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Lagrangian subspaces of symplectic vector spaces

  1. Jun 4, 2012 #1
    1. The problem statement, all variables and given/known data
    If [itex] (V,\omega) [/itex] is a symplectic vector space and Y is a linear subspace with [itex] \dim Y = \frac12 \dim V [/itex] show that Y is Lagrangian; that is, show that [itex] Y = Y^\omega [/itex] where [itex]Y^\omega[/itex] is the symplectic complement.

    3. The attempt at a solution

    This is driving me crazy since I don't think it should be that hard. If we try to go directly, the fact that [itex] \dim Y = \frac12 \dim V [/itex] implies that [itex] \dim Y^\omega = \frac12 \dim V [/itex]. At this point, it is actually sufficient to show that Y is either isotropic or co-isotropic, since dimensional arguments will give equality. However, I cannot see why this is the case.

    On the other hand, if we assume that [itex] Y \neq Y^\omega [/itex] then without loss of generality, we may assume there exists [itex] w \in Y \setminus Y^\omega [/itex] (otherwise, relabel [itex] Y [/itex] and [itex] Y^\omega [/itex]). The space [itex] W = \operatorname{span}(w) [/itex] is isotropic so [itex] W \subseteq W^\omega [/itex] and since [itex] W \subseteq Y [/itex] then [itex] Y^\omega \subseteq W^\omega [/itex]. I would like to put this together somehow, but I'm having trouble doing it.
     
  2. jcsd
  3. Jun 4, 2012 #2
    Nevermind. I've concluded that the question as phrased was unsolvable. Indeed, consider [itex] \mathbb R^4 [/itex] with the standard symplectic basis [itex] \{x_i,y_i\} [/itex]. The subspace generated by x1 and y1 is two-dimensional, but is in fact symplectic rather than Lagrangian. Hence the statement cannot be true in general.
     
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