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If $E$ is a complex vector bundle over a manifold $M$ then one defines the space of vector valued $p$-differential forms on them as $\Omega^p(M,E) = \Gamma ( \wedge ^p (T^*M) \otimes E) $

The connection be defined as the map , $\nabla: \Gamma(E) \rightarrow \Omega^1(M,E)$ satisfying $\nabla(fX) = df\otimes X+f\nabla X$ where $f \in C^{\infty}(M)$ and $X \in \Gamma(E)$

  • Can't the connection be thought of as an element of $\Omega^1(M,End(E))$? Difference of two connections though not a connection is such an element.

Curvature of the connection is defined as the map, $$R = \nabla \circ \nabla : \Gamma(E) \rightarrow \Omega^2(M,E)$$

Corresponding definition of "trace" is as a map $Tr: \Omega^p(M,End(E))\rightarrow \Omega^p(M)$ satisfying some natural conditions.

One can show that if $A,B \in \Omega^p(M,End(E))$ then $Tr[A,B]=0$

The crucial relationship from which a version of the Chern-Weil Theorem becomes almost immediately obvious is this,

$$dTr[A] = Tr[[\nabla,A]]$$

The argument for this begins by choosing a different connection say $\nabla'$ and seeing that, $$Tr[[\nabla',A]] = Tr[[\nabla,A]]$$

Hence the RHS of the desired equation is independent of the connection chosen and hence it can be evaluated for any connection to get the LHS. Using the fact that the a bundle is locally trivial one can choose the "trivial" connection and this should apparently yield $dTr[A]$

  • Here I can't see what is a "trivial connection". It would help if someone can write that down in local trivializing coordinates. And how for that does the evaluation of the Tr give a de-Rham derivative of the Tr. The appearance of the de-Rham derivative on the LHS looks very mysterious and thats what makes the Chern-Weil Theorem click! It would help if someone can give the intuition behind this.

    One defines the Chern form as $det(I + \frac{\sqrt{-1}}{2\pi}R)$

    • In this definition $I$ is the identity automorphism of $E$ but $R$ takes values in $\Omega^2(M,End(E))$. Then how is the "+" defined?

What is a good reference for this approach to Chern-Weil Theory in the language of connections and curvature? I found the notation of Kobayashi's lectures very old to relate to and Weiping's book is too terse and Milnor-Stasheff's book does it in the language of cohomology.

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  • $\begingroup$ Have you worked out the details for a U(1) connection on a complex line bundle? This case is easier but instructive. It is worth doing first. $\endgroup$
    – Deane Yang
    Commented Jun 22, 2010 at 19:35
  • $\begingroup$ Are you working from this book? books.google.com/books?id=T08AwbrdEPcC $\endgroup$
    – Steve Huntsman
    Commented Jun 22, 2010 at 19:36
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    $\begingroup$ I'll try to formulate a response to some of your specific questions if I have time later, but for now I'll just point you to the place where I learned this stuff: books.google.com/… Everything is worked out in great detail in chapters 17 and 18 (though the book only does the affine case). $\endgroup$
    – Paul Siegel
    Commented Jun 22, 2010 at 19:50
  • $\begingroup$ @Deane No. I at least got the impression that learning for vector bundles is easier first than for G-bundles. Anyway you have a good reference for that? @Steve Yes. Thats the Weiping's book I had in mind. $\endgroup$
    – Anirbit
    Commented Jun 22, 2010 at 20:17
  • $\begingroup$ I meant a 1-dimensional complex vector bundle. It's the connection that is U(1). $\endgroup$
    – Deane Yang
    Commented Jun 22, 2010 at 20:36

4 Answers 4

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(1) No, a connection is not a section of $\Omega^1(M,\mathrm{End}(E))$: a section would act tensorially and not satisfy the Leibniz rule. The connection is $\mathbb{C}$ linear and not $C^\infty(M,\mathbb{C})$ linear.

(2) Since you have a vector bundle over some manifold, by definition there is some complex vector space $V$ such that around a neighbourhood of some point $p$ in the base manifold $M$, $E$ splits locally as $U\times V$ (a bundle is by definition locally trivial), where $U$ is a domain in $\mathbb{R}^n$. A section of $E$ can be locally expressed as a $V$-valued function on $U$. The trivial connection is just given by component-wise partial derivation. (In coordinates, we can pick a basis of $V$ and use the standard coordinate on $\mathbb{R}^n$, then you just have a map from a domain in Euclidean space to Euclidean space, given by a collection of functions. The trivial connection acts on each of the functions like the exterior derivative.)

(3) The Chern form takes value in $H^*(M,\mathbb{R})$... if you accept that you can add objects in $H^2$ and $H^4$, why not the expression in the determinant? If you worry about such things, perhaps you'd be happier with the definition that the k'th Chern class is given by $(i / 2\pi)^k \sigma_k(\Omega)$, where $\sigma_k$ is the k'th symmetric polynomial acting on $\Omega$.

(4) Another book that may be useful is Morita's Geometry of Differential Forms.

