I've been watching some videos on differential forms, and I found this brilliant chap. What confuses me is at about 6:12, he calls the other component redundant, and I'm confused what allows him to do so.
Consider two 1-forms, $\omega,\tau\in\Omega^1(\mathbb{R}^2)$, and then consider their wedge \begin{align} \epsilon & = \omega \wedge \tau\\ &= \omega_\mu \tau_\nu dx^\mu \wedge dx^\nu \\ & = \epsilon_{\mu\nu}dx^\mu \wedge dx^\nu \end{align} using implicit summation. Expanding this sum, $$ \epsilon = \epsilon_{11}dx^1 \wedge dx^1 + \epsilon_{12}dx^1 \wedge dx^2 + \epsilon_{21}dx^2 \wedge dx^1 + \epsilon_{22}dx^2 \wedge dx^2, $$ but since these ($dx^\mu$) are $1$-forms, $dx^\mu \wedge dx^\mu = 0$, and $dx^\mu \wedge dx^\nu = - dx^\nu \wedge dx^\mu $, so we may cut out two terms and reduce the others to \begin{align} \epsilon & = \epsilon_{12} dx^1 \wedge dx^2 + \epsilon_{21} dx^2 \wedge dx^1 \\ & = (\epsilon_{12} -\epsilon_{21})dx^1 \wedge dx^2. \end{align}
Now, I know that this method (implicit summation) double counts, and hence the full expression reads $\epsilon = \frac{1}{2}\epsilon_{\mu\nu}dx^\mu \wedge dx^\nu$ in the end, but why can I say that $\epsilon_{12}=-\epsilon_{21}$? I know that for an antisymmetric matrix this must be true, but what allows me to say this just from the mathematics of wedge products, and the properties of the coefficients?
Sorry if this is quite a rudimentary question, I'm just trying to get my head around it.
Any $\epsilon_{\mu\nu}$ can be split into a symmetric part and an antisymmetric part: $$ \epsilon_{\mu\nu} = \underbrace{\frac12(\epsilon_{\mu\nu}+\epsilon_{\nu\mu})}_{\text{symmetric}} + \underbrace{\frac12(\epsilon_{\mu\nu}-\epsilon_{\nu\mu})}_{\text{antisymmetric}}. $$
When multiplied with the antisymmetric $dx^\mu \wedge dx^\nu$ only the antisymmetric part of $\epsilon_{\mu\nu}$ contributes. The symmetric part cancels. Therefore it's convenient to let $\epsilon_{\mu\nu}$ be antisymmetric.