Journal of Algebraic Combinatorics 5 (1996), 175-189
© 1996 Kluwer Academic Publishers. Manufactured in The Netherlands.
A Hecke Algebra Quotient and Some
Combinatorial Applications
C.K. FAN
Math Department, Harvard University, Cambridge, MA 02138
ckfan@math.harvard.edu
Received May 18. 1994; Revised April 25, 1995
Abstract. Let (W, S) be a Coxeter group associated to a Coxeter graph which has no multiple bonds. Let H be
the corresponding Hecke Algebra. We define a certain quotient H of H and show that it has a basis parametrized
by a certain subset Wc of the Coxeter group w. Specifically, Wc consists of those elements of W all of whose
reduced expressions avoid substrings of the form sts where s and t are noncommuting generators in 5. We
determine which Coxeter groups have finite Wc and compute the cardinality of Wc when W is a Weyl group.
Finally, we give a combinatorial application (which is related to the number of reduced expressions for w E Wc)
of an exponential formula of Lusztig which utilizes a specialization of a subalgebra of H.
Keywords:
1.
permutation, representation theory, non-commutative algebra, Lie theory, reductive group
Introduction
Let (W, S) be a Coxeter group whose associated Coxeter graph F is connected and has no
multiple bonds. Let I be the set of vertices so that S = {si}iEI. Let l(w), w e W, be the
smallest number n such that w is a product of n generators. Let H be the corresponding
Hecke algebra over Q[q, q - 1 ]. Denote by (T W ) W E W the standard basis. These satisfy the relations: (1) T W T W' = T w w ' , if w, w' e W, l(ww') = l(w) + l(w'), (2)Ts2 = (q - 1)Ts + q,
if s e S.
Let / be the two-sided ideal generated by the elements
where we have one such expression for each pair of non-commuting generators s, t e S.
Let H = H/I.
In the case where W is a Coxeter group of type An, this has been studied by Jones [7],
who attributes the notion to Temperley and Lieb [11]. For more details on this history, see
[6, p. 104].
Let Wc denote those elements of W whose reduced expressions avoid substrings of the
form sts where s and t are non-commuting generators in 5. In Section 2, we show that H
has a basis parametrized by Wc. In Section 3, we determine when Wc is finite. In Section 4,
we find explicit formulas for the cardinality of Wc when W is a Weyl group. Finally, in
Section 5, we give an application of an exponential formula of Lusztig to derive some
combinatorial identities.
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FAN
Throughout this paper, we shall use a theorem of Iwahori and Tits which says that any
reduced expression for w e W may be obtained from any other reduced expression for w
via a sequence of braid relations (see [3] for an explanation of braid relations and Coxeter
groups).
I wish to thank George Lusztig, without whom this paper would not exist.
(Added April 21, 1995: It has come to the author's attention that Stembridge [10] has
independently derived the formulas in section 4 using purely combinatorial methods and,
after hearing about the results of this paper, generalized the results of section 3 to the non
simply-laced case.)
2.
The Hecke algebra quotient H
Let (W, S) be as in the introduction. Depending on the context, we shall allow Tw to mean
Tw or its canonical projection in H.
Let Wc be as defined in the introduction. For example, in type A2, we have Wc =
{1, s, t, st, t s ] , where S = {s, t}.
Proposition 1 The elements {Tw } w e w c farm a basis for the algebra H.
The remainder of this section is devoted to a proof of Proposition 1.
Let V be a free Q[q, q-1]-module with basis (Xw)weWc.
Lemma 1 There exists an action of H on V with the property that for any w e Wc, we
have T W ( X 1 ) = Xw + (linear combination of X w ' , l ( w ' ) < l(w)).
Before proving the lemma, we show how the lemma implies the proposition.
First, we claim that {T w } w e W c spans H. Let Hc c H be the Q[q, q-1]-module spanned
by [T w ] w e W c . We proceed by induction on the length. If l(w) = 1, then w = 1 and T1 e Hc.
Suppose Tw e Hc for all w E W, l(w) < m. Choose w e W of length m. If w e Wc, then
Tw e Hc. Otherwise, there is some reduced expression si1,si2,si3 ··· Sip of w such that for
some p', we have si'p = sip'+2 and (Si'pSip'+1)3 = 1. Thus,
This last expression is a linear combination of Tw' with l(w') < l(w). By induction, we
have Tw e Hc. Since H is spanned by all Tw, w e W, we see that H = Hc.
