# Cohomology of vector bundles via Intersection Theory

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 $H_i(X, \mathbb{Q})$?

• Re: Q3. For surfaces $c_1^2$, $c_2$ (of the tangent bundles) are "independent variables" in the sense that one does not determine the other. Take a look at the book by Barth et. al. on surfaces. Re Q4: No, there is a map $CH_i\to H_{2i}$ which is usually not surjective (even after tensoring with $\mathbb{Q}$). – Donu Arapura Dec 12 '12 at 16:45
• In case it wasn't clear, $ch(T_X)$ determines the Chern numbers given above, and $c_{2}(X)$ is the Euler characteristic of surface. – Donu Arapura Dec 12 '12 at 16:56
• The Chern polynomial is additive in exact sequences of vector bundles, as is the Euler characteristic. However, the dimensions of individual cohomology groups are not, so there is no hope to recover these from the former. – ACL Dec 15 '12 at 9:03
• The Chern polynomial is a too rough invariant to detect continuous phenomena such as the existence of sections. A degree 0 line bundle on a curve has non zero sections if and only if it is trivial; it is semi-ample if and only if some power is trivial. – ACL Dec 15 '12 at 9:06
• On another note, chern classes are insensitive to adding copies of the trivial line bundle, which has global sections. So we can't detect global sections in complete generality. – LMN Dec 15 '12 at 20:49

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 $L$.

• The first chern class of $E$ is the divisor corresponding to $E$ (modulo rational = linear equivalence) (at least if $E$ has global sections). However, $\dim_k(\mathcal{O}_X(D))$ doesn't depend on the linear equivalence class of $D$. So it sounds like we should be able to recover the dimensions of cohom. groups in this case? – LMN Dec 12 '12 at 6:09
• That should be $dim_k H^0(\mathcal{O}_X(D))$. – LMN Dec 12 '12 at 6:11
• So, do you want to say that you consider Chern classes with values in the Chow ring rather than in the cohomology ring? It is better to specify this in the question. – Sasha Dec 12 '12 at 6:12
• Sasha, yes. You're right I should have specified that - it didn't even cross my mind. – LMN Dec 12 '12 at 6:14
• I added another example which shows that this does not help. – Sasha Dec 12 '12 at 6:17

(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?

• Will, are there specific examples you have in mind for (2)? – LMN Dec 12 '12 at 6:31

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.