[Symplectic geometry] Show that a submanifold is Lagrangian

In summary: Now, since ## v ## is a tangent vector to ## C ## at the base point of ## p_1 ##, we know that ## <\varphi_1 - \varphi_2, v> = 0 ## (since ## \varphi_1, \varphi_
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RedTachyon
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Homework Statement



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. In other words, you have to show that ## \omega |_L = 0 ## and ## dim\ L = \frac{1}{2} dim\ M ##

Homework Equations



## \alpha: M \to T^*M##
##\alpha^*\theta_M = \alpha## where ## \theta_M ## is Liouville's form ( ## d\theta_M = \omega_M ## )

The Attempt at a Solution



I'm trying to start with decomposing ## p \in L ## like that: ## p = df + \varphi ##, where ## \varphi \in TC^\circ ##, since ## p ## can differ from ## df ## only by ## \varphi:\ <\varphi,v> = 0\ \forall v \in TC ##. But I think there could be several possible ## \varphi ## for each point? There should be, since we know nothing about ## dim\ C##. I also tried doing something with the pullback ## P^*\theta_M ## where P would be a one-form that takes on the value ## p ## at ## \pi_M(p) ## but I can't really get any results.

Thanks in advance for any help!
 
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First, let's start by defining the submanifold ## L ## in a more explicit way. We can write ## L ## as the set of all points ## p \in T^*M ## that satisfy the following conditions:

1. ## \pi_M(p) \in C ## (i.e. the base point of ## p ## lies in ## C ##)
2. For all ## v \in T_{\pi_M(p)}C ##, the inner product ## <p, v> = <df, v> ## (i.e. the tangent vector to ## C ## at the base point of ## p ## is equal to the tangent vector to ## f ## at that same point)

Now, let's consider any two points ## p_1, p_2 \in L ##. We can write these points as ## p_1 = df_1 + \varphi_1 ## and ## p_2 = df_2 + \varphi_2 ##, where ## \varphi_1, \varphi_2 \in TC^\circ ## (the annihilator of ## TC ##). Now, let's consider the difference ## p_1 - p_2 ##:

$$
p_1 - p_2 = (df_1 + \varphi_1) - (df_2 + \varphi_2) = df_1 - df_2 + \varphi_1 - \varphi_2
$$

Notice that ## df_1 - df_2 ## is an exact form, since it is the difference of two exact forms (since ## f_1, f_2 ## are smooth functions, their differentials are exact forms). Therefore, we can write it as ## df_1 - df_2 = d\alpha ## for some one-form ## \alpha ##.

Now, let's consider the inner product of this difference with any tangent vector ## v \in T_{\pi_M(p_1)}C ##:

$$
<p_1 - p_2, v> = <d\alpha + \varphi_1 - \varphi_2, v> = <d\alpha, v> + <\varphi_1 - \varphi_2, v>
$$

Since ## d\alpha ## is an exact form, we can use the fact that the inner product is a symmetric bilinear form to rearrange the
 

1. What is a submanifold?

A submanifold is a subset of a manifold that is itself a manifold. It can be thought of as a smaller space contained within a larger space.

2. What does it mean for a submanifold to be Lagrangian?

In symplectic geometry, a Lagrangian submanifold is a submanifold that has a symplectic form (a mathematical structure that encodes information about the geometry of a manifold) that is closed. This means that the submanifold is "maximally symplectic" and has no holes or boundaries.

3. How can a submanifold be shown to be Lagrangian?

A submanifold can be shown to be Lagrangian by proving that its symplectic form is closed. This can be done by showing that the exterior derivative of the symplectic form is equal to zero.

4. What is the importance of a submanifold being Lagrangian?

Lagrangian submanifolds have important applications in physics, particularly in the study of classical mechanics and Hamiltonian systems. They also have important implications in symplectic topology, which is the study of the global properties of symplectic manifolds.

5. Can any submanifold be made into a Lagrangian submanifold?

No, not all submanifolds can be made into Lagrangian submanifolds. In fact, it is a highly non-trivial problem to find Lagrangian submanifolds in general. This is because not all manifolds have a symplectic form, and even when they do, not all submanifolds are Lagrangian with respect to that form.

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