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Jul 12, 2021 at 22:01 comment added Bertram Arnold To expand on Thomas's comment, you can think of the symbol of your operator as defining a compactly supported $K$-theory class on the cotangent bundle, which has an almost complex structure and therefore Poincaré duality for $K$-theory. Under this, the symbol defines a $K$-homology class on your manifold, whose pushforward to the point is the index. From general nonsense, this splits off a $\mathbb Z$-summand in even dimensions, but there's typically lots of things in the kernel, eg for $\mathbb{CP}^n$ with $n>1$. One can probably get a counterexample to your statement via this approach.
Jul 12, 2021 at 21:18 history edited Overflowian CC BY-SA 4.0
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Jul 12, 2021 at 11:25 comment added Thomas Rot I'm not too familiar with the symbol calculus. But the following is true: The space of Fredholm operators (say on a Hilbert space) of a given index is connected. There are a $\mathbb{Z}$ worth of connected components. All these components have the same homotopy type, namely those of $BO$.
Jul 12, 2021 at 10:44 history asked Overflowian CC BY-SA 4.0