Tensor Product

  • #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:

Answers and Replies

  • #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.
 
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  • #3
2,111
18
Hello. I'm going to be taking an exam on differential geometry on the Wednesday of next week. Let's see...

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].
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.

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]
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.
 
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  • #4
Thank you, sir.
 

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