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I'm trying to verify/show that:

$$T^*(F^\sigma) = (T^*F)^\sigma$$

I have that the alternating tensor, $A: L^k \to L^k$ is given by the fact that

$$AF = \sum_\sigma \mathop{\mathrm{sgn}} \sigma\, F^\sigma$$

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    What is $F^\sigma$? What is $T$? What have you tried?2012-03-20
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    I am a bit lost on how to go about it2012-03-20
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    F is a k-tensor on W (not necessarily alternating), and T is a linear transformation T: V--> W2012-03-20
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    Ok. No just compute: For $v_1, \ldots, v_k \in V$ we have $T^*(F^\sigma)(v_1, \ldots, v_k) = F^\sigma(Tv_1, \ldots, Tv_k) = \cdots$ ... can you see it?2012-03-20
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    it makes sense, but i can't say it polished2012-03-20
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    can i just say, by some linearity/interchanging2012-03-20
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    ... ${} = F(Tv_{\sigma(1)}, \ldots, Tv_{\sigma(k)}) = (T^*F)(v_{\sigma(1)}, \ldots, v_{\sigma(k)}) = (T^*F)^\sigma(v_1, \ldots, v_k)$. That's it (if I got your definitions right).2012-03-20
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    OK, but can you put a polished answer, so I can also reward you with points?2012-03-20
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    What's AB and can you also include the definitions of T* and (-)^sigma?2012-03-20
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    AB means "Allzeit bereit" - the german scout motto2012-03-20

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we have for every $v_1, \ldots, v_k \in V$ exploiting the definitions of $T^*$ and $(-)^\sigma$ given by \begin{align*} T^*F(v_1, \ldots, v_k) &= T(Fv_1, \ldots, Fv_k)\\ F^\sigma(w_1, \ldots, v_k) &= F(w_{\sigma(1)}, \ldots, w_{\sigma(k)}) \end{align*} that \begin{align*} (T^*F^\sigma)(v_1,\ldots, v_k) &= F^\sigma(Tv_1, \ldots, Tv_k)\\ &= F(Tv_{\sigma(1)}, \ldots, Tv_{\sigma(k)})\\ &= (T^*F)(v_{\sigma(1)}, \ldots, v_{\sigma(k)})\\ &= (T^*F)^\sigma(v_1,\ldots, v_k) \end{align*}

AB,