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\input { knots_ macros}
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\begin { document}
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\tableofcontents
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%\newpage
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%\input{myNotes}
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\section { Basic definitions \hfill \DTMdate { 2019-02-25} }
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\input { lec_ 1.tex}
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\section { Alexander polynomial \hfill \DTMdate { 2019-03-04} }
\input { lec_ 2.tex}
%add Hurewicz theorem?
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\section { \hfill \DTMdate { 2019-03-11} }
\input { lec_ 3.tex}
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\section { Concordance group \hfill \DTMdate { 2019-03-18} }
\input { lec_ 4.tex}
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\section { \hfill \DTMdate { 2019-03-25} }
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\input { lec_ 5.tex}
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\section { \hfill \DTMdate { 2019-04-08} }
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%
%
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$ X $ is a closed orientable four-manifold. Assume $ \pi _ 1 ( X ) = 0 $ (it is not needed to define the intersection form). In particular $ H _ 1 ( X ) = 0 $ .
$ H _ 2 $ is free (exercise).
\begin { align*}
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H_ 2(X, \mathbb { Z} ) \xrightarrow { \text { Poincar\' e duality} } H^ 2(X, \mathbb { Z} ) \xrightarrow { \text { evaluation} } \Hom (H_ 2(X, \mathbb { Z} ), \mathbb { Z} )
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\end { align*}
Intersection form:
$ H _ 2 ( X, \mathbb { Z } ) \times
H_ 2(X, \mathbb { Z} ) \longrightarrow \mathbb { Z} $ - symmetric, non singular.
\\
Let $ A $ and $ B $ be closed, oriented surfaces in $ X $ .
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\begin { proposition}
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$ A \cdot B $ doesn't depend of choice of $ A $ and $ B $ in their homology classes $ [ A ] , [ B ] \in H _ 2 ( X, \mathbb { Z } ) $ . \\
\noindent
If $ M $ is an $ m $ - dimensional close, connected and orientable manifold, then $ H _ m ( M, \mathbb { Z } ) $ and the orientation if $ M $ determined a cycle $ [ M ] \in H _ m ( M, \mathbb { Z } ) $ , called the fundamental cycle.
\begin { example}
If $ \omega $ is an $ m $ - form then:
\[
= [\omega ]
\]
\end { example}
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%$A \cdot B$ gives the pairing as ??
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\end { proposition}
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\section { \hfill \DTMdate { 2019-04-15} }
In other words:\\
Choose a basis $ ( b _ 1 , ..., b _ i ) $ \\
???\\
of $ H _ 2 ( Y, \mathbb { Z } $ , then $ A = ( b _ i, b _ y ) $ \\ ??\\ is a matrix of intersection form:
\begin { align*}
\quot { \mathbb { Z} ^ n} { A\mathbb { Z} ^ n} \cong H_ 1(Y, \mathbb { Z} ).
\end { align*}
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In particular $ \mid \det A \mid = \# H _ 1 ( Y, \mathbb { Z } ) $ .\\
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That means - what is happening on boundary is a measure of degeneracy.
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\begin { center}
\begin { tikzcd}
[
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column sep=tiny,
row sep=small,
ar symbol/.style =%
{ draw=none,"\textstyle #1" description,sloped} ,
isomorphic/.style = { ar symbol={ \cong } } ,
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]
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H_ 1(Y, \mathbb { Z} ) &
\times \quad H_ 1(Y, \mathbb { Z} )&
\longrightarrow &
\quot { \mathbb { Q} } { \mathbb { Z} }
\text { - a linking form}
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\\
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\quot { \mathbb { Z} ^ n} { A\mathbb { Z} } \ar [u,isomorphic] &
\quot { \mathbb { Z} ^ n} { A\mathbb { Z} } \ar [u,isomorphic] & \\
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\end { tikzcd}
$ ( a, b ) \mapsto aA ^ { - 1 } b ^ T $
\end { center}
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?????????????????????????????????\\
\noindent
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The intersection form on a four-manifold determines the linking on the boundary. \\
\noindent
Let $ K \in S ^ 1 $ be a knot, $ \Sigma ( K ) $ its double branched cover. If $ V $ is a Seifert matrix for $ K $ , then
$ H _ 1 ( \Sigma ( K ) , \mathbb { Z } ) \cong \quot { \mathbb { Z } ^ n } { A \mathbb { Z } } $ where
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$ A = V \times V ^ T $ , $ n = \rank V $ .
