Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm trying to define a mapping from $\mathbb{R}^4$ into $\mathbb{R}^3$ that takes the flat torus to a torus of revolution.

Where the flat torus is defined by $x(u,v) = (\cos u, \sin u, \cos v, \sin v)$. And the torus of revolution by $x(u,v) = ( (R + r \cos u)\cos v, (R + r \cos u)\sin v, r \sin u)$.

I think an appropriate map would be: $f(x,y,z,w) = ((R + r x)z, (R + r x)w, r y)$ where $R$, $r$ are constants greater than $0$.

But now I'm having trouble showing this is one-to-one.

share|improve this question
3  
Presumably the mapping is only going to be one-to-one when restricted to the torus? –  JSchlather Nov 30 '11 at 2:55

3 Answers 3

Compute the Jacobians of the two mappings. I'll call your first mapping $X(u, v)$ so as not to re-use the variable $x$. Let $R > r > 0$.

$$\frac{\partial X}{\partial(u, v)} = \left[ \begin{array}{cc} -\sin u & 0 \\ \cos u & 0 \\ 0 & -\sin v \\ 0 & \cos v \end{array} \right] $$

$$\frac{\partial f}{\partial(x, y, z, w)} = \left[ \begin{array}{cccc} rz & 0 & R + rx & 0 \\ rw & 0 & 0 & R + rx \\ 0 & r & 0 & 0 \end{array} \right] $$

The embedding in $\mathbb{R}^3$ is the composite $f \circ X$, whose Jacobian is the product of the Jacobians:

$$ \frac{\partial f}{\partial(x,y,z,w)}\frac{\partial X}{\partial(u,v)} = \left[ \begin{array}{cc} -rz\sin u & -(R + rx)\sin v \\ -rw\sin u & (R+rx)\cos v \\ r\cos u & 0 \end{array} \right] = \left[ \begin{array}{cc} -r\cos v\sin u & -(R + r\cos u)\sin v \\ -r\sin v\sin u & (R+r\cos u)\cos v \\ r\cos u & 0 \end{array} \right] $$

(I used sage notebook www.sagenb.org for my computations.)

Now all of the questions concerning 1-to-1 can be phrased in terms of rank and nullity of these matrices. For example, it easy to see the first and third matrices have rank 2 (so nullity=0) for all choices of $u, v$, (which shows both representations of the torus are immersive, meaning an appropriate restriction of domain will make those maps 1-to-1).

Now $f$ is 1-to-1 when restricted to $\mathrm{im} X$ (the flat torus) because $f \circ X$ is immersive and by restricting to $u, v \in [0, 2\pi)$, both $f$ and $f \circ X$ become 1-to-1. In other words, if $f( X(u,v) ) = f(X(u', v'))$, then $X(u,v) = X(u', v')$.

Hope this helps!

share|improve this answer

Shaun's answer is insufficient since there are immersions which are not 1-1. For example, the figure 8 is an immersed circle. Also, the torus covers itself and all covering maps are immersions. http://en.wikipedia.org/wiki/Immersion_(mathematics)

Your parametrization of the torus of rotation is the the same as in http://en.wikipedia.org/wiki/Torus You just have to notice that the minimal period in both coordinates of the $uv$ plane is the same $2\pi$ in the case of both the flat and rotated tori.

share|improve this answer

Do you need to explicitly give the map or are you asked to prove both sets have the same cardinilty?

If not, it is a good exercise to prove $\mathbb{R}$ and $\mathbb{R}^2$ are both the same size (Schroeder-Bernstein theorem - consider decimal expansions to show a surjection) then via induction and the fact both sets are subsets of $\mathbb{R}^n$ we see there is a bijection between them. Explicitly finding a bijection is not clear to me, perhaps it is to someone else.

share|improve this answer
4  
It doesn't look like you read the question very carefully... –  JSchlather Nov 30 '11 at 2:54
2  
It's possible that he only read the title of the question. –  Paul Nov 30 '11 at 2:57

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.