# Munkres wrote $(f^\sigma)^\tau=f^{\tau\circ\sigma}$ in “Analysis on Manifolds”. But I proved $(f^\sigma)^\tau=f^{\sigma\circ\tau}$.

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

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 $$(f^{\tau})^{\sigma}(v_1,\ldots,v_k)=f^{\tau}(v_{\sigma(1)},\ldots,v_{\sigma(k)})$$.
• It's better to think in the last step, to replace $f$ by $f^{\tau}$ in the first line. – 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}