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?)


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...

  • $\begingroup$ Do you restrict the rank of the bundle (for fixed degree of the Chern class)? If so, I think the problem is difficult. Here is another issue: the Chern classes do not uniquely determine a complex vector bundle up to equivalence. To see this, consider the case when the base manifold is an odd-dimensional sphere, so that the Chern classes are all zero. Since homotopy groups with odd degree of the classifying space can be nontrivial, the Chern classes are not sufficient to characterize complex vector bundles. $\endgroup$ May 4, 2015 at 15:42
  • $\begingroup$ No, I do not restrict the rank of the bundle. In fact, what sparkled this was Bott Residue formula, which holds when the form being integrated is a symmetric polynomial of (formal roots of) chern classes. So I asked my self, how many forms can we see as chern classes of a certain bundle so that I can apply the formula more often? Of course an higher dimensional bundle could make the other side of the formula more difficult. Your comment is interesting, I will think about it! $\endgroup$
    – Temitope.A
    May 4, 2015 at 15:58
  • $\begingroup$ For your edit, if you're doing complex geometry you could look at Clare Voisin's articles showing that the naive generalization of the Hodge conjecture to Kahler manifolds is false, that is, that the Chern classes of vector bundles or coherent sheaves are not enough to generate the rational classes: webusers.imj-prg.fr/~claire.voisin/Articlesweb/hodgeimrn.pdf $\endgroup$ May 4, 2015 at 19:59

4 Answers 4


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 $(n-1)!$.

  • $\begingroup$ Liviu, do you know a reference for the statement in the first sentence of your answer? Or can you suggest a proof? I can't recall seeing this statement anywhere in the past. $\endgroup$
    – Dan Ramras
    Jul 8, 2015 at 18:02
  • 2
    $\begingroup$ The proof is simple.Any complex line is the pullback of the tautological line bundle over $\mathbb{CP}^\infty$. Next you observe that $\mathbb{CP}^\infty$ is a $K(\mathbb{Z},2)$. $\endgroup$ Jul 8, 2015 at 19:02
  • 1
    $\begingroup$ Ah, great. I was thinking this had something to do with smoothness, whereas in fact it is just a statement about compact CW complexes (and now it is of course very familiar to me...). $\endgroup$
    – Dan Ramras
    Jul 8, 2015 at 19:24

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.


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.


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.

  • $\begingroup$ Do you have a reference for this result? $\endgroup$ Jul 22, 2020 at 16:35

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .