# Are tensor products of vector bundles “well-behaved”?

Do the "nice" properties of the tensor product of vector spaces always extend to tensor products of vector bundles? I'm working through Milnor-Stasheff and recently had to prove that the tensor product of line bundles is commutative, associative, behaves well under dualization, et cetera. For just about every step, I had to work with local trivializations to show that some very natural isomorphism of fibers was continuous on the total space. It seems like there should be slicker, categorical ways of proving these properties. If so, can we do this for other functors?

If you'd prefer, here's a formal variant of my question, in the language of Milnor-Stasheff $\S$3: Let $\mathcal{V}$ be the category of finite-dimensional real vector spaces and isomorphisms between them, and let $T: \mathcal{V}^n \to \mathcal{V}$ be a continuous (covariant) functor. Then if $\xi_i$ is a vector bundle over $B$ (for $1 \leq i \leq k$), then there is a well-defined vector bundle $\pi:T(\xi_1,\ldots,\xi_k) \to B$ with fibers $T(F_b(\xi_1),\ldots,F_b(\xi_k))$ and local trivializations $U \times T(\mathbb{R}^{n_1},\ldots,\mathbb{R}^{n_k}) \to \pi^{-1}(U)$. I ask:

(1) Associativity: If $T$ satisfies $T(U,T(V,W)) \cong T(T(U,V),W)$ for all $U,V,W \in \mathcal{V}$, then do we have $T(\xi_1,T(\xi_2,\xi_3))\cong T(T(\xi_1,\xi_2),\xi_3)$?

(2) Commutativity: If $T$ satisfies $T(V,W) \cong T(W,V)$ for all $V,W \in \mathcal{V}$, then do we have $T(\xi_1,\xi_2)\cong T(\xi_2,\xi_1)$?

(3) Induced maps: Given bundle maps $\ f_i :\xi_i \to \eta_i$, is there an induced bundle map $T(f_1,\ldots,f_k): T(\xi_1,\ldots,\xi_k)\to T(\eta_1,\ldots,\eta_k)$?

• "It seems like there should be slicker, categorical ways of proving these properties. " Yes. Work with sheaves of modules. And better don't argue with stalks or fibers - work with the universal properties ... the tensor product of sheaves of modules classifies bihomomorphisms as usual. Actually there is no need to specify the geometry mere, everything works in an arbitrary ringed topos, and is not really more complicated than the case of modules. – Martin Brandenburg Aug 6 '14 at 13:13
• @MartinBrandenburg: Thanks for the tip! I'll explore things along those lines. – Kyle Aug 6 '14 at 22:13
• @MartinBrandenburg: Thanks again. I've written down a correspondence of vector bundles over $X$ and locally free sheaves of $\mathcal{O}_X$-modules that respects isomorphisms. In particular, if $\Gamma(\xi)$ (resp. $\Gamma(\eta)$) is the sheaf of sections of $\xi$ (resp. $\eta$), then $$\Gamma(\xi \otimes \eta) \cong \Gamma(\eta \otimes \xi) \qquad \qquad (*)$$implies $\xi \otimes \eta \cong \eta \otimes \xi$. Should I be trying to prove ($*$)? From universal properties, it's clear that $\Gamma(\xi)\otimes \Gamma(\eta) \cong \Gamma(\eta)\otimes \Gamma(\xi)$. [continued...] – Kyle Aug 12 '14 at 18:45
• [continued...] In the case of (smooth?) manifolds, I think we have $\Gamma(\xi \otimes \eta) \cong \Gamma(\xi)\otimes \Gamma(\eta)$. That proves ($*$) in the (smooth?) manifold case, but not the general one. I think I could show ($*$) directly if I use the fact that the obvious map $\xi \otimes \eta \to \eta \otimes \xi$ is continuous, but this assumption makes a direct proof of "$\xi \otimes \eta \cong \eta \otimes \xi$" very easy. I'd appreciate any suggestions. – Kyle Aug 12 '14 at 18:45

Let $\mathcal{V}$ be the category of finite dimensional real vector spaces and isomorphisms of such (not all linear transformations, just isomorphisms, for maximum generality). Let $T : \mathcal{V}^n \to \mathcal{V}$ be a continuous functor of $n$ variables, i.e., a functor such that the induced map $T : \text{Iso}(U_1, V_1) \times \cdots \times \text{Iso}(U_n, V_n) \to \text{Iso}(T(U_1, \ldots, U_n), T(V_1, \ldots, V_n))$ is continuous with respect to the natural topology on $\text{Iso}(U,V)$ (the subspace topology induced from the space of all linear maps). Then there exists a bundle construction that does $T$ fibrewise.
If $T$ is smooth, it works for smooth vector bundles, etc. Also, since we only took isomorphisms, the distinction between covariant and contravariant functors is not important and this result also furnishes constructions such as the dual vector bundle.
• Thanks for your response, which is convenient since I'm working through Milnor-Stasheff. But does the above construction indicate when a map out of a space, such as the obvious map $\xi \otimes \eta\to \eta \otimes \xi$, is continuous? (Note that I edited my original question because I had accidentally typed $F_b(\xi)\otimes F_b(\eta)\to F_b(\xi \otimes \eta)$, which isn't too interesting a map because that's how the fibers of the tensor product are defined.) – Kyle Aug 5 '14 at 4:47