# Topology.Linear transformation

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

Verify that $T^*(S^{\delta})= (T^* (S))^{\delta}, \delta \in S_k.$

Here is what I did, but unfortunately it is wrong. Please, help.

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 $(T^*S)^\sigma(x)$

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Notation.........?? – user38268 Dec 23 '12 at 12:19
What is $L^k(W)$? Is this the space of $k$-linear maps $W^k\to \mathbb{R}$? – froggie Dec 23 '12 at 13:39

I'm going to go ahead and assume that $L^k(W)$ is the space of $k$-linear maps $W^k\to \mathbb{R}$ in this answer. If this is incorrect, my apologies (but you should really define your notation in your question). If I understand correctly, this is what is needed to solve the problem.
1. The symmetric group $S_k$ acts on $L^k(W)$ in the following way: if $\sigma\in S_k$ and $S\in L^k(W)$, then $S^\sigma$ is $k$-linear map $S^\sigma(w_1,\ldots, w_k) = S(w_{\sigma^{-1}(1)}, \cdots, w_{\sigma^{-1}(k)})$.
2. The linear map $T\colon V\to W$ defines a pullback map $T^*\colon L^k(W)\to L^k(V)$ given by $(T^*S)(v_1,\ldots, v_k) = S(Tv_1,\ldots, Tv_k)$.
Using these definitions, we want to show $T^*(S^\sigma) = (T^*S)^\sigma$. We do this by evaluating on a tuple $(v_1,\ldots, v_k)\in V^k$:$$T^*(S^\sigma)(v_1,\ldots, v_k) = S^\sigma(Tv_1,\ldots, Tv_k) = S(Tv_{\sigma^{-1}(1)},\ldots, Tv_{\sigma^{-1}(k)}),$$ and $$(T^*S)^\sigma(v_1,\ldots, v_k) = (T^*S)(v_{\sigma^{-1}(1)},\ldots, v_{\sigma^{-1}(k)}) = S(Tv_{\sigma^{-1}(1)},\ldots, Tv_{\sigma^{-1}(k)}).$$ This proves $T^*(S^\sigma) = (T^*S)^\sigma$.
The problem with your answer is that you've confused $L^k(W)$ with $W^{\otimes k}$. While these spaces are very related, they are not the same! What is true is that $L^k(W)$ is canonically isomorphic to $(W^{\otimes k})^*$, the dual space of $W^{\otimes k}$, which is in turn canonically isomorphic to $(W^*)^{\otimes k}.$ You could then turn your solution into a correct proof by working with $T^*\colon W^*\to V^*$ and the $k$-tensor products $(V^*)^{\otimes k}$ and $(W^*)^{\otimes k}$ instead of with $T$, $V$, and $W$.