Let $\mathfrak{g}=< X_{f_1},X_{f_2},...,X_{f_n}> $ be a lie algebra of contact symmetries , $\mathfrak{g}\subset Sym(E_\omega )$ , such that the orbit spact $ B_\mathfrak{g}=E_\mathfrak{g}/ \mathfrak{g}$ is smooth Then can we say $(B_\mathfrak{g}, C_\mathfrak{g}, \omega_\mathfrak{g})$ is contact manifold? where if $\pi :E_\mathfrak{g}\rightarrow B_\mathfrak{g}$ be the natural projection and $C_\mathfrak{g}([a])=\pi_*(C_\mathfrak{a}\cap T_a E_\mathfrak{g})$ and $[a]$ is orbit of $a\in E_\mathfrak{g}$ and $\omega_{\mathfrak{g}}=i_{X_{f_1}}\circ i_{X_{f_2}}\circ ...i_{X_{f_k}}(\omega)$ .Here $\omega\in \Omega^n(J^1M)$, and by sense of V.Lychagin we denote the monge ampere equation $\Delta_\omega=0$ by $E_\omega$ . Moreover $E_{\mathfrak{g}}$ is the set of solutions of $(f_1=0,f_2=0,....,f_k=0 )$ . .You can find this fact as theorem 12.3.1 link text page 258 In fact in this book the authors used of this fact and never proved it
Take the 2minute tour
×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.
