# Is Curvature Accumulative?

Is the curvature of GR accumulative? Can you integrate the curvature immediately around various points to find the curvature around a larger area? Thanks.

Matterwave
Gold Member
No. You can't go around adding the curvature tensors of different parts of the manifold, they don't exist in the same tangent space. Even if you parallel transported them or Lie dragged them or something, I would have no idea what this "addition" would even mean physically. Even if you just looked at the curvature scalar, which is a scalar function that you can actually integrate, I don't see any meaning in integrating it.

If you draw a circle, then the radius of curvature scalar at each point on the circle is R, the radius of curvature. What does it mean to add them together? You can integrate it and get 2πR2 but this means nothing...

lavinia
Gold Member
Is the curvature of GR accumulative? Can you integrate the curvature immediately around various points to find the curvature around a larger area? Thanks.

If you make this question more specific, it will be possible to answer it. Specify what curvature you mean. Write down the formula.

Ben Niehoff
Gold Member
No. You can't go around adding the curvature tensors of different parts of the manifold, they don't exist in the same tangent space. Even if you parallel transported them or Lie dragged them or something, I would have no idea what this "addition" would even mean physically.

Actually...the connection (viewed as a ##GL(n)##-valued 1-form) can be integrated along a path. One needs to use path-ordered integration. The result is a linear operator in ##GL(n)## which tells you how to parallel transport vectors along the path from one endpoint to the other. If you integrate around a closed path, then you end up in the same tangent space you started in, so the result is covariant: it is a ##GL(n)## matrix that tells you about parallel transport around the given loop.

The Riemann tensor (viewed as the curvature 2-form ##\Omega^a{}_b = \frac12 R^a{}_{bcd} \,dx^c \wedge dx^d##) cannot be integrated unless you soak up the extra indices somehow (because ##GL(n)## is non-Abelian and there isn't a way to define an ordering on surfaces). In 2 dimensions, there are no extra indices, and you can integrate the curvature directly; the curvature is proportional to the Euler class, and its integral is the Euler number of your manifold. In higher dimensions, you can write down combinations of products of Riemann with itself which are the Pontrjagin classes, and these can be integrated, giving you other topological invariants.

Even if you just looked at the curvature scalar, which is a scalar function that you can actually integrate, I don't see any meaning in integrating it.

The Ricci scalar measures the angular deficit per unit volume, so actually it's a perfectly sensible thing to integrate.

• deluks917
Matterwave
Gold Member
Actually...the connection (viewed as a ##GL(n)##-valued 1-form) can be integrated along a path. One needs to use path-ordered integration. The result is a linear operator in ##GL(n)## which tells you how to parallel transport vectors along the path from one endpoint to the other. If you integrate around a closed path, then you end up in the same tangent space you started in, so the result is covariant: it is a ##GL(n)## matrix that tells you about parallel transport around the given loop.

The Riemann tensor (viewed as the curvature 2-form ##\Omega^a{}_b = \frac12 R^a{}_{bcd} \,dx^c \wedge dx^d##) cannot be integrated unless you soak up the extra indices somehow (because ##GL(n)## is non-Abelian and there isn't a way to define an ordering on surfaces). In 2 dimensions, there are no extra indices, and you can integrate the curvature directly; the curvature is proportional to the Euler class, and its integral is the Euler number of your manifold. In higher dimensions, you can write down combinations of products of Riemann with itself which are the Pontrjagin classes, and these can be integrated, giving you other topological invariants.

The Ricci scalar measures the angular deficit per unit volume, so actually it's a perfectly sensible thing to integrate.

Huh, looks like I should read up more on this. Thanks for the corrections! :D

......
The Ricci scalar measures the angular deficit per unit volume, so actually it's a perfectly sensible thing to integrate.
I find myself thinking about the Einstein-Hilbert action whose equations of motion yield the field equations of GR. See: http://en.wikipedia.org/wiki/Einstein–Hilbert_action

There the Ricci curvature scalar is integrated against the volume form - a straightforward integration of the curvature. What do you suppose this means? What is the angular deficit per unit volume?