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Let $f: \mathbb{R} \rightarrow \mathbb{S}^{1} \times \mathbb{S}^{1}$ given by:

$f(t)=(\exp(it),\exp(qit))$.

I want to show that $f$ is an immersion. OK I know the definition: we compute its derivative and check it is injective.

My question is: can we view the map as $f(t)=(\cos(t),\sin(t),\cos(qt),\sin(qt))$ i.e $f: \mathbb{R} \rightarrow \mathbb{R}^{4}$ or do we have to introduce charts? I'm familiar with maps $g: \mathbb{R}^{n} \rightarrow \mathbb{R}^{m}$ but here we have the torus and the map in terms of complex numbers so I'm confused.

Can you please explain how to see $f$ is an immersion? What are the steps?

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2 Answers

up vote 2 down vote accepted

Let us use the definition, with local charts, to compute the differential of $f$ at $0$. We can choose a chart of the torus $\mathbb{T}$ as follows:

$\phi: (-\pi,\pi)\times(-\pi,\pi) \to \mathbb{T}$ defined as $\phi(\alpha,\beta) = (e^{i\alpha},e^{i\beta})$.

The domain of this chart is the open set $\mathbb{T} \setminus \{\{-1\}\times\mathbb{S}^1,\mathbb{S}^1\times\{-1\}\}$, which contains the point $f(0) = (1,1)$. The local form of $f$ is $\hat{f}(t) = \phi^{-1} \circ f =(t,qt)$.

It's differential at $0$ is, seen as a linear map from $\mathbb{R}$ to $\mathbb{R}^2$ $df|_0 = (1,q)$. This is indeed an injective map (for any $q$). Actually, by using the same local charts, you can compute the differential

You can similarly compute the differential at other points $t$.

You can also see $\mathbb{T}^2$ as an embedded submanifold of $\mathbb{R}^4$, then, you can simply compute the differential of the map as seen as a map from $\mathbb{R}$ to $\mathbb{R}^4$. This is injective. Than use the fact that the torus is embedded to conclude that the map f is injective.

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Are you familiar with $S^1$, seen as the set of unit complex numbers? In that case one can define a chart $\psi : (-\pi,\pi) \to \mathbb{S}^1\setminus\{-1\}$ by $\psi(\theta) = e^{i\theta}$. Heuristically, you're using an angular coordinate on the circle. The chart I'm using in the answer is just the chart obtained by taking the cartesian product of two such charts on $\mathbb{S}^1$. –  Luca Oct 5 '11 at 19:20
    
if we view the map as a map from $\mathbb{R}$ to $\mathbb{R}^{4}$ then its differential is given by $(-sin(t),cos(t),-qsin(qt),qcos(qt)$ right? why from this it follows that since the torus is embedded then the map is injective? –  user17182 Oct 5 '11 at 19:38
    
Let us call $i:\mathbb{T}^2 \to \mathbb{R}^4$ the embedding of the torus in $\mathbb{R}^4$. Since $i$ it's an embedding, its differential is injective. Now you're regarding the map $f$ as a map in $\mathbb{R}^4$, that is you're considering the map $F:=i \circ f : \mathbb{R} \to \mathbb{R}^4$. You computed the differential of this map, and you can see that it's injective everywhere, for any value of $q$. But $dF|_t = di|_{f(t)} \circ d\,f|_t$. Since $d\,F_t$ is injective, and $d\,i_{f(t)}$ is also injective, it follows that $d\,f_t$ is injective. –  Luca Oct 5 '11 at 19:45
    
@user17182, feel free to upvote and/or accept my answer :) –  Luca Oct 5 '11 at 20:02
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The map you mention is an immersion for all values of $q\in \mathbb R $ , rational or not.
Indeed , consider the first projection $p:S^1\times S^1\to S^1: (z,w)\mapsto z$.
Clearly $p \circ f:\mathbb R\to S^1:t\mapsto e^{it}$ is an immersion and so $f$ is an immersion.

Do you see why, quite generally, $v\circ u \quad immersion \Rightarrow u \quad immersion$ ?

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@user17182: Congratulations, perfect answer! –  Georges Elencwajg Oct 5 '11 at 20:05
    
Thank you for your help! –  user17182 Oct 5 '11 at 20:07
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