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

  • Thread starter petergreat
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  • #1
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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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  • #2
lavinia
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It's really a question about convention. Does such a metric have to be linear on each fiber?

symmetric bilinear form on each fiber
 
  • #3
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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)?
 
  • #4
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In other words, are we allowed to "curve" the base space only or the entire space?
 
  • #5
lavinia
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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.
 
  • #6
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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.
 
  • #7
lavinia
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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?
 
  • #8
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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?
 
  • #9
lavinia
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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.
 
  • #10
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Thanks! That's clear now.
 

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