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I am reading "Analysis on Manifolds" by James R. Munkres.

There is the following lemma in this book:

Lemma 27.3. Let $f$ be a $k$-tensor on $V$; let $\sigma, \tau \in S_k$.
(a) The transformation $f \to f^\sigma$ is a linear transformation of $\mathcal{L}^k(V)$ to $\mathcal{L}^k(V)$. It has the property that for all $\sigma, \tau$,
$$(f^\sigma)^\tau=f^{\tau\circ\sigma}.$$

But I "proved" $(f^\sigma)^\tau=f^{\sigma\circ\tau}$.
Please tell me my mistake in my proof?

My wrong "proof":
$f^\sigma(v_1, \cdots, v_k) := f(v_{\sigma(1)}, \cdots, v_{\sigma(k)}) = f(w_1, \cdots, w_k)$, where $w_i := v_{\sigma(i)}$.
$f^\tau(w_1, \cdots, w_k) := f(w_{\tau(1)}, \cdots, w_{\tau(k)}) = f(v_{\sigma(\tau(1))}, \cdots, v_{\sigma(\tau(k))}) = f^{\sigma\circ\tau}(v_1, \cdots, v_k).$
So, $(f^\sigma)^\tau=f^{\sigma\circ\tau}$.

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With $w_i=v_{\sigma(i)}$, let's write \begin{align} f^{\sigma}(v_1,\ldots,v_k)&=f(v_{\sigma(1)},\ldots,v_{\sigma(k)})=f(w_1,\ldots,w_k)\\ f^{\tau}(w_1,\ldots,w_k)&=f(w_{\tau(1)},\ldots,w_{\tau(k)})=f(v_{\sigma(\tau(1))},\ldots,v_{\sigma(\tau(k))})=f^{\sigma\circ\tau}(v_1,\ldots,v_k) \end{align} Your idea was to add $\tau$ as superscript at $f$ in the first line, as showed next: $$\color{red}{(f^{\sigma})^{\tau}(v_1,\ldots,v_k)=f^{\tau}(v_{\sigma(1)},\ldots,v_{\sigma(k)})}=f^{\tau}(w_1,\ldots,w_k)$$ and the red part is wrong! It must be $(f^{\tau})^{\sigma}(v_1,\ldots,v_k)=f^{\tau}(v_{\sigma(1)},\ldots,v_{\sigma(k)})$.

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  • $\begingroup$ Thank you very much, Fakemistake. $\endgroup$ – tchappy ha Mar 7 '20 at 3:25
  • $\begingroup$ It's better to think in the last step, to replace $f$ by $f^{\tau}$ in the first line. $\endgroup$ – Fakemistake Mar 13 '20 at 7:46
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$(f^\sigma)^\tau(v_1,\ldots,v_k)$ is $f^\sigma(v_{\tau(1)},\ldots,v_{\tau(k)})$, not $f^\tau(v_{\sigma(1)},\ldots,v_{\sigma(k)})$.

If we put $g=f^\sigma$ and $u_i=v_{\tau(i)}$, we should have \begin{aligned} (f^\sigma)^\tau(v_1,\ldots,v_k) &=g^\tau(v_1,\ldots,v_k)\\ &=g(v_{\tau(1)},\ldots,v_{\tau(k)})\\ &=\color{red}{f^\sigma(u_1,\ldots,u_k)}\\ &=f(u_{\sigma(1)},\ldots,u_{\sigma(k)})\\ &=f(v_{\tau(\sigma(1))},\ldots,v_{\tau(\sigma(k))})\\ &=f^{\tau\circ\sigma}(v_1,\ldots,v_k). \end{aligned}

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  • $\begingroup$ Thank you very much, user1551. Fakemistake answerd earlier than you, so I chose Fakemistake's answer. Thank you. $\endgroup$ – tchappy ha Mar 7 '20 at 3:26

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