TITLE: Inverting the Weyl Character Formula
QUESTION [9 upvotes]: 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?

REPLY [4 votes]: 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.