What does it mean, geometrically for two surfaces to be locally isometric?

I've heard it said that if two surfaces have the same first fundamental form, or if they're locally isometric, then they "behave metrically in the same way", or variants on that theme. The textbook I'm reading says that it's "intuitively clear" that the cylinder is isometric to the plane, because "by cutting a cylinder along a generator we may unroll it onto part of a plane". Well, couldn't I do something similar to cut and flatten out any surface onto a plane?

What is the geometric meaning of the existence or non-existence of a local isometry?

• Can you do it with a sphere? – Matthew Leingang Nov 16 '16 at 19:18
• @MatthewLeingang Locally, yes. I mean, globally I can't do it with a cylinder either. – Jack M Nov 16 '16 at 19:20
• @JackM You can't do it with a sphere locally. Try it! You'll run in to trouble while trying to make it flat: you'll have to stretch the surface of the small spherical bit, which does not make it an isometry any more. The whole point is to preserve distance while deforming. – Balarka Sen Nov 16 '16 at 19:50
• @BalarkaSen When you say "preserves distance", do you mean "preserves the distance of curves"? That there's no (local) smooth mapping from the sphere to the plane such that the image of any curve has the same length as the original curve? – Jack M Nov 16 '16 at 19:54
• @JackM An isometry preserves lengths of curves, yes. – Balarka Sen Nov 16 '16 at 20:07

Well, couldn't I do something similar to cut and flatten out any surface onto a plane?

Picture a sphere made of rubber (like a racquetball or tennis ball). Cut out a hunk of it and lay it on the table. Do all parts of the rubber touch the table?

Since the ball has some convexity to it, either the rubber is going to touch in the center and be bowed up around the edge, or touch around the edge and bubble up in the center.

Of course, you could squish the rubber and force it to lie flat against the table. But now you've changed the metric (you have literally stretched certain points of the rubber further away from each other).

So I think it's not true that any surface can be cut and flattened onto a plane.

This is a small-scale version of the problem of drawing accurate maps of the world. Cartographers tried to create flat representations of the surface of the Earth while preserving distance. As it turns out, this is impossible.

• I don't really see why stretching the piece of sphere is different to unrolling the cylinder. But I sort of take your point. – Jack M Nov 16 '16 at 19:45
• @JackM One doesn't stretch the material, the one does. Imagine a tennis ball made of paper, and a cylinder made of paper. The cylinder you can still unroll, since you dont have to strech the material. The tennis ball will tear as soon as you try to flatten it though (since paper doesn't stretch). This 'shows' that there is a fundamental difference between the two situations. – user2520938 Nov 16 '16 at 20:10
• @JackM Also, you have to realise that you're in the Riemannian category here. You seem to be thinking in the smooth category, in which case I agree that both unrolling the cylinder and flattening the sphere are just smooth maps. But the essential thing in the notion of an isometry is that there is more data: the metric. So while you seem to not find the notion of 'stretching' of fundamental importance, it really is, since we are viewing things in the Riemanning setting here, and hence the metric is a fundamental and essential piece of data. – user2520938 Nov 16 '16 at 20:13

If $S$ and $S'$ are two surfaces, and if $g$ is the metric on $S$ and $g'$ the metric on $S'$, $S$ is locally isometric to $S'$ (by $\phi: S \rightarrow S'$) in a neighborhood $V$ of $x \in S$, if: $$\forall y \in V, \forall h,h' \in TS_y,~g'_{\phi(y)}(D\phi_y(h),D\phi_y(h'))=g_y(h,h')$$ where $g_y$ is the metric at the point $y$ and $TS_y$ is the tangent space at $y$.

Maybe should I write "metric tensor":

https://en.wikipedia.org/wiki/Metric_tensor

• What do you mean by "the metric on $S$"? – Jack M Nov 16 '16 at 19:28

Of course we all know that a cylindrical surface can be unrolled onto a plane, hence is locally isometric to the euclidean plane ${\mathbb E}^2$. If differential geometry would not confirm this fact we would have given it up long ago.

But it is something different with a $2$-sphere: You cannot map it locally isometrically to ${\mathbb E}^2$. In order to prove this "geometrically" (i.e., without setting up a heavy differential geometric apparatus) we have to produce a certain invariant which is different for $S^2$ and ${\mathbb E}^2$. Such an invariant is, e.g., the circumference of a circle of radius $0<r\ll 1$. In ${\mathbb E}^2$ this circumference is $2\pi r$, but on a sphere of radius $1$ it is $2\pi\sin r$, whereby we measure $r$ as length of a geodesic arc on $S^2$. This discrepancy shows that it is impossible to map $S^2$ isometrically to ${\mathbb E}^2$ even locally.

• When you refer to a circle of radius $r$ "on a sphere", you're talking about a circle with radius $r$ as measured along the sphere, right? – Jack M Nov 16 '16 at 20:01