Cohomology of vector bundles via Intersection Theory - MathOverflow

Cohomology of vector bundles via Intersection Theory

2012-12-12T05:37:48Z

Let $X$ be a smooth projective variety over a fixed field $k$ (take $k = \mathbb{C}$ if necessary). For a vector bundle $E$ on $X$, $ch(E)$ will be in the Chow ring.

$\textbf{Question 1: }$ If $\mathcal{E}$ is a locally free sheaf of rank $n$ on $X$, (with associated vector bundle $E$) can one recover the dimensions of the sheaf cohomology groups $\dim_k H^i(X, \mathcal{E})$ from the total chern class $\textrm{ch}(E)$? How about just $\dim_k H^0(X, \mathcal{E})$? If not that, how about if $E$ is just a line bundle? Can we at least determine if $\mathcal{E}$ has global sections?

$\textbf{Question 2: }$ In the case $k = \mathbb{C}$, can one recover the dimensions of the singular cohomology groups $\dim_k H^i_{sing} (X, k)$ from total chern classes of various bundles? We can recover the Euler characteristic of $X$ as $\int_X c_n(T_X)$. In the case of curves, we can even recover the geometric genus (since this is a degenerate case: the Euler characteristic and geometric genus encode the same information). Can we recover the geometric genus of $X$ if $\dim X > 1$ from chern classes of various bundles?

$\textbf{Question 3: }$ Is there a good example to indicate the kind of information that $\textrm{ch}(T_X)$ carries about $X$ beyond it's Euler characteristic?

$\textbf{Question 4: }$ Colloquially, people refer to the Chow ring as giving a "homology theory". In the case $k = \mathbb{C}$, can one recover the usual (singular) homology groups $H_i(X,\mathbb{Z})$ from the Chow groups? If not, what about $H_i(X, \mathbb{Q})$?

Answer by Sasha for Cohomology of vector bundles via Intersection Theory

2012-12-12T05:48:17Z

About question 1. You can recover $\sum(-1)^i\dim H^i(X,E)$ by Riemann--Roch. But the individual cohomolgy groups cannot be recovered. For example, let $X$ be a curve of positive genus $g$ and $E$ a line bundle of degree $0$. If $E$ is generic it has $H^0 = H^1 = 0$, but for trivial bundle you have $\dim H^0 = 1$, $\dim H^1 = g$.

EDIT. Another example showing that the Chern classes with values in the Chow ring also don't help. Let $X = C \times P^1$ with $C$ being a curve of positive genus. Let $E = p^*L \oplus p^*L^{-1}$, where $L$ is a line bundle of degree zero on $C$ and $p:C \times P^1 \to P^1$ is the projection. Then $c_1(E) = 0$ and $c_2(E) = 0$ in the Chow ring. However, the dimension of the cohomology groups depend on $L$.

Answer by Will Sawin for Cohomology of vector bundles via Intersection Theory

2012-12-12T06:09:50Z

(2) Chern classes of all bundles generated by the tangent & cotangent bundles under all the obvious operations: sums, tensors, syms, wedgs, etc., are determined by the chern classes of the tangent bundle. But already in surfaces the chern classes of the tangent bundle are not sufficient to determine $H^1$.

(3) Algebraic cobordism?

Answer by Daniel Loughran for Cohomology of vector bundles via Intersection Theory

2012-12-15T11:06:26Z

The answer to Question $4$ is also no. Let $C$ be a smooth projective curve of genus $g$ over $\mathbb{C}$. Then the chow ring $A(C)$ of $C$ is isomorphic to $\mathbb{Z}[x]/(x^2)$, whereas of course the betti numbers of $C$ are $1,2g,1$. So the chow ring will not determine the homology in general.

However, for smooth projective varieties $X$ over $\mathbb{C}$ there is a cycle class map $A(X) \to H^*(X,\mathbb{Z})$ which is a ring homomorphism (this is all explained in the appendices of Hartshorne, though one needs to use $\ell$-adic cohomology over other fields). In particular algebraic cycles will give you cohomology classes. Determining which cohomology classes come from algebraic cycles is part of the Hodge conjecture.