# How to find lagrangian submanifolds.

I am quite confused on the definition of a lagrangian submanifold $L$ of a symplectic manifold $(M,\omega)$.

In particular, I read that $L \subset M$ is lagrangian iff the symplectic form field $\omega(x)$ evaluated on every point $p\in L$ gives zero. How is it possible that $\omega$ assumes the value zero on a subset $L$ of $M$, but still it is not degenerate (it is never zero) over the whole $M$?!

Furthermore, consider $M=\mathbb{R}^2$ with the standard symplectic form $\omega=dq\wedge dp$. It appears to me that there are no lagrangian submanifolds at all, since the symplectic form is everywhere constant and never zero. However, I am reading that for example all the submanifolds of the tipe $q=const$ are lagrangian, in this case.

Clearly, I am not understanding well the definition of Lagrangian submanifold.

Where am I going wrong?

The condition of being a lagrangian submanifold $L\subset M$ means that for every point $p\in L$ and every pair of tangent vectors $X,Y\in T_pL$, $$\omega_p(X,Y)=0\,.$$ This does not violate the non-degeneracy of $\omega$, since $T_pL$ is merely a subspace of $T_pM$. However, non-degeneracy implies that any submanifold satisfying this condition is of dimension at most $\frac12\dim M$. Lagrangian submanifolds are submanifolds satisfying the condition above of maximal possible dimension, i.e. $\dim L=\frac12\dim M$.

Another was of saying the same thing is that the pullback $i^*\omega$ of $\omega$ under the inclusion (smooth map) $i:L\hookrightarrow M$ of the $2$-form $\omega$ is zero: $$i^*\omega=0\,.$$

• Ok, now I understand. Thank you very much. The point is that since the dimension of $L$ is smaller than the dimension on $M$, $\omega(p)$ can be zero on $T_pL$ but still nonzero on $T_pM$, cause $T_pL$ is just a vector subspace of $T_pM$, preserving thus the nondegeneracy condition. I feel really stupid now, it was easy! – Federico Carta Dec 19 '14 at 18:49

In particular, I read that $L\subset M$ is lagrangian iff the symplectic form field $\omega(x)$ evaluated on every point $p\in L$ gives zero.

More precisely, the restriction of $\omega$ to $T_p L$ is zero. As an element of $\wedge^2 T_p M$, $\omega(p)$ is nonzero and nondegenerate.

For an example, consider $\mathbb{R}^2$ with the symplectic form $\omega=dx\wedge dy$. This is surely nondegenerate, but the 1-dimensional submanifold spanned by $L=\langle 1,0 \rangle$ (the $x$-axis) satisfies that $\omega$ vanishes on the tangent space of $L$.

It is, in fact, an exercise in linear algebra that a quadratic form of signature $(n,n)$ admits a decomposition of the vector space into two parts $V_1\oplus V_2$ s.t. the restriction of the form to $V_i$ is trivial. (This decomposition is not unique in general.) For example, the matrix $$\begin{pmatrix} 0 & 1\\ -1 & 0 \end{pmatrix}$$ is nondegenerate but the quadratic form it represents restricted to $\langle 1,0\rangle$ is just the zero in the upper left corner.

Nonetheless, there is in general not a simple answer to the question stated in the title of your question.