A kind of orthogonal subtorus - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T12:05:40Z http://mathoverflow.net/feeds/question/90342 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/90342/a-kind-of-orthogonal-subtorus A kind of orthogonal subtorus michael-grade83 2012-03-06T10:33:06Z 2012-04-19T20:22:00Z <p>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</p> <p><code>$S = \{x \in \mathbb{T}^n : k \cdot x = 0_{\mathbb{T}^n}\}$</code></p> <p>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.</p> <p>About the case $n=2$, setting $k = (k_1, k_2)$, $k_2 \neq 0$ I found as a possible candidate isomorphism</p> <p><code>$\mathbb{T} \to \mathbb{T}^2 : x \mapsto (k_2 x, \lfloor k_2 x \rfloor / k_2 - k_1 x)$</code></p> <p>where $\lfloor \;\rfloor$ is the floor function, however this seems too complicated in the general case, I hope you have useful tips, thanks!</p> http://mathoverflow.net/questions/90342/a-kind-of-orthogonal-subtorus/90351#90351 Answer by James Cranch for A kind of orthogonal subtorus James Cranch 2012-03-06T11:10:11Z 2012-03-06T11:10:11Z <p>It is known, in fact, that every compact connected abelian Lie group is isomorphic to a torus.</p> <p>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).</p> http://mathoverflow.net/questions/90342/a-kind-of-orthogonal-subtorus/93245#93245 Answer by Mikhail Borovoi for A kind of orthogonal subtorus Mikhail Borovoi 2012-04-05T19:36:13Z 2012-04-05T19:36:13Z <p>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 $e_1^\perp=\mathbb{T}^{n-1}$.</p>