What are the subgroups of $(S^1)^n$ isomorphic to the standard copy of $(S^1)^k$?

Question

Let $H$ be the set of all subgroups of the $n$-dimensional torus $(S^1)^n$ that are isomorphic by an element of $\operatorname{Aut}((S^1)^n)$, the set of continuous automorphisms of $(S^1)^n$, to the standard copy of $(S^1)^k$, where $1 \leqslant k \leqslant n$.

My question is:

Is $H$ in a bijective correspondence with $\operatorname{Hom}((S^1)^k, (S^1)^n)$, the set of continuous group homomorphisms from $(S^1)^k$ to $(S^1)^n$?

My thoughts - Given any automorphism $f$ of $(S^1)^n$ if we restrict it to $(S^1)^k$ then we get a member of $\operatorname{Hom}((S^1)^k, (S^1)^n)$. So sending $f ((S^1)^k)$ to $f \vert _{(S^1)^k}$, gives a map from $H$ to $\operatorname{Hom}((S^1)^k, (S^1)^n)$.

Can one define a map in the reverse direction? Would it then mean that any homomorphism from $(S^1)^k$ to $(S^1)^n$ is injective and can be extended to an automorphism of $(S^1)^n$?

Thanks!

Answer 1

Your argument in the paragraph "My thoughts…" is invalid. The map you define from $H$ to $Hom((S^1)^k,(S^1)^n)$, taking $f((S^1)^k)$ to $f \bigm|_{(S^1)^k}$, is not well-defined independent of $f$.

Just to be concrete, take the case $k=n$. Then the set $H$ has one element, namely the whole group $(S^1)^n$. But there are infinitely many continuous group homomorphisms from $(S^1)^n$ onto itself. In fact there are infinitely many which are isomorphisms: the group $Aut((S^1)^n)$ is an infinite group, isomorphic to $GL(n,\mathbb{Z})$.

Answer 2

Suggestion: Since $(S^1)^n$ is isomorphic to $\mathbb{R}^n/\mathbb{Z}^n$ and $(S^1)^k$ to $\mathbb{R}^k/\mathbb{Z}^k$, it may be helpful to seek for a correspondence between subgroups of $(S^1)^n$ and linear subspaces of $\mathbb{R}^n$.

Answer 3

Here is a reference which should help: Sidney A. Morris, Pontryagin Duality and the Structure of Locally Compact Abelian Groups (Cambridge University Press, Cambridge, 1977). I think you need to consider closed subgroups in your question: for instance you can get dense subgroups of $(S^1)^2$ which can be thought of as lines of irrational slope in the torus. There is also a nice treatment in a volume of Bourbaki.

There are extensions to countable products and sums in R. Brown, P.J. Higgins, and S.A. Morris, "Countable products of lines and circles: their closed subgroups, quotients and duality properties'', Math. Proc. Camb. Phil. Soc. 78 (1975) 19-32.