QUESTION 1:
So I know that if $\omega$ is an alternating $p$-form for odd $p$ on some vector space $V$, then $\omega\wedge\omega = 0$.
But...isn't the same true for any $p$? Ie, take for example $p = 2$, then any 2-form $\omega$ can be written as a wedge of 1-forms, ie, $\omega = \omega_1\wedge\omega_2$, then $$\omega\wedge\omega = \omega_1\wedge\omega_2\wedge \omega_1\wedge\omega_2 = -\omega_1\wedge\omega_1\wedge\omega_2\wedge\omega_2$$ ...which should be $0\wedge 0$, ie 0?
Alternatively, in general, isn't $(\omega_1\wedge\cdots\wedge\omega_p)(v_1,\ldots,v_p)$ defined as the determinant of the matrix $(\omega_i(v_j))_{i,j}$? Thus, if $\omega_n = \omega_m$ for some $n,m$, then the functional should be 0, since the matrix will always have linearly dependent rows, right?
For some reason every resource I've seen only states this result for odd $p$.
QUESTION 2:
On a compact genus $g$ Riemann surface $X$, there's a theorem that says that the space of holomorphic differential 1-forms has dimension $g$. (Does anyone have a good reference for a proof of this?) Let $\Omega^1(X)$ be this space, then it's stated in Diamond/Shurman's book "A First Course on Modular Forms", that the dual space $\Omega^1(X)^*$ is generated as an $\mathbb{R}$-vector space by the operators of integration along one of the 2g loops that give a basis for the 1st homology.
My question is - if you take any path $\gamma$ on $X$, why is the operator of integration over $\gamma$ a linear combination of integration over loops? (good references would be appreciated too)
