## Symplectic and contact manifolds

Let $(M,\omega)$ be a symplectic manifold. It said to be exact if $\omega=d\lambda$ for some one-form $\lambda$ on $M$.

(1) If $\omega=d\lambda$ is exact, then there is a canonical isomorphism between the v.f. and 1-forms. In particular, there exists a v.f. $X$ such that $\lambda=i_X\omega$. Then we have $\lambda(X)=\omega(X,X)=0$, and $L_X\lambda=i_X d\lambda+d i_X\lambda=i_X\omega +0=\lambda$, and $L_X\omega=d i_X\omega=d\lambda=\omega$.

(2) Suppose there exists a vector field $X$ on $M$ such that its Lie-derivative $L_X\omega=\omega$ (notice the difference with $L_X\omega=0$). Then Cartan’s formula says that $\omega=i_X d\omega+ di_X\omega=d\lambda$, where $\lambda=i_X\omega$. So $\omega=d\lambda$ is exact, and $L_X\lambda=i_Xd\lambda+di_X\lambda=i_X\omega+0=\lambda$.

A hypersurface $E\subset M$ is of restricted contact type if $\lambda|_E$ is a contact form on $E$. For example, let $h:M\to \mathbb{R}$ be a smooth function, $X_h$ be its Hamiltonian vector field, $E$ be a regular level set of $h$. Then $E$ is of restricted contact type if $\lambda(X_h)\neq 0$ on $E$.

Given a time-periodic Hamiltonian $H_t(x)=H(x,t)$ on $M$, its Hamiltonian v.f. $X_t$ is said to be Reeb-like if the one-form $\Lambda=\lambda-H_t dt$ defines a contact form on the $(2d+1)$-dimensional manifold $M\times S^2$. In this case one can compute the kernel $(u,a)\in\ker(d\Lambda)$:

$d\Lambda((u,a),(v,b))=\omega(u,v)-(b\cdot dH_t(u)-a\cdot dH_t(v))=0$ for all $(v,b)$.

This is equivalent to $dH_t(u)=0$, $a\cdot dH_t=\omega(u,\cdot)=i_u\omega$.
So $(u,a)=a(X_t,1)$. Then $\Lambda$ being a contact form on $M\times S^1$ is equivalent to that
$\lambda(X_t)-H_t\neq 0$.

Let $g$ be a Riemannian metric on $\mathbb{T}^2$, $K_g(x)$ be the Gauss curvature. Then Gauss-Bonnet Formula gives that $\int K_g d\mu=2\pi\cdot \chi(\mathbb{T}^2)=0$. So if $K_g\le 0$, then $K_g\equiv 0$ and hence $g$ is a flat metric. Hopf generalized this argument and proved that if $(\mathbb{T}^2, g)$ has no conjugate point, then $g$ is flat.

The starting point is the geodesic flow on $M=T_1\mathbb{T}^2$ and the induced Ricatti equation: $\dot u+u^2+K(t)=0$ along a geodesic $\gamma(t)$. For each $(x,v)\in M$, let $u(t;x,v,R)$ be the solution with $u(-R)=\infty$ (the geodesic variation focuses backward at time $-R$). By the assumption of non-conjugate point, $u(t;x,v,R)$ is defined for all $t\ge 0$. Consider the limit $u(t;x,v)$ as $R\to\infty$. This function describes the geodesic curvature of the unstable horocycle at $(x,v)$, and satisfies the following cocycle condition: $u(t+s;x,v)=u(t,\phi_s(x,v))$. Note that the limit function $u$ is measurable, but may not be continuous.

Then we integrate $\dot u+u^2+K(t)=0$ over $(t,x,v)\in [0,1]\times M$, and note that $\iint \dot u =\int u(1,x,v)-u(0,x,v) =0$ since the geodesic flow preserves the measure; $\iint K=\int K=2\pi \chi(\mathbb{T}^2)=0$.
Therefore, $\iint u^2=0$, $u\equiv 0$ and hence $K\equiv 0$: the metric $g$ is flat.

Riemannian manifold $(M,g)$, symplectic manifold $(M,\omega)$, (almost) complex manifold $(M,J)$, holomorphic tangent space $T^{(1,0)}M$. A Hermitian metric $h$ on a complex vector bundle $E$ over a smooth manifold $M$ is a smoothly varying positive-definite Hermitian form $h_x$ on each fiber $E_x$. A (almost) Hermitian manifold is a (almost) complex manifold with a Hermitian metric $h$ on its holomorphic tangent space $T^{(1,0)}M$. Every (almost) complex manifold admits a Hermitian metric.

The real part of $h$ defines a Riemannian metric $g$ on $M$, while the (minus) imaginary part of $h$ defines a (nondegenerate) 2-form $\omega$ on $M$, the fundamental form. Any one of the three forms $h$, $g$, and $\omega$ uniquely determine the other two: $g(X,Y)=\omega(X,JY)$, and $h=g-i\omega$.

If the fundamental class $\omega$ of a Hermitian manifold $(M,J,h)$ is closed, then it is called a Kahler manifold. In this case the form $\omega$ is called a Kahler form. A Kahler form is a symplectic form, and so Kahler manifolds are naturally symplectic.

An almost Hermitian manifold with $\omega$ closed is naturally called an almost Kahler manifold. Any symplectic manifold admits a compatible almost complex structure making it into an almost Kahler manifold. In holomorphic local coordinates $(z^a)$, any Kahler form can be written as $\omega=i\frac{\partial^2K}{\partial z^a \partial \bar z^b}dz^a\wedge dz^b=i\partial \bar{\partial}K$, where $K$ is a real function, the so-called Kahler potential.

As a compact manifold, $S^6$ admits no symplectic structure since any closed 2-form must be exact ($H_2(M)=0$). Note that $S^6$ has a well-known almost complex structure $J_0$, which comes from the octonions, when the 6-sphere is viewed as the unit sphere in the 7-space of imaginary octonions. This, however, is not integrable (by the non-associativity of the octonions). An open question is whether $S^6$ admits a complex structure.

Kodaira and Thurston constructed a manifold $M$ that admits a symplectic form $\omega$ and a complex structure $J$, which fails to be Kahler. In particular, $\omega$ and $J$ are not compatible. The construction is quite simple. Consider the direct product $G=\mathcal{H}\times \mathbb{R}$, where $\mathcal{H}$ is the 3D Heisenberg group, $\Gamma=\mathbb{Z}^3\times \mathbb{Z}$ be the integer lattice in $G$, and $M=\Gamma\backslash G$.

Let $\langle X_1, X_2, X_3, X_4\rangle$ be the natural basis of the Lie algebra $\mathfrak{g}$, and $\langle x_1, x_2, x_3, x_4\rangle$ be the natural basis of its dual $\mathfrak{g}^\ast$. Then only non-zero bracket is $[X_1,X_2]=X_3$, and the only non-zero differential is $dx_3=x_1\wedge x_2$.

Let $\omega=x_1\wedge x_4+ x_2\wedge x_3$. Clearly the corresponding left-invariant 2-form $\omega$ on $M$ is nondegenerate and closed. Note that the more natural 2-form $\sigma=x_1\wedge x_2+ x_3\wedge x_4$ is not closed.

Let $J$ be the almost complex structure with $J(X_1)=X_2$ and $J(X_3)=X_4$. Then the tensor $N_J(X,Y)=[X,Y]+J[JX,Y]+J[X,JY]-[JX,JY]$ vanishes. For example:

$N_J(X_1,X_2)=[X_1,X_2]+J[X_2,X_2]+J[X_1,-X_1]-[X_2,-X_1]=0$;

$N_J(X_1,X_3)=[X_1,X_3]+J[X_2,X_3]+J[X_1,X_4]-[X_2,X_4]=0$.

Therefore, the corresponding left-invariant almost complex structure $J$ on $M$ is integrable, and $(M,J)$ is a complex manifold. One can check that $J$ is not compatible with $\omega$: $\omega(X_1,JX_1)=\omega(X_1,X_2)=0$ (although compatible with $\sigma$).

Moreover, the first homology group $H_1(M)=(\pi_1(M))^{\text{ab}}=\mathbb{Z}^3$ and then the first Betti number $\beta_1(M)=3$. The odd-degree Betti numbers of any Kahler manifold are ever. So $M$ is not Kahler, and there is no compatible symplectic–complex structure on $M$.