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

It's really a question about convention. Does such a metric have to be linear on each fiber?

lavinia
Gold Member
It's really a question about convention. Does such a metric have to be linear on each fiber?

symmetric bilinear form on each fiber

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)?

In other words, are we allowed to "curve" the base space only or the entire space?

lavinia
Gold Member
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.

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.

lavinia
Gold Member
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?

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?

lavinia