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%\input{ink_diag}
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\begin { figure} [h]
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\fontsize { 20} { 10} \selectfont
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\centering {
\def \svgwidth { \linewidth }
\resizebox { 0.5\textwidth } { !} { \input { images/ball_ 4.pdf_ tex} }
\caption { Pushing the Seifert surface in 4-ball.}
\label { fig:pushSeifert}
}
\end { figure}
\noindent
Let $ X $ be the four-manifold obtained via the double branched cover of $ B ^ 4 $ branched along $ \widetilde { \Sigma } $ .
\begin { fact}
\begin { itemize}
\item $ X $ is a smooth four-manifold,
\item $ H _ 1 ( X, \mathbb { Z } ) = 0 $ ,
\item $ H _ 2 ( X, \mathbb { Z } ) \cong \mathbb { Z } ^ n $
\item The intersection form on $ X $ is $ V + V ^ T $ .
\end { itemize}
\end { fact}
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\begin { figure} [h]
\fontsize { 20} { 10} \selectfont
\centering {
\def \svgwidth { \linewidth }
\resizebox { 0.5\textwidth } { !} { \input { images/ball_ 4_ pushed_ cycle.pdf_ tex} }
\caption { Cycle pushed in 4-ball.}
\label { fig:pushCycle}
}
\end { figure}
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\noindent
Let $ Y = \Sigma ( K ) $ . Then:
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\begin { align*}
H_ 1(Y, \mathbb { Z} ) \times H_ 1(Y, \mathbb { Z} ) & \longrightarrow \quot { \mathbb { Q} } { \mathbb { Z} }
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\\
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(a,b) & \mapsto a A^ { -1} b^ { T} ,\qquad
A = V + V^ T.
\end { align*}
????????????????????????????
\begin { align*}
H_ 1(Y, \mathbb { Z} ) \cong \quot { \mathbb { Z} ^ n} { A\mathbb { Z} } \\
A \longrightarrow BAC^ T \quad \text { Smith normal form}
\end { align*}
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???????????????????????\\
In general
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%no lecture at 29.04
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\section { \hfill \DTMdate { 2019-05-20} }
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Let $ M $ be compact, oriented, connected four-dimensional manifold. If $ { H _ 1 ( M, \mathbb { Z } ) = 0 } $ then there exists a
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bilinear form - the intersection form on $ M $ :
\begin { center}
\begin { tikzcd}
[
column sep=tiny,
row sep=small,
ar symbol/.style = { draw=none,"\textstyle #1" description,sloped} ,
isomorphic/.style = { ar symbol={ \cong } } ,
]
H_ 2(M, \mathbb { Z} )&
\times & H_ 2(M, \mathbb { Z} )
\longrightarrow &
\mathbb { Z}
\\
\ar [u,isomorphic] \mathbb { Z} ^ n & & & \\
\end { tikzcd}
\end { center}
\noindent
Let us consider a specific case: $ M $ has a boundary $ Y = \partial M $ .
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Betti number $ b _ 1 ( Y ) = 0 $ , $ H _ 1 ( Y, \mathbb { Z } ) $ is finite.
