Talk:Kernel formation (Ex)

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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 \lambda + n \mu, where \lambda and \mu denote the longtitude and meridian of the solid torus. From this Heegard decomposition we obtain the kernel formation

\displaystyle (H_{(-1)}(\mathbb{Z}); F,G).

The first lagrangian F is spanned by the generator of H_2(S^1 \times D^2, S^1 \times S^1). This generator is exactly the meridian \mu. The same argument shows, that the second lagrangian is generated by the image of the meridian of complementary torus \psi_{\ast} \mu = m \lambda + n \mu. Certainly we have F \cap G = \{0\}. Furthermore it is easy to see that the index of F+G in \mathbb{Z} \times \mathbb{Z} is equal to m. Thus the kernel formation is trivial if and only if m=1. We have L(1,n) = S^3, thus we obtain trivial formations for S^3:

\displaystyle (H_{(-1)}(\mathbb{Z}),\mu \mathbb{Z}, (\lambda + n \mu) \mathbb{Z}).

For \mathbb{R}P^3 it is sufficient to notice that L(2,1) = \mathbb{R}P^3. Thus the associated formation is:

\displaystyle (H_{(-1)}(\mathbb{Z}); \mu \mathbb{Z}, (2 \lambda + \mu) \mathbb{Z}).
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