# Riemannian metrics and how spaces look

I thought I had a fairly good understanding of Riemannian metrics until I came across this exercise in Petersen's book.

Construct paper models of the Riemannian manifolds ($\mathbb{R}^2, dt^2 + a^2t^2d \theta ^2$). If $\alpha = 1$, this is of course the Euclidean plane, and when $\alpha < 1$, they look like cones. What do they look like when $\alpha >1$?

I fail to understand why changing the numbers that you assign to a pair of vectors in a tangent space (changing the Riemannian metric) would make a figure look different. If I give $\mathbb{R}^2$ Cartesian coordinates why would changing the inner product cause any difference in the way it looks? Is there some intuition behind Riemannian metrics that I am missing? Thank you.

• Isometrically embed these spaces in $\mathbb R^3$. Then they will "look" qualitatively different. As $\alpha$ changes, for instance, the angle of the cone will be different. I don't know if one can do so for $\alpha > 1$, but this is the first way one can measure what a Riemannian manifold 'looks like'. Other ways include looking at properties of the manifold, like its geodesics, curvature, etc, which should help you see what it "looks like" locally or globally. (This 'embed in $\mathbb R^3$ - what does the image look like?' will only work for a sparse few Riemannian manifolds.)
– user98602
Commented Jul 27, 2015 at 3:23
• Think of all the possible ways you could take the open unit disk and bend it to make whatever (smooth, contractible) surface your heart desires. They can look very different. And every one of them can be described by putting a metric on the open unit disk. The whole point of the metric is to be able to describe distances, and if you mess with all of the distances between points then you warp the space! I recommend reading my answer here to understand the motivation of metrics.
– anon
Commented Jul 27, 2015 at 6:01
• (Should have said distances and angles.) The metric is something you get to decide yourself, but only insofar as you get to decide the shape of the space you're imagining yourself. But keep in mind that given a space from the outset (say, a smooth surface embedded in $\Bbb R^3$ that makes for easy visualization) with all the distances/angles already defined, the metric is determined by this figure - you don't have any say in the matter. And conversely, this figure is determined by the metric (at least "intrinsically" anyway - angles and distances on it).
– anon
Commented Jul 27, 2015 at 6:10
• Suppose I have $\mathbb{S}^1 = \{(x,y)| x^2 + y^2 = 1\}$. I understand that the induced metric from $\mathbb{R}^2$ is $d \theta ^2$. However, if I just changed it to $2 d \theta ^2$ my circle still looks the same (It is the set I described above) but it is not isometrically embedded. I understand that the distance will be measured differently but it is still valid to speak of this set with this metric right? However if I wanted to embed it isometrically I would have to transform it somehow correct? Commented Jul 27, 2015 at 6:53
• Are you saying that the only metric such that the figure is isometrically embedded is determined once you specify the distances and angles? And in general how do people solve the problem "given a metric on a space find a figure that has this metric and is isometrically embedded in Euclidean space"? Commented Jul 27, 2015 at 6:56

The factor $a$ makes the length of each circle centered at the origin equal to $2\pi a r$, where $r$ is the radius.

### Case $a<1$: not enough circumference

To realize this case in practice, cut off a part of the circle, specifically $(1-a)$ part of it (in angle terms, $2\pi (1-a)$). This is well explained in wikiHow:

Note that this surface is inclined to close in on itself; this is a manifestation of positive curvature (concentrated in the vertex here).

### Case $a>1$: too much circumference.

Instead of removing a part of a circle, we need to add more of it. A way to achieve this is to take $2$, $3$, or more sectors of the kind shown above, and glue their edges so that a single surface is obtained. With $n$ sectors of angular size $\theta<2\pi$, you get $n\theta$ total angle, which can be any positive number you want.

The sectors (labeled $1,2,\dots,n$ and put in a stack) should be glued so that, say, the bottom end of 1st is glued to the top end of 2nd, the bottom end of 2nd to the top end of 3rd... and the bottom end of $n$ to the top end of $1$st. This last gluing step cannot actually be done in a three-dimensional space without creating self-intersection. You'll just have to imagine that the self-intersection isn't there: the surface gets around another part of itself via a detour into the 4th dimension. Here is an illustration from Wikipedia.

Unlike the first example, the surface wants to spread around (so much that there isn't enough room for it in 3D space); this is a manifestation of negative curvature.

• Thank you so much for the illustration. I was able to come to the conclusion of having more circumference but I was having a very hard time thinking of the picture. Commented Jul 29, 2015 at 0:28