# Inverting the Weyl Character Formula

The Weyl Character formula tells us how to write the character of a representation as a linear combination of integral weights. Since characters are invariant under the action of the Weyl group, $W$, we can write a character as a linear combination of $W$-symmetrized dominant integral weights. It is know that the representation ring of a Lie algebra is isomorphic to $\mathbb{C}$P$^W$ as a vector space, where $P$ is the weight lattice of some Cartan subalgebra.

So we have two bases for the same vector space: the $W$-symmetrized dominant integral weights and the the character basis. The Weyl character formula tells us how to go from the former to the latter. My question is: is there much known about the matrix of going from the latter to the former? I've gone through a few low rank examples, and many of the coefficients are coming out to be zero. Does anyone know of a reference for this question in general?

Addendum: Jim makes a good point. The Weyl character formula isn't really needed. Perhaps we should just say that the matrix from the weight basis to the character basis is precisely the matrix of weight multiplicities. From this point of view it is clear that the matrix will be "upper triangular" (since weight multiplicities are zero above the highest weight). Thus the inverse should also be upper triangular. So my modified question is there any way to interpret the coefficients of the inverse matrix as counting anything interesting? As the matrix is upper triangular, we can certainly give recursive formulas for the coefficients. Does anyone have any other insight?

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For type A, a combinatorial interpretation of the entries of the inverse matrix was given by O. Egecioglu and J.B. Remmel, A combinatorial interpretation of the inverse Kostka matrix, Linear Multilinear Algebra 26 (1990) 59-84. The formula involves a lot of cancellation which suggests that the entries are relatively small. Another interpretation of the entries was given by H. Duan, On the inverse Kostka matrix, J. Combinatorial Theory (A) 103 (2003), 363-376.

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In general Kostant's approach to weight multiplicities showed that these typically grow very large, bounded only by the Kostant partition function value for the difference between highest weight and lower dominant weight. But the inverse matrix is elusive. Richard's suggestion that entries might be relatively small in absolute value for type $A$ could carry over (at least qualitatively) to other Lie types, though it's hard to visualize the combinatorics for say type $E_8$. – Jim Humphreys Apr 3 '10 at 12:37
In fact, let $f(n)$ denote the absolute value of the largest entry of the inverse Kostka matrix for GL$(n,\mathbb{C})$, i.e., the maximum coefficient in the expansion of a monomial symmetric function $m_\lambda$ in terms of Schur functions, for all $\lambda\vdash n$. Then $f(n)$ is sequence A102462 of EIS. Beginning with $f(1)$, the first terms are 1, 1, 2, 3, 4, 6, 12, 20, 30, 60, 105, 168, 280, 504, 840, 1512, 2520, 5040, 9240, 15840, 27720, 55440, 102960, 180180, 360360, 675675, 1201200, 2162160, 4084080, 7351344, 12697776, 24504480, 46558512, 84651840. Thus they can be pretty big. – Richard Stanley Apr 4 '10 at 3:16

Let $m$ be any $W$-invariant compactly supported $\mathbb Z$-valued function on the weight lattice. Then to expand $m$ in the basis of weight multiplicity functions of irreps, first apply differencing operators $1 - shift_\beta$ in the directions of the negative roots $\beta$, obtaining a non-$W$-invariant function, and look at the values inside the positive Weyl chamber; the value at $\lambda$ will be the multiplicity of $V_\lambda$ in $m$.

(Proof: it works for $m=V_\mu$ by the Kostant multiplicity formula, essentially run in reverse. Then use linearity of the question and answer.)

Okay, now let's apply this to $m$ being $1$ at $W\cdot \lambda$, $0$ elsewhere. I don't have a complete answer to your question, but the following is clear: if $\lambda$ is deep in the Weyl chamber -- I think distance $3$ from all the walls should be good enough -- then the $1$s outside the positive Weyl chamber won't affect the final values inside. The $1$ inside, at $\lambda$, will only lead to contributions in a small area around $\lambda$; I think the answer will be $\sum_w (-1)^w V_{\lambda + \rho - w\cdot \rho}$.

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I think you meant $\sum_w (-1)^w V_{\lambda+w(\rho)-\rho}$: certainly any highest weights in the decomposition will be below $\lambda$. – Marc van Leeuwen Nov 23 '11 at 9:00

Thanks Bruce Westbury for reminding me that I wrote something about this in my younger days. Here's what I make of it today, which provides a "closed formula" of sorts, a bit along the line of Allen Knutson's answer.

