In ChernWeil theory, we choose an arbitrary connection $\nabla$ on a complex vector bundle $E\rightarrow X$, obtain its curvature $F_\nabla$, and then we get Chern classes of $E$ from the curvature form. A priori it looks like these live in $H^*(X;\mathbb{C})$, but by an argument that I don't really understand they're actually in the image of $H^*(X;\mathbb{Z})$ (which is where they're usually considered to live). Meanwhile, I've heard people say that whenever I see a arbitrary real constants that end up having to be integers I should wonder whether the AtiyahSinger index theorem is lurking in there somewhere. Is there anything to this wild guess?
4 Answers
It's true that in some sense the AtiyahSinger Index Theorem has led to some integrality results. The theorem states that for an elliptic operator on a compact manifold two numbers are equal. One of them, the "analytic index", is obviously an integer. The other one, the "topological index", which depends on the symbol of the operator, is obviously a rational number but not so obviously an integer.
On the other hand, you can argue that the integrality phenomena revealed in this way don't really have anything to do with the analysis.
There are two equivalent ways to define the topological index. There is the cohomological formula in terms of characteristic classes that live in rational cohomology. There is also the conceptually more fundamental $K$theory definition of the topological index, in which you let the symbol of the operator give you an element of the compactly supported $K^0$ of the cotangent manifold and then use Bott periodicity and the fact that $K$theory is a cohomology theory to get from here to an element of $K^0(point)\cong\mathbb Z$. The fact that the integer thus defined coincides with the rational number obtained from characteristic classes is a calculation involving the relationship between $K$theory and ordinary cohomology, and does not rely on the fact that there was a differential operator involved.

$\begingroup$ This seems really close to what I want. But just because proving that the two definitions of the tindex agree doesn't involve analysis  that doesn't mean that the actual index theorem doesn't care about the operator, right? $\endgroup$ Jul 11, 2011 at 15:26


$\begingroup$ Sorry, I was confusing myself. I guess what seemed weird was that in my question it's a quantity coming from analysis that ends up being integral, whereas in the index theorem it's a quantity coming from homotopy theory. Of course, that should've already tipped me off that this was a dead end in the first place. Your point, if I'm understanding correctly, is that since there's even an internaltohomotopytheory proof that the tindex is integral, then there's absolutely no way these two are related. That sounds reasonable enough to me. $\endgroup$ Jul 18, 2011 at 17:05
The integrality results coming out of the index theorem and that for the ChernWeil classes have a different flavour.
The ChernWeil classes are integral cohomology classes for any vector bundle on any manifold.
On the other hand, contemplate on, say, the Hirzebruch signature theorem (as a typical example for the integrality results coming from the index theorem). It implies that if $M$ is a closed oriented $4k$manifold, then a wellknown rational polynomial in the Pontrajgin classes of the tangent bundle in degree $4k$ is in integral cohomology. The statement is about very special bundles (tangent bundles) on very special manifolds, and it is only about the component in the top degree and it is easy to see that the statement fails for other vector bundles or away from the top degree.
I do not think you can derive the integrality of Chern classes from the index theorem (or from the less analytic integrality result that Tom Goodwillie describes).

$\begingroup$ The index theorem is a little more flexible than this  you can always twist your operator by a coefficient bundle, for example. But I agree that it is not flexible enough for ChernWeil theory. $\endgroup$ Jun 30, 2011 at 13:32

$\begingroup$ The bundle you twist with comes in via the Chern character, which is additive. This can be used to show the integrality of the Chern character on spehres, for example. But the (total) Chern class is not additive. $\endgroup$ Jun 30, 2011 at 16:02

$\begingroup$ Thank you. This was my original intuition, that the hypotheses of the index theorem seem too specialized to give this general result about Chern classes. $\endgroup$ Jul 11, 2011 at 15:38
The usual argument (that you had trouble with) is not so complicated: you prove that every vector bundle is the pullback of the ``universal'' vector bundle on a Grassmannian. Then you compute the Chern classes of the universal bundle, directly on the Grassmannian, and by a miracle they turn out to be integral. In fact, you can compute the relevant integrals explicitly by hand, so it turns out quite nicely. So the hard bit is showing that every vector bundle is the pullback of the universal bundle; see Milnor and Stasheff, p. 65. As for AtiyahSinger, that seems much more complicated and I don't see how to make use of it here.

2$\begingroup$ Actually, Milnor and Stasheff use the splitting principle to reduce the concrete computation you describe to the case of line bundles, which is probably a bit simpler. $\endgroup$ Jun 29, 2011 at 21:40
My first remark about this question is a little bit pithy  the standard cohomological index formula works only for even dimensional manifolds while ChernWeil theory is of course more general.
For a more substantive answer, let's look at the index formula itself:
$Index(D) = \int_{T^*M} ch(\sigma_D) Todd(TM \otimes \mathbb{C})$
Here $ch(\sigma_D)$ refers to the chern character of the "symbol class" in Ktheory of the elliptic operator $D$. My first observation is that this formula places special emphasis on the Todd class of the tangent bundle, and it seems to me that one would have to place similar emphasis on the Todd class in Chern  Weil theory in order to answer your question in the affirmative. I certainly do not claim that this is impossible  the Todd polynomial seems to take a privileged position in many places where it does not seem to belong  but I do not know how to do it.
Additionally, we are forced to ask what characteristic classes can be written in the form $ch(\sigma_D)$. As Tom Goodwillie points out, we will get an integrality result if we plug in any Ktheory class $x$ rather than a Ktheory class of the form $\sigma_D$ (although it may be that all classes in $K(T^*M)$ are symbol classes), but there are still constraints on which characteristic classes can appear. For example, we will only get classes in cohomology of even degree.
So there are a number of hints that the right hand side of the index formula is too specialized for ChernWeil theory. But the left hand side also helps answer your question. Here are two general principles in index theory that will help me explain:
 The index theorem for any operator can be deduced from the index theorem for a special class of operators on a special class of manifolds  the "spin$^c$ Dirac operators". These operators are completely determined by the geometry of the manifolds on which they live.
 The index depends continuously on the operator in an appropriate sense. In particular, if $D_t$ is a continuous family of operators then $Index(D_t)$ is constant in $t$.
Combining these two facts, we see that the integers which we can obtain from the index theorem depend only on homotopy theoretic information about the manifold and the coefficient bundle (if applicable). This is not true of all integers that can be obtained via ChernWeil theory, and it demonstrates that index theoretic invariants are much too crude for ordinary cohomology theory.

$\begingroup$ Thanks  your last paragraph is exactly the nail in the coffin of this idea that I was expecting. By the way, I've heard it said that whenever you see the Todd class it just means that you're looking at cohomology when you should be looking at Ktheory, and that the Todd class should be thought of as nothing more than a "correction factor". $\endgroup$ Jul 11, 2011 at 15:42

$\begingroup$ To be precise one has Thom isomorphisms in both Ktheory and De Rham cohomology, but they are not compatible with each other in the sense that $ch \circ Thom_{KT} \neq Thom_{DR} \circ ch$. The Todd class is precisely the "correction term" which makes this equation true. $\endgroup$ Jul 14, 2011 at 12:39