Talk:Kernel formation (Ex)
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Wpolitarczyk (Talk | contribs) (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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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. | 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 | From this Heegard decomposition we obtain the kernel formation | ||
| − | $$(H_{-1}(\mathbb{Z}),F,G).$$ | + | $$(H_{(-1)}(\mathbb{Z}),F,G).$$ |
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 $\psi_{\ast} \mu = m \lambda + n \mu$. | 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 $m=1$. | 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 $S^3$: | We have $L(1,n) = S^3$, thus we obtain trivial formations for $S^3$: | ||
| − | $$(H_{-1}(\mathbb{Z}),\mu \mathbb{Z}, (\lambda + n \mu) \mathbb{Z}).$$ | + | $$(H_{(-1)}(\mathbb{Z}),\mu \mathbb{Z}, (\lambda + n \mu) \mathbb{Z}).$$ |
For $\mathbb{R}P^3$ it is sufficient to notice that $L(2,1) = \mathbb{R}P^3$. Thus the associated formation is: | For $\mathbb{R}P^3$ it is sufficient to notice that $L(2,1) = \mathbb{R}P^3$. Thus the associated formation is: | ||
| − | $$(H_{-1}(\mathbb{Z}), \mu \mathbb{Z}, (2 \lambda + \mu) \mathbb{Z}).$$ | + | $$(H_{(-1)}(\mathbb{Z}), \mu \mathbb{Z}, (2 \lambda + \mu) \mathbb{Z}).$$ |
</wikitex> | </wikitex> | ||
Revision as of 10:19, 1 April 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:
