Inverse problem of Chern Classes

Question

For my graduate (master) thesis I am studying the theory of Chern Classes. As a possible personal development the only sensible idea I have so far, and which I frankly think is impossible, is to work on the inverse problem, i.e., given a class in cohomology, is there a vector bundle which has that class as one of its Chern classes? The following question is to understand what it is already known about this and if maybe I might focus my attention on a small sub-problem and if it is an idea worth telling my supervisor at all or just forget about it.

1. Has this problem been analysed before? Is it a sensible matter to tackle?
2. I know how to find Chern Classes by taking conjugate invariant symmetric polynomials of the curvature matrix of a connection on the manifold (so de Rham Cohomology). Is this approach better, worse, equivalent to studying line divisors, picard groups and related? (Is there any other approach to finding Chern classes?)

EDIT

Ok, the problem seems rather involved with fundamental questions. Can you suggest one or more sub-problem which I might be able to work on in 3 months? Maybe even pre-existing results over which I can elaborate which explicit computations, examples, counter-examples, generalizations...

Answer 1

For $c_1$ the problem is solved. $\newcommand{\bZ}{\mathbb{Z}}$ For any smooth manifold and any $c\in H^2(M,\bZ)$ there exists a smooth complex line bundle $L\to M$ such that $c_1(L)=c$.

By results of Thom, for any oriented manifold $M$, any $\alpha\in H_{n-4}(M,\bZ)$ is represented by an oriented submanifold.

On the other hand, for any $n\geq 7$, there exists an $n$-dimensional oriented manifold $M$ and a homology class $\alpha\in H_{n-4}(M,\bZ)$ such that the normal bundle of any submanifold representing $\alpha$ does not admit a $spin^c$-structure; see Theorem 3, page 9 of this paper.

If $\alpha^\dagger\in H^4(M,\bZ)$ denotes the Poincare dual of such an $\alpha$, then there exist no rank 2 complex vector bundle $E\to M$ such that $c_2(E)=\alpha^\dagger$.

If such a bundle existed, then the zero set of a generic section of $E$ will be an oriented submanifold $S$ of $M$ representing $\alpha$. The normal bundle of $S$ in $M$ is isomorphic to $E|_S$. In particular it admits $spin^c$ structures because it admits an almost complex structure.

Edit 1. A rather deep divisibility theorem shows that if $n\geq 3$ and $E\to S^{2n}$ is a complex vector bundle, then $c_n(E)\in H^{2n}(S^{2n},\bZ)$ is divisible by $(n-1)!$.

Answer 2

Suppose $M$ is a closed oriented $2n$-manifold which admits a complex spin ($\text{Spin}^c$) structure, which means that its second Stiefel-Whitney class $w_2(M) \in H^2(M, \mathbb{Z}_2)$ is the $\bmod 2$ reduction of a class $c_1(M) \in H^2(M, \mathbb{Z})$; this holds in particular if $M$ admits an almost complex structure or if $H^3(M, \mathbb{Z})$ has no $2$-torsion.

Then the Chern classes $c_k(V) \in H^{2k}(M, \mathbb{Z})$ of a complex vector bundle $V$ satisfy integrality conditions coming from the following version of the Hirzebruch-Riemann-Roch theorem, which is a consequence of the Atiyah-Singer index theorem: there is a rational characteristic class $\text{td}(M) \in H^{\bullet}(M, \mathbb{Q})$, the Todd class of $M$, given by

$$\text{td}(M) = e^{ \frac{c_1(M)}{2} } \hat{A}(M)$$

where $\hat{A}(M)$ denotes the $\hat{A}$ class of $M$, such that

$$\int_M \text{ch}(V) \text{td}(M) \in \mathbb{Z}$$

where $\text{ch}(V)$ denotes the Chern character of $V$ and $\int_M$ denotes the pairing of a class in $H^{\bullet}(M, \mathbb{Q})$ with the fundamental class $[M]$. Note that the integrand lives in even degrees and so this statement only has content for even-dimensional manifolds.

To get something easier to work with, suppose in addition that every component of $\text{td}(M)$ but the zeroth component $\text{td}_0(M) = 1$ and the top component $\text{td}_n(M)$ vanishes. Then the index theorem reads

$$\int_M \left( \text{ch}_n(V) + \text{td}_n(M) \right) \in \mathbb{Z}$$

and hence applying the theorem twice, once to the trivial vector bundle and once to $V$, an equivalent statement is that

$$\int_M \text{ch}(V) = \int_M \text{ch}_n(V) \in \mathbb{Z}.$$

When $n = 2$, so $M$ is a closed oriented $4$-manifold with a complex spin structure, the vanishing condition is that $\text{td}_1(M) = \frac{c_1(M)}{2} = 0$ vanishes (rationally). This follows, for example, if $M$ admits a spin structure ($w_2(M)$ vanishes) and the complex spin structure is chosen so that $c_1(M) = 0$. The integrality condition is then that

$$\int_M \text{ch}_2(V) = \int_M \frac{c_1(V)^2 - 2 c_2(V)}{2} \in \mathbb{Z}$$

or equivalently that $c_1(V)^2$ is even. If $M$ is simply connected and $w_2(M) = 0$ then in fact every class in $H^2(M, \mathbb{Z})$ has this property (see e.g. this blog post) but in general I think it's an extra condition.

If $M$ is stably frameable then all of its stable characteristic classes vanish, and in particular (with a complex spin structure in which $c_1(M) = 0$) all of the components of $\text{td}(M)$ except the zeroth one vanish, so $M$ satisfies the vanishing condition. In particular this is true if $M = S^{2n}$. Moreover, because $S^{2n}$ has vanishing cohomology in degrees between $0$ and $2n$, a straightforward computation shows that if $V$ is any complex vector bundle on $S^{2n}$, then

$$\text{ch}_n(V) = \frac{c_n(V)}{(n-1)!}.$$

The integrality condition is then that $c_n(V)$ is divisible by $(n-1)!$, as mentioned in Liviu Nicolaescu's answer.

Answer 3

There are definitely lots of ways of thinking about Chern classes.

Let me just make the following a bit imprecise statement:

Up to tensoring with $\mathbb Q$, a K-theory class (i.e. for compact spaces: a vector bundle up to stable equivalence) is the same as a collection of cohomology classes (i.e. Chern classes).

Edit: Integral results are much harder. Sometimes they are deduced from index theorems, where on one side, you have a characteristic number (the coefficients of the monomials in the Chern classes can be complicated, e.g. Bernoulli numbers show up regularly), and on the other side the index of an operator which must be an integer. This is one way to prove the result for even-dimensional spheres. This relates all three answers present at the moment.

Answer 4

For any smooth affine scheme $X$ of dimension $d$ over a field $k$ such that $(d-1)!$ is invertible in $k$, for any element $(\alpha_{d}\times\cdots\times\alpha_{1})\in CH^{d}(X)\times\cdots CH^{1}(X)$, there is a vector bundle $V$ such that $ch_{i}(V)=\alpha_{i}$ if and only if $\alpha_{i}\in (i-1)!ch^{i}(X).$ That, the condition mentioned in the answer of Qiaochu Yuan above is not only necessary but also sufficient.