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Tensor Product

  1. Dec 9, 2007 #1
    Hi, I have a quick question about a derivation that has annoyed me all day. I am trying to prove, from the definition of [itex]\otimes[/itex] that:

    [itex]T(\theta,\tau)=T^{i\alpha}(e_i,f_\alpha)[/itex]

    Where [itex]\theta \in V^*[/itex] and [itex]\tau \in W^*[/itex]

    [itex]T: V^* \times W^* \mapsto \mathbb{R}[/itex]

    and where [itex]V[/itex] has basis [itex]e_i[/itex] and [itex]W[/itex] has basis [itex]f_\alpha[/itex]. And where [itex]T^{i\alpha}=T(e^i,f^\alpha)[/itex].

    My lecturer only gave us the definiton for [itex]\otimes[/itex] where the it operated between elements of the dual basis, namely;

    [itex]f\otimes g (v,w)=f(v) \cdot g(w)[/itex]

    [itex]\forall v \in V, w \in W[/itex] and [itex]\forall f \in V^*,g \in W^*[/itex].

    So back to the question; here is my attempt:

    We wish to derive the following expression for [itex]T \in V^{**} \otimes W^{**}[/itex]:

    [itex]T(\theta,\tau)=T^{i\alpha}(e_i,f_\alpha)[/itex]

    Where [itex]\theta \in V^*[/itex] and [itex]\tau \in W^*[/itex]

    [itex]T: V^* \times W^* \mapsto \mathbb{R}[/itex]

    and where [itex]V[/itex] has basis [itex]e_i[/itex] and [itex]W[/itex] has basis [itex]f_\alpha[/itex].

    Okay so:

    Step 1: [itex]V^{**} \otimes W^{**} \simeq V \otimes W[/itex] so that [itex]T \in V \otimes W[/itex].


    Step 2: [itex]T(\theta,\tau)=T(\theta_i e^i, \tau_\alpha f^\alpha)=\theta_i \tau_\alpha T(e^i,f^\alpha)[/itex]

    And using the obvious notation:

    [itex]T(\theta,\tau)=T^{i\alpha}\theta_i \tau_\alpha[/itex]

    I am not sure where to go from here as I am unsure of the nature of [itex]T(\theta,\tau)[/itex], I mean is it the same as [itex]T(\theta(v),\tau(w))=\theta \otimes \tau(v,w)[/itex].

    I apologise if this is beneath all of you but this has really been bugging me.
     
    Last edited: Dec 9, 2007
  2. jcsd
  3. Dec 11, 2007 #2
    Okay I think I am just after getting this. After thinking hard about the problem I think I might have resolved it. The thing I was trying to derive was wrong to begin with!

    I have the problem wittled down to two final technicalities.

    The first line has changed to:

    [itex]T(\theta,\tau)=T^{i\alpha}(e_i\otimes f_\alpha)(\theta, \tau)[/itex]

    Where [itex]\theta \in V^*[/itex] and [itex]\tau \in W^*[/itex]

    [itex]T: V^* \times W^* \mapsto \mathbb{R}[/itex]

    and where [itex]V[/itex] has basis [itex]e_i[/itex] and [itex]W[/itex] has basis [itex]f_\alpha[/itex]. And where [itex]T^{i\alpha}=T(e^i,f^\alpha)[/itex].

    For [itex]\otimes[/itex] where it operates between elements of [itex]V,W[/itex] we have

    1. [Is this definition of [itex]\otimes[/itex] correct?];

    [itex]T(\theta, \tau)=v \otimes w (\theta, \tau)=v(\theta)\cdot w(\tau) \equiv \theta(v) \cdot \tau(w)[/itex]

    2. [Are these expressions indeed equivalent?]

    [itex]\forall v \in V, w \in W[/itex] and [itex]\forall \theta \in V^*,\tau \in W^*[/itex].

    So assuming that all previous results hold:

    [itex]T(\theta, \tau)= \theta(v) \cdot \tau(w)[/itex]

    Now we can just use linearity of the elements [itex]V^*,W^*[/itex] to get:

    [itex]T(\theta, \tau)= v^i w^\alpha \theta(e_i) \cdot \tau(f_\alpha)[/itex]

    Now we look at the quantity:

    [itex]e_i \otimes f_\alpha(\theta, \tau)=e_i \otimes f_\alpha(\theta,\tau)= e_i(\theta) \cdot f_\alpha(\tau)=\theta(e_i)\cdot \tau(f_\alpha)[/itex]

    by previous assumed-correct relations (1,2); so that

    [itex]T(\theta, \tau)=v^i w^\alpha \theta(e_i) \cdot \tau(f_\alpha)= v^i w^\alpha e_i \otimes f_\alpha(\theta, \tau)[/itex]

    Again by (1,2)

    Further, let us define, as usual [itex]T^{i\alpha}=T(e^i,f^\alpha)=v(e^i)\cdot w(f^\alpha)[/itex]

    Which, if (1,2) again hold gives;

    [itex]T^{i\alpha}= e^i(v) \cdot f^\alpha(w)[/itex]

    So from linearity again:

    [itex]T^{i\alpha}= e^i(v^j e_j) \cdot f^\alpha(w^\beta f_\beta)[/itex]

    And following the usual routine:

    [itex]T^{i\alpha}= v^i w^\alpha[/itex]

    And consequently;

    [itex]T(\theta, \tau)= v^i w^\alpha \theta(e_i) \cdot\tau( f_\alpha)[/itex]

    becomes:

    [itex]T(\theta, \tau)= v^i w^\alpha e_i \otimes f_\alpha(\theta, \tau)[/itex]

    finally;

    [itex]T(\theta, \tau)= T^{i\alpha} e_i \otimes f_\alpha(\theta, \tau)[/itex]

    [tex]\Box[/tex]

    Everything hinges on those two questions.
     
    Last edited: Dec 11, 2007
  4. Dec 11, 2007 #3
    Hello. I'm going to be taking an exam on differential geometry on the Wednesday of next week. Let's see...

    This looks good. [itex]T^{i\alpha}[/itex] are only real numbers, so what you wrote in the first post didn't make sense, but this does.

    I don't understand what those v and w are. In general a bilinear mapping

    [tex]
    T:V^*\times W^*\to \mathbb{R}
    [/tex]

    cannot be written as

    [tex]
    T=v\otimes w
    [/tex]

    where

    [tex]
    v:V^*\to\mathbb{R}
    [/tex]

    [tex]
    w:V^*\to\mathbb{R}
    [/tex]

    are linear. That is a special case.

    There's not much to prove. It's like this:

    [tex]
    T = T^{i\alpha} (e_i\otimes f_{\alpha})\quad\implies\quad T(\theta,\tau) = T^{i\alpha} (e_i\otimes f_{\alpha})(\theta,\tau)
    [/tex]

    :smile:

    Only thing that I can see, that needs some kind of proof, is that the numbers [itex]T^{i\alpha}[/itex] exist, so that the T can be written like that.

    The definition of [itex]\otimes[/itex] (in this context) is

    [tex]
    (v\otimes w)(\theta,\tau) = v(\theta)\; w(\tau)
    [/tex]

    where there is an usual product of real numbers on the right. This was your equation in the middle, but the other equations on left and right seem stranger.
     
    Last edited: Dec 11, 2007
  5. Dec 11, 2007 #4
    Thank you, sir.
     
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