Then the intersection form can be degenerated in the sense that:
\begin { align*}
H_ 2(M, \mathbb { Z} )
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\times H_ 2(M, \mathbb { Z} )
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& \longrightarrow
\mathbb { Z} \quad &
H_ 2(M, \mathbb { Z} ) & \longrightarrow \Hom (H_ 2(M, \mathbb { Z} ), \mathbb { Z} )\\
(a, b) & \mapsto \mathbb { Z} \quad &
a & \mapsto (a, \_ ) H_ 2(M, \mathbb { Z} )
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\end { align*}
has coker precisely $ H _ 1 ( Y, \mathbb { Z } ) $ .
\\ ???????????????\\
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Let $ K \subset S ^ 3 $ be a knot, \\
$ X = S ^ 3 \setminus K $ - a knot complement, \\
$ \widetilde { X } \xrightarrow { \enspace \rho \enspace } X $ - an infinite cyclic cover (universal abelian cover).
\begin { align*}
\pi _ 1(X) \longrightarrow \quot { \pi _ 1(X)} { [\pi _ 1(X), \pi _ 1(X)]} = H_ 1(X, \mathbb { Z} ) \cong \mathbb { Z}
\end { align*}
$ C _ { * } ( \widetilde { X } ) $ has a structure of a $ \mathbb { Z } [ t, t ^ { - 1 } ] \cong \mathbb { Z } [ \mathbb { Z } ] $ module. \\
$ H _ 1 ( \widetilde { X } , \mathbb { Z } [ t, t ^ { - 1 } ] ) $ - Alexander module, \\
\begin { align*}
H_ 1(\widetilde { X} , \mathbb { Z} [t, t^ { -1} ]) \times
H_ 1(\widetilde { X} , \mathbb { Z} [t, t^ { -1} ]) \longrightarrow \quot { \mathbb { Q} } { \mathbb { Z} [t, t^ { -1} ]}
\end { align*}
\begin { fact}
\begin { align*}
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& H_ 1(\widetilde { X} , \mathbb { Z} [t, t^ { -1} ]) \cong
\quot { \mathbb { Z} { [t, t^ { -1} ]} ^ n} { (tV - V^ T)\mathbb { Z} [t, t^ { -1} ]^ n} \; , \\
& \text { where $ V $ is a Seifert matrix.}
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\end { align*}
\end { fact}
\begin { fact}
\begin { align*}
H_ 1(\widetilde { X} , \mathbb { Z} [t, t^ { -1} ]) \times
H_ 1(\widetilde { X} , \mathbb { Z} [t, t^ { -1} ]) & \longrightarrow \quot { \mathbb { Q} } { \mathbb { Z} [t, t^ { -1} ]} \\
(\alpha , \beta ) & \mapsto \alpha ^ { -1} (t -1)(tV - V^ T)^ { -1} \beta
\end { align*}
\end { fact}
\noindent
Note that $ \mathbb { Z } $ is not PID. Therefore we don't have primer decomposition of this moduli. We can simplify this problem by replacing $ \mathbb { Z } $ by $ \mathbb { R } $ . We lose some date by doing this transition.
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\begin { align*}
& \xi \in S^ 1 \setminus \{ \pm 1\}
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\quad
p_ { \xi } =
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(t - \xi )(t - \xi ^ { -1} ) t^ { -1}
\\
& \xi \in \mathbb { R} \setminus \{ \pm 1\}
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\quad
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q_ { \xi } = (t - \xi )(t - \xi ^ { -1} ) t^ { -1}
\\
&
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\xi \notin \mathbb { R} \cup S^ 1 \quad
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q_ { \xi } = (t - \xi )(t - \overbar { \xi } )(t - \xi ^ { -1} )(t - \overbar { \xi } ^ { -1} ) t^ { -2} \\
&
\Lambda = \mathbb { R} [t, t^ { -1} ]\\
& \text { Then: } H_ 1(\widetilde { X} , \Lambda ) \cong \bigoplus _ { \substack { \xi \in S^ 1 \setminus \{ \pm 1 \} \\ k\geq 0} }
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( \quot { \Lambda } { p_ { \xi } ^ k } )^ { n_ k, \xi }
\oplus
\bigoplus _ { \substack { \xi \notin S^ 1 \\ l\geq 0} }
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(\quot { \Lambda } { q_ { \xi } ^ l} )^ { n_ l, \xi } &
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\end { align*}
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We can make this composition orthogonal with respect to the Blanchfield paring.