I'll deviate from the OP in writing $\Lambda$ for the weight lattice, $\Lambda^+$ for the set of dominant weights, and $X^\lambda$ for the basis elements of $\mathbf{Z}[\Lambda]$ with $\lambda\in\Lambda$, I'll use the dot-action $w\cdot\lambda=w(\rho+\lambda)-\rho$ and a related operator $J:\mathbf{Z}[\Lambda]\to\mathbf{Z}[\Lambda]$ sending $P\mapsto\sum_{w\in W}(-1)^{l(w)}w(X^\rho P)X^{-\rho}$ in general, and in particular $X^\lambda\mapsto\sum_{w\in W}(-1)^{l(w)}X^{w\cdot\lambda}$. Weyl's character formula says that the character $\chi_\lambda$ of $V_\lambda$ satisfies $$\chi_\lambda.J(1)=J(X^\lambda)\quad\text{for all \lambda\in\Lambda^+}$$ The left hand side is in fact also equal to $J(\chi_\lambda)$, since for any $P\in\mathbf{Z}[\Lambda]^W$ one has $$J(P)=\sum_{w\in W}(-1)^{l(w)}w(X^\rho P)X^{-\rho} = P \sum_{w\in W}(-1)^{l(w)}w(X^\rho)X^{-\rho} = P.J(1),$$ the second equality by $W$-invariance of $P$. Thus $\chi_\lambda$ and $X^\lambda$ have the same image by $J$.

Every $P\in\mathbf{Z}[\Lambda]$ is equivalent modulo $\ker(J)$ to a unique $P'\in\mathbf{Z}[\Lambda^+]$, i.e., a polynomial supported on the dominant weights. Concretely, define a $\mathbf{Z}$-linear operator $\alpha:\mathbf{Z}[\Lambda]\to\mathbf{Z}[\Lambda^+]$ by $\alpha(X^\mu)=0$ if $X^\mu\in\ker(J)$, which happens if $r\cdot\mu=\mu$ for some reflection $r\in W$, and otherwise $\alpha(X^\mu)=(-1)^{l(w)}X^{w\cdot\mu}$ where $w\in W$ is the unique element with $w\cdot\mu\in\Lambda^+$. Then $J(P)=J(\alpha(P))$ for all $P\in\mathbf{Z}[\Lambda]$. It follows from the above that $\alpha(\chi_\lambda)=X^\lambda$ for all $\lambda\in\Lambda^+$. In other words if $\chi:\mathbf{Z}[\Lambda^+]\to\mathbf{Z}[\Lambda]$ is the "character" map that linearly extends $X^\lambda\mapsto\chi_\lambda$, then $\alpha$ restricted to $\mathbf{Z}[\Lambda]^W$ defines the inverse "decomposition" map.

Now the decomposition of an orbit sum $m_\lambda=\sum_{\mu\in W(\lambda)}X^\mu$ is given by $\alpha(m_\lambda)=\sum_{\mu\in W(\lambda)}\alpha(X^\mu)$; since each $\mu$ gives at most one term, this shows that its decomposition involves at most as many irreducible factors as the size the $W$-orbit of $\lambda$.

If $\lambda$ is very large, it may be convenient to write $m_\lambda=\frac1s\sum_{w\in W}X^{w^{-1}(\lambda)}$ where $s$ is the size of the stabiliser in $W$ of $\lambda$, and using $\alpha(X^\mu)=\alpha((-1)^{l(w)}X^{w\cdot\mu})$ for any $\mu,w$, obtain $$\alpha(m_\lambda) =\frac1s\alpha\left(\sum_{w\in W}(-1)^{l(w)}X^{w\cdot(w^{-1}(\lambda))}\right) =\frac1s\alpha(X^\lambda J(1)).$$ If $\lambda$ is strictly dominant one has $s=1$, and if moreover $\lambda$ is far enough off the walls that $X^\lambda J(1)$ is entirely supported on dominant weights, then the right hand side simply becomes $X^\lambda J(1)$, which has $|W|$ distinct terms. This coincides with the expression that Allen Knutson guessed. However the requirement is rather stronger than he suggested: in type $A_n$ for instance, dominant weights can be represented by weakly decreasing $n+1$-tuples of integers, with $\rho=(n,n-1,\ldots,1,0)$, and the "off the walls" condition means that the successive entries for $\lambda$ decrease by at least $n+1$. In other words, in this case the simplified formula holds only if $\lambda-(n+1)\rho$ is dominant.

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The Weyl character is expressed as a quotient, which is elegant but does not easily tell you how to express the character as a nonnegative $\mathbb{Z}$-linear combination of $W$-orbits labelled by dominant integral weights. These coefficients are difficult to predict in most cases, as seen in Freudenthal's recursive formula for weight multiplicities. While the individual $W$-orbits can in principle be expressed as $\mathbb{Z}$-linear combinations of the characters, I don't expect interesting closed formulas in general.

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Would you call the formula given by the definition of the operation $\alpha$ in my answer a closed formula? – Marc van Leeuwen Dec 14 '11 at 15:30
I guess it qualifies as a closed formula, though my undefined term "interesting" may still be an obstacle for me. My concern here was just to downplay the role of the Weyl formula for this kind of combinatorial development, though of course it's always in the background. – Jim Humphreys Dec 14 '11 at 23:31

The LiE package http://www-math.univ-poitiers.fr/~maavl/LiE/ has implemented this and uses this idea quite a bit. There is some discussion of this in the manual although it may not be a complete answer.

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The most relevant points are in fact summarized in the diagram on page 95 (pdf: 96) of the LiE manual. See also my reply to this question, where $J$ corresponds to the alt_W_sum function of LiE, and the decomposition map $\alpha$ to the function alt_dom, inverse of the "full character" map $\chi$ corresponding to the LiE function Demazure. – Marc van Leeuwen Nov 23 '11 at 9:14