# The character of the alternating sum of the exterior powers of the standard representation

Let $$V$$ denote the standard representation of $$\mathfrak{S}_n$$. Based on by-hand calculations when $$n=4$$ and $$n=5$$, I suspect that $$\chi\left( \sum_{0\leq i < n} (-1)^i \textstyle{\bigwedge}^iV \right)(\tau) = \begin{cases} n, & \tau \textrm{ is an n-cycle} \\ 0, &\textrm{otherwise} \end{cases}$$ but I lack the techniques to prove this. Does anyone know a proof of or have a reference for this identity?

If $$V$$ is a representation of a group $$G$$ of degree $$d$$, $$g\in G$$ and $$p\geq0$$, then the character of $$\Lambda^pV$$ evaluated at $$g$$ is $$\chi_{\Lambda^pV}(g)=\sigma_p(\lambda_1,\dots,\lambda_m),$$ with $$\lambda_1,\dots,\lambda_d$$ the eigenvalues of $$\rho_V(g)$$ on $$V$$ listed with multiplicities and $$\sigma_p$$ the elementary symmetric function of degree $$p$$. It follows from this that if $$\zeta$$ is the character of the (virtual) representation $$\sum_{p\geq0}(-1)^p\Lambda^pV$$, then $$\zeta(g) = \sum_{p\geq0}(-1)^p\sigma_p(\lambda_1,\dots,\lambda_d) = \prod_{i=1}^d(1-\lambda_i).$$ If $$f_g=\det(t-\rho_V(g))\in\mathbb C[t]$$ is the characteristic polynomial of $$\rho_V(g)$$, we thus have $$\zeta(g) = f_g(1).$$

Now suppose that $$G=S_n$$, the symmetric group of degree $$n$$ and that $$V$$ is the standard, $$(n-1)$$-dimensional representation of $$G$$. If we add a copy of the trivial representation to get $$W:=\mathbb C\oplus V$$, this is isomorphic the permutation representation of $$S_n$$ on $$\mathbb C^n$$. If $$g$$ is a permutation and its disjoint cycles are of length $$r_1,\dots,r_k$$ (including all cycles of length $$1$$ corresponding to fixed points of $$f$$) then the characteristic polynomial of $$\rho_W(g)$$ is

$$\prod_{j=1}^k(t^{r_k}-1)$$

and it follows from this that the characteristic polynomial of $$\rho_V(g)$$ is

$$f_g=\frac{\prod_{j=1}^k(t^{r_k}-1)}{t-1},$$

which is a polynomial. This vanishes at $$1$$ unless $$k=1$$, that is, if $$g$$ is an $$n$$-cycle, and when it is such a cycle its value is clearly $$n$$.

• Beat me to it by 10 seconds! Sep 22, 2022 at 4:47
• :-) ${}{}{}{}{}$ Sep 22, 2022 at 4:47
• I had blanked on that identity regarding symmetric polynomials. Thanks!! Sep 22, 2022 at 5:28

Let $$S^1$$ be the unit circle in $$\mathbb C$$, let $$T=(S^1)^n$$ be the $$n$$-torus, and let $$U=\{(z_1,\dots,z_n)\in T:z_1\cdots z_n=1\}.$$ This is a subspace of $$T$$ homeorphic to an $$(n-1)$$-torus. The symmetric group $$S_n$$ acts on $$T$$ by permuting coordinates, and that action restricts to $$U$$, and passes on to the homology $$H_*(U)$$ (with complex coefficients). As an $$S_n$$-module $$H_*(U)$$ is the exterior algebra $$W=\Lambda^*V$$ of the standard representation of $$S_n$$, so the character of $$W$$ evaluated at a permutation $$\pi$$ is the Lefschetz number of the map $$\pi:U\to U$$, and is thus (see Lefschetz number equal to Euler Characteristic of Fixed Points) the Euler characteristic of the fixed point set of $$\pi$$ in $$U$$.

This fixed point set is a union of tori, all of the same dimension. That dimension is either zero or psotive — when it is positive, the Euler characeristic is zero, and when it is zero it is easy to see that the characteristic is $$n$$. You can probably guess for what permutations $$\pi$$ this happens.

• Wow! This is extremely slick. Sep 22, 2022 at 5:34
• I can't find a real reference to Lefschetz number = Euler characterstic, but this follows from the fact that both can be computed by intersecting something with a diagonal. Sep 22, 2022 at 5:36