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Let $\mathcal{D}:\Gamma(E)\to \Gamma(F)$ be an elliptic operator of order $k$. Where $E,F$ are $\mathbb{C}$-vector bundles over $X$, a compact smooth manifold.

In Atiyah-Singer "the index of elliptic operators I" pg 518, it is stated that:

the index of $\mathcal D$ as a Fredholm operator, depends just on the homotopy class of the symbol of $\mathcal D$, $\sigma(\mathcal D):SX\to Iso(E,F)$ in the space of continuous invertible symbols of order $k$.

I have no problem in showing that if two operators have homotopic symbols then they are also homotopic as Fredholm operators and therefore they have the same index. The converse seems to be harder to prove and I'm starting to suspect that it is false.

Q. Suppose that two elliptic operators $\mathcal{D_1},\mathcal {D_2}:\Gamma(E)\to \Gamma(F)$ of the same order have the same index. Are $\sigma(\mathcal{D_1}) $ and $\sigma(\mathcal{D_1})$ homotopic?

For example, if I am not wrong, when $\dim X$ is odd, then all elliptic operators have index zero. However if we consider $E=F$ to be a trivial bundle, then homotopy classes of symbols should correspond to homotopy classes of maps $[SX, GL(\mathbb{C},r)]$ and this set is not a singleton in general.

P.S. I have posted this on math.stackexchange.com but I did not get any answer or comment.

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    $\begingroup$ 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 $BO$. $\endgroup$
    – Thomas Rot
    Commented Jul 12, 2021 at 11:25
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    $\begingroup$ 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. $\endgroup$
    – Bertram Arnold
    Commented Jul 12, 2021 at 22:01

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