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?




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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.



Plugging $1-x$ into Schur polynomials
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  

<http://math.mit.edu/~rstan/ec/ch7supp.pdf>, we have  

 $$ \vartheta(s\_\lambda) =\sum\_{\mu\subseteq\lambda}
 \frac{f^{\lambda/\mu}}{|\lambda/\mu|!}\left( \prod\_{u\in\lambda/\mu}
 (t+c(u))\right)s\_\mu, $$  

 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
 $p=s\_\lambda$.


Item 75 is an improved version of Example I.3.10 on page 47 of
I. G. Macdonald, *Symmetric Functions and Hall Polynomials*,
second ed.


**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?


