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

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SUMMARY

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.

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

symmetric bilinear form on each fiber
 
lavinia said:
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?
 
petergreat said:
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.
 
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.

I'm talking about a vector bundle, so each fiber has a natural metric up to constant factor.
 
petergreat said:
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?
 
lavinia said:
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?
 
petergreat said:
Oops... You're right. But still, does it have to be a constant 2-tensor on each fiber?

On each fiber ,it is just a metric on the fiber viewed as a vector space.

Different fibers are different vector spaces and generally have different metrics - which means there is no natural way to compare these vector spaces or their metrics.
 
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Thanks! That's clear now.
 

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