# What is a Fourier decomposition of the index of an operator?

Consider a compact manifold $M$ equipped with some $S^1$-action, and let $E,F$ be vector bundles over $M$. Suppose further that a fixed elliptic operator $D:\Gamma(E)\to\Gamma(F)$ is preserved under the action (i.e. the induced action on $E,F$ commutes with $D$). Then its index $ind(D)=Ker(D)-Coker(D)$, as a virtual vector space, is also acted on by $S^1$.

The point now, which I just don't quite understand, is that this index can then be written as a finite sum of irreducible 1-dimensional representations $ind(D)=\sum_na_nL^n$, where $a_n\in\mathbb{Z}$ and $L^n$ is the representation $e^{i\theta}\mapsto e^{in\theta}$ of $S^1$ on $\mathbb{C}$.
What does this mean / what is this procedure / why is this true? I guess I'm just confused with the terminology and everything.

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I hope that I haven't misunderstood the setting, but $ker(D)$ and $coker(D)$ both have an S^1-action on it (I would assume it is at least continuous). But we know that all the finite dimensional continuous representations of $S^1$ are semisimple and the irreducible pieces are exactly your $L^n$.. So I suppose the meaning of the equality is that we agree to use additive notation for direct sum of $S^1$-modules, or we could be working in the $K$-group of representations etc. – user27126 Nov 16 '12 at 7:45
Ah OK, I located my original misstep. We're using the reformulation of modules $\leftrightarrow$ representations, and invoking Maschke's theorem. You could write this up in the answer box so that I could accept it. – Chris Gerig Nov 16 '12 at 7:56

$ker(D)$ and $coker(D)$ are both finite dimensional continuous representations of $S^1$. By Maschke's theorem we know that both of them can be split up as a direct sum of some copies of $L^n$, hence the result.