Is SO(n) a deformation retract of SO(n+1) for $n\ge 3$?

Question

I know that $SO(n)$ shares many topological invariants with $SO(n+1)$ for $n\ge 3$ (such as first fundamental group, first singular cohomology class over $\mathbb{Z}$, etc.). I also know that there is an inclusion $i: SO(n) \hookrightarrow SO(n+1)$ taking $A \mapsto \begin{bmatrix} A & 0 \\ 0 & 1 \\ \end{bmatrix}$. Is $SO(n)$ a deformation retract of $SO(n+1)$?

Edit: Perhaps this can be deduced from $SO(n+1)/SO(n)\cong \mathbb{S}^n$ altough I do not know how to do that.

Answer 1

It is not a deformation retract.

As you remark there is a map $SO(n+1)\rightarrow S^n$ that sends a matrix in $SO(n+1)$ to the first basis vector. As you can check, this map is a submersion and the fiber is $SO(n)$. Thus we have a fiber bundle

$$ SO(n)\rightarrow SO(n+1)\rightarrow S^n. $$

From this you can see that the "low dimensional" topology of $SO(n)$ stabilizes for large $n$. Namely there are long exact sequences in homotopy

$$ \ldots \rightarrow \pi_k(SO(n))\rightarrow \pi_k(SO(n+1))\rightarrow \pi_k(S^n)\rightarrow \pi_{k-1}(SO(n))\rightarrow \ldots $$ and a similar sequence in (co)homology called the Wang sequence. The latter sequence strongly uses the fact that the base of the fibration is a sphere, in general for a fiber bundle there is a more complicated relation: the Serre spectral sequence.

Anyway, returning to the long exact sequence above, we see that if $k<n-1$ then $\pi_{k+1}(S^n)\cong \pi_k(S^n)\cong 0$ and the long exact sequence breaks down to $$ 0\rightarrow \pi_k(SO(n))\rightarrow \pi_k(SO(n+1))\rightarrow 0 $$ i.e. the groups are isomorphic.

You can see that this cannot be the case for homotopy groups $k>>n$ as the homotopy groups of spheres are very complicated. Colloquially speaking: the homotopy groups of $SO(n)$ and $SO(n+1)$ differ by the homotopy groups of $S^n$.

Answer 2

Thomas's answer is fine (and I upvoted it), but here is a simpler one.

The dimension of $SO(n)$ is $n(n-1)/2$. Further, $SO(n)$ is orientable (as all Lie groups are) and compact. It follows that $H^{n(n+1)/2}(SO(n+1))\neq 0$, while $H^{n(n+1)/2}(SO(n)) = 0$. Thus, the two are not homotopy equivalent, so $SO(n+1)$ cannot deformation retract to $SO(n)$.