To show linear independence of the { T w } w E W c , we again proceed by induction on length.
Clearly, [ T 1 ] is a linearly independent set. Assume that {T w } w E w c ,l(w)<n is a linearly independent set. Suppose we have Z weWc cwTw = 0 where cw = 0 whenever l(w) > n.
Then Z weWc c w T w ( X 1 ) = Z wEWc,l(w)=n cwXw + Z wEWc,l(w)<n cwXw = 0. This implies
that cw = 0 for all w e Wc, l(w) = n. By the induction hypothesis, we must then have all
cw = 0. Therefore the {T w } w e w c form a basis of H and the proposition follows.
We prove Lemma 1.
A HECKE ALGEBRA QUOTIENT
177
We change (Hecke algebra) generators from Ts to TS = Ts + 1, for s e S. These
new generators satisfy the relations: (1) Ts2 = (q + 1)Ts, (2) TsTt = T i T S if st = ts, and
(3) TsTtTs - qTs = T t T S T t - qTt, if (st)3 = 1. (See [7, Section 11.6].)
The elements (*) are the elements T S T t T s — qTs.
Thus, H is a Q[q,q - 1 ]-algebra generated by the TS, s e S satisfying the relations:
(1) Ts2 = (q + 1)Ts, (2) T S T t = T t T S if st = ts, and (3) T s T t T s = qTs if (st)3 = 1.
Let Pn denote the following six hypotheses (all w are assumed to be in Wc):
(1)
(2)
(3)
(4)
(5)
T s ( X W ) is defined for all s e S, l(w) < n.
T s (T s (X w )) = (q + 1)T s X w , if l(w) < n - 1.
T s ( T t ( X w ) ) = T t (T s (X w )) whenever st = ts, and l(w) < n - 1.
T s ( T t ( T s ( X w ) ) ) = qT s (X w ) whenever (st)3 = 1 and l(w) <n-2.
Xw = TSt1 (··· (Tsip (X1 ))···) whenever Si1Si2Si3 ··· Sip is a reduced expression for w and
l(w) <n + 1.
(6) For any expression si1si2si3 ··· si m , m < n + 1 (not necessarily reduced), Tsi1 (··· (TSim
( X 1 ) ) ···) is a linear combination of Xw' with l(w') < m.
If Pn holds for all n, Lemma 1 follows. To see this, note that for w E Wc, it makes
sense to define TW = Tsi1 ··· TSin, where si1si2si3 ··· sin is a reduced expression for w. Then,
Tw = Tw + (linear combination of T W' , l(w') < l(w)). Therefore, by Pn, parts (5) and (6),
Lemma 1 follows.
Let s, t, u e S, t = u. Define T S ( X 1 ) = Xs. Define
Finally, define
One can verify that with these definitions, P2 is true.
Fix n > 2 and assume that Pk is true for 2 < k < n. We shall establish the statement Pn.
Let w e Wc, l(w) = n. We wish to define T s ( X w ) .
If sw E Wc, then define
If sw E Wc, we proceed as follows.
First, note that l(sw) > l(w). If not, we may write w = sw', with l(w) = 1 + l(w').
Since w e Wc, we must have w' = sw e Wc, a contradiction.
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Denote by supp(W) the set of generators u e S which appear in some (or any) reduced
expression for to.
We need the following lemma.
Lemma 2 With w and s as above, i.e. w e Wc and sw e Wc, there exists a unique
t e S such that every reduced expression of w may be parsed uniquely as follows: w =
w 1 tw 2 sw 3 , where l(w) = l(w 1 ) + l(wi) + l(w 3 ) + 2, (st)3 = 1, and s commutes with
every u e supp(w1) U supp(w2).
Assuming this lemma, we show how we define T s ( X w ) .
Let Si1Si2Si3 ···Sin be a reduced expresssion for w, and let w1, w2, w3, and t be as
in the lemma. Thus, for some integers f and g, we have w1 = si1si2si3 ···s i f _ 1 and
w2 = s if+1 s if+2 s if+3 ··· Sig-1. Furthermore, sif = t and sig = s.
