# Tensor Product

Hi, I have a quick question about a derivation that has annoyed me all day. I am trying to prove, from the definition of $\otimes$ that:

$T(\theta,\tau)=T^{i\alpha}(e_i,f_\alpha)$

Where $\theta \in V^*$ and $\tau \in W^*$

$T: V^* \times W^* \mapsto \mathbb{R}$

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

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

$f\otimes g (v,w)=f(v) \cdot g(w)$

$\forall v \in V, w \in W$ and $\forall f \in V^*,g \in W^*$.

So back to the question; here is my attempt:

We wish to derive the following expression for $T \in V^{**} \otimes W^{**}$:

$T(\theta,\tau)=T^{i\alpha}(e_i,f_\alpha)$

Where $\theta \in V^*$ and $\tau \in W^*$

$T: V^* \times W^* \mapsto \mathbb{R}$

and where $V$ has basis $e_i$ and $W$ has basis $f_\alpha$.

Okay so:

Step 1: $V^{**} \otimes W^{**} \simeq V \otimes W$ so that $T \in V \otimes W$.

Step 2: $T(\theta,\tau)=T(\theta_i e^i, \tau_\alpha f^\alpha)=\theta_i \tau_\alpha T(e^i,f^\alpha)$

And using the obvious notation:

$T(\theta,\tau)=T^{i\alpha}\theta_i \tau_\alpha$

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

I apologise if this is beneath all of you but this has really been bugging me.

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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:

$T(\theta,\tau)=T^{i\alpha}(e_i\otimes f_\alpha)(\theta, \tau)$

Where $\theta \in V^*$ and $\tau \in W^*$

$T: V^* \times W^* \mapsto \mathbb{R}$

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

For $\otimes$ where it operates between elements of $V,W$ we have

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

$T(\theta, \tau)=v \otimes w (\theta, \tau)=v(\theta)\cdot w(\tau) \equiv \theta(v) \cdot \tau(w)$

2. [Are these expressions indeed equivalent?]

$\forall v \in V, w \in W$ and $\forall \theta \in V^*,\tau \in W^*$.

So assuming that all previous results hold:

$T(\theta, \tau)= \theta(v) \cdot \tau(w)$

Now we can just use linearity of the elements $V^*,W^*$ to get:

$T(\theta, \tau)= v^i w^\alpha \theta(e_i) \cdot \tau(f_\alpha)$

Now we look at the quantity:

$e_i \otimes f_\alpha(\theta, \tau)=e_i \otimes f_\alpha(\theta,\tau)= e_i(\theta) \cdot f_\alpha(\tau)=\theta(e_i)\cdot \tau(f_\alpha)$

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

$T(\theta, \tau)=v^i w^\alpha \theta(e_i) \cdot \tau(f_\alpha)= v^i w^\alpha e_i \otimes f_\alpha(\theta, \tau)$

Again by (1,2)

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

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

$T^{i\alpha}= e^i(v) \cdot f^\alpha(w)$

So from linearity again:

$T^{i\alpha}= e^i(v^j e_j) \cdot f^\alpha(w^\beta f_\beta)$

And following the usual routine:

$T^{i\alpha}= v^i w^\alpha$

And consequently;

$T(\theta, \tau)= v^i w^\alpha \theta(e_i) \cdot\tau( f_\alpha)$

becomes:

$T(\theta, \tau)= v^i w^\alpha e_i \otimes f_\alpha(\theta, \tau)$

finally;

$T(\theta, \tau)= T^{i\alpha} e_i \otimes f_\alpha(\theta, \tau)$

$$\Box$$

Everything hinges on those two questions.

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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:

$T(\theta,\tau)=T^{i\alpha}(e_i\otimes f_\alpha)(\theta, \tau)$

Where $\theta \in V^*$ and $\tau \in W^*$

$T: V^* \times W^* \mapsto \mathbb{R}$

and where $V$ has basis $e_i$ and $W$ has basis $f_\alpha$. And where $T^{i\alpha}=T(e^i,f^\alpha)$.
This looks good. $T^{i\alpha}$ are only real numbers, so what you wrote in the first post didn't make sense, but this does.

For $\otimes$ where it operates between elements of $V,W$ we have

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

$T(\theta, \tau)=v \otimes w (\theta, \tau)=v(\theta)\cdot w(\tau) \equiv \theta(v) \cdot \tau(w)$
I don't understand what those v and w are. In general a bilinear mapping

$$T:V^*\times W^*\to \mathbb{R}$$

cannot be written as

$$T=v\otimes w$$

where

$$v:V^*\to\mathbb{R}$$

$$w:V^*\to\mathbb{R}$$

are linear. That is a special case.

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

$$T = T^{i\alpha} (e_i\otimes f_{\alpha})\quad\implies\quad T(\theta,\tau) = T^{i\alpha} (e_i\otimes f_{\alpha})(\theta,\tau)$$ Only thing that I can see, that needs some kind of proof, is that the numbers $T^{i\alpha}$ exist, so that the T can be written like that.

The definition of $\otimes$ (in this context) is

$$(v\otimes w)(\theta,\tau) = v(\theta)\; w(\tau)$$

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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Thank you, sir.