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 $p$.


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)

  • 1
    $\begingroup$ Not every 2-form can be written as a wedge product of 1-forms. For example, if $\omega_1,\dots,\omega_4$ are pointwise linearly independent 1-forms, then $\Omega=\omega_1 \wedge \omega_2+\omega_3 \wedge \omega_4$ cannot be so written, and in fact $\Omega \wedge \Omega = 2\omega_1 \wedge \omega_2 \wedge \omega_3 \wedge \omega_4 \neq 0$. $\endgroup$ – Micah Feb 5 '13 at 19:29
  • $\begingroup$ Oh of course, duh. Thanks! $\endgroup$ – oxeimon Feb 5 '13 at 20:35

I will only address question 1: Counter-example: Consider the following 2 form in $R^4$, with coordinates $w,x,y,z$

$$b=dx\wedge dz+dy\wedge dw$$


\begin{align*} b\wedge b &= (dx\wedge dz+dy\wedge dw)\wedge (dx\wedge dz+dy\wedge dw) \\ &=dx\wedge dz\wedge dy\wedge dw+dy\wedge dw\wedge dx\wedge dz \\ &=2*dx\wedge dz\wedge dy\wedge dw \end{align*}

which is not 0


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