Neumann boundary conditions on S^1/Z_2

[AntideSitter]

Hello everybody,

I've been puzzling over something (quite simple I assume).

Take S^1. Now consider the action of a Z_2 which takes x to -x, where x is a natural coordinate on the cylinder ( -1< x <1). Now we mod out by this action. The new space is an orbifold: smooth except at x=0. It is diffeomorphic to the line internal. We can take 0< x <1 as a coordinate on S^1/Z_2.

Take the set of functions on S^1, call it F_s. To find the set of functions on S^1/Z_2, we should restrict F_s to the the functions which are invariant under the Z_2 action. I.e. they should be symmetric under the reflection. These functions all have f '(0)=f '(1) = 0. These are Neumann boundary conditions on the boundary of S^1/Z_2.

Firstly, correct any errors I have made here. My question is: why must there be Neumann boundary conditions on functions on the line interval? Is this a generic property of manifolds with boundaries? Why are Dirichlet conditions not possible here? Is the singularity important in this example?

Comments welcome. Thank you!

 

[AntideSitter]

Just to correct myself: there are two singularities on S_1/Z_2 , at x=0,1. Strictly speaking we should use two coordinate patches. So there are singularities at each end of the line interval, just where my boundary conditions are.

 

[lavinia]

I am not sure what you are doing. Are you modding the unit disc out by reflection along the x axis?

 

[AntideSitter]

I'm modding out the circle (e.g. x^2 + y^2 =1) by reflection about the y-axis.

 

[blue_raver22]

I would like to clarify a few things about the concept of Neumann boundary conditions and their application in this specific example. Neumann boundary conditions are a type of boundary condition that specifies the derivative of a function at the boundary of a domain. In this case, the domain is S^1/Z_2, which is an orbifold obtained by taking the action of Z_2 on S^1 and modding it out.

In this example, the Z_2 action has created a singularity at x=0, which is why the functions on S^1/Z_2 must satisfy Neumann boundary conditions at this point. This is because the derivative of the function must be continuous at the singularity, which in this case means f '(0)=f '(1) = 0.

It is important to note that Neumann boundary conditions are not always required on manifolds with boundaries. In fact, on a smooth manifold with a boundary, the type of boundary condition will depend on the specific problem being studied. In some cases, Dirichlet boundary conditions (where the function itself must be continuous at the boundary) may be more appropriate.

In summary, the presence of Neumann boundary conditions on functions on S^1/Z_2 is a result of the specific Z_2 action and the resulting singularity at x=0. It is not a generic property of all manifolds with boundaries, and the type of boundary condition required will depend on the problem at hand. I hope this helps clarify any confusion. Thank you for your question!