# Is curvature possible for a 2D metric?

1. Aug 31, 2014

### space-time

I was recently trying to test something out with the Riemann tensor. I used only 2 dimensions for simplicity sake. As I was deriving the Riemann tensor, I noticed that it looked as if all of the elements were going to come out to be 0 (which they all did). Therefore, this coordinate system is flat space.

Then I realized something. I was working in 2D. The very definition of 2D alone essentially means flat planes. Therefore it would make sense for a 2D coordinate system to be flat space.

Just to make sure however, I must ask this: Is it possible for a 2 dimensional Riemann tensor to ever have a non zero element, or does the fact that it is 2D mean that there can never be any non zero elements (and therefore no curvature)?

2. Aug 31, 2014

### WannabeNewton

No this is incorrect. A 2-sphere is 2-dimensional and has non-zero Riemann curvature. If every 2-manifold was trivially a plane then the extremely elegant classification theorem of compact 2-manifolds would be a waste

3. Aug 31, 2014

### Orodruin

Staff Emeritus
Yes, it is perfectly possible for a 2-dimensional manifold to have a non-vanishing curvature tensor. The prime example would be a sphere embedded in three euclidean dimensions with the induced metric.

4. Aug 31, 2014

### A.T.

5. Aug 31, 2014

### Matterwave

You gotta go one more dimension down. A 1-D Riemann tensor is always 0. Any curve can only have extrinsic curvature and no intrinsic curvature.

6. Aug 31, 2014

### pervect

Staff Emeritus
Not so, for a counterexample, the previously mentioned surface of a sphere. But if you start with a plane, you'll get zero curvature, that much is true.
There are 16 components of the Riemann in 2d

$R_{ijkl}$ is zero whenever i=j or k=l. Since i,j,k and l can take only the values 0 and 1, that leaves only four non-zero components for the Riemann in 2d.

These four components will all have the same value in an orthonormal basis, there is really only one degree of freedom for (intrinsic) curvature in 2d.

7. Aug 31, 2014

### Matterwave

Adding on to pervect's reply a little, in general the number of independent components of the Riemann tensor is $n^2(n^2-1)/12$ where n is the dimension of the manifold. You can immediately see from this equation that for n=1 the Riemann has actually no independent components, it must vanish, and for n=2 the answer is 1.

8. Sep 1, 2014

### stevendaryl

Staff Emeritus
Wikipedia gives an expression for the simplification of $R_{ijkl}$ in the 2D case:

$R_{abcd} = K (g_{ac} g_{db} - g_{ad}g_{cb})$

where $K$ is a real number that varies from point to point and is equal to the Gaussian curvature.

9. Sep 1, 2014

### mal4mac

If you define the 2D space to have the geometry of the surface of a sphere would that be flat?

10. Sep 1, 2014

### robphy

No. On a sphere, ...
-the sum of the interior angles of a triangle won't add up to 180 degrees (you can make a triangle with three right angles).
-a triangle with one right angle won't satisfy the Pythagorean theorem.
-the circumference/radius ratio of a circle (generated by a fixed length string tied to a post) won't be 2*pi.

11. Sep 1, 2014

### mal4mac

OK so a sphere gives us a 2D space that isn't flat.

Extending that into one dimension (e.g. the great circle of a sphere) don't we get a 1D space that isn't flat?

12. Sep 1, 2014

### Staff: Mentor

No. Mark two points on a piece of thread, and no matter how the thread is laid out the distance between the two points is the same - and in one dimension that's all the geometry there is.

13. Sep 1, 2014

### ChrisVer

or in other words there is no difference between a circle of radius R and a line of length 2πR
you can cut the circle, extend it and make it a line. That's actually not the case for a sphere - you can't cut it and make a flat plane. That's because the sphere has curvature.

14. Sep 1, 2014

### mal4mac

Imagine a 3D sphere entering a 2D flatland space at constant speed. The inhabitants of flatland would see a circle - growing and then diminishing (this is in Abbot's book Flatland if memory serves...)

Now imagine the edge of a circle entering (i) my curved 1D space (ii) a straight 1D space. Assuming the circle moves into the 1D spaces at the same speed the "shadow line" would grow at different rates in the different 1D spaces.

Fun animation showing flat case:

http://www.rmcybernetics.com/science/physics/dimensions_1_dimensional_space.htm

Last edited: Sep 1, 2014
15. Sep 1, 2014

### Staff: Mentor

That's true, but that's not how intrinsic curvature (the kind described by the Riemann tensor and that matters in general relativity) is defined. Instead, it is defined by the geometrical relationships between points in the manifold itself - look at robphy's post above for examples of such relationships.

