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A subspace in a symplectic vector space could be one of two extremes: either symplectic (meaning the form is nondegenerate there) or Lagrangian. Or it could be something between the two, meaning a degenerate nonzero form.

What makes Lagrangian subspaces particularly interesting? For example, we consider the Lagrangian Grassmannian. But why not the symplectic Grassmannian?

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    $\begingroup$ IIRC, if $V, W$ are symplectic vector spaces and $f : V \to W$ a linear map, then the graph of $f$, as a subspace of $V \times W$, is Lagrangian iff $f$ is a symplectomorphism. $\endgroup$
    – Qiaochu Yuan
    Mar 22, 2019 at 19:22

3 Answers 3

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In symplectic linear algebra this is perhaps not completely clear. However, if one passes to symplectic geometry then the Lagrangean submanifolds indeed play a dominant role. This is in some sense Weinstein's Lagrangean creed: Every manifold $M$ is a Lagrangean manifold when viewed as zero section inside its cotangent bundle.

But also beyond this observation Lagrangean submanifolds show up in e.g. Hamilton-Jacobi theory, in completely integrable systems, and in quantization theory where they can be thought of semiclassical limit of quantum states (to some extend). Finally, in semiclassical analysis they turn out to be related to supports of pseudo-differential and Fourier integral operators.

Surprisingly, symplectic submanifolds turn out to be not that important. Instead, coisotropic submanifolds have an important role when it comes to phase space reduction (coisotropic reduction). Also in Poisson geometry, coisotropic submaifolds are in some sense the closest one can get to Lagrangean ones, a notion which no longer makes sense.

I hope this gives some inspiration why Lagrangean submanifolds are useful.

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    $\begingroup$ The answer for submanifolds has already been addressed multiple times on MO, while this current question is only about vector spaces. $\endgroup$
    – Chris Gerig
    Mar 22, 2019 at 21:50
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    $\begingroup$ mathoverflow.net/questions/60201/… and mathoverflow.net/questions/181920/… $\endgroup$
    – Chris Gerig
    Mar 23, 2019 at 14:52
  • $\begingroup$ @Chris Gerig You are right. I was misguided by the tag on Lagrangean submanifold first. So please take all a look at the questions linked by Chris above. $\endgroup$
    – Stefan Waldmann
    Mar 24, 2019 at 0:47
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The symplecticGrassmanian, as you define it, is just the complement of a codimension 1 subspace of the usual Grassmanian. Hence its geometric properties are very close to the original Grassmanian. However, the Lagrangisn Grassmanian has high codimension, so it’s properties are more new. Also, it is compact, which is useful for many purposes (e.g, making it a flag manifold).

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    $\begingroup$ To a certain kind of person, I think this is the right, though certainly not only, answer. The basic structure theory of the symplectic group, in particular its parabolic subgroups, demands you give special privilege to isotropic, and therefore Lagrangian, sub-spaces. I wonder if this can somehow be turned into an explanation for the privileged role that Lagrangian sub-spaces, and sub-manifolds, play in symplectic geometry, complementing the physical reasons that they are so important. $\endgroup$
    – Andy Sanders
    May 5, 2019 at 15:02
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    $\begingroup$ @AndySanders My answer was trying to focus on why we don't study the symplectic Grassmanian. No matter how interesting Lagrangian subspaces are, we would still study symplectic subspaces to if they were useful for something. I think their genericity largely explains why we do not. $\endgroup$
    – Will Sawin
    May 5, 2019 at 15:11
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If $L$ is a Langrangian in a symplectic vector space $V$, then $V$ is isomorphic to $L\oplus L^*$ with the standard symplectic form. This is a linear analogue of Weinstein's neighborhood theorem.

Recall that, given a vector space $L$ with dual $L^*$, the vector space $L\oplus L^*$ can be endowed with a symplectic form via the formula

$$ \omega_L((v,\xi),(v',\xi'))=\xi(v')-\xi'(v). $$ Note that $L$ and $L^*$ are Lagrangian subspaces of $L\oplus L^*$. Now let $L$ be a Lagrangian inside a symplectic vector space $(V,\omega)$. Then $L$ admits a vector space complement $L'$ that is also Lagrangian. This defines a bijection $\varphi:L'\rightarrow L^*$ via the formula $\varphi(v)(w)=\omega(v,w)$. Then the map $\Phi:V=L\oplus L'\rightarrow L\oplus L^*$ with $\Phi(v,v')=v+\varphi(v')$ is a symplectomorphism between $(V,\omega)$ and $(L\oplus L^*,\omega_L)$.

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    $\begingroup$ Something here confuses me. The map $\Phi$ depends on $\varphi$, which in turn uses $\omega$ in its definition. If you choose two different symplectic forms $\omega_1$, $\omega_2$ on $V$, with a common Lagrangian space $L$, then this construction would give you two different $\Phi_1$, $\Phi_2$. $\endgroup$
    – Kim
    Mar 24, 2019 at 14:57

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