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I was thinking about the following "obvious" construction the other day. Let $$ \mathbb{S}^{\infty}_{2}=\bigcup_{i=1}^{\infty}\mathbb{S}^{2i} $$ Then because we know that $\chi(\mathbb{S}^{2i})=2$ for all $i$, we should "expect" the above space to have Euler characteristic $2$ as well. But on the other hand we should have $\mathbb{S}^{\infty}_{2}\cong \mathbb{S}^{\infty}=\bigcup^{\infty}_{i=1}\mathbb{S}^{i}$ as subspaces of $\mathbb{R}^{\infty}$. And it is known that the later space is contractible via Whitehead's theorem. So not only the Euler characteristic is not $2$, but it should be $1$ instead. Even though strictly speaking it is not properly defined.

This posed a puzzling question for me from the index theory point of view, we know that the index of the Dirac operator on the even sphere modulo some constants should be the Euler characteristic. Naively I would expect that the index of $\mathcal{D}^{+}$ on the following space $$ \mathbb{S}^{2n}=\bigcup^{n}_{i=1}\mathbb{S}^{2i} $$ is continuous in some appropriate sense. But it is clear that the index must be one as $n\rightarrow \infty$ by the argument above. May I ask analytically what exactly "breaks down" at dimension infinity? Notice that obviously if we choose a different sequence of spheres, then we may get all $0$ or no limit at all. So I feel this argument must be flawed. The trouble is, I do not really know how the index of the Dirac operator changes when I modify the topological space by double suspension or infinite number of suspensions.

Motivation: I was thinking about how to extend the heat kernel proof of index theorem to infinite dimensional manifolds, but it is not clear to me how to interpret $Tr_{n/2}(h_{t})$ term when $n \rightarrow \infty$. So I come up with this wrong headed example hoping it would help to extract some information by letting $n\rightarrow \infty$. However, I am very surprised when I realized $\mathbb{S}^{\infty}$ is in fact contractible. I suspect this low level question is something others have thought about many, many years ago.

I know Bott periodicity, but it feels difficult to connect the big theorem with such a little example. I am not sure if Mazur's trick or Casero summation, Abel summation, Zeta function regularization, etc would be useful at here.

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    $\begingroup$ The Euler characteristic of a contractible space is (perfectly well-defined and equal to) $1$, not $0$. Roughly speaking the problem here is a failure of monotone convergence (thinking of the Euler characteristic as a kind of measure): the spheres only have homology in two places and when passing to the limit above, the homology in the top-dimensional place has escaped off to infinity. That's nothing too unexpected. $\endgroup$ Oct 19, 2014 at 4:12
  • $\begingroup$ Sorry! I forgot $\chi(*)=1$. Shall edit. $\endgroup$ Oct 19, 2014 at 4:16
  • $\begingroup$ @Qiaochu: I agree it is not too surprising from the topology side, but I do not know how to see it from the analysis side or make it rigorous enough so that I can compute more complicated spaces/operators. $\endgroup$ Oct 19, 2014 at 4:18
  • $\begingroup$ I don't really know what you have in mind on the analysis side. The index-theoretic description of the Euler characteristic involves constructing, one way or another, the Euler class of a compact oriented manifold, which lives in top cohomology. Where could the Euler class possibly live on an infinite-dimensional manifold? $\endgroup$ Oct 19, 2014 at 4:21
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    $\begingroup$ If you want something to regularize, you need more structure. For example, Witten's work relating the Dirac operator on loop space to modular forms uses the circle action in an essential way. $\endgroup$
    – S. Carnahan
    Oct 19, 2014 at 5:35

1 Answer 1

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At its core, the index theorem relates the index of an elliptic pseudo-differential operator to topological invariants of its symbol.

Assume that the operator acts between trivial rank $r$ complex vector bundles on a smooth $m$-dimensional manifold $M$ with trivial tangent bundle. (These assumptions are automatically satisfied in infinite dimensions if all your manifolds and bundles are Hilbert manifolds and bundles.) The symbol is then a map $\DeclareMathOperator{\GL}{GL}$ $\newcommand{\bC}{\mathbb{C}}$

$$S(TM)\cong M\times S^{m-1}\to \GL_r(\bC), $$

where $m=\dim M$, and $S(TM)$ is the unit tangent sphere bundle of $M$. The index depends only on the homotopy type of the above map.

Assuming you can come up with a workable definition of differential operator on a Hilbert manifold acting on sections of a Hilbert bundle you still have a problem because for an infinite dimensional Hilbert space the group $\GL(H)$ is contractible so, there exists only one homotopy class of maps $S(TM)\to \GL(H)$.

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