# Tori and metrics

I have been doing some reading on tori. What I can make out of it is that a torus can be equipped with different metrics -- locally Euclidean or as an embedded surface. It is said however that the torus with the locally Euclidean metric cannot be realized as an embedded surface. Why is this true and what is the metric as an embedded surface like? Why would we want the latter metric, since it seems to me the former is more natural?

Thanks.

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You have to be careful when you say "embedded metric". It depends on which space you embed it in. The flat torus (the locally Euclidean one) can be isometrically embedded in $\mathbb{R}^4$, but not in $\mathbb{R}^3$. –  Thomas Rot Mar 9 '12 at 11:07
I was browsing around, and apparently I was wrong. mathoverflow.net/questions/31222/… . –  Thomas Rot Mar 9 '12 at 12:24
The standard embedding en.wikipedia.org/wiki/Torus however is not flat in $\mathbb{R}^3$. The embedding of the torus as $S^1\times S^1\subset \mathbb{R}^2\times \mathbb{R}^2$ is flat I believe. –  Thomas Rot Mar 9 '12 at 12:27
The torus is compact. If it is smoothly embedded in $\mathbb{R}^3$ it must have at least one point of positive curvature. From this it can be deduced that it cannot be flat with the metric induced by the embedding. Depending on the background knowledge you have this might require some additional research/work to see this. –  user20266 Mar 9 '12 at 12:57
Usually we say tori instead of toruses. –  Mariano Suárez-Alvarez Mar 11 '12 at 3:38

## 2 Answers

To construct the isometrically embedded torus (e.g. the surface of a donut), first consider the circle of radius $r$ in the $yz$ plane whose center is distance $c$ from the origin along the $y$ axis. Given in polar coordinates, the parametric equation is $$v \mapsto (c + r\cos v, r\sin v)$$ We can obtain the torus by rotating this curve around the $z$ axis. There is a standard way to do this- namely the parametric equations are given by $$(u,v) \mapsto ((c+r\cos v)\cos u, (c+r\cos v)\sin(u), r\sin v )$$ It is therefore clear that we have written down parametric equations for a torus in $\mathbb R^3$. To find the metric that makes the torus an isometric embedding, we may calculate (if $s:\mathbb R^2 \to \mathbb R^3$ is the above map) $$ds\cdot ds = (c+r\cos v)^2du^2 + r^2dv^2$$ On the other hand, we could turn this construction around if we start abstractly with the set of points $M = S^1\times S^1$. Think of $M$ as a handful of clay waiting to be sculpted. Putting a metric on $M$ is a way of measuring lengths and angles, which is like taking the clay and sculpting it into a specific shape. The metric above makes $M$ look like a donut. But there are other metrics too, for instance we could consider $$du^2 + dv^2$$ i.e. we could put the euclidean metric on $M$. This make $M$ look like a flat piece of paper that has been rolled into a cylinder and then had the ends of the cylinder glued together. It is impossible to carry this out in $\mathbb R^3$ (e.g. there is no embedding $s:M \to \mathbb R^3$ so that $ds\cdot ds = du^2 + dv^2$). As mentioned in the comments, if it were possible to do this, then we could consider the smallest sphere in $\mathbb R^3$ that completely contains $M$. This sphere must touch $M$ at least once, and it is possible to show that any point at which this "minimal sphere" touches $M$ must be a point of positive curvature on $M$, but this is a contradiction since $M$ is flat and therefore has zero curvature (see Ted Shifrin's Differential Geometry book, Proposition 3.5).

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A $2$-torus is a topological space $X$ that is homeomorphic (one-one continuously equivalent) to the surface $S$ of a doughnut in three-space. For all practical purposes we may require that $X$ is diffeomorphic (one-one differentiably equivalent) to $S$.

A Riemannian metric on $X$ is a law that allows to measure the length of (continuously differentiable) curves on $X$. This law is expressed "in local coordinates" $(u_1,u_2)$ by means of a quadratic expression $ds^2=\sum_{i,k} g_{ik} du_i du_k$. For a curve $\gamma:\ t\mapsto u(t)$ $\ (a\leq t\leq b)$ its length is then given by $L(\gamma)=\int_a^b\sqrt{\sum_{i,k} g_{ik} \dot u_i \dot u_k}\ dt$. When $X=S$ with the metric "inherited from" ${\mathbb R}^3$ then the formulae from treble's answer apply.

There is a very deep theorem about Rienmannian $2$-tori in general. It says the following: If $X$ is a Riemannian $2$-torus then there is a lattice $\Lambda$ in ${\mathbb R}^2$ (with fundamental parallelogram $P$) and a function $\rho:{\mathbb R}^2\to{\mathbb R}_{>0}$, periodic with respect to $\Lambda$, such that $X$ can be regarded as ${\mathbb R}^2/\Lambda$ ("$P$ with parallel edges identified"), provided with the Riemannian metric $ds=\rho(z)|dz|$, where$|dz|:=\sqrt{|dx^2+dy^2}$.

If the function $g$ is actually a constant then we say that the metric on $X$ is "locally euclidean". But note that the global metric structure of $X$ depends also on the shape of $\Lambda$, so there is an infinity of different "locally euclidean" $2$-tori.

In the case of a "real" doughnut embedded in $3$-space the corresponding lattice $\Lambda$ is orthogonal, and the function $g$ is $\Lambda$-periodic, but not constant.

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