Another example worth considering is the two-dimensional surface of a cylinder. It doesn't appear at first glance to be flat, but it is - rolling a sheet of paper up into a cylinder doesn't stretch the paper or distort any of the geometrical relationships between points on the sheet of paper.

16. Sep 1, 2014

### ChrisVer

Since we are talking about the sphere, I have one question... Is it because of the sphere's curvature that the Rieamman sphere projection on $\textbf{C}^{2}$ has the pole of the sphere appearing at "$\pm ∞$"?

17. Sep 1, 2014

### mal4mac

18. Sep 1, 2014

### A.T.

A 1D curve can have extrinsic curvature, which depends on its embedding in higher dimensional space. For actual intrinsic curvature, that exists independently of any embedding you need at least 2D.

Last edited: Sep 1, 2014
19. Sep 1, 2014

### mal4mac

I don't quite understand that, I find it difficult to visualise a space without its embedding in a higher dimension. Wolfram says:

A 2D intrinsic curvature, "is detectable to the inhabitants of a surface and not just outside observers. An extrinsic curvature ... is not detectable to someone who can't study the three-dimensional space surrounding the surface on which he resides."
http://mathworld.wolfram.com/IntrinsicCurvature.html

Fair enough, but in the 1D case, surely you need to add there must be no observable intrusion from the embedding 2D space? (thinking of the circle passing through...)

Also, if an inhabitant of circular 1D space went for a walk and found himself back where he started couldn't he say the space was curved? It certainly isn't flat or he'd be walking to infinity. So shouldn't we also disallow "long walks" in the definition of extrinsic curvature?

20. Sep 1, 2014

### ChrisVer

You are mixing the Riemannian curvature with the Gaussian...

He could say that his space is closed, it doesn't have to be curved to be closed...

21. Sep 1, 2014

### ChrisVer

22. Sep 1, 2014

### Staff: Mentor

Everybody does. That's why we draw these embedding diagrams, but we can't allow them to mislead us into thinking that there are really is a higher embedding dimension. In particular, our four-dimensional spacetime is implicitly curved, it is not extrinsically curved in a fifth(!) dimension.
But with intrinsic curvature, there is no embedding space of higher dimension.

That circular 1D space does have extrinsic curvature, just as does the 2D surface of a cylinder that I mentioned a few posts back. But it does not have intrinsic curvature - it is flat but not unbounded.

23. Sep 3, 2014

### A.T.

How would he quantify the intrinsic curvature?

I think whether the space is bounded or not is more a question of topology. Intrinsic curvature is a matter of geometry.

Last edited: Sep 3, 2014
24. Sep 3, 2014

### stevendaryl

Staff Emeritus
An intuitive way to see that a cylinder or a torus can have zero intrinsic curvature is this:

A cylinder is a 2-D surface that can be described using coordinates in which one coordinate, y, ranges from -∞ to +∞, while another coordinate, x, wraps around, so that $(x+2\pi, y)$ is the same point as $(x,y)$.

This situation is locally indistinguishable from the situation in which $x$ ranges from -∞ to +∞, but everything (all functions of position) happen to be periodic; for every function $f$ of position, it happens to be the case that $f(x,y) = f(x+2\pi, y) = f(x+4\pi, y) = ...$. So a cylinder looks just like an infinite 2D plane in which things are periodic in the $x$ direction.

If you tried to do the same thing with the sphere, it wouldn't work. A function on the surface of a sphere $f(\theta, \phi)$ cannot be interpreted as a function of Cartesian coordinates with extra conditions about being periodic.

What does work is a torus (a shape like the surface of a doughnut, or bagel, depending on your culinary preferences). You can describe your location on a doughnut by a pair of angles $\theta, \phi$ such that $(\theta + 2\pi, \phi) = (\theta, \phi + 2\pi) = (\theta, \phi)$. So a torus looks just like a 2D plane in which things are periodic in two different directions.

In general, a way to characterize a 2D surface of arbitrary topology is by cutting it up into triangles. In terms of these triangles, the surface can be defined by:

• Within a triangle, describe things using coordinates that are approximately Cartesian. (The differences with Cartesian coordinates vanish in the limit as the size of the triangles goes to zero.)
• For each side of each triangle, specify which side of which other triangle it is identified with.
• For every pair of triangles that share a side, specify how tangent vectors crossing the boundary from one direction relate to tangent vectors crossing the boundary from the other direction.

The exact geometry of the 2D surface is specified by the limit as the lengths of the sides of the triangles go to zero. Similarly, a 3D space can be characterized by tetrahedrons, and a 4D space can be characterized by whatever the 4D analog of a tetrahedron is.

25. Sep 3, 2014