies that B and hence A{m} is wild.
n
Lemma 2.7. The algebra A{2,n}, n ≥ 2, is of finite representation type.
n
Proof. For n = 2,3 the statement follows from [13, Section 7]. The algebra A{2,n}, n ≥ 4,
n
is given by the following quiver with relations:
a u
1hh ((2 hh (( n uv = uab = abv = 0, vu = (ab)n−2, (2)
b v
where a = a , b = b , u = a ...a , v = b ...b . Note that these relations imply
1 1 n−1 2 2 n−1
(ab)n−1 = (ba)n = 0. The projective A{2,n}-module P(1) is injective, so we can replace
n
A{2,n} by A′ = A{2,n}/soc(P(1)) = A{2,n}/((ba)n−1), which has the same indecomposable
n n n
modules except P(1), see [18, Lemma 9.2.2]. So from now on we consider the algebra A′,
6
i.e. add the relation (ba)n−1 = 0 to (2). The algebra A′ has a simply connected covering
A, see [9], which is the category, given by the following quiver with relations (we show the
case n = 5, in the general case the arrow starting at n ends at 2 ):
k n−2+k
e
(cid:7)
(cid:7)
(cid:7)
.. .. v(cid:7)(cid:7) ..
. {{wwwbwww . (cid:3)(cid:3)(cid:7)(cid:7)(cid:7)(cid:7)(cid:7) .
10 a //20 u // n0
11111234 wwpwwpwwpwwpppppppppppppppppppppaaaabbbbpppppppppppppppppppppppp////////22221234(cid:0)(cid:0)(cid:2)(cid:0)(cid:0)(cid:2)(cid:0)(cid:0)(cid:1)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:2)(cid:2)(cid:1)(cid:1)(cid:2)(cid:2)(cid:1)(cid:1)(cid:2)(cid:2)vv(cid:2)(cid:2)(cid:1)(cid:1)(cid:2)(cid:2)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:2)(cid:1)(cid:1)(cid:2)(cid:2)(cid:2)(cid:2)(cid:2)(cid:2)vv(cid:2)(cid:2)(cid:2)(cid:2)uuu(cid:2)(cid:2)(cid:2)(cid:2)(cid:2)(cid:2)u(cid:2)(cid:2)(cid:2)(cid:2)(cid:2)(cid:2)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:2)(cid:2)(cid:0)(cid:0)(cid:2)(cid:2)(cid:0)(cid:0)(cid:2)(cid:2)(cid:0)(cid:0)(cid:2)(cid:2)vv(cid:2)(cid:2)(cid:0)(cid:0)(cid:2)(cid:2)(cid:0)(cid:0)(cid:0)(cid:0)(cid:2)(cid:2)(cid:0)(cid:0)(cid:2)(cid:2)////////nnnn4231
.. {{wwwbwww .. v(cid:7)(cid:7)(cid:7)(cid:7)(cid:7)..
. . (cid:7) .
(cid:7)
(cid:7)
(cid:3)(cid:3)(cid:7)(cid:7)
We omit the indices at the arrows a,b,u,v. They satisfy the same relations as in A′, which
are shown by the dotted lines. Consider the full subcategory B of A with the set of objects
m
S = {1 ,m ≤ k ≤ m+n−1; 2 ,m ≤ k ≤ m+n−2; n }. Let M be an A-module, N be
k k m m
its restriction to B , N = s K , where K are indecomposableeB -modules. It is well
m m i=1 i i m
knownthatevery K iscompletely determined bythesubset ofobjectsS =e {x|K (x) 6= 0}
i i i
L
and if 1 ∈ S , then 1 ∈/ S . Moreover, allK (x) with x ∈ S are one-dimensional and
m i m+n−1 i i i
all arrows between these objects correspond to the identity maps. Since uab = abv = 0,
K splits out of the whole module M whenever S ⊇ {2 ,2 }. Suppose that, for
i i m m+n−2
every integer m, N does not contain such direct summands. It implies that M(vu) = 0.
m
Therefore M can be considered as a module over A, where A is given by the following quiver
... n′ n′ ... n′ ...
v v v
(cid:15)(cid:15) (cid:15)(cid:15) (cid:15)(cid:15)
... 1 a // 2 b //1 a //2 b //... a //2 b //1 ...
u u u
(cid:15)(cid:15) (cid:15)(cid:15) (cid:15)(cid:15)
... n n ... n ...
with relations uv = uab = abv = (ab)n−2 = 0. One easily checks that any indecomposable
representation ofAisatmost ofdimension 2n−5. Hence, Aisrepresentation (locally)finite,
i.e. for every object x ∈ A there are only finitely many indecomposable representations M
7
with M(x) 6= 0. By [9], the algebra A{2,n} is representation (locally) finite as well, which
n
completes the proof.
