One interesting fact in symmetric function theory is that the power symmetric function $p\_n$ can be written as an alternating sum of hook Schur functions $s\_{\lambda}$:
$$
p\_n = \sum\_{k+\ell = n} (-1)^\ell s\_{k, 1^\ell}.
$$


A priori, all that is known is that $p\_n$ can be expressed as a sum of Schur functions $s\_{\lambda}$, with the sum ranging over all partitions of $n$, not just those of hook shape. The fact that the coefficients of all non-hook shapes are zero is quite interesting and most likely says something about representations of $S\_n$.


However, this property does not extend to other generalizations of the Schur functions. In particular, when $p\_n$ is expanded in the Jack basis $J\_{\lambda}$ of symmetric functions with coefficients in $\mathbb{Q}(\alpha)$, such a vanishing phenomenon no longer occurs. In particular, the coefficient of $J\_{2,2}$ in the expansion of $p\_4$ is non-zero (though, of course, it vanishes when $\alpha = 1$).


Is there a representation-theoretic or geometric explanation for why the expansion of power functions in the Schur basis has such a nice form? And is there any sort of generalization that could determine which polynomials in the $p\_n$ would have a similar property in one of the generalized bases, specifically the Jack basis? Though it may be too much to ask for, I would love to have a combinatorially explicit, algebraically independent set of polynomials in the $p\_i$ that generate the ring of symmetric functions in $\mathbb{Q}(\alpha)$ with the property that each generator is a sum of hook Jack symmetric functions only.



Why are the power symmetric functions sums of hook Schur functions only?
The coefficient of $s\_{\lambda}$ in $p\_{\mu}$ is (up to normalizing factors which I won't get right) the trace of the permutation of conjugacy class $\mu$ acting on the representation $Sp(\lambda)$ of $S\_n$. You are asking why the $n$-cycle acts with trace $0$ on every irrep except for the hooks (which are $\bigwedge^k \mathbb{Q}^{n-1}$, for various values of $k$.) Actually, a more general statement is true: If $\alpha/\beta$ is a skew shape of size $n$, then $\langle p\_n, s\_{\alpha/\beta} \rangle$ is $0$ unless $\alpha / \beta$ is a ribbon.


For a **combinatorial proof**, see Section 7.17 in Enumerative Combinatorics.


For a **more representation-theoretic proof**, see Proposition 8.2 in [Vershik-Okounkov](http://arxiv.org/abs/math/0503040). In summary: We want to evaluate the central element
$$z:= \sum\_{\mbox{$w$ an $n$ cycle}} w$$
in $\mathbb{Q}[S\_n]$ acting on $Sp(\lambda)$. Define $x\_k = (1k)+(2k)+\cdots + (k-1 k)$. Observe that the $x\_i$ commute, and that $z = x\_2 x\_3 \cdots x\_n$. Since the $x\_i$ commute we should expect them to be simultaneously diagonalizable, and indeed they are. 
V&O construct a basis $v\_T$ for $Sp(\lambda)$, indexed by standard Young tableaux of shape $T$, which is a simultaneous eigenbasis for all $x\_i$. In particular, $x\_i v\_T = 0$ if and only the element $i$ is on the main diagonal of the tableau $T$. (I draw tableaux in the English style, like [this](http://en.wikipedia.org/wiki/File%3aFerrer_partitioning_diagrams.svg), so for me the main diagonal runs from the upper left to the lower right.)


For $\lambda$ any non hook shape, there is a box one step down and to the right of the upper left corner. For any $T$, that box contains some $i>1$. Then $x\_i v\_T=0$ and hence $z v\_T=0$. So $z$ acts by $0$ on $Sp(\lambda)$ for any nonhook shape $\lambda$.


I thought about it a bit more, and here is what I think is the **most direct proof**. 
Define $z\_n$ to be the element $\sum\_{\mbox{$w$ an $n$ cycle}} w$ in the group algebra of $S\_n$. I shall prove the following statement:


**Lemma** If $V$ is a representation of $S\_n$ where $z\_n$ acts by $0$, then $z\_{n+1}$ acts by $0$ on $\mathrm{Ind}\_{S\_n}^{S\_{n+1}} V$. 


**Proof of Lemma:** We use the description of induction as a tensor product: $$\mathrm{Ind}\_{S\_n}^{S\_{n+1}} V \cong \mathbb{Q}[S\_{n+1}] \otimes\_{\mathbb{Q}[S\_n]} V,$$
the fact that $z\_{n+1}$ is central, and the identity $z\_{n+1} = x\_{n+1} z\_n$. (The notation $x\_{n+1}$ was defined in the previous proof.) Therefore, 
$$z\_{n+1} \cdot (w \otimes \vec{v}) = (z\_{n+1} w) \otimes \vec{v} = (w z\_{n+1}) \otimes \vec{v} = (w x\_{n+1} z\_n) \otimes \vec{v} = (w x\_{n+1}) \otimes (z\_n \vec{v})=0$$
for any $w \in S\_{n+1}$ and $\vec{v} \in V$, as desired. $\square$


Direct computation shows that $z\_4$ acts by $0$ on $Sp(2,2)$. So, inductively, $z\_n$ acts by $0$ on $\mathrm{Ind}\_{S\_4}^{S\_n} Sp(2,2)$. Therefore, $z\_n$ acts by $0$ on any irreducible summand of $\mathrm{Ind}\_{S\_4}^{S\_n} Sp(2,2)$. For $\lambda$ any non-hook, $Sp(\lambda)$ occurs as a summand of $\mathrm{Ind}\_{S\_4}^{S\_n} Sp(2,2)$ so $z\_n$ acts by $0$ on $Sp(\lambda)$. As discussed in the previous proof, this proves the claim.


