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Mathematical products

  1. Sep 5, 2014 #1
    why the metric tensor is the dot product of two basis (co-variants or contra-variants)
    if it's a tensor then the metric tensor should be written as the tensor product of two basis (basis of a vector) >>> why ???!!!
     
  2. jcsd
  3. Sep 5, 2014 #2

    HallsofIvy

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    ?? The metric tensor is NOT the dot product of any vectors. Where did you get the idea that it was?

    In your second sentence you say it is the tensor product of two vectors and ask "why"?

    Do you know what a "dot product" and a "tensor product" are?
     
  4. Sep 5, 2014 #3
  5. Sep 5, 2014 #4
  6. Sep 5, 2014 #5
    gij= bi.bj
     
  7. Sep 5, 2014 #6

    Fredrik

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    The metric tensor is a bilinear map ##g:V\times V\to\mathbb R##. The definition requires it to satisfy a few more conditions. Those other conditions are almost the same as the ones you see in the definition of "inner product", so you could say that g is "almost" an inner product.

    The bilinearity allows you to write
    $$g(u,v)=g(u^i e_i, v^j e_j) = u^i v^j g(e_i,e_j) = u^i v^j g_{ij},$$ where I have defined ##g_{ij}=g(e_i,e_j)##. These numbers are called the components of g, with respect to the ordered basis ##(e_i)_{i=1}^n##.

    The dual ordered basis to ##(e_i)_{i=1}^n## consists of the ##e^i\in V^*## such that ##e^i(e_j)=\delta^i_j##. Note that
    $$e^i(u)=e^i(u^je_j) =u^je^i(e_j)=u^j\delta^i_j =u^i.$$ This implies that we have
    $$g(u,v)=u^i v^j g_{ij} =e^i(u) e^j(v) g_{ij} = g_{ij} (e^i\otimes e^j)(u,v).$$ This implies that ##g=g_{ij}e^i\otimes e^j##.
     
  8. Sep 7, 2014 #7
    thank you very very much Fredrik ..
     
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