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I'm currently attempting to come up with an explicit parametrization of the $k$-torus into $\mathbb{R}^{k+1}$. I've been following the answer here but I'm struggling to prove injectivity.

Let $e_1,\ldots, e_{k+1}$ be the standard basis in $\mathbb{R}^{k+1}$ and $\varepsilon<1$. So according to the link, I should start with an element $v_1$ of length 1 in span$(e_1,e_2)$, i.e. $$ v_1(\theta_1)=\cos(\theta_1)e_1+\sin(\theta_2)e_2 $$ Then from there I should choose an element $v_2$ of length $\varepsilon$ in span$(v_1,e_3)$, i.e. $$ v_2(\theta_1,\theta_2)=\varepsilon\cos(\theta_2)v_1(\theta_1)+\varepsilon\sin(\theta_2)e_3 $$ and in general (for $1\leq j\leq k$), set $$ v_j(\theta_1,\ldots,\theta_j)=\varepsilon^{j-1}\cos(\theta_j)v_{j-1}(\theta_1,\ldots ,\theta_j)+\varepsilon^{j-1}\sin(\theta_j)e_{j+1}. $$ From this, the map $$ \Phi:(e^{i\theta_1},\ldots,e^{i\theta_{k}})\mapsto\sum_{j=1}^{k}v_j(\theta_1,\ldots,\theta_j) $$ is supposed to be the necessary embedding. This makes sense intuitively, but I can't seem to prove injectivity.

Things Tried

The only case I can make any progress is in the case $k=3$. I've tried doing induction and immitating this argument in the inductive step, but it doesn't seem to work.

If $k=3$, then $$\Phi(e^{i\theta_1},e^{i\theta_1})=\Phi(e^{i\theta_1'},e^{i\theta_1'})$$ implies $$ (1+\varepsilon\cos(\theta_2))v_1(\theta_1)+\varepsilon\sin(\theta_2)e_3=(1+\varepsilon\cos(\theta_2'))v_1(\theta_1')+\varepsilon\sin(\theta_2')e_3. $$ In this case the vectors being added together are orthogonal, so taking the norm-square of both sides gives $$ (1+\varepsilon\cos(\theta_2))^2+\varepsilon^2\sin(\theta_2)=(1+\varepsilon\cos(\theta_2'))^2+\varepsilon^2\sin(\theta_2'), $$ which after simplifying turns into $$ 2\varepsilon\cos(\theta_2)=2\varepsilon\cos(\theta_2), $$ so $\cos(\theta_2)=\cos(\theta_2')$. This combined with the immediate fact that $\sin(\theta_2)=\sin(\theta_2')$ shows that $\theta_2=\theta_2'$ (in $[0,2\pi)$). The rest follows from the fact that $v_1$ is injective.

Any help is greatly appreciated. Thanks in advance.

Blake
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  • Can you give some more background? Is there a reason you want $k+1$ space instead of just some Euclidean space? $\Bbb R^{2k}$ is particularly pleasant for this problem, for example. – Adam Hughes Jul 20 '18 at 03:23
  • @AdamHughes I probably didn't make it very clear. I'm attempting to embed the $k$-dimensional torus into $(k+1)$-dimensional Euclidean space, for any $k$. I'm pretty sure $k+1$ is the smallest dimension where this is possible. Does this clear up the confusion? – Blake Jul 20 '18 at 03:26
  • Right, I got what you were asking, I was just wondering why you cared about it being the smallest $k$, mostly because there is a ridiculously easy embedding if you're willing to use a few more dimensions. – Adam Hughes Jul 20 '18 at 03:52
  • Do you care about an explicit embedding, or just an existence proof? – Steve D Jul 20 '18 at 05:03
  • @SteveD Well I'm hosting a prep class for an upcoming qualifying exam and I'd like something intuitive. So if you have something very simple then I'd be happy to see it (after all beggers can't be choosers). $$

    $$ I was thinking this was very intuitive, up until the point where I couldn't prove it. I do believe I already have a proof, but it's weird and it actually shows there is an embedding from $S^k\times\mathbb{R}$ into $\mathbb{R}^{k+1}$, so I didn't like it.

     – Blake Jul 20 '18 at 05:08
  • Yeah I think I'm thinking of the same embedding you are. For example, for the 2-torus it sends $(\theta_1,\theta_2)$ to $(\cos\theta_1,e^{\sin\theta_1}\cos\theta_2,e^{\sin\theta_1}\sin\theta_2)$. It can be pretty readily generalized. – Steve D Jul 20 '18 at 05:15
  • @AdamHughes Do you visualise the usual torus as a subset of $\Bbb R^3$ or as a subset of $\Bbb R^4$? – Angina Seng Jul 20 '18 at 05:23
  • @Blake: With a bit of differential topology (the tubular neighborhood theorem), you can in fact show that any product of (positive-dimensional) spheres always embeds as a hypersurface in Euclidean space. – Ted Shifrin Jul 20 '18 at 23:22
  • @TedShifrin: there's a beautiful answer about how to do this explicitly: https://math.stackexchange.com/a/161313 – Steve D Jul 22 '18 at 00:53
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    @SteveD: I think this is an equivalent formulation. I guess, in fact, that it holds for any product of hypersurfaces in Euclidean space. No need to restrict to spheres. – Ted Shifrin Jul 22 '18 at 01:30

1 Answers1

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Here's a description of an embedding, which can be elaborated into an explicit formula if you like.

Take an embedding of $T^n$ into $\Bbb R^{n+1}$. We want to use this as a basis for one of $T^{n+1}=T^n\times S^1$ into $\Bbb R^{n+2}$.

Translate your $T^n$ so that it lies within the half-plane defined by $x_{n+1}>0$. Then let the embedding of $T^n$ be given by functions $p\mapsto (\phi(p),\psi(p))$ where $\phi(p)\in\Bbb R^n$ and $\psi(p)\in(0,\infty)$. Now the embedding of $T^{n+1}$ into $\Bbb R^{n+2}$ by $$(p,e^{it})\mapsto(\phi(p),\psi(p)\cos t,\psi(p)\sin t).$$ As $\psi(p)>0$ the last two coordinates determine $\psi(p)$ and $e^{it}$.

Angina Seng
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