# Classifying Map $p_{n,m}:BO(n) \times BO(m) \to BO(n+m)$

I have a question about a construction & its properties described in May's "A Concise Course in Algebraic Topology"(see page 190):

Let $$VB_n(-): (pcHoTop) \to (Set)$$ the functor which assigns to every paracompact topological space $$X$$ the set $$VB_n(X)$$ of real $$n$$-dimensional vector bundles.

We know that this functor is a homotopy functor and is representable by a classifying space $$BO(n)= G_n(\mathbb{R}^{\infty})$$. Therefore we have a natural equivalence $$VB_n(-) \cong [-,BO(n)]$$.

Consider following natural transformation $$\phi: VB_n(-) \times VB_m(-) \to VB_{n+m}(-)$$ given by the family of assignments $$\phi_X:VB_n(X) \times VB_m(X) \to VB_{n+m}(X), (E,F) \mapsto E \oplus F$$

$$E \oplus F$$ is the direct sum bundle defined as $$E \oplus F:= \{(v,w) \in E \times F \vert p_E(v)= p_F(w) \text{ for } p_E:E \to X, p_F:F \to X \}$$

Since $$VB_n(-)$$ is representable the natural transformation is nothing but a natural trafo $$[-,BO(n) \times BO(m)]=[-,BO(n)] \times [-,BO(m)] \to [-,BO(n+m)]$$.

By Yoneda lemma this natural trafo $$\phi$$ coincides uniquely with a morphism $$p_{n,m}':BO(n) \times BO(m) \to BO(n+m)$$

My question is how to see that this morphism $$p_{n,m}'$$ coming from $$\phi$$ coincides exactly with the morphism $$p_{n,m}$$ described concretely here:

namely $$p_{n,m}:BO(n) \times BO(m) \to BO(n+m)$$ is given by taking an $$n$$-plane $$N \subset \mathbb{R}^{\infty}$$ and an $$m$$-plane $$M \subset \mathbb{R}^{\infty}$$ and mapping it to $$N \oplus M$$. This gives indeed by definition a point of $$G_{n+m}(\mathbb{R}^{\infty})$$ after identification $$\mathbb{R}^{\infty} \cong \mathbb{R}^{\infty} \oplus \mathbb{R}^{\infty}$$.

Furthermore $$p_{n,m}$$ induces $$p_{n,m}^*$$ with $$p_{n,m}^*(\gamma_{n+m})= \gamma_{n} \times \gamma_{m}$$ where $$\gamma_n$$ is the universal $$n$$ bundle over $$BO(n)$$.

So the question is how to see that $$p_{n,m}'$$ and $$p_{n,m}$$ coincide?

It's because $$p_{n,m}$$ and $$p_{n,m}'$$ both classify the bundle $$\gamma_n\times\gamma_m$$ over $$BO(n)\times BO(m)$$, so it follows by the theory of classifying spaces that they are homotopic, i.e. they represent the same morphism in the homotopy category. You've pointed out that the concrete map $$p_{n,m}$$ classifies this bundle, I will argue formally that the mysterious $$p_{n,m}'$$ does as well.

Let $$\rho_n\colon BO(n)\times BO(m)\to BO(n)$$ and $$\rho_m\colon BO(n)\times BO(m)\to BO(m)$$ be projections. Any map $$f\colon X\to BO(n)\times BO(m)$$ is a product of maps $$(\rho_n\circ f) \times (\rho_m\circ f)$$, and in fact we have a natural isomorphism

$$(\rho_n)_*\times (\rho_m)_*\colon[-,BO(n)\times BO(m)] \cong [-,BO(n)]\times[-,BO(m)]$$

At the level of bundles, $$\rho_n^* V \cong V\times 0$$ and $$\rho_m^*W \cong 0\times W$$ where $$0$$ is the trivial bundle of rank $$0$$. Then our natural transformation $$\phi\colon VB_n\times VB_m \to VB_{n+m}$$ is represented by a homotopy class of maps $$p_{n,m}'\colon BO(n)\times BO(m) \to BO(n+m)$$

so that for any $$V, W$$ over $$X$$ with classifying maps $$c_V\colon X\to BO(n)$$ and $$c_W\colon X\to BO(m)$$, the map $$p_{n,m}'\circ (c_V\times c_W)$$ classifies $$V\oplus W$$.

So let's let $$X = BO(n)\times BO(m)$$, let $$V = \gamma_n\times 0$$, and let $$W = 0\times\gamma_m$$, which as above are classified by $$\rho_n$$ and $$\rho_m$$. Note that $$\rho_n\times \rho_m = id_{BO(n)\times BO(m)}$$. Then $$p_{n,m}'\circ(\rho_n\times \rho_m)=p_{n,m}'\colon BO(n)\times BO(m)\to BO(n+m)$$ classifies $$(\gamma_n\times 0)\oplus (0\times \gamma_m)\cong \gamma_n\times \gamma_m$$.

• Thank you for the answer. One aspect: Your argument is that $p_{n,m}$ and $p_{n,m}'$ induce the same bundle $\gamma_n\times\gamma_m$ via pullback. And you assumed that the "concrete map" $p_{n,m}$ does it. Unfortunately the author only postulates that. Intuitively that should hold since $\gamma_n\times\gamma_m$ lies in following commutive diagram below (I draw it as an answer since unfortunately a commutative diagram can't be mashed as a comment): Mar 28 '19 at 13:45

$$\require{AMScd} \begin{CD} \gamma_n \times \gamma_m @>{s_{n,m}} >> \gamma_{n+m} \\ @VVp \times pV @VVpV \\ BO(n)\times BO(m) @>{p_{n,m}}>> BO(n+m) ; \end{CD}$$

where $$s_{n,m}$$ maps by $$(x, v, y, w) \mapsto (x \oplus y, \tilde{v} +\tilde{w})$$ where $$\tilde{v}= (v,0)$$ and $$\tilde{w}= (0,w)$$.

Do you see an argument that this diagram also a pullback and not only a commutative square?

• The fibre of $p_{n,m}^*\gamma_{n+m}$ over the pair $(P_n, P_m)\in BO(n)\times BO(m)$ is the vector space $P_n\oplus P_m\subset\mathbb{R}^\infty\oplus \mathbb{R}^\infty\cong \mathbb{R}^\infty$, so the bundle map $\gamma_n\times\gamma_m \to p_{n,m}^*\gamma_{n+m}$ is tautological since $P_n\oplus P_m \cong P_n \times P_m$. Mar 28 '19 at 13:51
• Can this argument be generalized in the sense that when we have a commutative square as above with upper row a morphism of $n$-vector bundles then the left bundle fitting in the diagram is automatically isomorphic to the pullback bundle? I'm not sure where in your answer you used the property "tautological". the bundle map $\gamma_n\times\gamma_m \to p_{n,m}^*\gamma_{n+m}$ arises from universal property of pullback and we observe that it induces fiberwise an linear isomorphism. isn't this property always given when a bundle fits in such square where the upper morphism is iso on fibers? Mar 28 '19 at 23:01
• Yes, I think the fact that $\gamma_n\times \gamma_m \cong p_{n,m}^*\gamma_{n+m}$ should follow from the fact that $s_{n,m}$ already is a fibre-wise linear isomorphism. The universal property of the pullback gives a natural map $\pi\colon p_{n,m}^*\gamma_{n+m}\to \gamma_{n+m}$, and my "tautological" map $i\colon\gamma_n\times\gamma_m \to p_{n,m}^*\gamma_{n+m}$ is essentially just $s_{n,m}$, in the sense that $\pi\circ i = s_{n,m}$. Mar 29 '19 at 0:49