In Chern-Weil 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 Atiyah-Singer index theorem is lurking in there somewhere. Is there anything to this wild guess?
[Math] Can one use Atiyah-Singer to prove that the Chern-Weil definition of Chern classes are $\mathbb{Z}$-cohomology classes
at.algebraic-topologydg.differential-geometry
Related Solutions
The difference between f(1,g) and f(g,1) is generally an issue of whether mathematicians give preference to "domains" or "ranges" of maps.
Here is one way that you could think of this. I can write EG for a category whose objects are objects are elements of G, and where each pair of objects has a unique map between them. This category has an action of G on it, and you can ask about G-equivariant functors from this category to another category what has a G-action on it.
To define such a functor on the level of objects it suffices to define F(1), where 1 is the unit; equivariance forces us to define F(g) = g F(1). On the level of morphisms, however, we have to make a choice. The unique morphism g→h becomes a morphism F(g)→F(h), and to make such maps compatible with the G-action it suffices to make one of the following sets of choices:
- We could define maps fh:F(1)→F(h) for all h, and get all the other maps as g fh:F(g)→F(gh). To be a functor, we need this to satisfy the cocycle condition fgh = (g fh) fg.
- We could define maps dh:F(h)→F(1) for all h, and get all the other maps as g dh:F(gh)→F(g). To be a functor, we need this to satisfy the cocycle condition dgh = dg (g dh).
In group cohomology, H1(G,M) classifies splittings in the semidirect product of G with M, and the cocycle condition we get comes from our convention of writing this group as pairs (m,g) (which is in the same order as the exact sequence it fits into) and not (g,m). Similarly for H2(G,M).
I would say that I've hit nonstandard cocycle definitions several times because I've been too lazy to come up with sensible conventions about when I'm thinking about domains and ranges or trying to sweep it under the rug, especially when dealing with Hopf algebroids and cohomological calculations there.
I don't have a good answer for higher cocycle conditions other than saying that writing 2-cochains using f(g,1,h) is somehow more unusual than either of the other 2 choices because it's somehow derived from focusing on the "middle" object in a double composite of maps.
Why are you not happy with using Grassmanians as in:
http://en.wikipedia.org/wiki/Classifying_space_for_U(n) ?
A related approach that might interest you is given in Dupont's book
http://www.amazon.com/Curvature-Characteristic-Classes-Lecture-Mathematics/dp/3540086633
using simplicial manifolds. A simplicial manifold $X$ is a sequence of manifolds $\lbrace X_n \rbrace$ and various maps between them. From a simplicial manifold you can construct a topological space called its realisation. This is how you define $EG \to BG$. Although it isn't a manifold you can realise its topology using the finite dimensional spaces $X_n$ which is how you tie things back to the statement of Chern-Weil theory given in the question. In this example the simplicial space is also the one arising in the bar construction and Milnor's join construction of $EG \to BG$.
Best Answer
It's true that in some sense the Atiyah-Singer 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.