Ok the question is pretty dumb: suppose you have a torus $T^n=\mathbb{R}^n/\mathbb{Z}^n$ and a vector $\bar{v}=(v_1,\ldots,v_n)\in\mathbb{R}^n$.

Consider the torus $T_{\bar{v}}$ given by the closure of the one parameter group in $T^n$ generated by $\bar{v}$:

$T_{\bar{v}}=\overline{ \{t\cdot\bar{v}\mod\mathbb{Z}^n|\phantom{a}t\in\mathbb{R}\}}$

My questions are:

  1. what is the dimension of $T_{\bar{v}}$?
  2. How can i find a basis of vectors spanning the tangent space of $T_{\bar{v}}$ at the origin?

My guess for question 1. is $\dim T_{\bar{v}}=\dim_{\mathbb{Q}}\langle v_1,\ldots, v_n\rangle$, but i don't know what the answer to question 2. can be.


  • 1
    $\begingroup$ Nice question (& correct guess). Beware that the closure may not be connected (consider the point $(\sqrt{2},1/2)$ in $T^2$), so the torus you want is the identity component of the closure. Replacing $\overline{v}$ with a nonzero integral multiple is then harmless. Make integral change of coords and drop $n$ if necessary to get to case when $v_i$'s are lin. indep. over $\mathbf{Q}$. Then $\overline{v}$ generates dense subgp of $T^n$: if not get $T^1$-quotient killing it. Change coords (and pass to multiple) so quotient map is projection to 1st factor, contradicting lin. indep. hypothesis. $\endgroup$ – BCnrd May 12 '10 at 17:11
  • $\begingroup$ Why is the closure not connected? I mean, by definition my one-parameter group (call it G) is dense in the closure, and it is connected. If the closure consisted of 2 or more connected components A and B, and G is either in A or B. Say G is in A. but since A and B are both open and closed, the closure of G must be contained in A and cannot be A\cup B... Again, am i missing something? $\endgroup$ – CuriousUser May 12 '10 at 17:48
  • $\begingroup$ I was thinking more algebraically, using closure of subgroup generated by $\overline{v}$ rather than the 1-parameter subgroup passing through $\overline{v}$. That is, I made the same misreading as Ekedahl. The example I gave in my comment is a point generating a subgroup whose closure is the first circle factor and does not contain $\overline{v}$. Ekedahl's answer also shows that I missed the possibility that the $\mathbf{Q}$-span of the $v_i$'s contains 1, in which case the dimension drops by 1 from the guess; in fact, my example using $(\sqrt{2}, 1/2)$ demonstrates that! Mea culpa. $\endgroup$ – BCnrd May 12 '10 at 18:39
  • $\begingroup$ Thanks, i got what you meant it after I read Ekedahl's post! $\endgroup$ – CuriousUser May 12 '10 at 19:10

The key to solving both problems is the use of the following two facts: 1) Any closed subgroup of $T^n$ is the intersection of the kernels of characters of $T^n$, i.e., continuous group homomorphisms $T^n \rightarrow S^1$. 2) Any continuous homomorphism $T^n \rightarrow S^1$ is of the form $(\overline x)\mapsto e^{2\pi i x\cdot m}$ for a unique $m\in\mathbb Z^n$. Hence, the closure of $T_{\overline \nu}$ is the intersection of the kernels of the characters corresponding to $m$ for which $\nu\cdot m\in\mathbb Z$. Picking a basis $m_1,\dots,m_k$ of the group of such $m$ gives a surjection $T^n \rightarrow T^k$ for which $T_{\overline\nu}$ is the kernel ($T_\nu$ is not necessarily a torus as it might not be connected but it doesn't change anything). By the above this map is just given by an $n\times k$ integer matrix specified by $m_1,\dots,m_k$. The tangent map at the origin is then obtained by regarding this matrix as a real matrix and thus the tangent space of $T_{\overline \nu}$ is the null space of this matrix.

In particular this gives that the dimension of $T_{\overline \nu}$ is equal to $\dim_{\mathbb Q}\langle1,\nu_1,\nu_2,\ldots,\nu_n\rangle -1$. This is off by one from your guess if $1$ is in the span of of the $\nu_i$ but equal to it if it isn't.

[[Added]] I misread the question and the above is for the closed subgroup generated by $\overline\nu$ while the question was about the closure of the $1$-parameter subgroup generated by it. To answer the question everything works the same only the condition is that $r\nu\cdot m\in\mathbb Z$ for all real $r$ which gives $\nu\cdot m=0$ and indeed the dimension is $\dim_{\mathbb Q}\langle\nu_1,\nu_2,\ldots,\nu_n\rangle $.

  • $\begingroup$ I'm sorry, I'm a bit confused: suppose we're in $T^2$, and $v=(1,1/2)$. Then $T_v=S^1$, but according to your formula i should get $\dim T_v=1-1=0$. Is there something I'm missing? $\endgroup$ – CuriousUser May 12 '10 at 17:41
  • $\begingroup$ See amendment to the answer. $\endgroup$ – Torsten Ekedahl May 12 '10 at 18:12

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