Given a vector space $V$, and a $2$-tensor $w$ in the second exterior power $\Lambda^2 V$. Assume that $w \wedge w=0$. Why is $w$ decomposable? Thanks for your help!
Why is an alternating $2$-form decomposable if and only if its self-wedge vanishes?
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linear-algebra
abstract-algebra
2 Answers
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There is a canonical form: there is a basis $e_1,\dots,e_n$ of $V$ and a $k$ such that $w=e_1\wedge e_2+\dots +e_{2k-1}\wedge e_{2k}$. Notice that if $k>1$ then $w\wedge w$ contains the term $2e_1\wedge e_2\wedge e_3\wedge e_4$, hence $w\wedge w\neq 0$. We can conclude that if $w\wedge w=0$ then the canonical form is $w=e_1\wedge e_2$.
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0The existence of the canonical form is proven by taking a representation $w=e_1\wedge e_2 + ... + e_{2k-1}\wedge e_{2k}$ and noticing that if e_l could be expressed as a L.C. of smaller index vectors, we could put that term into earlier ones, contradicting minimality. Right? – 2012-07-17
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0I don't think this is true. If it was then the dim of $\Lambda^2 V$ would be k-1 when it is not. It is n choose 2. – 2013-04-21
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0If you can point out my mistake in my above comment I will remove my downvote. – 2013-04-21
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0@LeoSpencer: I don't know why you think that it should imply that the dimension is $k-1$ (perhaps this confused you (?): the basis $e_i$ depends on $w$) – 2013-05-13
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A proof not requiring a basis and canonical form: Evaluate the interior product with any vector $V$ for which $i_Vw\ne0$: $0=i_V(w\wedge w)=(2i_Vw)\wedge w$. So $w=(i_Vw)\wedge u$, for some 1-form $u$.
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0Are you sure that $2$ isn't a $0$ ? – 2017-12-11
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0Yes. $w$ is a 2-form, so $i_v$ distributes over $w\wedge w$ with a plus sign. – 2017-12-12
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0Ah, right! I see what you're doing, and it's quite nice. – 2017-12-12