Symplectic geometry

Symplectic geometry studies smooth manifolds equipped with a closed nondegenerate differential two-form.

The real symplectic group consists of matrices $A$ satisfying $A^{T}JA=J$. It is an embedded matrix Lie group of dimension $2n^2+n$ in size $2n$.

The map $F:M_{2n}(\mathbb{R})\to {\mathrm{Skew}}_{2n}(\mathbb{R})$ given by $F(A)=A^{T}JA$ has $Sp(2n,\mathbb{R})=F^{-1}(J)$. At a symplectic $A$,is surjective, so the regular level set theorem makes the symplectic group an embedded submanifold.

At the identity, the symplectic Lie algebra consists of $X$ satisfying $X^{T}J+JX=0$; tangent spaces elsewhere are its left translates.

On ${\mathbb{R}}^{2n}$ let $J_0$ be the standard complex structure, let ${\omega }_0$ be the standard symplectic form, and let $g_0(u,v)={\omega }_0(u,J_{0}v)$ be the Euclidean inner product. A linear map preserving any two of $J_0$, ${\omega }_0$, and $g_0$ preserves the third. Equivalently,

A Lagrangian subspace of a $2n$-dimensional symplectic vector space $(V,\omega )$ is an $n$-dimensional subspace $L$ on which $\omega$ vanishes.

The Lagrangian Grassmannian is the space of all Lagrangian subspaces of a symplectic vector space. The unitary group acts transitively on $\mathrm{LGr}({\mathbb{R}}^{2n})$, and the stabilizer of ${\mathbb{R}}^n$ is the orthogonal group $O(n)$, giving

The Maslov index assigns an integer to a loop of Lagrangian subspaces. Under $\mathrm{LGr}({\mathbb{R}}^{2n})\cong U(n)/O(n)$, it is the degree of the map induced by $A\mapsto \det (A{)}^2$.

The Maslov map is well defined because every matrix in $O(n)$ has determinant $\pm 1$. The loopmaps to $e^{2\pi it}$ and therefore has Maslov index one.

For a rank-$2n$ symplectic vector bundle $E\to B$, the Lagrangian Grassmannian bundle has fiber $\mathrm{LGr}(E_b)$ over $b$. A symplectic trivialization of $E$ induces a trivialization $\mathcal{L}Gr(E)\cong B\times \mathrm{LGr}({\mathbb{R}}^{2n})$.

If $E\to B$ is an oriented real plane bundle, then $\mathcal{L}Gr(E)=\mathbb{P}(E)$ is an oriented circle bundle. Its transition rotations have twice the angle of those of $E$, so

A symplectic manifold is an even-dimensional smooth manifold $M$ equipped with a closed nondegenerate differential two-form $\omega$.

A symplectic form is a closed differential form of degree two whose value on every tangent space is a nondegenerate alternating bilinear form. In canonical position-momentum coordinates it is ${\sum }_{i}d{q}_i\wedge d{p}_i$.

A symplectic surface is a two-dimensional symplectic manifold. Every embedded curve in it is a Lagrangian submanifold.

A symplectomorphism $\varphi :(M,\omega )\to (M^{\prime },{\omega }^{\prime })$ is a diffeomorphism satisfying ${\varphi }^{\ast }{\omega }^{\prime }=\omega$.

A symplectic isotopy is a smooth path ${\varphi }_t$ of symplectomorphisms beginning at the identity.

For a symplectic isotopy generated by $X_t$, its flux isIt vanishes for a Hamiltonian isotopy.

For a smooth function $H_t:M\to \mathbb{R}$, the Hamiltonian vector field is defined by ${\iota }_{X_{H_t}}\omega =-d{H}_t$, with the sign depending on convention.

A Hamiltonian function is a smooth function whose differential determines a Hamiltonian vector field.

A Hamiltonian isotopy is the flow of a time-dependent Hamiltonian vector field. Its time-one map is a Hamiltonian diffeomorphism.

A Lagrangian submanifold of a $2n$-dimensional symplectic manifold is an $n$-dimensional submanifold $L$ satisfying $\omega {|}_L=0$. The symplectic form identifies its normal bundle with its cotangent bundle.

A Lagrangian embedding is a smooth embedding whose image is a Lagrangian submanifold.

The cotangent bundle has the canonical one-form ${\lambda }_{(q,p)}(v)=p(d\pi (v))$ and the canonical symplectic form $d\lambda$ up to a conventional sign. Its zero section is a Lagrangian submanifold.

A neighborhood of a compact Lagrangian submanifold $L\subset (M,\omega )$ is symplectomorphic, by a map restricting to the identity on $L$, to a neighborhood of the zero section in $(T^{\ast }L,d\lambda )$, with the sign chosen to match the convention for the canonical form.

A Lagrangian $L\subset M$ is displaceable by a class of isotopies when some isotopy ${\varphi }_t$ in that class satisfies ${\varphi }_1(L)\cap L=\varnothing$.

For a compact orientable Lagrangian $L$, the identification $NL\cong T^{\ast }L$ givesIf $\chi (L)\ne 0$, no smoothly isotopic copy of $L$ can be disjoint from it.

A simple closed curve dividing a symplectic two-sphere into two regions of equal area cannot be displaced by a symplectic isotopy. Any disjoint image would lie in one complementary disc, yet it would still have to bound a disc of half the total area.

Let $M$ be compact and let ${\omega }_t$ be a smooth family of symplectic forms with constant de Rham cohomology class. Choose ${\sigma }_t$ smoothly so that ${\dot{\omega }}_t=d{\sigma }_t$, solveand let ${\varphi }_t$ be the flow of $X_t$. Thenso ${\varphi }_t^{\ast }{\omega }_t={\omega }_0$.

Any two smooth degree-$d$ hypersurfaces in ${\mathbb{CP}}^n$, equipped with the restricted Fubini-Study form, are symplectomorphic. Join them through the connected complement of the discriminant in the parameter space, use the resulting smooth family to identify the fibers, and apply Moser's trick to the cohomologous restricted forms.

The degree-$d$ Fermat hypersurface in ${\mathbb{CP}}^n$ is

The Fermat hypersurface $z_0^d+\cdots +z_n^d=0$ is preserved by multiplying each coordinate by a $d$th root of unity. Projective scalar multiplication is trivial, so the effective diagonal group isThese transformations preserve the Fubini-Study form.

A submanifold $S\subset (M,\omega )$ is symplectic when $\omega {|}_{TS}$ is nondegenerate.

The symplectic normal bundle of $S$ is the symplectic orthogonal complement $(TS{)}^{\omega }\subset TM{|}_S$. It is a symplectic vector bundle and is naturally isomorphic to the ordinary normal bundle.

If a symplectomorphism between compact symplectic submanifolds lifts to an isomorphism of their symplectic normal bundles, then it extends to a symplectomorphism between neighborhoods. Thus the germ of a symplectic neighborhood is determined by the restricted form and the symplectic normal bundle.

A symplectic sum removes tubular neighborhoods of symplectic submanifolds whose normal bundles have opposite Euler classes and glues their boundaries by a fiber-reversing identification.

A symplectic sphere of self-intersection $-4$ can be rationally blown down by taking the symplectic sum with ${\mathbb{CP}}^2$ along the sphere and a smooth conic of self-intersection $+4$. Equivalently, its disk-bundle neighborhood is replaced by the rational ball ${\mathbb{CP}}^2\setminus \nu (Q)$ whose boundary is the lens space $L(4,1)$.