Nontriviality of Godbillon-Vey class

On $G=\SL(2,\bR)$, with $TG\isom G\times\gerg$, ($\gerg$ - Lie slgebra of $G$ = traceless matrices) take the foliation given by the subbundle $E$ generated by the left invariant vector fields corresponding to

$$X= \left( \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array}\rig... ...\left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right),$$

with

$$[X, H]= \left( \begin{array}{cc} 0 & -1 \\ 0 & 0 \end{array... ...ft( \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array}\right)=-2X.$$

The third basis element is

$$Y= \left( \begin{array}{cc} 0 & 0 \\ 1 & 0 \end{array}\right),$$

with

$$[Y, H]=2Y,\;\; [X, Y]=H.$$

Take the dual basis $\{\zeta, \eta, \chi\}$ of $\gerg^*$ and extend them as left-invariant 1-forms. Then $\eta$ defines $\sF$ (i.e. $E=\ker\eta$). One has

$$d\chi = a\chi\land\zeta + b\chi\land\eta + c\zeta\land\eta,$$

$$b=d\chi(H, Y)=-\chi([H, Y])=2\chi(Y)=0$$

$$c=d\chi(X, Y)=-\chi([X, Y])=-\chi(H)=-1$$

$$a=d\chi(H, X)=\chi([X, H])=-2\chi(X)=0,$$

hence

$$d\chi=-\zeta\land\eta.$$

Similarly

$$d\zeta=-2\chi\land\zeta,$$

$$d\eta=2\chi\land\eta.$$

The last implies

$$\alpha=4\chi\land d\chi=-4\chi\land\zeta\land\eta.$$

The form $\alpha$ drops down to $M=\Gamma\setminus G$ for any $\Gamma$ cocompact giving a volume form, hence

$$[\alpha_{\Gamma}]=\mbox{ generator of }\HH^3(M; \bR).$$

More precisely, let $\Sigma_g$ be the Riemann surface of genus $g\geqslant 2$. Then its universal cover is the upper half plane

$$\bH=\SL(2,\bR)/\SO(2),$$

on which $\Gamma=\pi_1(\Sigma_g)$ acts by Mobius transformation

$$\Gamma\subset \PSL(2, \bR),\;\;z\mapsto\frac{az+b}{cz+d}.$$

Let $\tilde{\Gamma}$ be the double cover of $\Gamma$. Then $\tilde{\Gamma}$ is cocompact. Morover $M\isom S^1\Sigma_g$ (unit tangent bundle), hence

$$[\alpha_{\Gamma}]([M])=4\int_{S^1\Sigma_g}\zeta\land\eta\land\chi= 4\pi \int_{\Sigma_g}\zeta\land\eta=4\pi \Area(\Sigma_g)=$$

$$=-4\pi\int_{\Sigma_g}Kd\sigma=-8\pi^2(2-2g).$$