Is there a classification of closed subgroups $H\le SO(n)$ such that the inclusion $H\to SO(n)$ is trivial on all homotopy groups?
This happen e.g. when the group $H$ is finite. Are there other examples?
$\begingroup$
$\endgroup$
asked Apr 24, 2017 at 13:38
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
5
Well, the identity component $H^0$ of $H$ would have to be abelian, since, otherwise, it would have a compact simple component, and the induced map on $\pi_3$ would then be nontrivial.
If $H^0$ has positive dimension and is abelian, then what you are asking is whether $\pi_1(H^0)\to\pi_1(\mathrm{SO}(n))\simeq \mathbb{Z}_2$ is trvial or not. This does happen, of course. For example, if $H^0=S^1$ and it represents the zero element in $\pi_1(SO(n))\simeq \mathbb{Z}_2$, then the induced map on homotopy is trivial in $\pi_1$ and hence in all higher cases as well. This first happens for $n=4$, as there is such an $S^1$ sitting in $\mathrm{SO}(4)$ (in fact, a countably distinct family of them). It also happens for all $n\ge 4$.
As far as classification goes: Given $n$, there will be a countable number of codimension $1$ subtori $T^{r-1}$ in a maximal torus $T^r\subset\mathrm{SO}(n)$ (where $r$, the rank is the greatest integer in $n/2$) such that the inclusion $T^{r-1}\hookrightarrow T^r$ induces an inclusion $\pi_1(T^{r-1})\subset \pi_1(T^r)$ so that all the elements of $\pi_1(T^{r-1})$ are zero in $\pi_1(\mathrm{SO}(n))$. Then the abelian groups that you want are the ones that are conjugate to a subgroup of such a sub-maximal torus.
answered Apr 24, 2017 at 14:20
-
$\begingroup$ For a precise description, one has to determine the case of the inclusion of the maximal torus $T$; if $f_*:\pi_1(T)\to\pi_1(SO(n))$ is trivial then all tori work. Otherwise for a comprehensive description, we need to make this map $f_*$ explicit; then for a subtorus $S$ of $T$ one just checks if the composite map $\pi_1(S)\to\pi_1(T)\to\pi_1(SO(n))$ is trivial. $\endgroup$– YCorCommented Apr 24, 2017 at 14:46
-
$\begingroup$ However, it's never the case that the homomorphism is trivial for a maximal torus. The homomorphism $\pi_1(T^r)\to\pi_1(SO(n))\simeq\mathbb{Z}_2$ is always surjective. $\endgroup$ Commented Apr 24, 2017 at 14:49
-
$\begingroup$ OK. So there's only one possible description: indeed, for a "standard" copy of $SO(2)$, the induced homomorphism should be always trivial or always nontrivial (since they are all conjugate), and the first case is excluded since the maximal torus is generated by finitely many such tori images of standard embeddings. Probably this amounts to saying that there is only one homomorphism from $\pi_1(T)$ onto the cyclic group of order 2 that is invariant under the normalizer of $T$. $\endgroup$– YCorCommented Apr 24, 2017 at 15:08
-
$\begingroup$ Thank you! I am not sure what codimension 1 subtori are for. Here is how I think about this. Identify the maximal torus with $\mathbb R^n/\mathbb Z^n$ and let $L\le \mathbb Z^n$ be the index $2$ subgroup that is the kernel of the epimorphism onto $\pi_1(SO(n)$. Given a subgroup $G\le L$ consider its span $\mathbb RG$ in $\mathbb R^n$. Then $\mathbb RG/G$ is a subtorus iff $\mathbb RG\cap L=G$. This I think gives a description of all subgroups $H^0$ inside the maximal torus: $\pi_1(H^0)$ goes to zero under the inclusion into $\pi_1(SO(n))$ iff $H^0=\mathbb RG/G$ with $\mathbb RG\cap L=G$. $\endgroup$ Commented Apr 24, 2017 at 15:25
-
$\begingroup$ @IgorBelegradek: Yes, this is equivalent to that I was describing. $\endgroup$ Commented Apr 24, 2017 at 19:06