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The type-A nilpotent pyramid, good grading and Whittaker character

Definition

Work over C. Let λ=(0≤p1≤⋯≤pn) be a partition of N≥0 with transpose q1≥⋯≥ql, where l=pn for a nonempty row list and l=0 for an empty row list. Zero-length rows have no boxes. Let g=glN(C) with matrix units ei,j in the standard basis v1,…,vN (Classical complex matrix Lie algebras). Number the boxes of the diagram of λ by 1,…,N down columns from the left, and for a box i write row(i),col(i) for its row and column, L(i) for the box immediately to its left (or ∅ if absent), and R(i) for the box immediately to its right (or ∅ if absent). Define the nilpotent endomorphism e=∑i: L(i)≠∅eL(i),i∈g,evi=vL(i), and declare deg⁡ei,j=col(j)−col(i) for all i,j, so that g=⨁r∈Zgr with gr=span⁡{ei,j:col(j)−col(i)=r}. Put p=⨁r≥0gr,h=g0,m=⨁r<0gr, and for rows 1≤i,j≤n define the shift si,j=pj−pi if i≤j and si,j=0 if i≥j, together with the centralizer elements ei,j;r=∑row(h)=i, row(k)=jcol(k)−col(h)=reh,k(si,j≤r<pj). Finally let χ ⁣:m→C be χ(x)=tr⁡(xe), where the trace is the trace of the N×N matrix.

The verification below records the properties of this data that are used throughout this page. e∈g1 has Jordan type λ; p is a parabolic subalgebra with Levi factor h≅glq1⊕⋯⊕glql and nilradical ⨁r>0gr, while m is the nilradical of the opposite parabolic; the elements ei,j;r with si,j≤r<pj form a basis of the centralizer ge={X∈g:[X,e]=0}; the box set J={(a,b)∈{1,…,N}2:col(a)≤col(b), R(b)=∅} is in bijection with K={(i,j,r):1≤i,j≤n, si,j≤r<pj} via (a,b)↦(row(a),row(b),col(b)−col(a)), and ξa,b:=∑t=0col(a)−1eLt(a),Lt(b)=erow(a),row(b);col(b)−col(a); and χ is a Lie algebra character, χ([m,m])=0. This grading is the type-A pyramid good grading in the degree-one convention: ad⁡e:gr→gr+1 is injective for r<0 and surjective for r≥0. For e≠0 it differs from the standard Dynkin grading, in which e has degree 2 rather than 1. For e=0 all boxes lie in one column or there are no boxes, and the pyramid grading and the trivial Dynkin grading both concentrate g in degree zero. If N=0, take gl0=0: the box set and both centralizer index sets are empty, e=0, p=h=m=0, and χ=0. These conventions also permit an empty row list.

Facts & Assumptions

Given: A partition λ=(0≤p1≤⋯≤pn) of N with transpose q1≥⋯≥ql, the diagram boxes with row, column, L and R data, the nilpotent e=∑eL(i),i, the grading deg⁡ei,j=col(j)−col(i) and the subspaces p,h,m, and the elements ei,j;r and χ.

[F1]

Matrix units satisfy eh,kek,l=eh,l and eh,kek′,l=0 for k≠k′, and the trace is tr⁡(eh,k)=δhk (Classical complex matrix Lie algebras).

[F2]

Jordan strings: a nilpotent endomorphism is described by its strings w,Nw,N2w,…; the Jordan type is the partition of the string lengths, and the centralizer of a nilpotent of type μ has dimension ∑h,lmin⁡(μh,μl). Indeed its string-pair block is Hom⁡C[t](C[t]/tb,C[t]/ta); the image of 1 is any vector killed by tb, a subspace of dimension min⁡(a,b) (Every finite-dimensional nilpotent endomorphism has a basis of Jordan strings, Nilpotent endomorphisms and their nilpotency index).

[F3]

The trace form tr⁡(XY) on glN(C) is symmetric and invariant, and the degree grading is multiplicative: if x∈gr and y∈gs then xy∈gr+s, and tr⁡(x)=0 for x∈gr with r≠0 (Trace forms are symmetric and invariant, Classical complex matrix Lie algebras).

[F4]

The enveloping algebra and representations of g (Universal enveloping algebra, Representations of Lie algebras).

Proof

technique · direct

Given: The pyramid data of the Definition.

1.1F1F2F4algebra

The endomorphism e acts on the box basis by evi=vL(i) and evi=0 for boxes with L(i)=∅; hence the vi in one row form a Jordan string of length pi, because left shifts run from the rightmost box of the row to the first. Consequently e is nilpotent of Jordan type λ=(p1,…,pn), e∈g1, and the number of columns is l=pn when the row list is nonempty, with l=0 for the empty row list. Zero-length rows contribute no Jordan strings; for N=0 all box-indexed sums and vector spaces are empty or zero.

1.2F1algebra

Parabolic structure. The degree defines a Z-grading of the associative algebra: deg⁡(eh,kek′,l)=deg⁡eh,k+deg⁡ek′,l when k=k′ by [F1], and [gr,gs]⊆gr+s. Hence p=⨁r≥0gr is a subalgebra, its degree-zero part h consists of the matrices eh,k with col(h)=col(k), which is exactly the block-diagonal subalgebra with blocks of sizes q1,…,ql, so h≅glq1⊕⋯⊕glql; the strictly positive part ⨁r>0gr is a nilpotent ideal and m=⨁r<0gr is the nilradical of the opposite parabolic, with g=m⊕h⊕⨁r>0gr.