We define Y1 = Tsi1 (··· (Tsic-1 (T s (X W 2 W 3 ))) ···). By the uniqueness of w1, w2, and w3,
and the induction hypothesis Pn-1 parts (1) and (6), this is a well-defined construct. Note
that Y\ is a linear combination of X'w with l(w') < n — 1. (This follows from the induction
hypothesis Pn-1 parts (5) and (6).) Also, note that l(w 2 W 3 ) = l(w 2 ) + l(w3).
We claim that Y\ does not depend on i. Since every reduced expression of w can be
obtained from any other via a sequence of commutation relations (recall that w e Wc),
it suffices to show that Y1 = Yl, where i' is obtained from i by switching two adjacent
coordinates ih, ih+1 of i where sih, and Sih+1 commute.
Let w'1, w'2, w'3, and t be associated to i' as in the lemma.
If f < h, then w'1 = w1 and w'2w'3 = w2w3. Thus, YI = YI' because both are defined by
the same expression.
If h < f — 1, then w'k = wk, but the expressions which define Yi and Yi' differ by the
switching of Tsih and Tsih+1. In this case, the induction hypothesis Pn-1 parts (3) and (6)
imply Yi = Yi'.
The remaining two cases, namely h = f — 1 and h = f, are similar. We treat the
case where h = f - 1. In this case, Yi' = Tsi1 (··· (Tsif-2 (T s (Xsi f-1w2w3 ))) ···). Using the
induction hypothesis Pn-1 parts (3), (5), and (6), this can be seen to be the same as Yi.
Because Y\ is independent of i, we can define
for any choice of reduced expression for w.
We now prove Lemma 2.
Let si1si2si3 • • • sin be a reduced expression for w. Since w e Wc, every reduced expression
for w may be obtained from this one via a sequence of commutation relations. Therefore
every reduced expression of w is of the form SiP(1)siP(2)SiP(3) • • • siP(n) where P is a permutation
of the letters 1,... ,n.
We remark that if Sik does not commute withsSi1,with k < l, then P - 1 (k) < P - 1 (l).
Let si0 = s. Since sw E Wc, we can apply a sequence of commutation relations to
Si 0 Si 1 Si 2 • • • Sin and obtain some expression Si z(0) Si z(1) si z(2) • • • siz(n), where a is a permutation
of the letters 0 , . . . , n, and there exists some m where siz(m) = siz(m+2) and (Siz(m)Siz(m+1))3 = 1.
A HECKE ALGEBRA QUOTIENT
179
We claim that z(m) = 0. First note that by the remark, z (m + 1) = 0 and z(m + 2) = 0.
(There are non-commuting generators to the left of these.) If z ( m ) = 0, then by the
remark, we can move si0 back to its original place. That is, if we set z = z -1 (0), then
ssiz(0)siz(1)siz(2) ··· Siz(z-1)Siz(z+1)Siz(z+3) ··· Siz(n) is a reduced expression for sw. However,
this yields a reduced expression for w which is plainly not in Wc, a contradiction. Therefore
a(m) = 0.
We also have Siz(m+2) = s and Si z(m+1) = t for some t E S, such that (st) 3 = 1 .
Let f = z(m + 1) and g = z(m + 2). Notice that, by the remark, f and g are
uniquely determined by i. Indeed, sig = s is characterized by being the first occurrence of
s in Si1Si2Si3 • • • sin (reading from left to right), and sif = t is the only generator occurring
before sig which does not commute with s. (If more than one such element existed, the
reduced expression obtained via a would not be possible.) Now set w1 = si1si2si3 • • • sif-1,
w2 = s if+1 s if+2 s if+3 • • · Sig-1, and w3 = sig+1sig+2sig+3 • • • sin.
All that remains to show is that t is independent of the choice of reduced expression for w.
Let si'1si'2si'3 • • • si'n be another reduced expression for w. Let t' be the unique generator which
does not commute with s and which occurs before the first occurrence of s (which must
necessarily occur) in this reduced expression. From the remark, since t appears before the
first occurrence of s in si1si2si3 • • • sin, we must have that t appears before the first occurrence
of s in si'1si'2si'3 • • • Si'n as well. Therefore, we must have t' = t.
The lemma follows.
By construction, it follows that Pn parts (1), (5), and (6) hold.
