Symplectic Basics II : The Lagrangian Grassmanian

Before we proceed with Lagrangian stuff, I should really talk about a fundamental property of (finite dimensional, real!) symplectic vector spaces:

proposition 1: They’re always even dimensional!


Let (E,\sigma) be a symplectic vector space of dimension m.  I claim then that is even.  Choose some basis of E, so that we get a matrix representative A of \sigma, i.e., \sigma(x,y) = x^T A y for all x, y \in E.  Since \sigma is alternating, A is skew-symmetric, giving the relation A^T = -A.  Hence,

det(A) = det(A^T) = det(-A) = (-1)^m det(A).

So, if m is odd, we must have det(A) = 0.  But, since \sigma is non-degenerate, det(A) \neq 0, which forces m to be even in the above equation.  Done!

end proof.

A quick application of the “rank-nullity” theorem for a subspace W \subseteq E applied to \sigma implies \dim W + \dim W^\perp = m = 2n; hence, all Lagrangian subspaces of E are of dimension n = \frac{1}{2} \dim E.  Similarly, if \dim W = n and W \subseteq W^\perp, then W is Lagrangian.

Of course, from plain old linear algebra, we know that, given any Lagrangian subspace \lambda_0 of E, there exists an n-dim subspace \lambda_1 \subseteq E such that E = \lambda_0 \oplus \lambda_1.  The question now is:

proposition 2: Can the complementary subspace \lambda_1 be chosen such that \lambda_1 is also Lagrangian?


Given \lambda_0 Lagrangian, choose an isotropic space \rho \subseteq E such that \lambda_0 \cap \rho = \{0\} (\lambda_0 is a proper subspace, so we can always at least choose a line that satisfies this property, and lines are always isotropic). If \rho^\perp \neq \rho,  \rho^\perp is not a subset of \lambda_0 + \rho.  Indeed, if it were, then we’d have (by “symplectic duality” given by regarding \rho^\perp as the annihilator of \rho w.r.t. \sigma) \lambda_0 \cap \rho^\perp \subseteq \rho.  Since \rho^\perp \cap \lambda_0 \subseteq \rho \cap \lambda_0 = \{0\} (so \rho^\perp \cap \lambda_0 = \{0\}), we contradict the assumption that \rho is isotropic: i.e., we assumed \dim \rho^\perp > 0.  Thus, \rho + \lambda_0 does not contain \rho^\perp.  Choose then some x \in \rho^\perp \backslash (\lambda_0 + \rho).  Then, \rho + \langle x \rangle is an isotropic subspace (as \sigma is bilinear and alternating) satisfying (\rho + \langle x \rangle) \cap \lambda_0 = \{0\}.  Arguing by induction on \dim \rho, we get our Lagrangian space \lambda_1.  Perfect.

end proof.

 Why is this relevant?

Recall the space of n-dim subspaces of E, called the Grassmannian and denoted G(E,n).  It is a smooth manifold of dimension n^2.  My main goal is to introduce the Lagrangian Grassmannian \Lambda (E) of Lagrangian planes in E, a closed (smooth) submanifold of G(E,n) of dimension \frac{n(n+1)}{2}. Also, it’s compact! (implicitly, this is proposition 3)


Let \lambda \in \Lambda(E).  I claim that there is a bijection between the spaces \Lambda_\lambda(E) := \{ \mu \in \Lambda(E) | \lambda \cap \mu = \{0\} \} (which is open in the subspace topology of \Lambda(E) in G(E,n)) and the space of (real) quadratic forms on \mathbb{R}^n (a real vector space of dimension \frac{n(n+1)}{2}).

Set W = \lambda \oplus \lambda^* (where \lambda^* := \text{Hom}_\mathbb{R}(\lambda,\mathbb{R}) is the algebraic dual of \lambda), equipped with the standard symplectic form \omega((x,\xi);(x',\xi')) := \langle x', \xi \rangle - \langle x,\xi' \rangle.  With respect to \omega, the subspace \lambda \cong \lambda \oplus \{0 \} is Lagrangian (follows immediately from the definition of \omega and the assumption \lambda \in \Lambda(E)).  By proposition 4 from the last post, we know there exists a symplectic map \psi : (E,\sigma) \to (W,\omega) with \psi(\lambda) = \lambda \oplus \{0\}.  So, we might as well work inside W.  Let \mu \in \Lambda_\lambda(W).  Then, (it’s an easy exercise to show) that \mu is the “graph” of a (unique!) linear map A : \lambda^* \to \lambda, that is, we can write \mu = \{ (Ay^*,y^*) | y^* \in \lambda^*\} (this is actually very similar to how you construct the smooth atlas on the ordinary Grassmannian manifold; check it out!).

Since \mu \in \Lambda_\lambda(W) \subseteq \Lambda(W), \omega|_{\mu} = 0.  Using the fact that \mu is the graph of the matrix A (suppose we’ve chosen a basis), this tells us

\langle Ay_2^*,y_1^* \rangle = \langle y_2^*, Ay_1^* \rangle

i.e., A is a symmetric matrix.  With a basis fixed for E (giving a basis of W by taking the dual basis for \lambda^*), we know that the collection of n \times n real symmetric matrices is linearly isomorphic to the space of real quadratic forms on \mathbb{R}^n (cf. the wiki page on quadratic forms, or whatever linear algebra reference you hold dear), which has the desired dimension.  Since this map is linear, it’s (a fortiori) smooth.  Compactness follows trivially from the fact that G(E,n) is compact and \Lambda(E) is closed in G(E,n).  The only thing left to check is that the transition maps between different charts are smooth.  Screw that; left to the reader.

end proof.

In the up and coming posts, we generalize our “symplectic vector spaces” to get symplectic manifold: these are smooth (real, at least for us) manifolds equipped with a closed, non-degenerate 2-form that gives each tangent space the structure of a symplectic vector space.  Neat, right?  Will there be some sort of “standard” symplectic manifold, like we saw last post (cf. example 1)?

(The answer is yes).



Published by brianhepler

I'm a third-year math postdoc at the University of Wisconsin-Madison, where I work as a member of the geometry and topology research group. Generally speaking, I think math is pretty neat; and, if you give me the chance, I'll talk your ear off. Especially the more abstract stuff. It's really hard to communicate that love with the general population, but I'm going to do my best to show you a world of pure imagination.

2 thoughts on “Symplectic Basics II : The Lagrangian Grassmanian

  1. Did you use the fact that your vector space was real? Seems like even dimension should hold over any field (though I get a little shaky about alternating forms over fields of char = 2);

    1. Explicitly, no, I didn’t (other than avoiding char =2 nonsense). You CAN do this stuff for complex vector spaces, grabbing material from Hermitian forms and almost complex structures. I didn’t read too much into those situations, so I didn’t want to talk about them here without knowing all the details!

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