1
$\begingroup$

I'm interested in showing the existence of a generating function. Explicitly:

Suppose $\Lambda\subset T^*M\times T^*M$ is a Lagrangian submanifold. Consider the projection $\pi:(x_1,\xi_1,x_2,\xi_2)\mapsto(x_1,\xi_2)$. If $\pi|_\Lambda$ is a diffeomorphism, how might I show that there exists a generating function $G(x_1,\xi_2)$ such that $\Lambda=\{(x_1,\frac{\partial{G}}{\partial{\xi_2}},-\frac{\partial{G}}{\partial{x_1}},\xi_2)\}$?

$\endgroup$

1 Answer 1

1
$\begingroup$

When $(M_1,\omega_1)$ and $(M_2,\omega_2)$ are symplectic manifolds then we endow $M_1\times M_2$ with the symplectic form $\omega_1\ominus\omega_2:=\pi_1^\ast\omega_1-\pi_2^\ast\omega_2,$ (where $\pi_i:M_1\times M_2\to M_i$ denotes the projection on the $i$-th factor.)

Let $\Lambda$ be an arbitrary Lagrangian submanifold of $M_1\times M_2,\omega_1\ominus\omega_2),$ and $i:\Lambda\to M_1\times M_2$ the inclusion map.
Fixed $p\in\Lambda$, for any primitive $\theta$ of $\omega_1\ominus\omega_2$ in a neighborhood of $p,$ there exists a smooth local function $S$ around $p$ such that $dS=i^\ast\theta.$ (Because of Poincaré Lemma and $0=i^\ast(\omega_1\ominus\omega_2)=i^\ast d\theta=d(i^\ast\theta).$)
Such a function $S$ is called generating function for $\Lambda,$ and, being only locally defined, it depends on the choice of $\theta.$


To be more specific:
In your context $\omega_i=d\xi_i\wedge dx_i$ is the canonical $2$-form on $M_i=\mathbb{R}^{2n},$ for $i=1,2,$ and $\Lambda$ is the image of $i:(u,f(u,v),g(u,v),v)\in\mathbb{R}^{2n}\to(u,v)\in\mathbb{R}^{4n}.$
As remarked above the local generating functions for $\Lambda$ are sensitive to our choice of the local primitive $\theta_i$ of $\omega_i.$

Let us choose $\theta_1=\xi_1 dx_1$ and $\theta_2=-x_2d\xi_2.$
Then the corresponding generating function is determined (up to a constant) by $dS=i^\ast(\theta_1\ominus\theta_2)=fdu+gdv,$ i.e.: $f=\frac{\partial S}{\partial u},\ g=\frac{\partial S}{\partial v}.$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.