We shall use the following immediate consequence of Lemma 2.
Lemma 3 Let w e Wc, s e S. If sw e Wc, then we may write w = w 1 tsw 2 , where
l(w) = l ( w 1 ) + l(w 2 ) + 2, s commutes with all u e supp(w 1 ),and(st) 3 = 1. Furthermore,
if w = w'1t'sw'2 is another such expression, then t = t'.
We now check Pn, part (2).
Let w e Wc,l(w) = n - 1.
Case 1. Suppose sw e Wc. If w < sw, then TS(TS(XW)) = (q + 1 ) X s w and (q + 1)
T s ( X w ) = (q + 1 ) X S W . If w> sw, then TS(TS(XW)) = (q + 1) 2 X W and (q + 1 ) T s ( X w ) =
(q + 1 ) 2 X W .
Case 2. Suppose sw E Wc. Write w = w 1 tsw 2 , as in the lemma.
We have T s (T s (X w )) = T s ( q T w 1 ( T s ( X W 2 ) ) ) = q(q + 1 ) T W 1 ( T s ( X W 2 ) ) . On the other hand,
we have (q + 1)TS(XW) = (q + 1 ) ( q T W 1 ( T s ( X W 2 ) ) ) .
Thus, Pn, part (2) is established.
We now check Pn, part (3).
Let r, s e S commute and pick w e Wc, l(w) = n - 1.
Case 1. Suppose rsw e Wc. We have four possibilities depending on whether w is
shortened or lengthened by r and s. The proof for each case is similar. We consider only
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the case where r w < w and sw > w. We have T s ( T r ( X w ) ) = Ts((q + 1 ) X w ) = (q + 1 ) X s w .
On the other hand, T r (T s (X w )) = Tr(Xsw) = (q + 1 ) X s w .
Case 2. Suppose rw e Wc, but sw e Wc. Write w = w 1 tsw 2 , as in the lemma. If
rw < w, we have T s (T r (X w )) = Ts((q + 1 ) X W ) = q(q + 1)Tw1 (T s (X w2 )). On the other
hand, T r ( T s ( X w ) ) = T r (qT W 1 (T s (X w 2 ))) = q(q + 1 ) T W 1 ( T X ( X W 2 ) ) . Note that if rw < w,
then rw 1 sw 2 < w 1 sw 2 , since t = r. The case rw > w follows similar lines.
Case 3. Suppose rw, sw e Wc. We have two possibilities. Either we can write w =
w 1 trsw 2 where t does not commute with r or s but both r and s commute with all u e
supp(w1), or we can write w = w 1 tsw 2 vrw 3 , where t does not commute with s, v does
not commute with r, but s commutes with all w E supp(w 1 ) and r commutes with all
u E supp(w 1 tsw 2 ). We treat only the latter case.
We have then, Ts(Tr(Xw)) = T s (qT W 1 t S W 2 (T r (X W 3 ))) = q 2 T W 1 (T s (T W 2 (T r (X w 3 )))).
To
arrive at this, we have made extensive use of induction hypothesis P n - 1 . On the other hand,
T r (T s (X w )) = T r (qT W 1 (T s (T W 2 v r w 3 (X 1 )))) =
q 2 T W 1 (T s (T w 2 (T r (X w 3 )))).
This exhausts the possibilities and establishes Pn, part (3).
The establishment of Pn, part (4) follows similar lines. Let w e Wc, l(w) = n — 2. Fix
s,r E S such that (sr)3 = 1.
Case 1. Suppose l(rsw) < l(sw) < l(w). Note that l(srsw) = l(w) — 1. We must then
have w = srw' where l(w) = l(w') + 2. Wefind,T s ( T r ( T x ( X w ) ) ) = (q + 1)T s (T r (X s r w ' )) =
q(q + 1)T s (T r (X w ' )). On the other hand, qT s (X w ) = q ( q + 1 ) ( X w ) = q(q + 1)T s (T r (X w ,)),
as desired.
Case 2. Suppose l(sw) < l(w), l(rsw) = l(w). Here, w = sw' where l(w) = l(w') + 1.
Then, T s (T r (T s (X w ))) = (q + 1)T s (T r (X sw' )) = (q + 1)T s (T r (T s (X w ' ))) = q(q + 1)T s (X w ,),
by Pn-1 part (4). On the other hand, qT s (X w ) = q(q + 1 ) ( X w ) as desired.
Case 3. Suppose l(sw) > l(w) and both sw, rsw e Wc. Note that necessarily, l(rsw) >
l(sw). We have Ts(Tr (TS(XW))) = Ts(Xrsw) = qT s (X w ) as desired.
Case 4. Suppose l(sw) > l(w), sw e Wc, and rsw e Wc. In this case, we see that we may
write w = rw 1 . This follows since by lemma 3, sw = w'1trw'2. According to Lemma 3, t is
the unique generator occurring to the left of r which fails to commute with r, but this is just
s. By the remark on ordering, we see that sw = sw'1rw'2 = srw'1w'2. We take w1 = w'1w'2.
We then find T s ( T r ( T s ( X w ) ) ) = T s (qT r (X W 1 )) = q(T s (X r w 1 )), as desired.
Case 5. Suppose sw E Wc. In this case, write w = w1tsw2 as in the lemma. T s (X w ) =
qT w1 (T s (X w2 )) = qT s (T W 1 (X W 2 )). Recall that s commutes with all u e supp(w1). Since
this last expression is a linear combination of Xw' with l(w') < n — 3, we can write
T s (T r (qT s (T w1 (X w2 )))) = q 2 T s (T W1 (Xw2)), which is just qTs(Xw), as desired.
Thus, Pn, part (4) is established.
This completes the proof of Proposition 1.
A HECKE ALGEBRA QUOTIENT
3.
181
Classification of finite Wc
By En we mean the extended E series which is defined for n > 5 and consists of a string
of n — 1 nodes which comprise a graph of type A n - 1 , along with an additional node which
is connected by a single edge to the third node of the string counting from one end. Note
that E9 = E8.
Proposition 2 If W is of type An, Dn, or En, then Wc is finite. Otherwise, Wc is infinite.
Proof: Suppose that F has a subgraph of type An for some n > 2. Denote by S0, s1,
s 2 , . . . ,sn the corresponding generators such that si and Sj commute unless i — j =
±1 mod n + 1. Let w = s0s1s2 • • • V We see that w' e Wc for any t > 0.
Therefore, if Wc is finite, F has no loops.
Suppose that F has a subgraph of type Dn for some d > 4. Label the associated generators
as follows:
Let w = S 1 S 2 S 3 • • • S n-4 b 1 b 2 S n-4 S n-5 S n-6 • • • S1a1a2. We see that wt e Wc for any t > 0.
Therefore, if Wc is finite, F can have at most one branch point, and this branch cannot
have more than three arms.
Suppose that F has a subgraph of type E6. Label the associated generators as follows:
Let w = ma2a1b1mb2b1c1mc2C1a1. One can check that w' has the property that between
any two occurrences of a generator s in wt, there occur two generators which do not commute
with s. Therefore, w' e Wc for any t > 0.
Therefore, if Wc is finite and F has a branch point, then the three arms cannot all extend
a distance of two or more from the branch point.
Suppose that F has a subgraph of type E7. Label the associated generators as follows:
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Let w = ma 2 a 1 b 1 ma 3 a 2 a 1 c 1 mb 1 c 2 c 1 ma 1 C 3 C 2 c 1 . One can check that wt has the property
that between any two occurrences of a generator s in wt, there occur two generators which
do not commute with s. Therefore, wt e Wc for any t > 0.
Therefore, if Wc is finite and F has a branch point, then no two arms can extend a distance
of three or more from the branch point.
The above considerations eliminate all graphs except those of type An, Dn, or En. Since
W itself is finite in case An, Dn, E6, E7, and E8, it suffices to check the case En, n > 8.
Assume F = En, n > 8. We shall proceed by induction on n. Let L denote the longest
possible length of a reduced expression in Wc for the Coxeter group of type En-1. Label
the generators as follows:
We group all generators of type ak together into one family called a. Similarly, define
families B and G. Note that elements from different families commute.
Lemma 4 Let w E Wc. In any reduced expression for w, we claim that:
(1) Between any two occurrences of b1, there must be at least two occurrences of m.
(2) Between any two occurrences of m, there must be generators from at least two of the
three families a, B, and G.
(3) For any four consecutive occurrences of m, at least one consecutive pair of these m's
must be separated by generators other than a\.
