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Apr 24, 2017 at 19:06 comment added Robert Bryant @IgorBelegradek: Yes, this is equivalent to that I was describing.
Apr 24, 2017 at 18:00 vote accept Igor Belegradek
Apr 24, 2017 at 15:25 comment added Igor Belegradek 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 $\mathbb RG\cap L=G$.
Apr 24, 2017 at 15:08 comment added YCor 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 $T$.
Apr 24, 2017 at 14:49 comment added Robert Bryant 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.
Apr 24, 2017 at 14:46 history edited Robert Bryant CC BY-SA 3.0
Added a remark about the classification.
Apr 24, 2017 at 14:46 comment added YCor 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.
Apr 24, 2017 at 14:20 history answered Robert Bryant CC BY-SA 3.0