Show that a submanifold is lagrangian The problem is as follows:
Let $ (M, \omega_M) $ be a symplectic manifold, $ C \subset M $ a submanifold, $ f: C \to \mathbb{R} $  a smooth function. Show that $ L = \{ p \in T^* M: \pi_M(p) \in C, \forall v \in TC <p, v> = <df, v> \} $ is a langrangian submanifold.
I'm trying to start with a single point $x \in C$ for which I'll have a number of points $p_{xi} = df(x) + \varphi_i$, where $\varphi_i \in T_xC^\circ$. How many $\varphi_i$ there are depends on $dim\ C$ and, intuitively, it should make the dimensions "work", or make it so that $dim\ L = dim\ M = \frac{1}{2} dim\ T^*M$, but I don't really know how to show it precisely. And then there's the matter of showing that $\omega|_L = 0$ - if we restricted ourselves to $df(x)$, it would be simple, but there's still the $\varphi$ part that I'm uncertain how to handle.
 A: One approach is to work in coordinates.  Let $k = \dim(C)$ and $n = \dim(M)$.  Write the usual symplectic structure on $T^*M$ as $(T^*M, \omega)$.  To show that $L \subset T^*M$ is Lagrangian, we must show that: (a) $\dim(L) = n$, and (b) $\omega|_L = 0$.
Since we are working with a submanifold $C \subset M$, we can choose "slice" coordinates $(x^1, \ldots, x^k, x^{k+1}, \ldots, x^n)$ for an open set $U \subset M$.  In other words, the map $(x^1, \ldots, x^n) \colon U \to \mathbb{R}^n$ identifies $C \cap U \to \mathbb{R}^k \times 0$.
In general, a covector $p \in T^*M$ has the form
$$p = \xi^1\,dx^1|_x + \cdots + \xi^k \,dx^k|_x + \xi^{k+1}dx^{k+1}|_x + \cdots + \xi^n\,dx^n|_x,$$
where $x = \pi_M(p) \in M$.  Note that $p \in T^*M$ is specified by $2n = \dim(T^*M)$ numbers: the basepoint $x \in M$ gives $n$, and the components $(\xi^1, \ldots, \xi^n)$ give $n$.
However, a typical element $p \in L \subset T^*M$ takes the form
$$p = \frac{\partial f}{\partial x^1}(x)\,dx^1|_x + \cdots + \frac{\partial f}{\partial x^k}(x)\,dx^k|_x + \xi^{k+1}\,dx^{k+1}|_x + \cdots + \xi^n\,dx^n|_x,$$
where now $x = \pi_M(p)$ lies in $C \subset M$.  Notice that $p \in L$ is specified by $n = k + (n-k)$ numbers: the basepoint $x \in C$ gives $k$, and the components $(\xi^{k+1}, \ldots, \xi^n)$ give $n-k$.  This illustrates that $\dim(L) = n$.
(To make this completely precise, one should use this idea to define a coordinate chart $L \to \mathbb{R}^k \times \mathbb{R}^{n-k}$.)
Now, notice that we not only chose coordinates $(x^1, \ldots, x^n)$ on an open set $U \subset M$, but we've also described coordinates $(\widetilde{x}^1, \ldots, \widetilde{x}^n, \widetilde{\xi}^1, \ldots, \widetilde{\xi}^n)$ on the open set $\widetilde{U} = \pi^{-1}(U) \subset T^*M$.  In these coordinates, the usual symplectic form on $T^*M$ is
$$\omega = \sum d\widetilde{x}^i \wedge d\widetilde{\xi}^i =  (d\widetilde{x}^1 \wedge d\widetilde{\xi}^1 + \cdots + d\widetilde{x}^k \wedge d\widetilde{\xi}^k) + (d\widetilde{x}^{k+1} \wedge d\widetilde{\xi}^{k+1} + \cdots + d\widetilde{x}^n \wedge d\widetilde{\xi}^n).$$
Using this, one can show that $\omega|_L = 0$.  Details are left to you.  (Hint: Show that $d\widetilde{\xi}^j|_L = 0$ for $1 \leq j \leq k$, while $d\widetilde{x}^j|_L = 0$ for $k+1 \leq j \leq n$.)

Note: The original question statement states that $(M, \omega_M)$ is a symplectic manifold.  However, one only needs that $M$ is a smooth manifold, while $T^*M$ is given its usual symplectic structure.
