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Let $k$ be a natural number and $F$ be $\mathbb C$ or $\mathbb R$.

What conditions should be imposed on a topological space that it was true that a $k$-dimensional vector bundle defined over $F$ on this space is determined by its characteristic classes (Chern or Stiefel-Whitney and Pontrjagin classes)?

For example, in the case of complex line bundles it is enough to have $H^1(X,\mathbb C)=0.$ This comes from the exponential sequence which doesn't exist in higher dimension.

Perhaps for $k>1$ this question is difficult to answer in general, so I'll be more specific.

Is it true for compact simply-connected oriented smooth 4-manifolds that a real 4-dimensional vector bundle is determined by its Stiefel-Whitney and Pontrjagin classes? This would give a partial answer to my yesterday's question (Tangent bundles of exotic manifolds) since in this case characteristic classes of tangent bundle give a purely topological information (signature, Euler number and diagonal of the intersection form mod $2$).

It seems well known that over complex projective space (and even over Grassmannian) in each dimension a complex vector bundle is determined by its total Chern class. Is it true in the real case (or in the mixed case of complex bundles over real Grassmannians or vice versa)? Can we say something about another algebraic varieties?

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    $\begingroup$ I found that the question about 4-manifolds is almost covered by Dold-Whitney Theorem: "If two oriented 4-plane bundles over an oriented 4-manifold have the same second Stiefel-Whitney class $w_2$, Pontryagin class $p_1$ and Euler class $e$, then they must be isomorphic". $\endgroup$
    – Misha
    Aug 19, 2013 at 10:36
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    $\begingroup$ You're thinking of holomorphic complex line bundles. Topological complex line bundles are always completely determined by their first Chern class. $\endgroup$
    – Qiaochu Yuan
    Jul 15, 2014 at 5:28
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    $\begingroup$ In the topological case the exponential sequence doesn't say what you think it does. For $H^n(X, \mathbb{C})$ the cohomology is taken with coefficients in $\mathbb{C}$ with the usual, rather than discrete, topology, and this vanishes because $\mathbb{C}$ is contractible. So here the exponential sequence tells you that $H^n(X, \mathbb{C}^{\times}) \cong H^{n+1}(X, \mathbb{Z})$ for all $n$ ($\mathbb{C}^{\times}$ also has the usual, rather than discrete, topology). $\endgroup$
    – Qiaochu Yuan
    Jul 15, 2014 at 7:47
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    $\begingroup$ If instead you want both $\mathbb{C}$ and $\mathbb{C}^{\times}$ to have the discrete topology, then $H^n(X, \mathbb{C})$ is the thing you expect but $H^1(X, \mathbb{C}^{\times})$ no longer classifies line bundles; instead it classifies flat line bundles. $\endgroup$
    – Qiaochu Yuan
    Jul 15, 2014 at 8:40
  • $\begingroup$ Said another way, it's important to distinguish between the sheaf of holomorphic complex-valued functions (only available in the presence of a complex structure), the sheaf of continuous complex-valued functions, and the sheaf of locally constant complex-valued functions. $\endgroup$
    – Qiaochu Yuan
    Jul 15, 2014 at 18:15

2 Answers 2

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(Let's start with the complex case.)

Basically there are two issues here.

1) First of all, Chern classes are stably invariant, i.e. can't distinguish stably isomorphic vector bundles. So the question becomes

Do Chern classes determine an element in the group $\tilde K^0(X)$ of vector bundles on $X$ modulo stable equivalence?

(Example. Since $\pi_4(U(2))=\pi_4(S^1\times S^3)=\mathbb Z/2$ there is a non-trivial 2-bundle on $S^5$. But of course $\tilde K^0(S^5)=0$ so this bundle is stably trivial — in particular, it has zero Chern classes.)

2) This 'stable' variant of the question is (still very nontrivial but) much more tractable (essentially because K-theory is a generalized cohomology theory etc). The Chern character gives an isomorphism $K^0(X)\otimes\mathbb Q\to H^{\text{even}}(X;\mathbb Q)$. So

if $K(X)$ has no torsion, an element of $K(X)$ is completely determined by its Chern classes.

And if $H(X)$ is torsion-free so is $K(X)$ (this follows from AHSS). That's true e.g. for complex Grassmannians and flag manifolds.

(Example. The if $L\in\tilde K^0(\mathbb RP^6)$ is the complexified tautological bundle, $4L$ is a non-trivial element, i.e. a stably non-trivial bundle, with trivial Chern classes.)

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Is it true for compact simply-connected oriented smooth 4-manifolds that a real 4-dimensional vector bundle is determined by its Stiefel-Whitney and Pontrjagin classes?

No. For example, the tangent bundle of $S^4$ is stably trivial (because the outward pointing normal gives a trivialization of its normal bundle in $\mathbb{R}^5$; true more generally of all spheres) but not trivial (because its Euler class is nontrivial; true more generally of all even spheres), so its Stiefel-Whitney and Pontrjagin classes vanish but it is not isomorphic to a trivial bundle.

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