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I'm trying to find a smooth degree 1 map from the torus $T^2 = S^1 \times S^1$ to the $2$-sphere $S^2$.

My first thought was to use the two coordinates $(\theta_1,\theta_2)$ to map onto the usual spherical polar coordinates on the sphere - however I can't quite get that to work because it seems that one of these coordinates would cover the sphere twice and so give a degree $2$ map?

Alternatively I thought about using the winding number. So in general if we embed circles into $\mathbb{R}^3$ by the maps $f_1,f_2: S^1 \to \mathbb{R}^3$ we can then define a map $F:T^2 \to S^2$ by:

$$F(\theta_1,\theta_2) = \frac{f_1(\theta_1) - f_2(\theta_2)}{\| f_1(\theta_1) - f_2(\theta_2) \|}$$

However I'm now struggling to make suitable choices for $f_1$ and $f_2$, my guess is that I'd want to make the circles cross over each other once but I'm struggling to visualise this and write down a map so I would appreciate any help.

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    $\begingroup$ Your spherical coordinates idea actually gives a map of degree zero, see for example Why spherical coordinates is not a covering?. One approach to your question is: If you represent a torus as a square with boundary identifications, can you see a way of further collapsing the boundary of the square to get as torus? $\endgroup$ – Andrew D. Hwang May 12 '16 at 11:49
  • $\begingroup$ Degree zero because this map wouldn't be surjective? It misses out the north pole? $\endgroup$ – Wooster May 12 '16 at 12:03
  • $\begingroup$ I guess if we then further identify all the sides of the square we would have a sphere - is that map obviously smooth though? $\endgroup$ – Wooster May 12 '16 at 12:04
  • $\begingroup$ Degree zero because "the outside of the torus wraps one way the the inside wraps the other way". :) For your second comment (which is Michael's answer), you're right that you'll need to find a smooth implementation. (The topological picture is likely to be helpful; I don't have formulas off the top of my head.) $\endgroup$ – Andrew D. Hwang May 12 '16 at 12:09
  • $\begingroup$ Okay great, yes I see! Thanks. I wonder if it is possible to do this with the linking map I wrote down? $\endgroup$ – Wooster May 12 '16 at 12:17
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Let $(U, \varphi)$ be a chart on $T^2$, i.e. $U$ is an open subset of $T^2$ and $\varphi : U \to \mathbb{R}^2$ is a homeomorphism. Let $V$ be the open subset of $U$ such that $\varphi|_V : V \to B(0, 1)$ is a homeomorphism, i.e. $V = \varphi^{-1}(B(0, 1))$. Then $\varphi(\overline{V}) = \overline{\varphi(V)} = \overline{B(0, 1)}$ and $\varphi(\partial V) = \partial\varphi(V) = \partial B(0, 1) = S^1$.

Note that the quotient $\overline{B(0, 1)}/S^1$ is homeomorphic to $S^2$; let $\psi : \overline{B(0, 1)}/S^1 \to S^2$ be a homeomorphism. The composite $\psi\circ\varphi|_{\overline{V}} : \overline{V} \to S^2$ maps $\partial V$ to a single point, call it $p$, and $(\psi\circ\varphi|_{\overline{V}})|_V = \psi\circ\varphi|_V$ is a homeomorphism from $V$ to $S^2\setminus\{p\}$.

Now define $f : T^2 \to S^2$ by

$$f(x) = \begin{cases} \psi(\varphi(x)) & x \in \overline{V}\\ p & x \not\in \overline{V}. \end{cases}$$

Then $f$ is a continuous map. Furthermore, it has degree one .

More generally, we can use the same technique to construct a degree one map from any closed, connected, orientable $n$-manifold to $S^n$.

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  • $\begingroup$ Hello @Michael Albanese. Is $f$ injective?. Can you show me a continuous function from $S^2\rightarrow T^2$ please?, thanks $\endgroup$ – Joe May 5 '17 at 18:52
  • $\begingroup$ @Joe: No it isn't (consider two points in the complement of $\overline{V}$). There are many maps $S^2 \to T^2$, for example constant maps. Up to homotopy, these are the only ones. $\endgroup$ – Michael Albanese May 5 '17 at 19:12
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Here's a way to construct a degree one map.

Consider the standard CW complex for the torus consisting of a $0$-cell, two $1$-cells, and one $2$-cell. If we forget about the $2$-cell, what remains is $S^1\vee S^1$. That is $T^2 = (S^1\vee S^1)\cup_{\varphi} e^2$ where $e^2$ is the $2$-cell and $\varphi$ is the attaching map $\varphi : \partial e^2 \to S^1\vee S^1$.

Now consider the space $T^2/(S^1\vee S^1)$ where all the points of $S^1\vee S^1$ are identified with one another. This quotient space has the CW structure $e^0\cup_{\tilde{\varphi}} e^2$ where $e^0$ is a $0$-cell (i.e. a point) and $\varphi : \partial e^2 \to e^0$. We see that the quotient is $S^2$ and so we can view the quotient map as a map $f : T^2 \to S^2$.

