Loring Tu's "An Introduction to Manifolds", Proposition 3.12

Question

As I tried to prove the following proposition by myself:

I got this:

$$(\tau(Af))(v_1,\dots,v_k)=(Af)(v_{\tau(1)},\dots,v_{\tau(k)})=\sum_\sigma (\operatorname {sgn} \sigma) f(v_{\sigma (\tau(v_1)),\dots,\sigma(\tau(v_k))})=\sum_\sigma (\operatorname {sgn} \sigma)(\sigma \tau) f(v_1,\dots,v_k),$$ which is not the same as the proof in the book claims ($\sigma \tau \ne \tau \sigma$). Is this a typo, or did I miss something?

Of course both mine and his partial results give the same answer, but I was wondering whether I (or Tu) applied definitions incorrectly. In my understanding, his first equality does not follow directly from the definition.

(Note that $\sigma f(v_1,\dots,v_k)=f(v_{\sigma(1)},\dots,v_{\sigma(k)})$ by definition.)

Answer 1

$\DeclareMathOperator{\sgn}{sgn}$

I think that you and the book are both correct. You have

$$\sum\limits_{\sigma}( \sgn \sigma)(\sigma \tau) f$$

The expression in the sum does not change if you replace every $\sigma$ in the sum with $\sigma \tau^{-1}$:

$$\sum\limits_{\sigma} \sgn(\sigma \tau^{-1})( \sigma) f$$

Now $\sgn \sigma \tau^{-1} = \sgn \sigma \sgn \tau^{-1} = \sgn \sigma \sgn \tau$, so you can factor out $\sgn \tau$:

$$\sgn \tau \sum\limits_{\sigma} (\sgn \sigma)\sigma f = \sgn \tau Af$$

Answer 2

$\DeclareMathOperator{\sgn}{sgn}$The confusion is in your second equality where you say $$ (Af)(v_{\tau(1)}, \dots, v_{\tau(k)}) = \sum_\sigma (\sgn \sigma)(\sigma f)(v_{\tau(1)}, \dots, v_{\tau(k)}) = \sum_\sigma (\sgn \sigma)f(v_{\sigma(\tau(1))}, \dots, v_{\sigma(\tau(k))}). $$

Actually, following the definitions gives $(\sigma f)(v_{\tau(1)}, \dots, v_{\tau(k)}) = f(v_{\tau(\sigma(1))}, \dots, v_{\tau(\sigma(k))})$. Note that Tu uses this same step in the proof of Lemma 3.11.