(4) Between any two occurrences of c n - 4 , there must be an occurrence of two m's separated
by generators of type a and B only.
Proof: Note that b\ commutes with all other generators except m. Therefore, if there
are two occurrences of b\ separated by one or fewer occurrences of m, we may perform a
series of commutations until either b1b1 or b1mb1 appears. Then either our expression is
not reduced, or w E Wc. This proves (1).
Suppose we have two occurrences of m separated by generators from family G only.
Since m commutes with all but one member of G, namely c1, there must be two occurrences
of c1 between these m's. Two consecutive occurrences of c1 must be separated by at least
two occurrences of c2. Of these, we can find two consecutive occurrences of c2. These must
be separated by at least two occurrences of c3, etc. We continue this argument until we have
two consecutive occurrences of cn_4 which must be separated by at least two occurrences
of generators which do not exist, an impossibility. A similar argument applies to the other
families. This proves (2).
Suppose we have four consecutive occurrences of m, such that between any two consecutive m's there exists an occurrence of a1. We may arrange that a\ is the only generator of
type a between the middle consecutive pair of m's by commuting a2 as necessary. If this
is not possible, it means there is an occurrence of a2 surrounded by a1 's which contradicts
w E Wc. Since this lone a] can be commuted to be adjacent to either of the two middle
A HECKE ALGEBRA QUOTIENT
183
m's, a2 must be the first generator of family a which appears reading from the second m
to the left, and a2 must be the first generator of family a which appears reading from the
third m to the right. In this case, we may commute these a 2 's so that they lie between the
two middle m's. But this allows us to form an occurrence of a 2 a 1 a 2 , contradicting w e Wc.
This proves (3).
For (4), we proceed in a similar way as in the proof of (2). If we have two occurrences
of c n-4 , we can find two consecutive such occurrences. These must be separated by at least
two occurrences of c n _ 5 . Of these, we pick two consecutive occurrences, and so on, until
we arrive at two consecutive occurrences of m separated by no member of family G. This
proves (4).
This completes the proof of Lemma 4.
n
Now assume Wc is infinite. We shall expose a contradiction.
Note that any element in Wc of length p(L + 1), p a positive integer, must involve at
least p occurrences of c n - 4 , by definition of L. Removing the vertex corresponding to m
from F results in a Coxeter graph of type A1 x A2 x A n - 4 . The corresponding Coxeter
group has a longest element, say, of length L0. Thus, any element in Wc of length p(L 0 +1)
must involve at least p occurrences of m.
Since we are assuming that Wc is infinite, there must occur elements in Wc with reduced
expression of arbitrary length. Let w e Wc be an element of length greater than 2(L + 1)
(L0 + 1). Then any reduced expression for w must have 2(L + 1) occurrences of m.
Between the first and last of these occurrences of m, there must occur at least 2 occurrences
of c n - 4 . By Lemma 4, part (4), between these two occurrences of c n _ 4 there occurs two
m's separated by generators from families a and B, but not G. By construction, there occur
four consecutive m's of which the middle two m's are precisely the aforementioned m's.
From now on, we shall refer only to these four m's.
By Lemma 4, part (1), the first and third consecutive pairs of m's are not separated by
any occurrence of b1, since b1 already occurs between the middle pair of m's. By Lemma 4,
part (2), we must then have the first and third consecutive pairs of m's separated by generators from both families a and G, and only these families. By Lemma 4, part (3), between
some consecutive pair of these four m's there cannot occur a1. The middle pair must be
separated by an occurrence of a1 because these two m's must be separated by at least two
of a1 and b1 and b1 cannot occur twice (by Lemma 4, part (1)). Thus, between the first
or third pair of m's there cannot occur a1. Between the pair which excludes a1, there must
be two occurrences of c1. Grouping together the generators of family G between this pair
yields an element of the Wc associated with a graph of type An_4 with two occurrences
of c1. The argument for Lemma 4, part (2) shows that this is impossible. Proposition 2
follows.
D
4. Some explicit formulas
In this section, let Cn = 1(2nn), the nth Catalan number. We adopt the convention that
(nk) = 0 whenever k e {0, 1, 2
n}.
