# Do you know an elegant proof for this expression for a Schur function?

I know that the identity $$s_\mu = \sum_{\mu-\lambda \text{ is a horizontal strip}} \;\sum_{\alpha\vdash|\lambda|} \frac{\chi^\lambda_\alpha}{z_\alpha} \prod_i(p_i-1)^{a_i}$$ holds. Here $$\alpha=1^{a_1}2^{a_2}\dotsb$$ in exponential notation; in other words, $$a_i$$ is the number of times that $$i$$ occurs in $$\alpha$$.

I know this because I can show that it is essentially equivalent to the formula I.5 of Garsia and Goupil EJC 2009.

Note that the equality of top-degree terms is the well-known expansion of a Schur function in terms of power sum symmetric functions, so it suffices to show that all lower degree terms vanish.

I feel that his simple-looking formula should have a direct and elegant proof. Is there one?

I would like to suggest an interpretation using super symmetric functions. These are symmetric functions that are symmetric in two sets of variables $$\{x_i\}$$ and $$\{y_j\}$$ separately. They satisfy the property that setting a single $$x_i$$ variable to equal $$z$$ and a single $$y_j$$ variable equal to $$z$$ gives a polynomial independent of $$z$$, in addition to several other conditions (see Chapter 1, Section 3, Exercise 23 of Macdonald's book). Note that since it's independent of $$z$$, it's the same as just setting $$x_i = y_j = 0$$, which amounts to considering the same super symmetric function in a smaller set of variables.

A concrete description of supersymmetric functions is as follows. The supersymmetric functions are the image of the map $$\Lambda \to \Lambda \otimes \Lambda$$ given by $$(\mathrm{Id} \otimes S) \circ \Delta$$, where $$\Lambda$$ is the ring of symmetric functions, $$\Delta$$ is the usual comultiplication, and $$S$$ is the usual antipode (which sends the power sum $$p_i(y)$$ to $$-p_i(y)$$). The variables in the second factor (i.e. $$\{y_j\}$$) are the "super variables". Because the map is injective, the supersymmetric functions are abstractly isomorphic to $$\Lambda$$, but they are convenient for performing certain specialisation tricks. But let me give an example: $$\sum_{r=0}^\infty e_r(x/y) t^r = \frac{\prod_i(1+x_i t)}{\prod_j (1+y_j t)}.$$ These are the elementary super symmetric functions. You can see that if you set a single $$x_i = z$$ and a single $$y_j = z$$, then a factor of $$1+zt$$ cancels in the numerator and denominator, giving an expression independent of $$z$$ as stated above. Another example is that $$p_i(x/y) = p_i(x) - p_i(y)$$.

Now we're in a situation where we can explain the identity. Clearly $$s_{\mu}(x) = s_\mu(x,1/1)$$ by the specialisation property of super symmetric functions. But the right hand side can be obtained by first setting an $$x_i$$ equal to 1, then passing from $$\Lambda$$ to super symmetric functions, then setting a $$y_j=1$$ and all other $$y$$'s to zero (that setting $$x_i=1$$ commutes with passing to super symmetric functions isn't totally obvious, but I'm omitting it for the sake of brevity). The upshot of this is that $$s_\mu(x,1) = \sum_\lambda s_\lambda(x) s_{\mu/\lambda}(1)$$, and $$s_{\mu/\lambda}(1)$$ is zero unless $$\mu/\lambda$$ is a horizontal strip. This gives the outer sum in your expression. Then we recognise that second part of the operation replaces $$p_i(x)$$ with $$p_i(x/1) = p_i(x) - p_i(1) = p_i(x)-1$$, which recovers your expression, where I have taken for granted that $$s_\lambda(x) = \sum_{\alpha \vdash |\lambda|} \frac{\chi_\alpha^\lambda}{z_\alpha}\prod_i p_i^{a_i}.$$

• I'm not sure if I'm allowed to follow-up your answer with a question. In any case what happens if you apply your supersymmetric construction in the case when $\Lambda$ is the tensor algebra $T(V)$ (viewed as a Hopf algebra)? Nov 15 '20 at 17:58
• I don't know how to interpret the "super" tensor algebra when defined with the Hopf structure analogously to above. A more analogous construction seems to be to consider the map $T(V) \to T(V\oplus V)$ induced by $v\mapsto (v,-v)$. (Note that $T(V\oplus V) \neq T(V)\otimes T(V)$.) The image of an element of $T(V)$ under this map has the property that if you set the element corresponding to $(e_i,0)$ equal to the element corresponding to $(0,e_i)$, the output is independent of the actual value. (This is like the cancellation above, but you must use the same index variable and super variable.) Nov 15 '20 at 23:37

Here is another argument. Let $$\chi^{\mu/k}$$ denote the skew character of the symmetric group $$\mathfrak{S}_n$$ corresponding to the skew shape $$\mu/k$$. Then by Pieri's rule, $$\sum_{\mu-\lambda\ \mathrm{is\ a\ horizontal\ strip}} \sum_{\alpha\vdash|\lambda|}\frac{\chi^\lambda_\alpha}{z_\alpha} \prod_i p_i^{a_i} = \sum_k \sum_{\alpha\vdash n-k} \frac{\chi^{\mu/k}_\alpha}{z_\alpha}p_\alpha$$ $$\qquad = \sum_k s_{\mu/k}.$$ Let $$s_k^\perp$$ denote the linear operator taking $$s_\lambda$$ to $$s_{\lambda/k}$$. Let $$\psi$$ denote the linear operator taking $$f$$ to $$\sum_k s_k^\perp f$$. Let $$\theta$$ denote the algebra automorphism taking $$p_i$$ to $$p_i-1$$.

I claim that $$\psi$$ and $$\theta$$ are inverses (so in fact $$\psi$$ is an algebra automorphism). By linearity it suffices to show that $$\psi p_\lambda = \prod_i (p_{\lambda_i}+1).\ \ \ (*)$$ Now it is not hard to see that $$s_k^\perp p_\lambda$$ is equal to $$\sum p_\nu$$, where $$\nu$$ is obtained from $$\lambda$$ by removing a set of parts (regarding equal parts as distinguishable) summing to $$k$$. From this (*) is immediate.

We therefore have $$\left.\sum_k s_{\mu/k}\right|_{p_i\to p_i-1} = \theta\psi s_\mu = s_\mu,$$ as desired.