Are tensor products of vector bundles "well-behaved"?

470 Views Asked by At

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 $T(f_1,\ldots,f_k): T(\xi_1,\ldots,\xi_k)\to T(\eta_1,\ldots,\eta_k)$?

1

There are 1 best solutions below

3
On

A proof analogous to the one for tensor products will establish the following more general result that you can find in Milnor and Stasheff's Characteristic Classes:

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.