Is the following true?
Every injective matrix $m \times m-1$ with integral coefficients induces an injective mapping $\mathbb{T}^{m-1}=\mathbb{R}^{m-1}/\mathbb{Z}^{m-1} \to \mathbb{T}^m=\mathbb{R}^m/\mathbb{Z}^m$ and every inclusion $\mathbb{T}^{m-1} \to \mathbb{T}^m$ is represented by such a matrix.
I know for sure that such a matrix, call it $A$, induces a well defined map $\mathbb{T}^{m-1} \to \mathbb{T}^m$ because it sends integral vectors to integral vectors. It is not obvious to mme why the quotient map so obtained should be injective.
Edit: I will motivate further my question and add some tags. I do not know much about tori from an algebraic-geometrical point of view, in particular I do not know what "invariant factors" are. This question arise as I am trying to comprehend (part of) the proof of the Atyiah's theorem about the convexity of the image of the moment map for an Hamiltonian torus action, you can find this at page 169 of this.
Two passages are unclear to me:
- "Choose an injective matrix $A \in \mathbb{Z}^{m \times m-1}$..." this should determine a subtorus inside $\mathbb{T}^m$, but I don't get why.
- "Any $p_0, p_1$ in $M$ can be approximated arbitrarily closely by point $p′_0$ and $p′_1$ with $\mu(p′_1)−\mu(p′_0) \in \ker A^\intercal$ for some injective matrix $A \in \mathbb{Z}^{m \times m-1}$" and again this matrix is supposed to represent the inclusion of a subtorus, but I don't get why this can be done.
In order for the quotient map to be injective, you would need that the preimage under the matrix $A$ of an integral vector is an integral vector. But, this is not necessarily the case.
Consider that
$$\begin{bmatrix}1 & 1 \\ 1 & -1 \\ 0 & 0\end{bmatrix} \begin{bmatrix}1/2 \\ 1/2\end{bmatrix} = \begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}.$$