# Wedge product of forms

Let $\alpha = y^2 dx + dy \in \Omega^1(R^2)$, $\beta = xy dx \wedge dy \in \Omega^2(R^2)$. Is $\alpha \wedge \beta = 0$?

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## 1 Answer

Yes. Since $\dim \mathbb{R}^2=2$ and $\alpha\wedge\beta$ is a $3$-form, it must be $0$. In general, if $M$ is a $n$-dimensional manifold, then for every $k>n$, all the $k$-forms are by definition $0$.

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