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Definition of tensor product

  1. May 27, 2015 #1
    I'm relatively new to differential geometry and would like to check that this is the correct definition for the tensor product of (for simplicity) two one-forms [itex]\alpha,\;\beta\;\;\in V^{\ast} [/itex]: [tex](\alpha\otimes\beta)(\mathbf{v},\mathbf{w})=\alpha (\mathbf{v})\beta (\mathbf{w})[/tex] where [itex]\alpha\otimes\beta\;\;\in V^{\ast}\otimes V^{\ast}[/itex] and [itex]\mathbf{v},\;\mathbf{w}\;\;\in V[/itex].
    Given this, is it correct to write, [tex](dx^{\mu}\otimes dx^{\nu})(\mathbf{v},\mathbf{w})=dx^{\mu}(\mathbf{v})dx^{\nu}(\mathbf{w})=V^{\mu}W^{\nu}[/tex]
    where we have expressed [itex]\mathbf{v}=V^{\mu}\partial_{\mu}[/itex] and [itex]\mathbf{w}=W^{\nu}\partial_{\nu}[/itex] in terms of a coordinate basis [itex]\lbrace\partial_{\mu}\rbrace[/itex] for [itex]V[/itex], and [itex]\lbrace dx^{\mu}\otimes dx^{\nu}\rbrace[/itex] is a coordinate basis for [itex]V^{\ast}\otimes V^{\ast}[/itex] (with [itex]\lbrace dx^{\mu}\rbrace[/itex] a basis for [itex]V^{\ast}[/itex]). As such, if we express [itex]\alpha[/itex] and [itex]\beta[/itex] in terms of the coordinate basis [itex]\lbrace dx^{\mu}\rbrace[/itex] as [itex]\alpha = \alpha_{\mu}dx^{\mu}[/itex] and [itex]\beta = \beta_{\nu}dx^{\nu}[/itex], respectively, we have [tex](\alpha\otimes\beta)(\mathbf{v},\mathbf{w})=\alpha (\mathbf{v})\beta (\mathbf{w})=\alpha_{\mu}\beta_{\nu}dx^{\mu}\otimes dx^{\nu}(\mathbf{v},\mathbf{w})=\alpha_{\mu}\beta_{\nu}V^{\mu}W^{\nu}[/tex].

    Would this be correct at all?
     
  2. jcsd
  3. May 27, 2015 #2

    Orodruin

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    Yes.
     
  4. May 27, 2015 #3
    Excellent, thanks.
     
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