I am trying to construct an isometric immersion of $T^n := S^1\times\ldots \times S^1$ into $\mathbb{R}^{2n}.$
I am considering The flat Torus, i.e., on each $S^1$ I am considering the metric induced by $\mathbb{R}^2$ and on $T^n$ the product metric.
The way I am trying to obtain this is to define the map:
$$f : T^n \to \mathbb{R}^{2n} $$
$$f(z_1,\ldots,z_n) := (a_1,b_1,\ldots,a_n,b_n),$$
where $z_j \equiv (a_j,b_j) \equiv a_j + ib_j$ and $|z_j| = 1.$
Here $i$ is the imaginary unit.
Once I show that this map is an immersion the claim follows pulling-back the metric of $\mathbb{R}^{2n}.$
Is it ok?
I am not sure about this argument once I can not give an clear statement that $f$ is an immersion without calculating on coordinates charts.
Best Answer
$\newcommand{\dd}{\partial}$Your $z_{j}$ notation is potentially vexing because the $z_{j}$ are unit complex numbers rather than real numbers. You might instead pick local coordinates $\theta_{j}$ on the torus so that $z_{j} = \exp(i\theta_{j})$ for each $j$, and define $$ f(\theta_{1}, \dots, \theta_{n}) = (\cos\theta_{1}, \sin\theta_{1}, \dots, \cos\theta_{n}, \sin\theta_{n}). $$ It's elementary to show that the (vector-valued) partial derivatives $df(\theta)(\mathbf{e}_{j}) = \dd f/\dd \theta_{j}$ form an orthonormal set at each point, so that $f$ is a local isometry (and, particularly, is an immersion).