rt.representation theory - Inverting the Weyl Character Formula

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 $lambdainLambda$, I'll use the dot-action $wcdotlambda=w(rho+lambda)-rho$ and a related operator $J:mathbf{Z}[Lambda]tomathbf{Z}[Lambda]$ sending $Pmapstosum_{win W}(-1)^{l(w)}w(X^rho P)X^{-rho}$ in general, and in particular $X^lambdamapstosum_{win W}(-1)^{l(w)}X^{wcdotlambda}$. Weyl's character formula says that the character $chi_lambda$ of $V_lambda$ satisfies
$$
chi_lambda.J(1)=J(X^lambda)quadtext{for all $lambdainLambda^+$}
$$
The left hand side is in fact also equal to $J(chi_lambda)$, since for any $Pinmathbf{Z}[Lambda]^W$ one has
$$
J(P)=sum_{win W}(-1)^{l(w)}w(X^rho P)X^{-rho}
= P sum_{win W}(-1)^{l(w)}w(X^rho)X^{-rho} = P.J(1),
$$
the second equality by $W$-invariance of $P$. Thus $chi_lambda$ and $X^lambda$ have the same image by $J$.

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

Now the decomposition of an orbit sum $m_lambda=sum_{muin W(lambda)}X^mu$ is given by $alpha(m_lambda)=sum_{muin W(lambda)}alpha(X^mu)$; 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 $m_lambda=frac1ssum_{win W}X^{w^{-1}(lambda)}$ where $s$ is the size of the stabiliser in $W$ of $lambda$, and using $alpha(X^mu)=alpha((-1)^{l(w)}X^{wcdotmu})$ for any $mu,w$, obtain
$$
alpha(m_lambda)
=frac1salphaleft(sum_{win W}(-1)^{l(w)}X^{wcdot(w^{-1}(lambda))}right)
=frac1salpha(X^lambda J(1)).
$$
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.