184
Proposition 3
FAN
For the simply laced Weyl groups, the cardinality of Wc is given by:
We remark that for type An, the above result is well-known as the number of 321avoiding permutations (permutations P such that there does not exist a < b < c such that
P ( a ) > P(b) > P(c)), see for example [2, Section 2], and also [7, Section 11].
Proof: By Proposition 1, Wc parametrizes a basis of H. When W is a Weyl group, this
algebra is semi-simple since it is the quotient of a semi-simple algebra. Thus, the cardinality
of Wc is the sum of the squares of the dimensions of those representations of H which factor
through to H. These, in turn, are in one to one correspondence with representations of W
on which the elements (*) (interpreted as elements of the group algebra of W) act as 0.
Equivalently, this condition is given by: (ResWA2x, 1/A2)A2 = 0, where A2 is some (or any)
parabolic subgroup of W of type A2. We call this restriction property (R).
Note that any two parabolic subgroups of type A2 are conjugate. Because F is connected,
this statement follows from the following observation: If {s, t, u} C S generate a subgroup
of type A3 where su = us, then the parabolic subgroup (s, t) can be conjugated to (t, u)
using the element sutsut.
Case 1. The representations of An are parametrized by diagrams with n + 1 squares. Let
h = n + 1. The ones which satisfy the restriction property (R) are precisely those with two
or fewer columns (where we take the single column diagram to be the sign representation.)
Using the hook length formula for dimension, we have:
Case 2. Let n > 4.
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185
The irreducible representations of W of type Dn arise from certain pairs of diagrams.
For details of this correspondence, we refer the reader to [8, Sections 4.5,4.6]. Briefly, let
r + r = n, r > r > 0. Let x1 (resp. X2) be the character of an irreducible representation
E1 (resp. E2) of the symmetric group Sr (resp. S r ). There are two natural ways in which
W is a subgroup of W'n, the Weyl group of type Bn, though the choice of the imbedding is
immaterial for our purposes, so we fix one. The symmetric group Sn is naturally a quotient
of Wn. Thus, E1 ® E2, a representation of Sr x Sr, can be lifted to a representation E1,2
of W'r x W'r. Inducing this latter representation to W'n and then restricting to W gives us a
representation E, with character x, of W. If E1 and E2 are not given by the same diagram,
then E is irreducible. Otherwise, £ is a direct sum of two irreducible representations of the
same dimension.
We have the following formulae:
Condition (R) requires:
186
FAN
If r > 2, this last expression can be 0 if and only if x1(S) = -X1(1) and X2(s) =
—X 2 (1). Because any normal subgroup of the symmetric group which contains all simple
transpositions must be the whole group, we see that this is possible if and only if E\ and
E2 are the respective sign representations of Sr and Sr.
If r = 1, we must have X 1 (S) = — X1(1). Therefore, E\ is the sign representation and
E2 is the trivial representation.
If r = 0, we must have x1(1) + 3x1(S) + 2x 1 (st) = 0, which means that E1 is parametrized
by a diagram with two or fewer columns.
Note that if E\ and £2 are given by the same diagram, that is, if r = r and E1 and
E2 are both sign representations, then the two irreducible summands of E both satisfy the
restriction condition (R) because if one did not, neither could E.
We compute
By inspection, this formula is valid also when n = 2 and n = 3.
Case 3. For E6, E7, and E8 we consulted [1] to determine which representations contain
an A2 fixed vector. We shall follow the notation in [8] for the irreducible representations.
For E6, the representations are 1'p, 6'p, 15'q, and 20'p.
For E7, the representations are 1'a, 7a, 15a, 21b, 27'a, and 35'b.
For E8, the representations are 1'x, 8'z, 35'x, 50'x, and 84'x.
This completes the proof of Proposition 3.
5.
n
Application of an exponential formula of Lusztig
In this section, we assume that (W, S) is the Weyl group of a reductive algebraic group G
defined over C. Let g be the Lie algebra of G. We continue to assume g is simple of type
A, D, or E. Fix a Cartan subalgebra h and a root decomposition of g. Let R be the set
of roots. Choose a set of simple roots P = {ai}iei. Let ht (P), P E R denote the sum of
the coefficients of P when written as a linear combination of simple roots. Here, si e S
corresponds to the reflection in the root ai. Let w0 E W be the unique element for which
2l(w0) = card(R). For each i e I, fix some Ei E Gai, Ei = 0. Let u+ be the nilpotent
subalgebra of g generated by the Ei and let U+ = exp(u+).
