Classifying space level description of the tensor product of vector bundles?

Question

Let's say $\xi: X \to BO(n), \eta : X \to BO(m)$ are two vector bundles over $X$. If I want to take the sum of these two vector bundles, then at the level of classifying spaces, I have the map $\oplus : BO(n) \times BO(m) \to BO(n+m)$ given by taking the (interwoven) direct sum of subspaces (for me $BO(n)$ is the set of $n$ planes in $\mathbb{R}^\infty$), and I can use my two vector bundles to get a map $\xi \times \eta : X \to BO(n) \times BO(m)$ which when I post compose with $\oplus$ gives me a map $X \to BO(n+m)$ that represents $\xi \oplus \eta$.

I was wondering about the analogous situation for tensor products of vector bundles. Namely, is there a map $\otimes : BO(n) \times BO(m) \to BO(nm)$ that when precomposed with $\xi \times \eta$ yields $\xi \otimes \eta$?

Answer 1

Yes. Given real vector bundles $E\rightarrow X$ and $F\rightarrow Y$ of rank $n,m$, respectively, you can form the external tensor product $E\widehat\otimes F\rightarrow X\times Y$. This is the real vector bundle of rank $n\cdot m$ whose fiber over $(x,y)\in X\times Y$ is the tensor product $$(E\widehat\otimes F)_{(x,y)}=E_x\otimes F_y.$$ In fact we can also construct this in another way. Namely, if $\pi_X,\pi_Y:X\times Y\rightarrow X$ are the projections onto each of the factors, then $$E\widehat\otimes F\cong (\pi_X^*E)\otimes(\pi_Y^*F)$$ where the right-hand side is the ordinary tensor product.

Notice moreover that if $X=Y$, and $\Delta:X\rightarrow X\times X$ is the digonal map, then $$\Delta^*(E\widehat\otimes F)\cong E\otimes F$$ as bundles over $X$.

Now, to get a universal example you can just apply the consturctions to the universal $n$- and $m$-plane bundles. That is, if $E(k)\rightarrow BO(k)$ denotes the universal $k$-plane bundle, then the external tensor product $$E(n)\widehat\otimes E(m)\rightarrow BO(n)\times BO(m)$$ classifies all external tensor products. In more detail, if $f:X\rightarrow BO(n)$ and $g:Y\rightarrow BO(m)$ classify the respective bundles $E,F$, then $(f\times g)^*(E(n)\widehat\otimes E(m))\cong E\widehat\otimes F$ as bundles over $X\times Y$.

Now the universal example is itself a real bundle of rank $n\cdot m$, so itself admits a classifying map $$\rho_{n,m}:BO(n)\times BO(m)\rightarrow BO(mn)$$ such that $$\rho_{m,n}^*E_{mn}\cong E(n)\widehat\otimes E(m).$$ The composite $\rho_{m,n}(f\times g)$ then classifies $E\widehat\otimes F\rightarrow X\times Y$.

Answer 2

So the lame answer is yes because of Yoneda's lemma. Tensor product (like direct sum) is clearly a natural transformation of the relevant functors, and so corresponds to some map between the representing objects.

The constructive answer is a little harder than the direct sum case. The direct sum case basically relies on picking an isomorphism $\mathbb{R}^ \infty \oplus \mathbb{R}^ \infty \rightarrow \mathbb{R}^ \infty $, your so-called interweaving.

The analogous process for tensor product is finding an isomorphism $\mathbb{R}^ \infty \otimes \mathbb{R}^ \infty \rightarrow \mathbb{R}^ \infty$. An explicit isomorphism here is harder to describe, but one exists because they have the same dimension.

An interesting question is whether or not there is a model of the spaces $BO(n)$ (i.e. a space that classifies the same thing) such that the tensor product and direct sum maps are commutative and associative on the nose (rather than up to homotopy). If you only care about direct sum, this is definitely possible for homotopical reasons, but I'm not sure what happens when you add in tensor product.