petergreat
- 266
- 4
It's really a question about convention. Does such a metric have to be linear on each fiber?
A Riemannian metric on a vector bundle does not require linearity on each fiber, as each fiber is treated as a separate vector space with its own distinct metric. There is no natural Euclidean metric on a fiber, and different fibers cannot be directly compared due to their unique metrics. The discussion emphasizes that while a fiber may have a natural metric up to a constant factor, this does not imply a uniform metric across fibers. Additionally, metrics can be inherited from the ambient manifold when dealing with submanifolds.
PREREQUISITESMathematicians, differential geometers, and students of geometry interested in the properties of vector bundles and Riemannian metrics.
petergreat said:It's really a question about convention. Does such a metric have to be linear on each fiber?
lavinia said:symmetric bilinear form on each fiber
petergreat said:Does it have to preserve the natural Euclidean metric up to a constant factor in each fiber (which is a vector space)?
lavinia said:not sure what you mean but each fiber is a vector space with a metric defined on it. Different fibers have there own separate metric and there is generally no way to compare them among different fibers.
There is generally no natural Euclidean metric on a fiber.
If you have a submanifold of another manifold then its tangent and normal bundles inherit a metric from the ambient manifold.
petergreat said:I'm talking about a vector bundle, so each fiber has a natural metric up to constant factor.
lavinia said:no. There is no natural metric. Why do you think that? Can you give me a proof?
petergreat said:Oops... You're right. But still, does it have to be a constant 2-tensor on each fiber?