Exterior 2-form, 1-form, Hodge star operator.

Question

In $\mathbb{R}^{2n}$ with coordinates $x_1, x_2, \dots, x_{2n}$, consider an exterior 2-form$$\eta = \sum_{k=1}^n x_{2k-1} \wedge x_{2k}.$$Given a 1-form $\alpha = \sum_{i=1}^{2n} a_ix_i$, what is the 1-form$$\beta = \star\left(\alpha \wedge \underbrace{\eta \wedge \dots \wedge \eta}_{n-1}\right)?$$Many thanks in advance.

Answer 1

We have$$\eta^{n-1} = (n-1)! \sum_{j=1}^n x_1 \wedge \dots \wedge x_{2n} \text{ (}x_{2j - 1} \wedge x_{2j} \text{ is missing).}$$Then$$x_{2j - 1} \wedge \eta^{n-1} = (n-1)! \sum_{j=1}^n x_1 \wedge \dots \wedge x_{2n} \text{ (}x_{2j} \text{ is missing)}$$and$$x_{2j} \wedge \eta^{n-1} = (n-1)! \sum_{j=1}^n x_1 \wedge \dots \wedge x_{2n} \text{ (}x_{2j-1}\text{ is missing).}$$Hence, $\star(x_{2j-1} \wedge \eta^{n-1}) = (n-1)!x_{2j}$ and $\star(x_{2j} \wedge \eta^{n-1}) = -(n-1)!x_{2j-1}$.

Therefore,\begin{align}\beta &= \star(\alpha \wedge \eta^{n-1}) = \sum_{j=1}^n (a_{2j-1} \star (x_{2j-1} \wedge \eta^{n-1}) + a_{2j} \star (x_{2j} \wedge \eta^{n-1}))\\ &= (n-1)! \sum_{j=1}^n (-a_{2j} x_{2j-1} + a_{2j-1} x_{2j}).\end{align}Note that these formulas also hold for $n=1$. In this case, $\star(a_1x_1 + a_2x_2) = a_1x_1 - a_1x_2$.