Alternating sum of symmetric and exterior powers vanishes

Let $V$ be a vector space with some extra structure (maybe to be general, an object of an abelian tensor category?), where we can form tensor products and exterior powers $\Lambda^i V$ and symmetric powers $\text{Sym}^i V$. I believe the following always holds for $n > 0$: $$\sum_{i=0}^n (-1)^i\Lambda^i V \otimes \text{Sym}^{n-i}V = 0$$ where I take the sum as a "virtual" object; another way to say this is just that $$\sum_{i=0}^{[n/2]} \Lambda^{2i} V \otimes \text{Sym}^{n-2i}V = \sum_{i=0}^{[(n-1)/2]} \Lambda^{2i+1} V \otimes \text{Sym}^{n-2i-1}V.$$

For example, if $V$ is a representation of a finite group, you can use character theory to reduce to the analogous identity for the elementary symmetric and complete homogeneous polynomials (see Wikipedia) applied to the eigenvalues of a group element. But it'd be nice to have a higher-level proof that applied more generally.

• I remember that it's the same thing as the relation between elementary and complete homgeneous symmetric polynomials $\sum_{i=0}^n (-1)^i e_i h_{n-i}$ (where $e_i$ is the sum of all products of $i$ distinct variables and $h_i$ is the sum of all products of $i$ variables), but I can't remember whether it's just an analogy or whether one can be deduced from the other. Apr 5 '17 at 21:46
• Probably we can argue by expressing the products using Schur functors in the context of your question, Schur polynomials in the context of my previous comment, and the two should be equivalent. So I guess the higher-level proof would be that the relation is correct whenever Schur functors make sense. But there's too much handwaving in what I wrote for me to dare make it into an answer. :-) Apr 5 '17 at 21:53

If $V$ is a vector space, then you always have a Koszul complex
$$\cdots \to \bigwedge^i V \otimes Sym(V) \to \bigwedge^{i-1} V \otimes Sym(V) \to \cdots \to V \otimes Sym(V) \to Sym(V)$$
which has no homology except at the end where it's a copy of the ground field in degree $0$ (this complex is naturally graded where $$\bigwedge^i V \otimes Sym^k V$$ has degree $i+k$).
The identity you mention is the Euler characteristic of the degree $n$ piece being $0$. The Koszul complex is equivariant for the natural action of $GL(V)$. If $V$ happens to have more structure, great: it will likely be compatible, for example a finite group acting on $V$ gives a subgroup of $GL(V)$.