This is a [cross-post from math.se](https://math.stackexchange.com/questions/383358/expression-of-basis-vectors-of-permutation-modules-in-different-bases?noredirect=1#comment824882_383358), because I did not get any answer there:


Write $[n]:=\{1,\ldots,n\}$. For a partition $\lambda\vdash n$, I will write $[\lambda]$ for the Specht module that corresponds to $\lambda$, i.e. the complex vector space spanned by all standard tableaux of shape $\lambda$.


Consider the complex vector space $V\_k$ which has as a basis the $k$-element subsets of $[n]$. The symmetric group $S\_n$ acts on $V\_k$ in the was that $g\in S\_n$ maps a basis vector $A\subseteq[n]$ to the basis vector $g(A)=\{ g(i) \mid i\in A \}$. Let us assume $k\le \frac n2$. It is known that 
$$V\_k = \bigoplus\_{j=0}^k~ [n-j,j]$$
is the decomposition of $V\_k$ into irreducibles. See, for instance, Example 7.18.8 in Stanley's book *Enumerative Combinatorics* (Volume 2). 


**Question:** Given any $j\le\frac n2$ and a standard tableau $T$ of shape $\lambda=(n-j,j)$, I would like to know how to express the corresponding basis vector $v\_T$ of $[\lambda]$ in the canonical basis of $k$-element subsets of $[n]$.


**Thoughts:** For $k=1$, this is rather easy. For the only standard tableau $T$ on $\lambda=(n)$, the vector $v\_T=\sum\_{i=1}^n \{ i \}$ is just the sum of all $1$-element subsets of $[n]$. On the other hand, for a standard tableau $T$ of $(n-1,1)$, let $a\_T$ be the number in the single box in row $2$. Then it can be checked that $v\_T=\{1\}-\{a\_T\}$. However, I already have trouble with $V\_2$. There is an obvious $S\_n$-invariant inclusion
\begin{align\*}
\phi\_1: V\_1 &\longrightarrow V\_2 \\\\
\{i\} &\longmapsto \sum\_{j\ne i} \{ i,j \}
\end{align\*}
and the $[n-2,2]\subseteq V\_2$ must be isomorphic to $V\_2/\mathrm{im}(\phi\_1)$, but I didn't get much further. 



Expression of basis vectors of permutation modules in different bases.
Maybe I'm mistaken but I think the answer is quite easy if you use the language of Specht module, that is realize all these in polynomials. More precisely, you are working in the subspace of $C[x\_1,\dots,x\_n]$, where no variables are squared. In this space any standard tableau is realized as the product of the Vandermonde determinant of its columns. Then if I understand correctly your question, it just amount to expand such a product on the monomials. Does it makes sense ? 


