Homotopy fiber of $BSO(n) \to BSO$

Question

By definition there is an inclusion of $BSO(n)$ into $BSO$. I am asking for the homotopy fiber of this inclusion. Actually I would like to know what is the first non trivial homotopy group of that fiber.

Answer 1

It is the infinite Stiefel manifold

$$SO/SO(n)=colim_k(SO(n+k)/SO(n))$$

and I'm afraid there isn't a whole lot more to say.

Now the good news is that for each $k\geq 0$ the canonical projection $p:SO(n+k+1)\rightarrow S^{n+k}$ induces a map $SO(n+k+1)/SO(n)\rightarrow S^{n+k}$ with fibre $SO(n+k)/SO(n)$ so that we have a fibration sequence

$$SO(n+k)/SO(n)\rightarrow SO(n+k+1)/SO(n)\rightarrow S^{n+k}.$$

Since $SO(n+1)/SO(n)\cong S^n$ we infer that the homotopy groups of $SO/SO(n)$ are trivial below degree $n$, and in degree $n$ we have, from the above fibration, a surjection

$$\rho:\pi_n S^n\rightarrow \pi_n(SO(n+2)/SO(n))$$

where the right-hand group is already the stable group,

$$\pi_n(SO(n+2)/SO(n))\cong \pi_n(SO/SO(n))$$

and these groups were calculated by G. Paechter, here. In fact this group is never trivial, so is in fact the first non-zero homotopy group of $SO/SO(n)$.

The kernel of $\rho$ is generated by the image of the composite $\pi_{n+1}S^{n+1}\xrightarrow{\Delta} \pi_nSO(n+1)\xrightarrow{p} \pi_nS^n$ where the first map is the connecting homomorphism in the long exact sequence of $SO(n+1)\rightarrow SO(n+1)$, and you might like to play around with this if you want to make an explicit calculation yourself.

Now the bigger picture is detailed in the link that Mike Miller has kindly provided in the comments details, where the mod 8 periodicity of the homotopy groups of $SO$ is discussed. The group you are interested in will depend on the mod $8$ value of $n$ (there are some hiccups for low values of $n$, caused by the exceptional isomorphisms).