Lemma 2.8. The algebra A{n−1,n}, n > 3, is tame.
n
Proof. For q = n−1 the algebra A{q,n} is given by the following quiver with relations
n
u a
c 991 hh ((q hh ((n cn = ab = uv = 0, vu = cn−2, cv = vba, uc = bau,
v b
wherec = b a , a = a , b = b , u = a ...a , v = b ...b . Theprojective moduleP(1)
1 1 q q n−2 1 1 n−2
is also injective, hence, using [18, Lemma 9.2.2] as it was done in the proof of Lemma 2.7,
we can replace A by A′ = A/soc(P(1)) = A/(cq). Let M be an A′-module. Choose a basis in
M(1) so that the matrix C = M(c) is in the Jordan normal form, or, further,
q
M(c) = J ⊗I ,
i mi
i=1
M
where J is the nilpotent Jordan block of size i× i and I is the identity matrix of size
i mi
m ×m (here m is just the number of Jordan blocks of size i). Thus
i i i
0 I 0 ... 0 0
m
0 0 I ... 0 0
m
 
J ⊗I = ......................
i m
0 0 0 ... 0 I 
m
 
0 0 0 ... 0 0 
 i×i
 
(here i×i means i boxes times i boxes, each of size m ). Choose bases in M(q) and M(n)
i
such that the matrices A = M(a) and B = M(b) are of the form
0 I 0 0
0 0 0 I 0
0 0 0 I
0 0 0 0 I  
A =  , B = 0 0 0 0 ,
0 0 0 0 0
0 0 0 0
0 0 0 0 0  
  0 0 0 0
   
 
where the vertical (horizontal) stripes of A are of the same size as the horizontal (re-
spectively, vertical) stripes of B, and I is the identity matrix; we do not specify these
sizes here. Set r = nq/2; it is the number of the horizontal and vertical stripes in C.
Then M(u) and M(v) can be considered as block matrices: M(u) = U = (Uij) and
k 5×r
M(v) = V = (Vk) , where k = 1,...,5 correspond to the k-th horizontal stripe of B;
ij r×5
i = 1,...,q, j = 1,...,i, and the stripe (ij) corresponds to the j-th horizontal stripe of
the matrix J ⊗ I in the decomposition of C. The conditions uc = bau and cv = vba
i mi
imply that for i > 1 the only nonzero blocks Uij and Vk can be
k ij
Uii and Ui,i−1 = Uii
k 1 5
Vk and V5 = V1.
i1 i2 i1
8
Moreover, we also have U11 = V1 = 0. Changing bases in the spaces M(x), x = 1,q,n, so
5 11
that the matrices A,B,C remain of the same form, we can replace U and V respectively
by T−1US and S−1VT, where S,T are invertible matrices of the appropriate sizes such
that SA = AS and TU = UQ, QV = VT for an invertible matrix Q. We also consider
S and T as block matrices: S = (Sij) and T = (Tk) with respect to the division of
st r×r l 5×5
A,B,C. Then the conditions above can be rewritten as follows:
• Sij can only be nonzero if i−j < s−t or i−j = s−t, s ≤ i;
st
• Sij = Sij′ if t−j = t′ −j′;
st st′
• T is block triangular: Tk = 0 if k < l, and T1 = T5;
l 1 5
• all diagonal blocks Sij and Tk are invertible.
ij k
Especially, for the vertical stripes Uii and for the horizontal stripes U of the matrix U the
k
following transformations are allowed:
1. Replace Uii by UiiZ.
2. Replace U by ZU , where k = 2,3,4.
k k
3. Replace U and U respectively by ZU and ZU .
1 5 1 5
4. Replace Uii by Uii +UjjZ, where j < i.
5. Replace U by U +U Z, where k < l.
k k l
Here Z denotes an arbitrary matrix of the appropriate size, moreover, in the cases 1–3 it
must be invertible. One can easily see that, using these transformations, one can subdivide
all blocks Uii into subblocks so that each stripe contains at most one nonzero block, which
k
is an identity matrix. Note that the sizes of the horizontal substripes of U and U must
1 5
be the same. Let Λii and Λ be respectively the sets of the vertical and the horizontal
k
stripes of these subdivisions. Note that all stripes Uij must be subdivided respectively
to the subdivision of Uii and recall that Ui,i−1 = Uii. Especially, there is a one-to-one
1 5
correspondence λ 7→ λ′ between Λ and Λ .
