A kind of orthogonal subtorus - MathOverflow

A kind of orthogonal subtorus

2012-03-06T10:33:06Z

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!

Answer by James Cranch for A kind of orthogonal subtorus

2012-03-06T11:10:11Z

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).

Answer by Mikhail Borovoi for A kind of orthogonal subtorus

2012-04-05T19:36:13Z

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 $e_1^\perp=\mathbb{T}^{n-1}$.