1.3F2algebra

The centralizer basis. First [e,eh,k]=eL(h),k−eh,R(k) for boxes h,k, with a term omitted when the relevant neighbour is absent (this is [e,eh,k]=∑d(eL(d),deh,k−eh,keL(d),d) and eL(d),deh,k=δd,heL(d),k, while eh,keL(d),d=δL(d),keh,d and L(d)=k holds exactly for d=R(k) in the same row). Now write ei,j;r=∑cehc,kc with hc the box of row i in column c and kc the box of row j in column c+r, the sum over c with 1≤c≤pi and 1≤c+r≤pj; here r≥si,j≥0 by the defining range, so this index set is exactly c=1,…,min⁡(pi,pj−r). Then [e,ei,j;r]=∑c(ehc−1,kc−ehc,kc+1) where a term is omitted when the neighbour does not exist; shifting the second sum by c↦c+1 and using R(kc)=kc+1 (which exists exactly when c+r<pj) shows that the two sums run over the same index set {c:c≥2, c≤min⁡(pi,pj−r)}, because for every c in the defining range one has c+r≤pj; hence the commutator is 0. The ei,j;r are linearly independent because they are sums over pairwise disjoint matrix units, and their number is ∑i,j#{r:si,j≤r<pj}=∑i,j(pj−max⁡(0,pj−pi))=∑i,jmin⁡(pi,pj), which equals dim⁡ge by [F2]. Hence they form a basis of ge.

2.1step 1.3algebra

The bijection J→K. For (a,b)∈J put i=row(a), j=row(b) and r=col(b)−col(a). Because b is rightmost, col(b)=pj; hence r=pj−col(a)≥0, r<pj because col(a)≥1, and r≥pj−pi because col(a)≤pi. Thus si,j=max⁡(0,pj−pi)≤r<pj, so the map lands in K. The triple determines b as the rightmost box of row j and a as the box of row i in column pj−r, proving injectivity. Conversely, if (i,j,r)∈K, then r<pj gives 1≤pj−r, while r≥max⁡(0,pj−pi) gives pj−r≤pi. Thus those two boxes exist, their columns satisfy col(a)≤col(b), and they give the required preimage. If either row has length zero, the index range in K is empty, consistently with the absence of such a box pair.

2.2F3step 1.2algebra

The character. For x,y∈m one has [x,y]∈[m,m]⊆g≤−2 by step 1.2, and e∈g1, so [x,y]e∈g≤−1 and tr⁡([x,y]e)=0 by [F3]. Hence χ([m,m])=0, that is, χ is a Lie algebra character on m: for x,y∈m, χ([x,y])=tr⁡([x,y]e)=0.

2.3F1F3step 1.1step 1.3algebra

The good-grading conditions. By step 1.3, the centralizer ge has a homogeneous basis ei,j;r with r≥si,j≥0, so it has no negative-degree part. The kernel of ad⁡e:gr→gr+1 is ge∩gr, hence this map is injective for r<0. The trace pairing is nondegenerate between gr and g−r: each matrix unit ea,b pairs with eb,a, and matrix units of other total degree have zero trace. Invariance gives tr⁡([e,x]y)=−tr⁡(x[e,y]), so the transpose of the map on degree r is, up to sign, ad⁡e:g−r−1→g−r. For r≥0 the latter is injective by the negative-degree result; finite-dimensional duality therefore makes the original map surjective. These are the good-grading conditions for degree-one e. They remain valid when e=0, because all nonzero spaces then have degree zero.

3.1step 1.3step 2.1algebra

The identity ξa,b=erow(a),row(b);col(b)−col(a). Left shifts of a run through the boxes of row i in columns col(a),col(a)−1,…,1, and the corresponding left shifts of b run through the boxes of row j in columns col(b),col(b)−1,…,col(b)−col(a)+1; all these exist because col(b)−col(a)+1≥1 and the shifts stop while the row i box exists. The pairs (row(h),row(k)) obtained are exactly the pairs (i,j) with column difference col(b)−col(a) and col(h) ranging over 1,…,col(a); by step 2.1 these are exactly the pairs indexing erow(a),row(b);col(b)−col(a). Hence the two finite sums coincide, and the two displayed families are the same basis of ge.

4.1F1step 1.1step 1.3step 2.3algebra∎

Dynkin normalization and the zero boundary. If e≠0, step 1.1 puts this nonzero element in pyramid degree 1, whereas the standard Dynkin grading defined by an sl2-triple has [h,e]=2e and puts it in degree 2; these two gradings therefore differ. There is no assertion that every Dynkin matrix degree is even: on Jordan strings of lengths one and two the standard h-weights are 0 and 1,−1, so matrix units between these strings have odd weights. If e=0, no row has length greater than one, since a row of length at least two contributes a distinct nonzero left-shift matrix unit. Hence all boxes lie in the first column or there are none, every matrix unit has pyramid degree zero, and the zero triple gives the trivial Dynkin grading as well. If N=0, there are no matrix units or centralizer indices; the parabolic, Levi, opposite nilradical and trace character are zero. Thus the conventions and good-grading conditions cover repeated row lengths, zero-length rows and the empty pyramid without an exclusion.

Remarks

  • The numbering of boxes down columns from the left makes each row a Jordan string of e; this is why the row lengths pi (weakly increasing) are the Jordan block sizes, while the column heights qj are the ranks' increments and index the Levi blocks of h.

  • The shift si,j=max⁡(0,pj−pi) counts the overlap min⁡(pi,pj) of the two rows i,j through the identity #{r:si,j≤r<pj}=pj−max⁡(0,pj−pi); this count is used again in the loop-mode index sets of the finite W-algebra items.

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