Talk:Kernel formation (Ex)
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(Created page with "<wikitex>; The lens space $L(m,n)$ is a union of two solid tori glued along the boundary. The gluing diffeomorphism $\psi \colon T^2 \to T^2$ is such that $\psi_{\ast}\mu = m ...")
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(Created page with "<wikitex>; The lens space $L(m,n)$ is a union of two solid tori glued along the boundary. The gluing diffeomorphism $\psi \colon T^2 \to T^2$ is such that $\psi_{\ast}\mu = m ...")
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Revision as of 00:09, 29 March 2012
The lens space is a union of two solid tori glued along the boundary. The gluing diffeomorphism is such that , where and denote the longtitude and meridian of the solid torus. From this Heegard decomposition we obtain the kernel formation
The first lagrangian is spanned by the generator of . This generator is exactly the meridian . The same argument shows, that the second lagrangian is generated by the image of the meridian of complementary torus . Certainly we have . Furthermore it is easy to see that the index of in is equal to . Thus the kernel formation is trivial if and only if . We have , thus we obtain trivial formations for :
For it is sufficient to notice that . Thus the associated formation is: