Let $V $ be a vector space. Let $\xi$ be a non zero $1$ form. I want to show that if some $k$ form $\omega$ satisfies $\xi \wedge \omega=0$ then it is of the type $\omega=\xi \wedge \eta$. I have been able to prove this in case $\xi$ is one of the $dx_i$ but I am unable to prove it when it is a sum of these. Any hints will be appreciated. Thanks.
Proving that a form is of the type $\xi \wedge \eta$
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$\begingroup$
vector-spaces
differential-forms
1 Answers
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If $\xi$ is a non-zero one form, you can complete $\{\xi\}$ into a basis of $\Lambda^1(V)$ and write $\omega$ in this basis, which brings you back to the case $\xi = dx_i$.
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0thank you I just realised how silly the question is – 2017-01-15
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2Don't worry, often we already solved the problem but we just don't realize we did. It happens often to me as well ... – 2017-01-15