In Hatcher's book on Vector Bundles he states (page 14) that

It is routine to verify that the tensor product operation for vector bundles over a fixed base space is commutative, associative, and has an identity element, the trivial line bundle. It is also distributive with respect to direct sum.

The gluing functions for the bundle $E_1 \otimes E_2$ are the tensor product of the two matricies $g^1_{\beta \alpha}(x)$ and $g^2_{\beta \alpha}(x)$

So for the tensor product for vector bundles to be commutative, wouldn't we require that $g^1_{\beta \alpha}(x) \otimes g^2_{\beta \alpha}(x) = g^2_{\beta \alpha}(x) \otimes g^1_{\beta \alpha}(x)$?

This is certainly not true - the best we can say is that they are permutation equivalent

(Since $g_{\beta \alpha}(x):U_\alpha \cap U_\beta \to GL_n(\mathbb{R})$)

What am I missing?


1 Answer 1


The desired statement about tensor products is that $E_1 \otimes E_2$ is isomorphic to $E_2 \otimes E_1$, not that it is equal to $E_2 \otimes E_1$.

If $E'$ and $E''$ are vector bundles given by transition functions $g'_{ij}$ and $g''_{ij}$, then they will be isomorphic iff these transition functions determine the same element in the (sheaf) cohomology set $H^1(X,\operatorname{GL}_n)$. This means two things: (i) one is allowed to pass to a refinement of the covering, and (ii) one regards two sets of transition functions on the same covering as being equivalent if there exists $\lambda_i: X \rightarrow \operatorname{GL}_n$ such that

$g''_{ij} = \lambda_i^{-1} g'_{ij} \lambda_j$:

one says that $g'$ and $g''$ are cohomologous.

This is all rather abstract, but if you look back at what you've already written, you should find that you've explained exactly what functions $\lambda_i$ to take to show that your two sets of transition functions are cohomologous (in this case it is not necessary to refine the cover), hence the corresponding vector bundles are isomorphic.

Added: I just looked at Hatcher's notes/prebook, and he does not introduce the cohomological perspective here. So probably there's a better way to look at it. How about this: one can argue Atiyah-style (i.e., from his book on K-theory) that there is a natural isomorphism of finite dimensional vector spaces $V \otimes W \rightarrow W \otimes V$ (just switch the factors!), so this will extend to an isomorphism of the tensor product of vector bundles. (This seems a lot simpler, but the OP phrased things in terms of transition functions, and I like that perspective a lot. It requires nonabelian sheaf cohomology to really be done right, but -- what can I say? -- in my line of work nonabelian cohomology of sheaves -- on Grothendieck topologies, no less -- is fairly ubiquitous anyway.)

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    $\begingroup$ Hi! Could you explain the sheaf cohomology in more detail? I certainly know the material but I am not familiar with the terminology. Glad to know what you had written. $\endgroup$
    – Kerry
    May 25, 2011 at 5:35
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    $\begingroup$ @user7887: I'm afraid it would take me too long to expose the connection between fiber bundles and sheaf cohomology in detail. I'll tell you where I learned it from: $\S 1.3$ of Hirzebruch's Topological Methods in Algebraic Geometry. I recommend it highly. $\endgroup$ May 25, 2011 at 5:43
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    $\begingroup$ thanks for that, I had suspected it was something like this. I agree if you use Atiyah's 'continuous functor $T$' that is is clear, but I was trying to prove what Hatcher says. Maybe we have different definitions of routine! (Or perhaps he was referring to using the construction on the fibers - $h_1 \otimes h_2:p^{-1}_1(U) \otimes p^{-1}_2(U) \to U \times (\mathbb{R}^{n_1} \otimes \mathbb{R}^{n_2})$ and then use the isomorphism $\mathbb{R}^{n_1} \otimes \mathbb{R}^{n_2} \to \mathbb{R}^{n_2} \otimes \mathbb{R}^{n_1})$ $\endgroup$
    – Juan S
    May 25, 2011 at 6:28
  • $\begingroup$ L.Clark: Thank you greatly for the reference. $\endgroup$
    – Kerry
    May 25, 2011 at 11:32
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    $\begingroup$ As a minor comment, there is a bit more than "naturality of the isomorphism": namely, under the tensor product, vector bundles form a symmetric monoidal category (which includes various coherence axioms for these various isomorphisms). $\endgroup$ May 25, 2011 at 12:41

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