petergreat
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It's really a question about convention. Does such a metric have to be linear on each fiber?
The discussion revolves around the nature of a Riemannian metric on a vector bundle, focusing on whether such a metric must be linear on each fiber, how it relates to the Euclidean metric, and the implications of curving the base space versus the entire space.
Participants do not reach consensus on whether a Riemannian metric must be linear on each fiber or whether there is a natural metric on the fibers. Multiple competing views remain regarding the nature of metrics on vector bundles.
Participants express uncertainty about the definitions and assumptions regarding metrics on fibers and the comparison of different fibers' 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?