i try to show that one can define a metric on all tensor bundles with the riemannian metric.
I have really no idea, how i can show this. Here are the definition, which i know:
A Riemannian Metric on a Manifold M is a map f which assigns to each point p this: which assigns to each point p a positive definite symmetric bilinearform.
And the tensor bundles are of the form:
Ok, meanwhile i'm not sure what the word "metric" mean. A metric on all tensor bundles. Is it a metric in the usual sense, that is a metric space?
Or is it a metric like the riemannian metric is. I.e. a positive definite symmetric bilinearform ??
Feb 6th 2011, 09:26 PM
Consider a vector space V with an inner product defined. The inner product induces a canonical isomorphism between V and V*: . So we can identify V and V* by this isomorphism. Then we can consider only tensor products .
Define an inner product on as follows:
Feb 6th 2011, 10:46 PM
Excuse me this was a bad mistake of me! In europe it is very early in the morning! ;-)
Feb 6th 2011, 10:53 PM
why did you say that I was considering only for even ranks?
Feb 6th 2011, 11:13 PM
Excuse me! You are absolutely right. I did a mistake. I have forgotten that the inner product takes two elements of the tensor bundle therefor you are right!
Now i have to show that this construction is indeed continuous. Why do we have this property?
Feb 7th 2011, 12:27 AM
Since the induced inner product is (smoothly) expressible by the original one g, and g is smooth, we're done.
You can write down the exact expression to be sure.
Feb 7th 2011, 08:50 AM
Thank you again!
I think they are continuous, since our g is the product of the original one g (which is smooth).
So now we have defined a smooth inner product on each Tensor space of the form (r,s).(with r,s fixed).
What does this construction mean? What can i do with it? Or better: What do we need it for?
Do you know perhaps what this meaning of this construction is?