So, now with the preliminaries on connections and curvature established, and the Chern classes summarized, it’s time to see how they connect with one another. Namely, we want to say that, given a complex vector bundle, we can compute the Chern classes in de Rham cohomology by picking a connection — any connection — on it, computing the curvature, and then applying various polynomials.

We shall start by warming up with a special case, of a line bundle, where the algebra needed is easier. Let be a smooth manifold, a complex line bundle. Let be a connection on , and let be the curvature.

Thus, is a global section of ; but since is a line bundle, this bundle is *canonically* identified with . (Recall the notation that is the bundle (or sheaf) of smooth -forms on the manifold .)

Proposition 1 (Chern-Weil for line bundles)is a closed form, and the image in is times the first Chern class of the line bundle .

*Proof:* Let us suppose that we have an open cover of such that each finite intersection of elements of this cover is either empty or contractible; we can do this by choosing a Riemannian metric on , and then taking geodesically convex neighborhoods.

The line bundle is described by nonvanishing, complex-valued continuous functions , satisfying the usual cocycle condition. To compute the first Chern class, we take the family of functions (this can be done, since each intersection is contractible!), and take their Cech 2-coboundary. That is, for a triple , we consider the *integer*

these integers, for varying, form a Cech 2-cocycle (integer-valued), which is the first Chern class, by *definition* of the connecting homomorphisms in sheaf cohomology.

Now let us try to understand where the curvature lives in de Rham cohomology, and first that it is actually a closed 2-form. Over each , the bundle is trivial, and we have chosen an isomorphism of it with the trivial bundle (this is what choosing the amounted to), so we have a canonical frame over (so ). The transition from to over is given by

This is the definition of the local trivializations.

Now, we know that the connection form is a simply a 1-form for each (as it’s a one-by-one matrix) such that , and the transition rule is, by what we saw last time,

The curvature form is given by, locally,

because is a 1-form (and not a large matrix), so .

With these preliminaries established, we can figure out what is happening. Note first that (2) implies that is a closed 1-form, and consequently (which is obtained by gluing the together) is a closed 1-form itself. We next need to figure out where maps to in . To do this, we need to unwind the de Rham isomorphism, and use (1). Namely, the de Rham isomorphism came from the sheaf-theoretic resolution

So, if we have a global closed 2-form , and we want to figure out where it goes in (as a Cech 2-cocycle), we need to start by lifting over each : that is, we need to find 1-forms such that .

Then, we need to form the associated 1-cocycle , which is an element of . Then we have to unapply , and take the 2-coboundary of this. This is how the de Rham isomorphism works.

So, let’s do it. Locally we can lift to . So we can take the . The differences are given by , by the transition rules. Now we have to unapply and take the 2-coboundary of this.

But then we get precisely the differences which were used to define the first Chern class.

August 21, 2011 at 9:07 am

[…] like to finish the series I started a while back on Chern-Weil theory (and then get back to exponential […]

August 10, 2012 at 6:51 pm

Thanks for this nice proof of Chern-Weil for line bundles. I have a puzzlement: I too have found the factor to be 1/(2*pi*i). That is, the Chern class is this factor times the de Rham class of the curvature form. Many others get (-1) times this factor and I have not been able to discover the source of the discrepancy.

August 11, 2012 at 8:25 am

I think the conventions regarding signs of curvature sometimes vary, but at the end, what you want is that the Chern class of the line bundle on (that is, the dual of the tautological one) should correspond to the cohomology class on dual to a point. More generally, the first Chern class of any line bundle should be dual to the vanishing locus of a generic section (I believe this is how algebraic geometers think of it).

August 11, 2012 at 10:59 am

Thanks so much for your rapid response. However I think it is a matter of your being correct and many others being wrong. I don’t think it’s a matter of convention in this case, since it is two cohomology classes that are being compared. I have tracked the discrepancy to the tricky transition equation (1) which you get right and for instance, Griffiths and Harris seem to get wrong, mirabile dictu: see their corresponding equation on page 141. Does $g_{\alpha\beta}$ transform from $\alpha$ to $\beta$ or the other way around? This seems to be the issue and affects the sign. I think you have got it right. (I have seen a few others who also get it with the unsigned factor.) It seems odd, but perhaps this is an error that has been inherited from time immemorial and just get repeated.

August 11, 2012 at 11:35 am

PS: On looking further at G&H: if one accepts their analogue (on p.141) of your equation (1), then it is not hard to show that the connection matrix (and therefore the curvature matrix and the DR class it determines) should get a minus sign, which gives your result.

August 12, 2012 at 6:25 pm

Griffiths-Harris is amazing, but in a book that long there are bound to be small mistakes around — though certainly far fewer than abound on this blog! My feeling is that you should ask someone who has thought more carefully about differential geometry than me about this sign discrepancy. I’ll take another look at their equations, though.