# Plugging $1-x$ into Schur polynomials

I have a symmetric Laurent polynomial $f$ in $k$ variables expressed as a linear combination of Schur polynomials. I'd like to know what happens when I make the substitution $p(x_1,\ldots,x_k)\mapsto p(1-x_1,\ldots,1-x_k)$ and re-expand in the Schur basis. Is there some nice combinatorial description of what happens when you do this?

That's the whole question, but I'll say more about where this comes from in case it's helpful or you're just interested. The symmetric polynomial is an element of the equivariant K-group $K_0^{GL(V)}(\mathrm{Hom}(V,W))$ where $V$ is some $k$-dimensional vector space and $W$ is some $n$-dimensional vector space (so $\mathrm{Hom}(V,W)$ is just a $kn$-dimensional affine space as a variety). This is isomorphic to the ring of symmetric Laurent polynomials in $k$ variables.

I know that my K-class is a linear combination of modules supported on the subvariety of $\mathrm{Hom}(V,W)$ consisting of non-full-rank matrices. I have an explicit but complicated description of this subspace as an ideal in the ring of symmetric polynomials, but after making the $1-x$ substitution, the ideal is exactly spanned by $s_\lambda$ where $\lambda$ doesn't fit in a $k\times(n-k)$ rectangle. So answering the question above would be enough to give a set of $\binom nk$ explicit linear equations cutting out this ideal, which is what I'm after.

Let $t$ be an indeterminate. Let $\vartheta:\Lambda \rightarrow \Lambda[t]$ be the specialization (homomorphism) defined by
$$\vartheta(p_k)=t+\sum_{i=1}^k {k\choose i}p_i,$$
where $p_i$ is a power sum summetric function. According to item 75 of
$$\vartheta(s_\lambda) =\sum_{\mu\subseteq\lambda} \frac{f^{\lambda/\mu}}{|\lambda/\mu|!}\left( \prod_{u\in\lambda/\mu} (t+c(u))\right)s_\mu,$$
where $c(u)$ denotes the content of the square $u$ of the skew shape $\lambda/\mu$, and $f^{\lambda/\mu}$ is the number of standard Young tableaux of shape $\lambda/\mu$.
If we restrict to $n$ variables and set $t=n$, then for any symmetric function $f$, $$\vartheta(f)=f(x_1+1,\dots,x_n+1),$$ as may be seen by setting $f=p_k$. We need only send $x_i$ to $-x_i$ (also a homomorphism) to get an answer to the question when $p=s_\lambda$.
Addendum. Here is a sketch of the proof. Both sides are polynomials in $t$, so it suffices to prove the result for $t=n\in\{1,2,\dots\}$. Since $$\vartheta(s_\lambda)(x_1,\dots,x_n) = s_\lambda(x_1+1,\dots,x_n+1),$$ we get (using notation from Chapter 7 of Enumerative Combinatorics, vol. 2) \begin{eqnarray*} s_\lambda(x_1+1,\dots,x_n+1) & = & \frac{a_{\lambda+\delta}(x_1+1,\dots,x_n+1)} {a_\delta(x_1+1,\dots,x_n+1)}\\ & = & \frac{a_{\lambda+\delta}(x_1+1, \dots,x_n+1)}{a_\delta(x_1,\dots,x_n)}. \end{eqnarray*} By expanding the entries of $a_{\lambda+\delta}(x_1+1,\dots,x_n+1)$ and using the multilinearity of the determinant we get (see Example I.3.10 mentioned above) $$s_\lambda(x_1+1,\dots,x_n+1)=\sum_{\mu\subseteq\lambda} d_{\lambda\mu}s_\mu,$$ where $$d_{\lambda\mu} = \det\left( {\lambda_i+n-i\choose \mu_j+n-j} \right)_{1\leq i,j\leq n}.$$ We can factor out factorials from the numerators of the row entries and denominators of the column entries of the above determinant. These factorials altogether yield $\prod_{u\in\lambda/\mu}(n+c(u))$. What remains is exactly the determinant for $f^{\lambda/\mu}/|\lambda/\mu|!$ given by Corollary 7.16.3 of Enumerative Combinatorics, vol. 2, and the proof follows. Is there a more conceptual proof that doesn't involve the evaluation of a determinant?