# What does it mean by a Riemannian metric on a vector bundle?

by petergreat
Tags: riemannian metric, vector bundle
 P: 270 It's really a question about convention. Does such a metric have to be linear on each fiber?
P: 1,716
 Quote by petergreat It's really a question about convention. Does such a metric have to be linear on each fiber?
symmetric bilinear form on each fiber
P: 270
 Quote by lavinia symmetric bilinear form on each fiber
Does it have to preserve the natural Euclidean metric up to a constant factor in each fiber (which is a vector space)?

 P: 270 What does it mean by a Riemannian metric on a vector bundle? In other words, are we allowed to "curve" the base space only or the entire space?
P: 1,716
 Quote by petergreat Does it have to preserve the natural Euclidean metric up to a constant factor in each fiber (which is a vector space)?
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 metric on the tangent space of the ambient manifold.
P: 270
 Quote by lavinia 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.
I'm talking about a vector bundle, so each fiber has a natural metric up to constant factor.
P: 1,716
 Quote by petergreat I'm talking about a vector bundle, so each fiber has a natural metric up to constant factor.
no. There is no natural metric. Why do you think that? Can you give me a proof?
P: 270
 Quote by lavinia no. There is no natural metric. Why do you think that? Can you give me a proof?
Oops... You're right. But still, does it have to be a constant 2-tensor on each fiber?