If curvature were an exact differential

[Mike2]

If curvature were an exact differential so that the derivative of the curvature were zero, then is it possible to solve for the metric?

 

[Mike2]

I'm wondering what happens if you put the gaussian curvature in Stoke's theorem. So the curvature would have to be an exact differential. This might be physically important since the curvature of spacetime has a connection with energy and therefore mass, etc.

 

[lethe]

If curvature were an exact differential so that the derivative of the curvature were zero, then is it possible to solve for the metric?

the curvature is exact on a Calabi-Yau

 

[Mike2]

the curvature is exact on a Calabi-Yau

Interesting, I didn't know that. Thank you.

Are the 3 expanded dimensions part of the Calabi-Yau manifolds?

 

[lethe]

Interesting, I didn't know that. Thank you.

Are the 3 expanded dimensions part of the Calabi-Yau manifolds?

the expanded 3 dimensions? huh? what expanded 3 dimensions? I have no idea what you are asking, so i guess i will just say some stuff about Calabi-Yaus

A Calabi-Yau manifold is a Kähler manifold whose first Chern class vanishes. by a theorem of Yau, this implies that the Ricci tensor also vanishes. Calabi-Yaus are compact.

In general, we do not know how to find the metric for a Calabi-Yau, so the answer to the original question "can we solve for the metric if the curvature is exact" is "no".

 

[Mike2]

the expanded 3 dimensions? huh? what expanded 3 dimensions? I have no idea what you are asking, so i guess i will just say some stuff about Calabi-Yaus

If I remember right, space-time is suppose to be divided between 3 expanding spatial dimensions, 1 time dimension, and 6 compactified spatial dimensions. I wasn't sure whether the expanding 3 dimension were part of the Calabi-Yau manifold or not. It sounds to me from your answers below that it does not.

A Calabi-Yau manifold is a Kähler manifold whose first Chern class vanishes. by a theorem of Yau, this implies that the Ricci tensor also vanishes. Calabi-Yaus are compact.

In general, we do not know how to find the metric for a Calabi-Yau, so the answer to the original question "can we solve for the metric if the curvature is exact" is "no".

I was talking in theory can we find the metric given an exact curvature. Or whether there was some math that tells us that there is not sufficient information to get the metric given an exact differential. Perhaps we also need some boundary conditions.

 

[Haelfix]

Mike, I think you are confusing a few things here.

Curvature as a word is meaningless, unless you specify a connection and a context (is it extrinsic or intrinsic).

A metric space then may or may not be possible given your choice of connection on some topology.

I think what you are thinking of is the notion of gaussian curvature, and other related things like mean curvature, etc. This typically pressupposes a specific connection and an associated metric. Then you can play around with things.

there's other ways to think about 'curvature' and the term is often used differently here, for instance you can specify a topological invariant, like the Chern class.. And this kinda tells you a little something about how your space acts independantly of the connection. Make sure that you keep the definitions in context.

 

[Mike2]

What curvature was it that uncurled as the universe expanded? Was it the curvature described by general relativity? Was it the gaussian curvature, what?

I do appeciate your responses, thanks.