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I am interested in the following piece of data appearing in the cohomological Atiyah-Singer theorem. My reference is "The index of elliptic operators. III" by Atiyah and Singer.

Let $D:\Gamma(E)\to\Gamma(F)$ be an elliptic operator where $E\to X$ and $F\to X$ are complex vector bundles over a closed manifold $X$. The principal symbol of $D$ is a bundle isomorphism $\pi^\ast E\to\pi^\ast F$, where $\pi$ is the restriction of the projection $T^\ast X\to X$ to the complement of the zero section.

Given a Riemannian metric on $X$, the Thom space $\operatorname{Th}_X$ of $T^\ast X\to X$ is the topological space arising as the one-point compactification of the unit ball subbundle of $T^\ast X$. Using the above bundle isomorphism (and the obvious extension across the zero section), one extends $\pi^\ast E-\pi^\ast F$ to an element of $K(\operatorname{Th}_X)$. The Chern character gives an element of cohomology on the Thom space, and the cohomological Thom isomorphism gives an element of cohomology on $X$.

Now, the de Rham theorem says that this final cohomology element can be represented by a smooth differential form.

My question is:

  • can such a smooth form be prescribed in differential-geometric terms from the principal symbol, the (choice of) Riemannian metric on $X$, and perhaps a choice of hermitian bundle metrics and metric-compatible connections on $E\to X$ and $F\to X$?

This seems non-obvious since the Thom space does not seem to have a natural smooth structure [edit: see the comment below - it is not even a topological manifold], and so the use of the Chern character in the above presentation doesn't seem immediately amenable to the Chern-Weil approach. But, given the context and the conclusion, it seems unnatural to be forced to reach into the theory of topological characteristic classes.

If I understand correctly, the answer to my question is "essentially yes" according to Quillen's article "Superconnections and the Chern character," which identifies such a differential form via a choice of "superconnection" on $\pi^\ast E\oplus\pi^\ast F.$ However, Quillen's article seems to be answering a more general question, and has nothing to do with, for instance, the Thom space. Can one make use of the more special situation above to give a simpler answer than Quillen's?

I asked the same question on math.stackexchange with no response https://math.stackexchange.com/questions/3721536/chern-weil-theory-in-the-cohomological-atiyah-singer-theorem

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    $\begingroup$ I can't comment on most of the post, but re: "the Thom space does not seem to have a natural smooth structure" -- in general it is not a smooth manifold at all. The Thom space of a trivial bundle $\underline{\mathbb C}^r\to X$ is homeomorphic to $\Sigma^{2r}X$. Suspension shifts the cohomology groups of $X$ in a way that even mod 2 Poincaré duality usually does not hold, so the Thom space cannot even be a topological manifold. $\endgroup$
    – Arun Debray
    Jun 24, 2020 at 17:40
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    $\begingroup$ When $D$ is a Dirac-type operator (i.e., an odd symmetric first-order differential operator $D$ on a $\mathbb{Z}_2$-graded Hermitian vector bundle $E$ such that $D^2 = -g^{ij}\partial_i\partial_j + \text{lower order terms}$ for some Riemannian metric $g$), isn’t this exactly what the heat kernel proof for Atiyah–Singer gives you? Berline–Getzler–Vergne is basically the canonical reference for this, but there are many others besides. $\endgroup$
    – Branimir Ćaćić
    Jun 24, 2020 at 19:32
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    $\begingroup$ I'm not sure that I understand you, because the heat kernel proof gives you the index of a Dirac-type operator $D$ on a $\mathbb{Z}_2$-graded Clifford module bundle $E \to (M,g)$ as $\int_M \hat{A}(M,g) \wedge \mathop{ch}(E/\mathbb{S})$, where $\hat{A}(M,g)$ is the $\hat{A}$-genus form of $(M,g)$ and $\mathop{ch}(E/S)$ is the relative Chern character of the Clifford module bundle $E$, which is explicitly computable, Chern-Weil-style, from the curvature of the Quillen superconnection corresponding to $D$ and the Riemannian curvature of $(M,g)$ $\endgroup$
    – Branimir Ćaćić
    Jun 24, 2020 at 20:35
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    $\begingroup$ Indeed, when $(M,g)$ is spin and $E = S \otimes W$ is a twisted spinor bundle, $\operatorname{ch}(E/S)$ is just the Chern character of $W$. $\endgroup$
    – Branimir Ćaćić
    Jun 24, 2020 at 20:37
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    $\begingroup$ just one more comment, you can interpret the cohomology and K-theory of the Thom space as cohomology of the cotangent bundle with proper support along the projection to the base. So you can try to represent it by differential forms that decay fast at the vertical direction hopefuly. $\endgroup$
    – S. carmeli
    Jun 24, 2020 at 21:54

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