5 1
We make the respective subdivision of the blocks of the matrix V, too. The condition
UV = 0 implies that, whenever the λ-th vertical stripe of U is nonzero (λ ∈ Λii), the λ-th
horizontal stripe of V is zero. The conditions VU = Cq can be rewritten as
I if (i,j,s,t) = (q,1,q,q),
V Ust =
ij
(0 otherwise.
It implies that there are no zero vertical stripes in the new subdivision of Uq,q. Moreover,
if λ ∈ Λii, µ ∈ Λ , and the block Vλ is nonzero, then the µ-th vertical stripe of U is zero if
k µ
i 6= q; if i = q this stripe contains exactly one non-zero block, namely, Uµ = I. We denote
λ
9
ii
by Λ and Λ the set of those stripes from Λii and Λ , which are not completely defines
k k
by these rules. Let λ ∈ Λ , λ′ be the corresponding element of Λ . If the blocks Uµ and
5 1 λ
Uµ′ are both nonzero, write µ ∼ µ′. Note that there is at most one element µ′ such that it
λ′
holds, and µ′ 6= µ.
ii
One can verify that the sets Λ and Λ can be linearly ordered so that, applying the
k
transformations of the types 1–5 from above, we can replace a stripe Vλ by Vλ + Vλ′Z
with λ′ < λ and a stripe Vµ by Vµ + ZVµ′, where λ′ < λ, µ′ < µ for any matrix Z (of
the appropriate size). We can also replace Vλ by VλZ, where Z is invertible, and replace
simultaneously Vµ and Vµ′, where µ′ ∼ µ, by ZVµ andZVµ′ (if µ′ does notexist, just replace
V by ZV ) with invertible Z. Therefore, we obtain a special sort of the matrix problems
µ µ
considered in [8], which is known to be tame. Hence, the algebra A{q,n} is tame as well.
n
Proof of Theorem 2.1. Lemma 2.7 and Lemma 2.2 imply Theorem 2.1(i). The statement
of Theorem 2.1(iii) follows from Theorem 2.1(i) and Theorem 2.1(ii) using the Tame and
Wild Theorem. Hence we have to prove Theorem 2.1(ii) only.
It is known, see for example [13], that A has finite representation type for n ≤ 3, is
n
tame for n = 4, and is wild for all other n. This, in particular, proves Theorem 2.1(ii) for
n ≤ 4.
If n ≥ 6 then from Lemma 2.6 it follows that if AX is tame then X ⊂ {2,3,n−1,n}.
n
FromTheorem2.1(i)weknowthatX 6⊂ {2,n}. FromLemma2.3itfollowsthat{3,n−1} 6⊂
X and{3,n} 6⊂ X. FromLemma2.4itfollowsthat{2,n−1} 6⊂ X. This leaves usthecases
X = {n−1,n}, {n−1}, {2,3} and {3}. In the first two cases AX is tame by Lemma 2.8.
n
The algebra A{2,3}, n ≥ 3, is given by the following quiver with relations:
n
a s ab = ts,
1jj **2 jj ** 3
(st)n−2 = 0,
b t
where a = a , b = b , s = a , t = b . For n ≥ 5 this algebra is tame as a quotient of the
1 1 2 2
classical tame problem from [26]. Hence A{3} is tame as well.
n
For n = 5 Lemma 2.5 implies that AX is wild if X ⊃ {3,4}, Lemma 2.3 implies that
n
AX is wild if X ⊃ {3,5}, and Lemma 2.4 implies that AX is wild if X ⊃ {2,4}. Above
n n
we have already shown that the algebras A{2,3} is tame, and hence A{3} is tame as well.
5 5
Finally, that the algebras A{4,5} and A{4} are tame follows from Lemma 2.8. This completes
5 5
the proof.
3 Proof of Theorem 1.1
We briefly recall the structure of ∞H1. We refer the reader to [5, 28, 20, 24] for details. By
λ µ
[5, Theorem 5.9], the category ∞H1 is equivalent to the block O of the BGG category O,
λ 0 λ
[6]. Let O(W,G) denote the basic associative algebra, whose module category is equivalent
to O . The simple modules in O are in natural bijection with the cosets W/G (under
λ λ
this bijection the coset G corresponds to the dominant highest weight). For w ∈ W let
10