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.


Thanks!



sub-tori of a torus, generated by 1-dimensional subgroup
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 $.


