# One-to-one mapping from $\mathbb{R}^4$ to $\mathbb{R}^3$

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.

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

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!

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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.

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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.

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