# Wedge product with a non-degenerate form

Let $\alpha$ be a non-degenerate form in $\Lambda^k(V)$ for some vector space $V$, $\dim V = n$. (Here non-degenerate means that if $x\in V$ is nonzero, then $(y_1 , ... , y_{k-1}) \mapsto \alpha(x , y_1 , ... , y_{k-1})$ is a nonzero form). Is it true that if $\beta$ is a nonzero $\ell$-form for $\ell + k\leq n$, then $\alpha \wedge\beta \not=0$? Is this true under stricter conditions on $\beta$ (e.g. $\beta$ also non-degenerate)?

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## 1 Answer

Let us take $n=4$ and $$\alpha:=e^1\wedge e^2+e^3\wedge e^4\text{ and }\beta:=e^1\wedge e^3+e^2\wedge e^4.$$ Both $\alpha$, $\beta$ are non-degenerate forms but $$\alpha\wedge\beta=0.$$

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What you are asking is essentially related to the notion of exterior annihilators and divisibility. You may refer to the book "The Pullback Equation for Differential Forms" by Csato-Dacorogna-Kneuss for more information in this direction. –  Saugata Nov 18 '12 at 15:18