Neumann boundary conditions on S^1/Z_2by AntideSitter Tags: boundary, conditions, neumann, s1 or z2 

#1
Apr3010, 10:37 PM

P: 7

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! 



#2
Apr3010, 10:48 PM

P: 7

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.




#3
May110, 08:37 AM

Sci Advisor
P: 1,716

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




#4
May110, 04:32 PM

P: 7

Neumann boundary conditions on S^1/Z_2
I'm modding out the circle (e.g. x^2 + y^2 =1) by reflection about the yaxis.



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