As $f|_{T^2\setminus(S^1\vee S^1)} : T^2\setminus(S^1\vee S^1) \to S^2$ is a homeomorphism (it identifies the $2$-cell of $T^2$ with the $2$-cell of $S^2$), we see that $f$ is a degree one map $T^2 \to S^2$.


The above argument can be used to show that any connected closed smooth $n$-manifold $M$ admits a degree one map to $S^n$. Using Morse theory, one can show that $M$ admits a CW complex structure with only one $n$-cell, i.e. $M = M^{(n-1)}\cup_{\varphi}e^n$ where $M^{(n-1)}$ is the $(n-1)$-skeleton of $M$. Then $M \to M/M^{(n-1)} \cong S^n$ is a degree one map.

We needed to assume $M$ was smooth above in order to use Morse theory. However, for $n \neq 4$, every closed connected $n$-manifold has such a CW complex structure with one $n$-cell, even if it isn't smoothable, so all such manifolds admit a degree one map to $S^n$. If $n = 4$, it is not known whether such a CW complex structure exists (in fact, it is not even known if every closed four-manifold has a CW complex structure). See this MathOverflow question for more details.

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  • $\begingroup$ Thank you for your answer - unfortunately I am not familiar with CW complexes but I'm sure this will be of interest to those who are! $\endgroup$ – Wooster May 12 '16 at 12:46
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Take lat/lon coordinates on the sphere: latitude $\phi$ from 0 to $\pi$, longitude $\theta$ from $o0$ to $2\pi$.

Take polar coordinates $\alpha, \beta$ on the torus, both running from $0$ to $2\pi$.

Let $$ a = \frac{\alpha - \pi}{\pi} \\ b = \frac{\beta - \pi}{\pi} $$ be coordinates with the range $-1$ to $1$ on the torus.

Now consider $$ f(a, b) = \begin{cases} \sqrt{1 + \frac{\min(|a|, |b|)^2}{\max(|a|, |b|)^2}} & \text{either $a$ or $b$ nonzero}\\ 0 & \text{a = b = 0} \end{cases}. $$ This is the length of a ray through $(a, b)$ hitting the edge of the square $S = \{(x, y) \mid -1 \le x,y \le 1 \}$.

Define $$ g(a, b) = \frac{1}{f(a, b)} (a, b) $$ This maps the square $S$ to the unit disk $D$ (essentially by "radial stretching").

Finally, define $$ h: D \to S^2 : (u, v) \mapsto (\pi\sqrt{u^2+v^2}, atan2(u, v) ) $$ where points of $S^2$ are represented in $(\phi, \theta)$ coordinates.

Now the map $h \circ g$ takes $(a, b)$ coordinates on the torus to $(\phi, \theta)$ coords on the sphere, and you've got your map.

Shorthand: take the square $S$, which is the domain for $(a, b)$ coordinates on the torus, and send its whole boundary to the south pole, and its center to the north pole.

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  • $\begingroup$ This is great, thank you. So this is a smooth realisation of Andrew D. Hwang's suggestion. I would still be interested to know if it is possible to use the winding map or what goes wrong in that case? $\endgroup$ – Wooster May 12 '16 at 12:48
  • $\begingroup$ I have no idea about the winding map. But the thing I've provided may well not be smooth; it is, however, continuous. $\endgroup$ – John Hughes May 12 '16 at 15:01
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Your idea about using linking numbers is certainly on the right track. Consider the Hopf link. If we think of the circles as being embedded in two perpendicular planes then one can understand the map \begin{align} S^1 \times S^1 &\to S^2 \\ (x,y) &\mapsto \frac{x-y}{\|x - y\|} \end{align} by drawing the images of a circle for each $x$ fixed. If you think about this for a little while you realize that this can be thought of as placing the sphere in the interior of the torus and then flattening the torus onto the sphere. At this point one can see that the map generically has 3 or 1 preimages and wherever it has 1 preimage, it is orientation preserving so the map has degree 1.

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Toroidal coordinates are generally reconstructed from Cartesian, cylindrical, or spherical coordinates. The problem is the skewness of the toroidal basis vectors. Without redefining the isomorphism, you can just select regions where the sphere to torus mapping is defined. Why do you need to do this? Toroidal eigenfunctions and harmonics exist. Maybe you want to improve these. A nice idea is to look at spherical modes and then map them to toroidal modes. This could be useful in toroidal structures like transformers with fluid, tokamaks, and pipes with toroidal flow. I can show you my calculations, but you might find research into the Gamma function and toroidal coordinates interesting. The current problem is the modulation of the cosine and hyperbolic cosecant. It seems mathworld cannot get the sign of the cosine correct. You have stumbled onto an unknown problem. Follow this link for the quick and dirty mapping. https://en.m.wikipedia.org/wiki/Toroidal_coordinates Good luck.

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