1
$\begingroup$

I was reading this question, and the first answer says:

There is a canonical form: there is a basis $e_1,\cdots,e_n$ of $V$ and a $k$ such that $w=e_1 \wedge e_2 + \cdots + e_{2k−1} \wedge e_{2k}$.

Why is this true? I wasn't able to come up with a proof. My knowledge of exterior algebra is the most basic. If I understand correctly, vectors $e_{i_1} \wedge e_{i_2}$ form the basis of $\bigwedge^2(V)$, but wouldn't that mean that every $w \in \bigwedge^2(V)$ could be written as $a_1(e_1 \wedge e_2) + a_2(e_1 \wedge e_3) + \cdots$, meaning that not only $e_k \wedge e_{k + 1}$ terms are present, but also $e_k \wedge e_{k + p}$? I'm very confused about this.

$\endgroup$
  • $\begingroup$ This is a standard result concerning symplectic forms. I don't have reference in English, but I think most advanced linear algebra books would cover this topic. $\endgroup$ – Cave Johnson Dec 11 '17 at 5:11
  • $\begingroup$ (5.3) in Keith Conrad, Bilinear forms says this in the language of matrices. $\endgroup$ – darij grinberg Dec 11 '17 at 5:16
  • $\begingroup$ @darijgrinberg thank you for the reference! I already know that for any alternating bilinear form, there is a symplectic basis of the vector space, but I struggle to see the relation between alternating bilinear forms and second exterior power (it's the dual space of the vector space of alternating bilinear forms, right?). I don't understand how to get the statement in question from symplectic basis... $\endgroup$ – wouldnotliketo Dec 11 '17 at 5:46
  • $\begingroup$ Writing this up is a pain in the ass. Basically, take a matrix that translates the standard basis into the symplectic basis. Either this matrix or its inverse will send $w$ to $e_1 \wedge e_2 + \cdots + e_{2k-1} \wedge e_{2k}$. $\endgroup$ – darij grinberg Dec 11 '17 at 5:56
  • $\begingroup$ @darijgrinberg thank you, but unfortunately I'm still unable to understand. Why does this matrix (or its inverse, which is it?) send $w$ to the representation in question? Would you maybe know the book where I could look this up if expanding this idea is too time-consuming for you at the moment? $\endgroup$ – wouldnotliketo Dec 11 '17 at 6:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.