$\require{AMScd}$ I am interested in understanding better why the sphere subfiberbundle of the tautological n-plane bundle over $BU(n)$ is (up to homotopy) the homotopy fibration $BU(n-1) \hookrightarrow BU(n)$.

One way of seeing that this is true: The disk bundle of $BU(n)$ is the bundle $EU(n) \times_{U(n)} \mathbb{C^n}$ and the total space of the sphere subbundle is $EU(n) \times_{U(n)} U(n)/U(n-1)$. By the basechange homomorphism this is $EU(n)/U(n-1) \cong BU(n-1)$. This definitively shows that the sphere bundle of the tautological bundle is a map $BU(n-1) \hookrightarrow BU(n)$.

One question is left unanswered here:

Question 1): How can I show that the diagram $\begin{CD} EU(n)/U(n-1) @>{p}>> EU(n)/U(n)\\ @VVV @VVV \\ BU(n-1)@>{B(inclusion)}>> BU(n) \end{CD}$

homotopy commutes.

Now there is an easy way of seeing that there is a sphere bundle $S^{2n-1} \hookrightarrow BU(n-1) \to BU(n)$. Namely one has a fibration sequence $U(n-1) \to U(n) \to S^{2n-1} \to BU(n-1) \to BU(n)$.

Question 2) Is there a way of directly using this fibration sequence without going to the process above to show that this is the sphere bundle of the tautological bundle over $BU(n)$? Put another way, why is it true that $\Omega$ applied to this fibration is the fibration $U(n-1) \hookrightarrow U(n) \to S^{2n-1}$

Question 2 is another way of asking question 1 since I know that the top map in the diagram is the sphere bundle of the tautological bundle, and the last map in the written fibration sequence is the delooping functor applied to the inclusion $U(n-1) \to U(n)$.

If this question requires an explicit construction of the delooping functor, the one I am familiar with is the geometric realization functor applied to the classifying functor of a category(in this case of a groupoid).

I suspect that the answer to my second question(and hence the first) will not require an explicit construction: I expect using ,for instance, that the only $A-\infty $ H-space structure on $U(n)$ is the multiplication on $U(n)$.

  • $\begingroup$ Is this the right diagram in Q1? What is the map $p$? $\endgroup$ – Tyrone Jul 6 '17 at 12:58
  • $\begingroup$ Thanks! I should be quotienting by $U(n)$. The map then is the quotient map. $\endgroup$ – wonderwoman Jul 6 '17 at 13:39

Since $U(n−1)$ acts freely on $EU(n)$ (through the inclusion $U(n−1)\hookrightarrow U(n)$, it follows that $EU(n)/U(n−1)\simeq BU(n−1)$, and you can use this to define the classifying space $BU(n−1)$, since these objects are only defined up to homotopy. The map $B(inc.)$ in your diagram is then just the induced map of quotient spaces, and the diagram commutes strictly by definition.

Now taking the strict fibres of the vertical maps in the diagram gives the first map in the sequence $U(n)\hookrightarrow U(n-1)\rightarrow S^{2n-1}$ and since the bottom row of the diagram is a delooping of this fibration, it follows that the homotopy fibre of

$B(inc):BU(n-1)=EU(n)/U(n-1)\rightarrow EU(n)/U(n)=BU(n)$

is $S^{2n-1}$.

You can use the bar construction to generate a classifying space functor. For a group $G$ the geometric realisation $EG=B(\ast,G,G)$ is contractible, carries a free right $G$-action and maps canonically to $BG=B(\ast,G,\ast)$ to give a fibration $EG\rightarrow BG$ with fibre $G$. Moreover $EG/G\cong B(\ast,G,\ast)$ since the action is free.

Now the map $B(inc):BU(n-1)\rightarrow BU(n)$ is given functorially as the quotient of the map $EU(n-1)\rightarrow EU(n)$ that is the geometric realisation of the simplicial map which is the $n$-fold product of $(inc):U(n-1)\rightarrow U(n)$ in degree $n$. The map $B(inc)$ factors $EU(n-1)/U(n-1)\rightarrow EU(n)/U(n-1)\rightarrow EU(n)/U(n)$ with the first map a homotopy equivalence since $EU(n-1)$, $EU(n)$ are contractible. The second map is the one you want explicitly.

  • $\begingroup$ Thanks! Can you see my answer/response to your answer? It was too long to put into comments. Kind Regards, Wonderwoman $\endgroup$ – wonderwoman Jul 6 '17 at 14:31
  • $\begingroup$ Ok, I think it's clearer what you're asking now. I've updated my original answer with some comments that I hope are helpful. $\endgroup$ – Tyrone Jul 6 '17 at 16:03
  • $\begingroup$ Thanks this is great! $\endgroup$ – wonderwoman Jul 7 '17 at 0:01

This is a response to Tyrone's answer since it is too long to fit into comments.

Thanks so much for the time you put into the answer! You didn't answer my question though :).

There are three things you point out, 2 of which I had already shown and one that is false.

I know that $EU(n)/U(n-1)$ is a model for $BU(n-1)$.

You also show $EU(n)/U(n-1) \to EU(n)/U(n)$ is a sphere bundle(I show that it is in fact the sphere bundle of the canonical bundle).

The third thing you assert is that the diagram defines the meaning of $B(inc.)$. It doesn't because this does not define functor $B$ on the category of groups with group homomorphisms. Given a group homomorphism $G \xrightarrow{\phi} H$ with a nontrivial kernel, $EH/G$ is not a model for $BG$ because the action of $G$ on $EH$ is not free. Thus one cannot define $B(\phi)$ to be the map $EH/G \to EH/H$.

  • $\begingroup$ Although it can be made perfectly functorial, it isn't necessary to define it as such. The ad hoc definition of $B(inc.)$ suited the ad hoc definition of $BU(n-1)$ (and indeed $BU(n)$, since we assumed no knowledge of how $EU(n)$ had been supplied). The definition of $B(inc.)$ I gave loops to the required inclusion so up to homotopy is perfectly well defined. Rather than show that the sphere bundle was that of the canonical bundle (which you knew) I showed it was in fact the space $U(n)/U(n-1)\cong S^{2n-1}$, which I thought was the connection you wanted. Did I misunderstand your question? $\endgroup$ – Tyrone Jul 6 '17 at 14:59
  • $\begingroup$ Yes :). How do you show that your definition of $B(inc.)$ is the same as the 'perfectly functorial' definition up to homotopy.(say, using the definition of the functor using simplicial sets at the end of my question) $\endgroup$ – wonderwoman Jul 6 '17 at 15:09

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