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Math Help - Tensor bundle

  1. #1
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    Tensor bundle

    Hello,

    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:
    f(p):T_{p}M\;\times\;T_{p}M->\mathbb{R} which assigns to each point p a positive definite symmetric bilinearform.

    And the tensor bundles are of the form:
    T_{r,s}(M):=\bigsqcup_{m\in M}(T_{m}M)_{r,s}
    with (T_{m}M)_{r,s}:=T_{m}M\otimes...\otimes T_{m}M\otimes (T_{m}M)^{*}\otimes...\otimes (T_{m}M)^{*}

    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 ??


    Regards
    Last edited by Sogan; February 6th 2011 at 01:47 PM.
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  2. #2
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    Consider a vector space V with an inner product \langle ,  \rangle defined. The inner product induces a canonical isomorphism \phi between V and V*: (\phi(v))(w) = \langle v, w \rangle . So we can identify V and V* by this isomorphism. Then we can consider only tensor products V_{r,0}.
    Define an inner product on V_{r,0} as follows:
    \langle v_1 \otimes v_2 \otimes ... \otimes v_r,  w_1 \otimes w_2 \otimes ... \otimes w_r \rangle = \langle v_1, w_1 \rangle \langle v_2, w_2 \rangle ... \langle v_r,  w_r \rangle
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  3. #3
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    Excuse me this was a bad mistake of me! In europe it is very early in the morning! ;-)

    Regards
    Last edited by Sogan; February 6th 2011 at 11:15 PM.
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  4. #4
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    why did you say that I was considering only for even ranks?
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  5. #5
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    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?

    Regards
    Last edited by Sogan; February 6th 2011 at 11:25 PM.
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  6. #6
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    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.
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  7. #7
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    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?

    Regards
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