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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

[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.
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