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Multi-linear algebra Dual basis

  1. Jun 28, 2012 #1
    {(a_i)_j} is the dual basis to the basis {(e_i)_j}
    I want to show that
    ((a_i)_1) \wedge (a_i)_2 \wedge.... \wedge (a_i)_n ((e_i)_1,(e_i)_2,...,(e_i)_n) = 1

    this is exercise 4.1(a) from Spivak. So my approach was:

    \BigWedge_ L=1^k (a_i)_L ((e_i)_1,...,(e_i)_n) = k! Alt(\BigCross_L=1^k (a_i)_L)((e_i)_1,...,(e_i)_n)= k! Alt(T)((e_i)_1,...,(e_i)_n) = k!(1/k! Sum _ {permutations σ} sgn σ T ((e_i)_σ (1),...,(e_i)_σ (n))

    where T = \BigCross_L=1^k (a_i)_L

    So there is already a result on what T ((e_i)_1,...,(e_i)_n) is. 1 if all the sub-indices agree, and 0 otherwise. My question is..... is T ((e_i)_σ (1),...,(e_i)_σ (n)) any different?

    I'm assuming that in the one dimensional case we would say that T acts on one element in a linear fashion.... but I'm kinda confused by the idea of having several arguments.....

    Otherwise,....is there an easier approach to the solution?
     
  2. jcsd
  3. Jun 28, 2012 #2
    Okay.... I have since figured out the solution .... the real question then becomes why is the second equals sign true (below):

    ×_L=1^k (a_i)_L ((e_i)_σ(1),.... (e_i)_σ(k)) = ∏_L=1^k ((a_i)_L)(e_i)_σ(L) = 1 if and only if is identity and 0 otherwise.

    where × denotes multiple(indexed) tensor products. And if this second equals sign is true then can I have this view for all tensor products? Namely, is a tensor product of k-parts operating on k-arguments equal to the product of each "part" acting on its corresponding argument (with the same index)?
    Can I always hold that view of a tensor? Are there tensors where this is more obvious and others ... not so much?
     
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