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  • $\begingroup$ It is not clear how you can add elements of two different homology classes. They are two different vector spaces! Similar is the confusion with the "+" in the definition of the Chern Form. May be I am missing something obvious. Kindly explain. $\endgroup$
    – Anirbit
    Commented Jun 23, 2010 at 7:00
  • $\begingroup$ Don't treat $H^*$ as a collection of vector-spaces. Instead look at it as a commutative algebra. Let $\mu$ be a 2k-form, and $\nu$ a 2l-form, we can define a product $\mu\nu = \mu\wedge\nu$ which is a 2(k+l)-form. The product commutes for obvious reasons. So the determinant acts on linear transformations of $E$ with coefficients in the commutative algebra of even-degree differential forms. So you should take the identity to be $1\otimes I$, where $1$ is the constant function in $\Lambda^0(T^*M)$ (0-forms, aka functions), and $I$ the identity transform on $E$. $\endgroup$
    – Willie Wong
    Commented Jun 23, 2010 at 11:17
  • $\begingroup$ ...or rather, the identity transform on the fibre $V$ of $E$ over $M$. $\endgroup$
    – Willie Wong
    Commented Jun 23, 2010 at 11:19
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    $\begingroup$ It is rather hard to clarify more. The idea is that the determinant can be defined for a matrix whose coefficients is in some algebraic structure where addition and multiplication is defined. Observe that $\Lambda^{2k}(T^*M)$ is each a vector space. And there are finitely many of them (if $2k > dim(M)$, the space is trivial). So we can form the direct sum $\Lambda^* = \oplus_{k} \Lambda^{2k}(T^*M)$ to obtain a vector space. Each element of $\Lambda^*$ can be decomposed into a sum of a bunch of 2k forms, and when you add two of them you add the ones of the corresponding degrees. $\endgroup$
    – Willie Wong
    Commented Jun 25, 2010 at 11:24
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    $\begingroup$ Note that a question which you failed to ask is: even if the identity were tensored against something in $\Lambda^2(T^*M)$, you will still need to define a product between two elements therein for the determinant to make sense. I think you need to go a review some basics: (i) linear algebra over commutative rings [any decent algebra book should do] and (ii) the algebra of differential forms [Morita like I mentioned above, or Madsen & Tornehave's From Calculus to Cohomology also has a description]. $\endgroup$
    – Willie Wong
    Commented Jun 25, 2010 at 11:35
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Was going to put this as a comment on Willie's answer, but it was getting pretty long:

The idea is that, while a connection is not a section of $\Omega^1(M,End(E))$- one may view it as a map $\Gamma(E)\to \Gamma(T^*M \otimes E)$- that is, if you give a connection a section, it will give you a peculiar creature which eats up vector fields and spits out what look like sections of E as the $T^*M$ tensor factors go to scalars.

In reality these things that look like sections are morally our connection's choice of lift of the vector field to E (- the confusing factor being that because E is a vector space, it is its own tangent space). Now, as already mentioned above a connection is globally $\mathbb{C}$-linear, but not locally (that is not $C^\infty(M)$-linear) as would be reqired to have $\nabla \in \Gamma(\Omega^1(M,End(E))) $.

This is because any candidate for a lift must perform the dirty deed of differentiating the section along your vector field (this is what the Leibniz rule condition is about). With this in mind, our first draft of a connection would just be $\Theta=$ an extended $d_{DeRham}$-taking sections to their derivatives tensored with the duals of the directions in which they are differentiated (a so called 'trivial' connection)- but there is still room for manoevre and we can add to $\Theta$ a 1-form $A \in \Gamma(\Omega^1(M,End(E)))$ with coefficients in $End(E)$ and still get something that satisfies our conditions.

In fact for any connection $\nabla$ we may write $\nabla= \Theta +A$ for some $End(E)$ valued 1-form A. Now in $\nabla \circ \nabla$ 'the $\Theta \circ \Theta $ bit' will go straight to zero for the same reason $d^2=0$ in the DeRham complex and you will have yourself just an $End(E)$ valued 2-form left (again, strictly, an $End T_{\Gamma (x)}E$ valued 2-form, but who's counting?)- as you said, the curvature $R \in \Omega ^2 (M,E)$.

So (and this is the actual content of what I was going to put in the comment- the rest was just me getting carried away) what is an $End(E)$ valued 2-form when it's at home?

It's something that takes in pairs of vector fields and spits out an element of $End(E)$

$\iff$ It's something that takes in pairs of vector fields and spits out a $Rank(E) \times Rank(E)$ matrix

$\iff$ It's something that takes in pairs of vector fields and spits out the entries of a $Rank(E) \times Rank(E)$ matrix

$\iff$ it is a $Rank(E) \times Rank(E)$ matrix with entries in $\Omega^2 (M)$

So we can add the identity without fear, and we can sensibly take the determinant since even forms commute when we multiply them.

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Classical references: the little book by Chern "Complex manifolds without potential theory" and Wells "Differential analysis on complex manifolds".
Quick hints: the trivial connection is $\nabla=d$; trace is an algebraic operation, the exterior derivative a differential one, hence they commute.

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I suggest taking a second look at Milnors book. I found his treatment of chern classes in one of the appendices really clear and enlightening. I don't have acces to his book at the moment and think he doesn't really state the Chern Weil theorem, but that should only be a small step away. You could also take a look at chapters 5.3 to 5.5 in Moritas "Geometry of Differential Forms". I think page 196 in particular might be enlightening. Chapter 6 is devoted to Chern-Weil theory for principal bundles.

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