Exterior product of n copies of 2-form

Question

I have a problem with calculating exterior product of differential forms. Here is the problem: Let $\omega$ be a 2-form in $\mathbb{R}^{2n}$ given by $\omega=dx_{1}\wedge dx_{2}+dx_{3}\wedge dx_{4}+...+dx_{2n-1}\wedge dx_{2n}$. Calculate $\omega\wedge\omega\wedge...\omega$, where is $n$ $\omega's$.

Answer 1

The situation here is not difficult to understand, and the desired calculation not too difficult to perform, if one simply bears in mind and exploits a few basic properties of the exterior algebra of forms, several of which will be developed here; for rest, we rely on general knowledge.

First, some notation. We are working in $\Bbb R^{2n}$; thus for $1 \le j \le n$ we define the $2$-forms $\theta_j$ by

$\theta_j = dx_{2j - 1} \wedge dx_{2j}; \tag{1}$

then we see that the $\theta_j$ satisfy

$\theta_k \wedge \theta_k = 0, \tag{2}$

and

$\theta_j \wedge \theta_k = \theta_k \wedge \theta_j, \tag{3}$

for all $1 \le j, k \le n$. (2) holds since $\theta_k \wedge \theta_k$ is a wedge product containing two factors of each $dx_{2k - 1}$ and $dx_{2k}$, and $dx_j \wedge dx_j = 0$ for all $j$; (3) follows from the fact that forms of degree $2$ always commute; see this Wikipedia page for more information. For $1 \le m \le n$ set

$\omega_m = \sum_1^m \theta_i; \tag{4}$

then

$\omega = \omega_n = \sum_1^n \theta_i. \tag{5}$

We also define, again for $1 \le m \le n$,

$\Omega_m = \theta_1 \wedge \theta_2 \wedge . . . \wedge \theta_m. \tag{6}$

Using the above notation and properties, we compute first $\omega_2 \wedge \omega_2 = \omega_2^{(2)}$, where adopt the notation $\beta^{(k)}$ for the $k$-fold wedge product of any form $\beta$ with itself. We have

$\omega_2^{(2)} = (\theta_1 + \theta_2) \wedge (\theta_1 + \theta_2) = \theta_1^{(2)} + \theta_1 \wedge \theta_2 + \theta_2 \wedge \theta_1 + \theta_2^{(2)} = 2\theta_1 \wedge \theta_2 = 2\Omega_2, \tag{7}$

where we have used (2) and (3) in simplifying (7). Next we compute $\omega_3^{(3)}$, using the fact that

$\omega_{m + 1} = \omega_m + \theta_{m + 1}, \tag{8}$

for $1 \le m \le n- 1$. Thus,

$\omega_3^{(3)} = (\omega_2 + \theta_3)^{(3)} = (\omega_2 + \theta_3)^{(2)} \wedge (\omega_2 + \theta_3). \tag{9}$

Before proceding further, we pause to make note of the following useful fact: the algebraic maneuvers required to manipulate polynomial expressions in the $\theta_j$ are identical to those permitted in any commutative $\Bbb R$-algebra, or even any $R$-algebra over a commutative ring $R$, for that matter, by virtue of (3). Exploiting this information can somewhat smooth the performance many of our calculations. In any event, and bearing these observations in mind, returning to (9) we see

$(\omega_2 + \theta_3)^{(2)} \wedge (\omega_2 + \theta_3)$ $= (\omega_2^{(2)} + 2\omega_2 \wedge \theta_3 + \theta_3^{(2)}) \wedge (\omega_2 + \theta_3) = (\omega_2^{(2)} + 2\omega_2 \wedge \theta_3) \wedge (\omega_2 + \theta_3)$ $= (\omega_2^{(3)} + 2\omega_2^{(2)} \wedge \theta_3 + \omega_2^{(2)} \wedge \theta_3 + 2\omega_2 \wedge \theta_3^{(2)}) = 3\omega_2^{(2)} \wedge \theta_3. \tag{10}$

In establishing (10), we have used (3) more than once to eliminate terms containing $\theta_3^{(2)}$, and also

$\omega_2^{(3)} = 0, \tag{11}$

which follows from (7), viz.

$\omega_2^{(3)} = \omega_2^{(2)} \wedge \omega_2 = 2\Omega_2 \wedge \omega_2 = 0, \tag{12}$

which in its turn follows readily from (2), (5) and (6) since each term of $\Omega_2 \wedge \omega_2 = 0$ contains a multiple factor $\theta_i$ for $i = 1 \; \text{or} \; 2$. Combining (7), (9) and (10) we see that

$\omega_3^{(3)} = 6\Omega_2 \wedge \theta_3 = 6 \Omega_3, \tag{13}$

since for all $m$, $1 \le m \le n - 1$ we have

$\Omega_{m + 1} = \Omega_m \wedge \theta_{m + 1}, \tag{14}$

which follows readily from (6).

Inspection of (7) and (13) leads us to form the inductive hypothesis that

$\omega_m^{(m)} = m!\Omega_m \tag{15}$

holds for $1 \le m \le n$; we have already validated the base cases $m = 2, 3$ by direct calculation. So suppose (15) holds for some $m$, $1 \le m < n$. We have

$\omega_{m + 1}^{(m + 1)} = (\omega_m + \theta_{m + 1})^{m + 1} = (\omega_m^{(m +1)} + (m + 1)\omega_m^{(m)} \wedge \theta_{m + 1} + \beta \wedge \theta_{m + 1}^{(2)}) \tag{16}$

for some form $\beta$ of degree $2m - 2$. (16) follows from our previous observation that the ring of even dimensional forms, being commutative, permits the same algebraic manipulations as any commutative ring; (16) is a simple application of the binomial theorem, known to hold in such rings, to the present situation. The $2m -2$ form $\beta$ is simply the coefficient, as a polynomial in $\omega_m$ and $\theta_{m + 1}$, of $\theta_{m + 1}^{(2)}$ in such a binomial expansion. In any event, we have that

$\beta \wedge \theta_{m + 1}^{(2)} = 0 \tag{17}$

by virtue of (3). Furthermore,

$\omega_m^{(m + 1)} = \omega_m^{(m)} \wedge \omega_m = m!\Omega_m \wedge \omega_m \tag{18}$

by virtue of (15), and

$\Omega_m \wedge \omega_m = 0 \tag{18}$

by (2), since each term in its expression as a polynomial in the $\theta_j$ contains factors of the form $\theta_k \wedge \theta_k$. Thus, by (14) and (15) we are left with

$\omega_{m + 1}^{(m + 1)} = (m + 1)\omega_m^{(m)} \wedge \theta_{m + 1} = (m + 1)!\Omega_m \wedge \theta_{m + 1} = (m + 1)!\Omega_{m + 1}; \tag{19}$

taking the case $m = n - 1$ specializes (19) to the required result,

$\omega^{(n)} = \omega_n^{(n)} = n!\Omega_n = n!\Omega. \tag{20}$

QED.