Verify the expression.

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December 19th 2012, 06:13 AM
vercammen

Verify the expression.

$T:V \rightarrow W$ is a linear transformation and $S \in L^k (W).$

Is it true that $T^*(S^{\delta})= (T^* (S))^{\delta}, \delta \in S_k$ ?
December 20th 2012, 12:14 AM
chiro

Re: Verify the expression.

Hey vercammen.

This might be a dumb question, but is W just a matrix? Also what does raising W to the delta do with regards to the map?
December 20th 2012, 03:46 AM
vercammen

Re: Verify the expression.

As far as I understood U,W are vector fields, S in a k-tensor on W (can be written using basis); S = sum of all possible k-tuples.
December 20th 2012, 11:46 AM
chiro

Re: Verify the expression.

So is delta an Einstein summation?
December 20th 2012, 01:16 PM
vercammen

Re: Verify the expression.

just a permutation, I guess
December 20th 2012, 01:25 PM
chiro

Re: Verify the expression.

The reason I ask is that if it is just a summation, then it should hold (the identity that is).

The reason has to do with distributivity of matrix multiplication,
December 20th 2012, 01:35 PM
vercammen

Re: Verify the expression.

I asked a professor, he told me it's just the matter of using the definitions...
December 22nd 2012, 02:35 PM
vercammen

Re: Verify the expression.

Here is what I did, but unfortunately it was graded as incorrect.

On the k-tensor powers the induced map is $T:V^{\otimes k}\to W^{\otimes k}$ which on pure tensors is $T(v_1\otimes v_2\otimes \ldots \otimes v_k)=T(u_1)\otimes T(u_2)\otimes \ldots T(u_k)$

If $\sigma$ is a permutation, then $(u_1\otimes \ldots u_k)^\sigma=u_{\sigma^{-1}(1)}\otimes \ldots \otimes u_{\otimes k}$ and thus we can immediately verify the identity

$T(u_1\otimes \ldots u_k)^\sigma=T((u_1\otimes \ldots u_k)^\sigma)$

because both sides will equal $T(u_{\sigma^{-1}(1)})\otimes \ldots T(u_{\sigma^{-1}(k)})$

Because pure tensors span the k-tensor power space, we conclude that

$T(v^\sigma)=T(v)^\sigma$ (1)

Now let's get back to the problem.

By definition, $T^*(S^\sigma)(x)=S^\sigma(T(x))$ which is $S(T(x))^\sigma$

Now by two consecutive applications of (1), $S(T(x))^\sigma=S(T(x)^\sigma)=S(T(x^\sigma))$ and that is, by definition, $(T^*S)(x^\sigma)$ - which again by definition equals $(T^*S)^\sigma(x)$