I'm trying to write the proof of the following proposition.
Proposition: if $f \in A_k(V)$ and $g \in A_l(V)$ then $f \wedge g = (-1)^{kl} g \wedge f$.
Part 1 of the proof
– Part 2 of the proof
$A_k(V)$ and $A_l(V)$ are the spaces of the alternating k-tensors and l-tensors, respectively, in a vector space V of finite dimension.
I understand the identidy $A(f\otimes g)=sgn(\tau)A(g\otimes f)$ and the necessity of changing the indexes that $\tau$ makes , but, i don't have the insight to calculate $sgn(\tau)$
$$ \tau = \left[
\begin{array}{c}
1&\cdots&l&l+1 &\cdots&l+k\\
k+1&\cdots&k+l&1 &\cdots&k
\end{array}
\right] .$$
The images are from the book Introduction to Manifolds from Loring W. Tu. Any recommendations are welcome.
Best Answer
From KReiser's explanations, here's my answer to this question:
We want to write $\tau$ as product of inversions in order to calculate $sgn(\tau)$. To do this, we have to turn $\tau$ into the identity permutation of $S_{k+l}$. So we must to move the elements $1,...,k$ across the elements $k+1,...,k+l$.
Starting by the element 1, we multiply $\tau$ on the left by the $l$ transpositions $((l+k) \ \ 1),...,((k+1) \ \ 1)$, respectively, to obtain
$$ ((k+1) \ \ 1)\cdots((k+l) \ \ 1) \ \tau = \left[ \begin{array}{c} 1&\cdots&l&l+1 &\cdots&l+k\\ 1&\cdots&k+l-1&k+l &\cdots&k \end{array} \right] .$$
Doing the same process to $2,...,k$, that is, moving $2,...,k$ across $k+1,...,k+l$ we have
$$ ((k+1) \ \ k)\cdots((k+l) \ \ k) \cdots ((k+1) \ \ 1)\cdots((k+l) \ \ 1) \ \tau = id .$$
Multiplying both sides of the equation above on the left by the corresponding transpositions we get
$$ \tau= \underbrace{ \underbrace{ ((k+l) \ \ 1) \cdots ((k+1) \ \ 1) }_{l\text{ times}} \cdots \underbrace{((k+l) \ \ k)\cdots((k+l) \ \ k)}_{l\text{ times}} }_{k \text{ times}}.$$
And therefore,
$$ sgn(\tau)= \underbrace{ (-1)^{l} \cdots (-1)^{l}}_{k\text{ times}} = (-1)^{kl} .$$
The notation is a bit overwhelmed, but I finally got the insight on how to calculate the permutation signal simply by switching positions.