Let $V$ be a finite dimensional vector space. Let $ \omega \in \Lambda^k(V^*) $ and $\eta \in \Lambda^l(V^*)$. Then
$\omega \wedge \eta = (-1)^{kl} \eta \wedge \omega$
I am trying to understand the reason for $ (-1) ^{kl} $ and i have understood upto the following
If $I$ and $J$ are increasing multi indices then ,
$ \epsilon^I \wedge \epsilon^J = sign \tau \epsilon^J \wedge \epsilon^I $
where $ \tau $ is the permutation that sends $IJ$ to $JI$
Then $ sign \tau= (-1)^{kl} $ is what I cannot understand.
I tried working out in the simple case of $k=2$ and $l= 3$
Then the number of transpositions required for (1,2,3,4,5) to (3,4,5,1,2) is only 4 and the result of each transposition is given below.
(4,2,3,1,5) (4,5,3,1,2) (3,5,4,1,2) (3,4,5,1,2)
There were only 4 transpositions required, while according to the text, it must require $kl=6$ transpositions. Where am i going wrong?
I don't think your text says that kl=6 is required. The important thing is NOT whether 6 is the lowest number of transpositions possible but if the number is even or odd. Your text is probably showing you a systematic way of rearranging $k$ terms $l$ times. Like this
$\{\bar{1},\bar{2},3,4,5\}\to$$\{\bar{1},3,\bar{2},4,5\}\to$$\{\bar{1},3,4,\bar{2},5\}\to$$\{\bar{1},3,4,5,\bar{2}\}$
Now we have rearranged our first term. We have used 3 transpositions to move $\bar{2}$. Let's continue with $\bar{1}$
$\to\{3,\bar{1},4,5,\bar{2}\}$$\to\{3,4,\bar{1},5,\bar{2}\}$$\to\{3,4,5,\bar{1},\bar{2}\}$
Not surprisingly this required another $l=3$ transpositions.
There are $k$ terms being transposed $l$ times, hence $kl=2\cdot3=6$ transpositions all in all.
If you start by moving $\bar{1}$ you will end up with $8$ transpositions instead.