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My question is as follows.

Do the Chern classes as defined by Grothendieck for smooth projective varieties coincide with the Chern classes as defined with the aid of invariant polynomials and connections on complex vector bundles (when the ground field is $\mathbf{C}$)?

I suppose GAGA is involved here. Could anybody give me a reference where this is shown as detailed as possible? Or is the above not true?

Some background on my question:

Let $X$ be a smooth projective variety over an algebraically closed field $k$. For any integer $r$, let $A^r X$ be the group of cycles of codimension $r$ rationally equivalent to zero. Let $AX=\bigoplus A^r X$ be the Chow ring.

Grothendieck proved the following theorem on Chern classes.

There is a unique "theory of Chern classes", which assigns to each locally free coherent sheaf $\mathcal{E}$ on $X$ an $i$-th Chern class $c_i(\mathcal{E})\in A^i(X)$ and satisfies the following properties:

C0. It holds that $c_0(\mathcal{E}) = 1$.

C1. For an invertible sheaf $\mathcal{O}_X(D)$ on $X$, we have that $c_1(\mathcal{O}_X(D)) = [D]$ in $A^1(X)$.

C2. For a morphism of smooth quasi-projective varieties $f:X\longrightarrow Y$ and any positive integer $i$, we have that $f^\ast(c_i(\mathcal{E})) =c_i(f^\ast(\mathcal{E}))$.

C3. If $$0\longrightarrow \mathcal{E}^\prime \longrightarrow \mathcal{E} \longrightarrow \mathcal{E}^{\prime\prime} \longrightarrow 0$$ is an exact sequence of vector bundles on $X$, then $c_t(\mathcal{E}) = c_t(\mathcal{E}^\prime)c_t(\mathcal{E}^{\prime\prime})$ in $A(X)[t]$.

So that's how it works in algebraic geometry. Now let me sketch the complex analytic case.

Let $E\longrightarrow X$ be a complex vector bundle. We are going to associate certain cohomology classes in $H^{even}(X)$ to $E$. The outline of this construction is as follows.

Step 1. We choose a connection $\nabla^E$ on $E$;

Step 2. We construct closed even graded differential forms with the aid of $\nabla^E$;

Step 3. We show that the cohomology classes of these differential forms are independent of $\nabla^E$.

Let us sketch this construction. Let $k= \textrm{rank}(E)$. Let us fix an invariant polynomial $P$ on $\mathfrak{gl}_k(\mathbf{C})$, i.e. $P$ is invariant under conjugation by $\textrm{GL}_k(\mathbf{C})$.

Let us fix a connection $\nabla^E$ on $E$. We denote denote its curvature by $R^E = (\nabla^E)^2$. One shows that $$R^E \in \mathcal{C}^\infty(X,\Lambda^2(T^\ast X)\otimes \textrm{End}(E)).$$ That is, $R^E$ is a $2$-form on $X$ with values in $\textrm{End}(E)$. Define $$P(E,\nabla^E) = P(-R^E/{2i\pi}).$$ (This is well-defined.)

The Chern-Weil theorem now says that:

The even graded form $P(E,\nabla^E)$ is a smooth complex differential form which is closed. The cohomology class of $P(E,\nabla^E)$ is independent of the chosen connection $\nabla^E$ on $E$.

Choosing $P$ suitably, we get the Chern classes of $E$ (by definition). These are cohomology classes. In order for one to show the equivalence of these "theories" one is forced to take the leap from the Chow ring to the cohomology ring.

How does one choose $P$? You just take $P(B) = \det(1+B)$ for a matrix.

Motivation: If one shows the equivalence of these two theories one gets "two ways" of "computing" the Chern character.

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1 Answer 1

up vote 15 down vote accepted

See this question.

Also, read Milnor-Stasheff or Hatcher's book "Vector Bundles and K-Theory". In particular, Milnor-Stasheff and Hatcher prove that there is a unique "theory of Chern classes" for complex vector bundles over topological spaces satisfying axioms totally analogous to the C0, C1, C2, C3. Milnor-Stasheff constructs Chern classes using the Thom isomorphism theorem; Hatcher constructs them using the Leray-Hirsch theorem. Hatcher's construction (in topology) is essentially the same as Grothendieck's construction (in algebraic geometry). I think if you study the two constructions (Hatcher's and Grothendieck's) carefully, their equivalence should follow fairly easily. I did this once a while ago.

I don't think you need any GAGA theorem. I think you just need the fact that there is an analytification functor.

Appendix C of Milnor-Stasheff then proves the equivalence with the Chern-Weil theory.

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1  
Don't I also need an "algebrafication" functor? That is, something that goes the other way around. –  Ariyan Javanpeykar Mar 31 '10 at 19:43
4  
No. The point is that we want to prove that the different definitions of Chern classes agree for an algebraic vector bundle, whether we consider it as an algebraic vector bundle or as a topological vector bundle. If we have a topological vector bundle, there is no canonical way to consider it as an algebraic vector bundle. –  Kevin H. Lin Mar 31 '10 at 20:14
    
Kevin, your answer looks like the right one to me, but I'm sure you didn't mean to claim that one can construct Chern classes using Steenrod squares! –  Tim Perutz Mar 31 '10 at 21:23
    
Tim, you're right. Thanks. I confused Chern with Stiefel-Whitney. I've fixed it. –  Kevin H. Lin Apr 1 '10 at 0:27

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