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](https://doi.org/10.37236/85).


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?



Do you know an elegant proof for this expression for a Schur function?
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
$$
s\_\lambda(x) = \sum\_{\alpha \vdash |\lambda|} \frac{\chi\_\alpha^\lambda}{z\_\alpha}\prod\_i p\_i^{a\_i}.
$$


