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Math Help - Verify the expression.

  1. #1
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    Verify the expression.

    T:V \rightarrow W is a linear transformation and S \in L^k (W).


    Is it true that T^*(S^{\delta})= (T^* (S))^{\delta}, \delta \in S_k?
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  2. #2
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    Re: Verify the expression.

    Hey vercammen.

    This might be a dumb question, but is W just a matrix? Also what does raising W to the delta do with regards to the map?
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  3. #3
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    Re: Verify the expression.

    As far as I understood U,W are vector fields, S in a k-tensor on W (can be written using basis); S = sum of all possible k-tuples.
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  4. #4
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    Re: Verify the expression.

    So is delta an Einstein summation?
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  5. #5
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    Re: Verify the expression.

    just a permutation, I guess
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  6. #6
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    Re: Verify the expression.

    The reason I ask is that if it is just a summation, then it should hold (the identity that is).

    The reason has to do with distributivity of matrix multiplication,
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  7. #7
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    Re: Verify the expression.

    I asked a professor, he told me it's just the matter of using the definitions...
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  8. #8
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    Re: Verify the expression.

    Here is what I did, but unfortunately it was graded as incorrect.


    On the k-tensor powers the induced map is T:V^{\otimes k}\to W^{\otimes k} which on pure tensors is T(v_1\otimes v_2\otimes \ldots \otimes v_k)=T(u_1)\otimes T(u_2)\otimes \ldots T(u_k)

    If \sigma is a permutation, then (u_1\otimes \ldots u_k)^\sigma=u_{\sigma^{-1}(1)}\otimes \ldots \otimes u_{\otimes k} and thus we can immediately verify the identity

    T(u_1\otimes \ldots u_k)^\sigma=T((u_1\otimes \ldots u_k)^\sigma)

    because both sides will equal T(u_{\sigma^{-1}(1)})\otimes \ldots T(u_{\sigma^{-1}(k)})

    Because pure tensors span the k-tensor power space, we conclude that

    T(v^\sigma)=T(v)^\sigma (1)

    Now let's get back to the problem.

    By definition, T^*(S^\sigma)(x)=S^\sigma(T(x)) which is S(T(x))^\sigma

    Now by two consecutive applications of (1), S(T(x))^\sigma=S(T(x)^\sigma)=S(T(x^\sigma)) and that is, by definition, (T^*S)(x^\sigma) - which again by definition equals (T^*S)^\sigma(x)
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