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 $\sigma, \tau$,

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, $(f^\sigma)^\tau=f^{\sigma\circ\tau}$.


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 $(f^{\tau})^{\sigma}(v_1,\ldots,v_k)=f^{\tau}(v_{\sigma(1)},\ldots,v_{\sigma(k)})$.

  • $\begingroup$ 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

$(f^\sigma)^\tau(v_1,\ldots,v_k)$ 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}

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.