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Here $\mathbb{T}^n := (\mathbb{R} / \mathbb{Z})^n$ is the topological group of the n-dimensional torus and $k \in \mathbb{Z}^n$ is a non-null vector, I'm working about the subgroup

$S = \{x \in \mathbb{T}^n : k \cdot x = 0_{\mathbb{T}^n}\}$

where $\cdot$ is the scalar product. I think that $S$ is isomorphic, as topological group, to $\mathbb{T}^{n-1}$, but I could not prove it.

About the case $n=2$, setting $k = (k_1, k_2)$, $k_2 \neq 0$ I found as a possible candidate isomorphism

$\mathbb{T} \to \mathbb{T}^2 : x \mapsto (k_2 x, \lfloor k_2 x \rfloor / k_2 - k_1 x)$

where $\lfloor \;\rfloor$ is the floor function, however this seems too complicated in the general case, I hope you have useful tips, thanks!

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Consider the subgroup $N:=\langle k\rangle\subset \mathbb{Z}^n$. There exists a basis $f_1,\dots,f_n$ of $\mathbb{Z}^n$ such that $uf_1$ is a basis of $N$, where $u\in \mathbb{Z}$, $u>0$, see Vinberg, A Course in Algebra, Thm. 9.1.5, or Lang, Algebra, 3d ed., Thm. III.7.8. Changing, if necessary, $f_1$ to $-f_1$, we may think that $k=uf_1$, $u>0$. If you assume that your vector $k$ is not divisible by any positive integer different from 1, you obtain that $k=f_1$. Let $e_1,\dots,e_n$ be the standard basis of $\mathbb{Z}^n$. If follows easily that $S:=k^\perp=f_1^\perp$ is isomorphic to $e_1^\perp=\mathbb{T}^{n-1}$.

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  • $\begingroup$ I gave the theorem number in Vinberg's book in 3d Russian edition. I hope it is relevant for the English edition. $\endgroup$
    – Mikhail Borovoi
    Commented Apr 5, 2012 at 19:41
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It is known, in fact, that every compact connected abelian Lie group is isomorphic to a torus.

Your construction has a danger of producing disconnected subgroups, however: this occurs when the coordinates of $k$ are all multiples of some fixed $i>1$. (You can see this even in small dimensions).

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  • $\begingroup$ Well, so I can argue as follows: $S$ is closed because is the pre-image of the closed set $\{0_{\mathbb{T}^n}\}$ by the continuous map $\mathbb{T}^n \to \mathbb{T} : x \mapsto k \cdot x$, then $S$ is compact because $\mathbb{T}^n$ is compact. But how I can show that $S$ is connected when $k \neq ih$ for all $i =2,3,\ldots$ and $h \in \mathbb{Z}^d$? $\endgroup$
    – user21706
    Commented Mar 6, 2012 at 12:33
  • $\begingroup$ I thought that if $P : \mathbb{R}^n \to \mathbb{T}^n$ is the standard projection then $S=\bigcap_{n \in \mathbb{Z}} S_n$, where $S_n := P(\{x \in \mathbb{R}^n : k \cdot x = n\})$. Now $\{x \in \mathbb{R}^n : k \cdot x = n\}$ is connected because is an affine hyperplane, then $S_n$ is connected because is the continuous image of a connected set. Remains to prove something like $\bigcap_{n \in \mathbb{Z}} S_n \neq \emptyset$ $\endgroup$
    – user21706
    Commented Mar 6, 2012 at 13:10
  • $\begingroup$ Maybe I have concluded: if $k = (k_1, \ldots, k_n) \in \mathbb{Z}^n$ with $\gcd(k_1, k_2, \ldots, k_n) = 1$ then for the generalized Bezout's lemma exists $m \in \mathbb{Z}^n$ such that $k \cdot m = 1$, then for any $x \in \mathbb{R}^n$ such that $k \cdot x = n \in \mathbb{Z}$ we have $k \cdot (x + m) = n + 1$ and $P(x) = P(x + m) \in S_n \cap S_{n+1}$. Can someone check my reasoning? $\endgroup$
    – user21706
    Commented Mar 6, 2012 at 13:32

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