## Proof of Formua for Interchanging Two Forms

Theorem
Let
$\omega^p = \sum_{i_1 <... i_p f_ ...f_ dx_ ...> \[\omega^q = \sum_{j_1 <... j_q g_ ...g_ dx_ ...> The exterior product of \[\omega^p$
and
$\omega^q$
i
$\omega^p \wedge \omega^q = \sum f_{{i_1} ...{i_p}} g_{{j_1} ...{j_q}} dx_{i_1} \wedge ... \wedge dx_{i_p} \wedge dx_{j_1} \wedge ... \wedge dx_{j_q}$
(1)
We can reverse the order of
$\omega^p , \: \omega^q$
to give
$\omega^p \wedge \omega^q =(-1)^{p+q} \omega^q \wedge \omega^p$

Proof
The summation in (1) includes the factors
$dx_{i_1} \wedge ... \wedge dx_{i_p} \wedge dx_{j_1} \wedge ... \wedge dx_{j_q}$

Interchanging consecutive
$dx_k$
changes the sign of the whole expression so moving
$dx_{j_1}$
past the
$p$
terms to the left introduces a factor
$(-1)^p$
and we get
$dx_{i_1} \wedge ... \wedge dx_{i_p} \wedge dx_{j_1} \wedge ... \wedge dx_{j_q}=(-1)^p dx_{j_1} \wedge dx_{i_1} \wedge ... \wedge dx_{i_p} \wedge dx_{j_2} \wedge ... \wedge dx_{j_q}$

Moving each of the
$dx_{j_k}$
past the
$p$

$dx{i_m}$
introduces a factor
$(-1)^p$
and since there are
$q$

$dx_{j_m}$
to move, we introduce a factor
$(-1)^{pq}$
when this is finished hence
$\omega^p \wedge \omega^q =(-1)^{p+q} \omega^q \wedge \omega^q$