# Why are the power symmetric functions sums of hook Schur functions only?

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.

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+1 for the last question ("representation-theoretic or geometric explanation" seems awfully subjective to me: in this subject, any proof of a statement can be regarded as either explanatory or completely technical and unenlightening depending of one's viewpoint). See also Theorem 8.1 in arxiv.org/abs/math/0503040v3 for a generalization to skew shapes. –  darij grinberg Feb 14 '14 at 16:22
Nice! It never occurred to me that the first column of the character table of Sn (class {n}) equals 0 for irreps whose partitions are not hook shaped and (-1)^l for hook-shaped partitions (k,1^l). It's like a sister to the well-known fact that the characters of class {1^n} equal the tableaux-count. –  Wouter M. Dec 29 '14 at 13:46

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. 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, 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.

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"Yu" is a father's initial, not a segment of the last name. –  darij grinberg Feb 14 '14 at 17:27
I was wondering who this "Yu" was. Incidentally, remember you have the power to fix such things... –  Ben Webster Feb 14 '14 at 17:44
Oops. I was wondering myself "why does no one give Yu credit for this?" –  David Speyer Feb 14 '14 at 18:41
"For $\lambda$ any non-hook, $Sp(\lambda)$ occurs as a summand of $\mathrm{Ind}_{S_4}^{S_n} Sp(2,2)$" -- I just realized that I don't know why this holds for $\lambda$ a non-ribbon skew shape. –  darij grinberg Apr 27 '14 at 18:01
OK, I see (by repeated skewing of Schur functions) why this holds if $\lambda$ is a connected skew shape which is not a ribbon. If $\lambda$ is not connected, this is not generally true (e.g., $\lambda$ could be a disjoint union of two boxes), so I guess another argument would be necessary here. Still, nice proof in the straight-$\lambda$ case! –  darij grinberg Apr 27 '14 at 19:01