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.

Advertisements
Post a comment or leave a trackback: Trackback URL.

Leave a Reply

Please log in using one of these methods to post your comment:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: