# Verify the expression.

• Dec 19th 2012, 06:13 AM
vercammen
Verify the expression.
$\displaystyle T:V \rightarrow W$ is a linear transformation and $\displaystyle S \in L^k (W).$

Is it true that $\displaystyle T^*(S^{\delta})= (T^* (S))^{\delta}, \delta \in S_k$?
• Dec 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?
• Dec 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.
• Dec 20th 2012, 11:46 AM
chiro
Re: Verify the expression.
So is delta an Einstein summation?
• Dec 20th 2012, 01:16 PM
vercammen
Re: Verify the expression.
just a permutation, I guess
• Dec 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,
• Dec 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...
• Dec 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 $\displaystyle T:V^{\otimes k}\to W^{\otimes k}$ which on pure tensors is $\displaystyle T(v_1\otimes v_2\otimes \ldots \otimes v_k)=T(u_1)\otimes T(u_2)\otimes \ldots T(u_k)$

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

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

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

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

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

Now let's get back to the problem.

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

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