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Let $C$ be a smooth projective curve and let $C^{(n)}$ be its $n$th symmetric power.

Let $E$ be a $S_n$-equivariant vector bundle over the Cartesian power $C^n$. Suppose that $E$ descends to a vector bundle $\tilde{E}$ over the symmetric power $C^{(n)}$.

Then there are two things I can do:

  1. I can compute the $S_n$-equivariant index of $E$ to get an element $V \in K_{S_n}(\text{pt})$. Then I can take the dimension of the $S_n$-invariant part $V^{S_n}$ to get a number.

  2. I can compute the ordinary index of $\tilde{E}$ over $C^{(n)}$, which is a number.

Do these two numbers agree?

Of course I can ask this same question for a more general setting, but this is the setting that I care about right now; I'll be happy with answers for both this setting or a more general one. I am also interested in any references that talk about this kind of "take-equivariant-index-and-then-take-invariant-part" procedure; I haven't really been able to find anything about this in the literature.

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1 Answer 1

up vote 2 down vote accepted

If I interpret your question correctly, the answer should be positive. In stack-theoretic terms, you have a bundle on the quotient stack $[C^n/S_n]$. This gives an element of the equivariant K-theory $K_{S_n}(C^n)$. In case 1, you push forward from $K([C^n/S_n]) = K_{S_n}(C^n)$ to the representation ring $K([\mathop{\rm Spec}k/S_n]) = K_{S_n}(\mathop{\rm Spec}k)$, then to $K(\mathop{\rm Spec}k) = \mathbb Z$. In the second you push forward to $K(C^{(n)})$, then to $K(\mathop{\rm Spec}k)$. Since pushforward in K-theory is functorial, the two operations coincide.

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