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Math Help - Product Basis?

  1. #1
    Super Member Showcase_22's Avatar
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    Product Basis?

    I have this theorem:

    Theorem 33.2: Let \{e_i \} and \{ e^i \} be dual bases for V and V^*. Then the set of tensor products:

    \{ e_{i_1} \otimes \ldots \otimes e_{i_p} \otimes e^{j_1} \otimes \ldots \otimes e^{j_q}, \ i_1, \ldots , i_p,j_1, \ldots , j_q=1, \ldots , N \}

    forms a basis for \vartheta_q^p(V) (the set of all tensors of order (p,q) on a vector space V).
    The proof starts by proving that the set is linearly independent. It does this as follows:

    "We shall prove that the set of tensor products is a linearly independent generating set for \vartheta_q^p(V). To prove that the set is linearly independent, let

    A^{i_1, \ldots, i_p}_{j_1, \ldots , j_q} e_{i_1} \otimes \ldots \otimes e_{i_p} \otimes e^{j_1} \otimes \ldots \otimes e^{j_q}=0

    where the RHS is the 0 tensor".

    Firstly, what is A^{i_1, \ldots, i_p}_{j_1, \ldots , j_q} e_{i_1}? Is it a linear map going from V to V^* with the bases outlined in the question?

    Secondly, how will this prove that the set is linearly independent?
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  2. #2
    MHF Contributor
    Opalg's Avatar
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    Quote Originally Posted by Showcase_22 View Post
    I have this theorem:



    The proof starts by proving that the set is linearly independent. It does this as follows:

    "We shall prove that the set of tensor products is a linearly independent generating set for \vartheta_q^p(V). To prove that the set is linearly independent, let

    A^{i_1, \ldots, i_p}_{j_1, \ldots , j_q} e_{i_1} \otimes \ldots \otimes e_{i_p} \otimes e^{j_1} \otimes \ldots \otimes e^{j_q}=0

    where the RHS is the 0 tensor".

    Firstly, what is A^{i_1, \ldots, i_p}_{j_1, \ldots , j_q} e_{i_1}? Is it a linear map going from V to V^* with the bases outlined in the question?

    Secondly, how will this prove that the set is linearly independent?
    A^{i_1, \ldots, i_p}_{j_1, \ldots , j_q} is a scalar, for each set of indices \{i_1, \ldots, i_p,j_1, \ldots , j_q\}. You should read this expression as \sum A^{i_1, \ldots, i_p}_{j_1, \ldots , j_q}(e_{i_1} \otimes \ldots \otimes e_{i_p} \otimes e^{j_1} \otimes \ldots \otimes e^{j_q}). The summation convention is being used, which means that you are meant to sum (from 1 to N) over each index that appears both as a superscript and as a subscript. So that expression simply denotes a linear combination of the given tensors. To prove that they are linearly independent, you put a linear combination of them equal to 0, and then you are aiming to show that each coefficient A^{i_1, \ldots, i_p}_{j_1, \ldots , j_q} is zero.
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