# Integration of Ricci Scalar Over Surface

1. Apr 12, 2015

### darida

Does this integration of Ricci scalar over surface apply in general or just for compact surfaces?

∫RdS = χ(g)

where χ(g) is Euler characteristic.
And could anybody give me some good references to prove the formula?

2. Apr 13, 2015

### Ben Niehoff

On a non-compact surface, the integral might fail to converge. Take hyperbolic space, for example.

3. Apr 13, 2015

### wabbit

This might still be true when the integral converges though ?

4. Apr 13, 2015

### Ben Niehoff

I'm not sure.

Another way to look at this is an order-of-limits issue. Consider a modified integral for computing the Euler number of manifolds-with-boundary. This integral will contain a boundary term (sorry, I don't remember what it looks like). Then you can take any non-compact surface, and just cut a piece out of it, adding a boundary and making the integral finite. Perform the integral first, and then take the limit as the boundary goes to infinity.

I think this will give you an Euler number of 1 for the infinite plane. It is homologous to the disk, and boundaries add 1 to the Euler number.

5. Apr 13, 2015

### wabbit

For a non compact surface, I was more thinking of compacification with a point/curve at infinity and hopefully making use of R=0 at infinity... But you're right the boundary term can make this whole business tricky.

6. Apr 13, 2015

### lavinia

In the non-compact case where the integral does converge, the total curvature can be less than 2πχ(S).

The flat plane has zero total curvature 0 but has Euler characteristic 1.

For a compact manifold with boundary one needs to add on the total geodesic curvature of the bounding curve.