p-adic analog of the Kazhdan-Lusztig Hypothesis

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia

Last update: 15 April 2014

Notes and References

This is an excerpt of the paper p-adic analog of the Kazhdan-Lusztig Hypothesis by A. V. Zelevinskii. Terrestrial Physics Institute, Academy of Sciences of the USSR. Translated from Funktsional'nyi Analiz i Ego Prilozheniya, Vol. 15, No. 2, pp. 9-21, April-June, 1981. Original article submitted November 27, 1980.

2. Contiguity Theorem

2.1. A Partial Order on M(𝒥) (see [Zel1980, Sec. 7]). Let a,bM(𝒥) be graded partitions. We say that b is obtained from a by an elementary operation if b is obtained from a by the deletion of a pair of connected segments Δ1 and Δ2 and addition in its place of the segments Δ=Δ1Δ2 and Δ=Δ1Δ2 (if Δ1Δ2=, then we add only the segment Δ); more formally, b=a-χΔ1-χΔ2+χΔ+χΔ, where χΔM(𝒥) is the characteristic function of the singleton {Δ}𝒥. We write b<a if b can be obtained from a by a series of elementary operations. It is easily seen that this definition turns M(𝒥) into a partially ordered set, where the elements from different M(𝒥)φ are incomparable with each other.

Theorem 7.1 of [Zel1980] asserts that the multiplicity mb,a(ρ) is nonzero if and only if ba; in addition, ma,a(ρ)=1.

2.2. Contiguity Theorem. Xb=abXa for each bM(𝒥). In other words, "ba" and "XaXb" are equivalent.

The remaining part of Sec. 2 is devoted to the proof of this theorem.

2.3. Proof of the Implication baXaXb. We can assume that b can be obtained from a by an elementary operation. By the same token, the whole problem reduces to the case where a=χΔ1+χΔ2 and b=χΔ1Δ2+χΔ1Δ2, where Δ1 and Δ2 are connected segments in . Let TXa. This means that V has a basis that consists of the vectors ξiVi (iΔ1) and ηiVi (iΔ2) such that Tξi=ξi+1 and Tηi=ηi+1 (we assume that ξe(Δ1)+1=0 and ηe(Δ2)+1=0). Let Δ1Δ2 (see Sec. 1.2). Let us consider the operator T0E(V) such that T0ξi=ηi+1 for b(Δ2)-1ie(Δ1) and T0 is equal to 0 on the remaining basis vectors. It is easily verified that the operator T+ε·T0 belongs to Xb for ε0. Therefore TXb, which was required to be proved.

2.4. Correspondence ada. With each graded partition a we associate a finite function da:𝒥+ by setting da(Δ)=ΔΔa(Δ) for Δ𝒮. Obviously, a is uniquely regenerated by da. Namely, for each Δ=[i,j]𝒥 we set Δ+=[i,j+1], +Δ=[i-1,j], and +Δ+=[i-1,j+1]. In this notation we have a(Δ)=da (Δ)-da (Δ+)- da(+Δ) +da(+Δ+) ,Δ𝒥.

The following proposition shows the origin of the function da.

Proposition. For each TE(V)=Eφ and each segment ΔT=[i,j]T𝒥T, define the operator TΔ:ViVj as the composition ViVi+1Vj. If TXa, then da(Δ)=rkTΔ for all Δ𝒥 if [in particular, da({i})=dimVi=φ(i) for i].

The proof is obvious. Let us observe that this proposition gives an elementary proof of the fact that aM(𝒥) is uniquely determined by the operator T (see Sec. 1.8).

2.5. Proof of the Implication XaXbba. By virtue of the preceding proposition, the inclusion XaXb implies that da(Δ)db(Δ) for all Δ𝒥. Therefore, it remains to prove the following proposition.

Combinatorial Proposition. Let a,bM(𝒥) be graded partitions. The following conditions are equivalent:

(1) ba;
(2) da(Δ)db(Δ) for all Δ𝒥, moreover, da({i})=db({i}) for i.

2.6. Proof of the Combinatorial Proposition. In fact, the implication (1) (2) has already been proved [we have shown that (1) XaXb (2)]. It can also be easily proved directly: If b=a-χΔ1-χΔ2+χΔ+χΔ (see Sec. 2.1), then we easily see that db=da+χ(+Δ+ΔΔ) (if Δ=, i.e., say, b(Δ2)=e(Δ1)+1, then +Δ+ is defined as the segment [e(Δ1),b(Δ2)]).

Fundamental Lemma. Let d:𝒥L+ be a nonzero finite function such that d({i})=0 for all i, and suppose that the function c:𝒥 is defined by the equality c(Δ)=d(Δ)-d(+Δ)-d(Δ+)+d(+Δ+) (Δ𝒮). Then there exist connected segments Δ1 and Δ2 such that c(Δ1)<0, c(Δ2)<0 and d(Δ)>0 for any segment Δ such that +(Δ1Δ2)+ΔΔ1Δ2.

We deduce the implication (2) (1) from this lemma. Let a and b satisfy condition (2) and suppose that ab. Let us apply the lemma to d=db-da (so that c=b-a). Let Δ1 and Δ2 be the segments whose existence is asserted in the lemma. We have a(Δ1)>0 and a(Δ2)>0, so that a=a-χΔ1-χΔ2+χΔ+χΔM(𝒥). Therefore, a>a, and the partitions a and b satisfy the condition (2). By means of obvious induction, we can assume that ab, i.e., a>b, which was required to be proved.

2.7. Proof of the Fundamental Lemma. Step 1. Reduction to the case where d([i,i+1])>0 for a certain i. If d([i,i+1])=0 for all i, then the function d-:𝒥+, defined by the equality d-(Δ)=d(Δ+), satisfies the conditions of the fundamental lemma. The validity of the statement of the lemma for d follows obviously from its validity for d-. If d-([i,i+1])=0 for all i, then we consider the function (d-)-, etc. Since d is nonzero, we finally arrive at a function d such that d([i,i+1])>0 for a certain i, and it is sufficient to prove the assertion of the lemma for d. Thus, we can assume that d([i,i+1])>0 for a certain i; without loss of generality, we will assume that d([0,1])>0.

Step 2. We will often use the following identity, which can be directly verified:

(*) If Δ0 and Δ are two segments of such that Δ0=[i0,j0]Δ=[i,j], then Δ0ΔΔ c(Δ)=d(Δ0) -d([i0,j+1]) -d([i-1,j0]) +d(+Δ+).

First of all we apply (*) to the segments Δ0={0} and Δ=[i,0], where i0. We get ii0 c([i,0]) =-d([0,1]) -d([i-1,0]) -d([i-1,1]). (1) Taking limit as i-, we see that i0c([i,0])=-d([0,1])<0. Hence there exists a k0 such that c([i,0])0 for k<i0, and c([k,0])<0. Again applying (1), we see that d([i,1]) d([0,1])>0 forki0. (2)

In the same manner, there exists an l1 such that c([1,j])0 for 1j<l, and c([1,l])<0; in addition, d([0,j]) d([0,1])>0 for1jl. (3)

Step 3. We consider the following statement for j=1,2,,l: d(Δ)>0 for[0,1] Δ[k,j]. (Aj) Let us observe that (A1) is valid by virtue of the inequality (2).

Let us now suppose that the assertion of the fundamental lemma is not fulfilled for our function d. Starting from this supposition, we prove the validity of all (Aj) by means of induction on j.

Let (Aj) be valid. We prove (Aj+1), i.e., that d([i,j+1])>0 for ki0. By virtue of (3), we can assume that i<0. Let us apply (*) to the segments Δ0=[0,1] and Δ=[i+1,j], and rewrite the obtained equation in the form d([i,j+1])= Δ0ΔΔ c(Δ)+ { d([0,j+1])- d([0,1])+ d([i,1]) } . (4) By virtue of (2) and (3), the expression within the curly brackets is positive. It remains to verify that c(Δ)0 for Δ0ΔΔ. Let us suppose that this is not so, i.e., c(Δ)<0 for a certain Δ such that Δ0ΔΔ. Taking (Aj) into account, we see that the segments Δ1=[k,0] and Δ2=Δ satisfy the conclusion of the fundamental lemma, which contradicts our supposition.

Thus, we have proved the statement (Aj) for all j[1,l]; in particular, (Al) is valid. But this means that the segments Δ1=[k,0] and Δ2=[1,l] satisfy the conclusion of the lemma; the last statement is a contradiction!

The fundamental Lemma is proved and, with it, Proposition 2.5 and Theorem 2.2 are also proved.

2.8. Remarks. a) The results of this section have "nongraded" analogs. With each partition λ of a number n (see Sec. 1.1) we associate the variety Xλ of the nilpotent (n×n)-matrices that have λ(k) Jordan cells of dimension k. The analog of Theorem 2.2 describes the contiguity of the cells Xλ. The resulting partial order relation on the partitions is well known (see, e.g., [LVi1973]). The analog of the elementary operation (see Sec. 2.1) is the replacement of the pair {k,l} with kl>0 by the pair {k+1,l-1} in the partition λ, and the analog of the function da is defined by the equality dλ(k)=mkλ(m). The analog of Proposition 2.5 is well known; its proof is substantially simpler than that in the graded case.

b) The proofs of the combinatorial results about the relation in [Zel1980] are greatly simplified with the help of Proposition 2.5. Let us observe, e.g., the following result (see [Zel1980, 9.10]): Each set M(𝒥)φ has a unique minimal element a=amin(φ). Proof: The corresponding function da is defined by the equality da(Δ)=minnΔφ(n).

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