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Andrius Kulikauskas

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Why symplectic manifolds are even dimensional

The elements of a symplectic group/manifold preserve a (symplectic) 2-form {$s(x,y)$} which is bilinear and antisymmetric: {$s(x,y)=-s(x,y)$}. When we write this out in terms of a matrix:

{$s(x,y) = {x^a}{s_{ab}}{y^b} = {x^T}Sy$}

then {$s_{ba} = -s_{ab}$}. In particular, {$s_{aa}=0$}. The matrix S is invertible iff {$det(S)\neq 0$}.

{$det(S)= \sum_{\sigma \in S_n} sgn(\sigma)a_{1\sigma(1)}\dots a_{n\sigma(n)} $}

In this sum, note that we get zero terms whenever {$i = \sigma(i)$}. For each remaining term, we have an involution which matches:

{$a_{1\sigma(1)}\dots a_{n\sigma(n)}$} and {$a_{\sigma(1)1}\dots a_{\sigma(n) n} = -a_{1\sigma(1)}\dots -a_{n\sigma(n)} = (-1)^n a_{1\sigma(1)}\dots a_{n\sigma(n)} $}

The terms add to zero if {$(-1)^n = -1 $}, which is the case when {$n$} is odd. Thus {$n$} is even.

Exercise4


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Puslapis paskutinį kartą pakeistas 2019 sausio 19 d., 18:20
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