Deriving T(\theta,\tau) from Definition of $\otimes$

[cathalcummins]

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.

 

[cathalcummins]

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.

 

[jostpuur]

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

<br /> T:V^*\times W^*\to \mathbb{R}<br />

cannot be written as

<br /> T=v\otimes w<br />

where

<br /> v:V^*\to\mathbb{R}<br />

<br /> w:V^*\to\mathbb{R}<br />

are linear. That is a special case.

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

<br /> 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)<br />



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

<br /> (v\otimes w)(\theta,\tau) = v(\theta)\; w(\tau)<br />

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.

 

[cathalcummins]

Thank you, sir.