The following exponential formula was conjectured by Lusztig [9, Section 11.4] and
proven in [4].
A HECKE ALGEBRA QUOTIENT
187
Proposition 4 Choose ( i 1 , . . . , i N ) e IN so that Si 1 Si 2 si 3 • • • SiN is a reduced expression
for W0. Let hk = ht(si 1 si 2 si 3 • • • sik-1 (aik)). For each i e I, let ni = Zik=i hk. Then
For any w E W, we denote by J(w) the set of i for which si1si2si3 • • •sin is a reduced
expression for w.
For each z e W, we shall define a map fz : W ->U.
Fix w e W and pick some j E J(w).
For each 1 < k < l(w), let hk = ht(s j 1 • • • sjk-1 (ajk)). Let n = l(z).
Now define
We claim that this definition is independent of the choice of j. The proof can be achieved
using the theorem of Iwahori and Tits referred to in the introduction. We omit the proof.
We remark that in the case where w e Wc, this will follow from Proposition 5.
We define
for any choice of j e J(w).
Note that for w e Wc, the number of times a given generator occurs in a reduced expression is independent of the reduced expression. Therefore, it makes sense to define
cz = 1ni1..niz f z ( w 0 ) , z e Wc,i e J(z). Here, the ni are as defined in the statement of
Proposition 4.
Proposition 5 Proposition 4 is equivalent to the set of equations:
where z e Wc.
Proof:
Let U+ be the universal enveloping algebra of u+. Define
Let Ei denote the image of Ei in U+. The Serre relation 2EiEjEi = E21E2 + E2E21 implies
that EiEjEi = 0 in U+ whenever (siSj)3 = 1.
FAN
188
We now return to the algebra H. Let es = qTs. We see that H is the algebra generated
by the e s , s e S with the following relations: (1) e2s- = q(q + 1)e s , (2) e s e t = e t e s , if
st = ts, and (3) e s e t e s = q3es, if (st)3 = 1. Let HQ[q] be the Q[q]-subalgebra of H
generated by the es. As a Q[q]-module, this is free with basis ew = ql(w)Tw, w e Wc. Let
H0 = HQ[q] ®Q[q] C where q acts as 0 on the Q[q]-module C. This is the the specialization
of HQ[q] to q = 0 with base field extended to C.
By construction, we have an algebra isomorphism C : H0 = U+ where C(e Si ) = Ei.
For any i = ( i 1 , . . . , in) e In, denote by £1 the monomial Ei1 • • • Ein. By convention,
when n = 0 we set E1 = 1. For each z e Wc, choose i(z) e J(z).
n
Lemma 5
The set of monomials ( E i ( z ) | z E Wc} form a basis for the algebra U+.
Proof: Via the isomorphism C, this is equivalent to showing that {e w } wEWc . is a basis of
H0. Since HQ[q] is free over Q[q], we see that dim C(H0) = dimQ [q] HQ [q] . The lemma
follows.
n
Lemma 6 There exists a unique injective group homomorphism i : U+ -> (U+)* such
that exp(aE i ) -> 1 +aEi where a e C.
Proof: Because U+ is unipotent, exp : u+ -> U+ is a bijection. Observe that U+ =
C ® rad(U+). Therefore exp : rad(U + ) -> 1 + rad(U+) C (U+)* is an injection into U+.
By the Poincare-Birkoff-Witt theorem, the natural map - : u+ -> U+ is injective because
+
U is obtained from U+ by reducing modulo elements of homogeneity 2.
Using functoriality of the exponential map between nilpotent Lie algebras and unipotent
groups, we get the desired map i. One can check that i(exp(aE i )) = 1 + aEi and the lemma
follows.
D
By Lemma 6, Proposition 4 holds if and only if the formula holds in U+ via the homomorphism i.
Direct computation reveals that the formula becomes
when interpreted as a formula in U+.
By Lemma 5, this equality can occur if and only if the various coefficients are equal,
whence the proposition.
References
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A HECKE ALGEBRA QUOTIENT
189
4. C.K. Fan and G. Lusztig, "Factorization of certain exponentials in Lie groups," to appear in Tribute to
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