\vspace { 0.5cm} \\
Historical remark:
\begin { itemize}
\item John Milnor, \textit { On isometries of inner product spaces} , 1969,
\item Walter Neumann, \textit { Invariants of plane curve singularities}
%in: Knots, braids and singulari- ties (Plans-sur-Bex, 1982), 223– 232, Monogr. Enseign. Math., 31, Enseignement Math., Geneva
, 1983,
\item András Némethi, \textit { The real Seifert form and the spectral pairs of isolated hypersurfaceenumerate singularities} , 1995,
%Compositio Mathematica, Volume 98 (1995) no. 1, p. 23-41
\item Maciej Borodzik, Stefan Friedl
\textit { The unknotting number and classical invariants II} , 2014.
\end { itemize}
\vspace { 0.5cm}
Let $ p = p _ { \xi } $ , $ k \geq 0 $ .
\begin { align*}
\quot { \Lambda } { p^ k \Lambda } \times
\quot { \Lambda } { p^ k \Lambda } & \longrightarrow \quot { \mathbb { Q} (t)} { \Lambda } \\
(1, 1) & \mapsto \kappa \\
\text { Now: } (p^ k \cdot 1, 1) & \mapsto 0\\
p^ k \kappa = 0 & \in \quot { \mathbb { Q} (t)} { \Lambda } \\
\text { therfore } p^ k \kappa & \in \Lambda \\
\text { we have } (1, 1) & \mapsto \frac { h} { p^ k} \\
\end { align*}
$ h $ is not uniquely defined: $ h \rightarrow h + g p ^ k $ doesn't affect paring. \\
Let $ h = p ^ k \kappa $ .
\begin { example}
\begin { align*}
\phi _ 0 ((1, 1))=\frac { +1} { p} \\
\phi _ 1 ((1, 1)) = \frac { -1} { p}
\end { align*}
$ \phi _ 0 $ and $ \phi _ 1 $ are not isomorphic.
\end { example}
\begin { proof}
Let $ \Phi :
\quot { \Lambda } { p^ k \Lambda } \longrightarrow
\quot { \Lambda } { p^ k \Lambda } $
be an isomorphism. \\
Let: $ \Phi ( 1 ) = g \in \lambda $
\begin { align*}
\quot { \Lambda } { p^ k \Lambda }
\xrightarrow { \enspace \Phi \enspace } &
\quot { \Lambda } { p^ k \Lambda } \\
\phi _ 0((1, 1)) = \frac { 1} { p^ k} \qquad & \qquad
\phi _ 1((g, g)) = \frac { 1} { p^ k} \quad \text { ($ \Phi $ is an isometry).}
\end { align*}
Suppose for the paring $ \phi _ 1 ( ( g, g ) ) = \frac { 1 } { p ^ k } $ we have $ \phi _ 1 ( ( 1 , 1 ) ) = \frac { - 1 } { p ^ k } $ . Then:
\begin { align*}
\frac { -g\overbar { g} } { p^ k} = \frac { 1} { p^ k} & \in \quot { \mathbb { Q} (t)} { \Lambda } \\
\frac { -g\overbar { g} } { p^ k} - \frac { 1} { p^ k} & \in \Lambda \\
-g\overbar { g} & \equiv 1\pmod { p} \text { in } \Lambda \\
-g\overbar { g} - 1 & = p^ k \omega \text { for some } \omega \in \Lambda \\
\text { evalueting at $ \xi $ : } \\
\overbrace { -g(\xi )g(\xi ^ { -1} )} ^ { >0} - 1 = 0 \quad \contradiction
\end { align*}
\end { proof}
????????????????????\\
\begin { align*}
g & = \sum { g_ i t^ i} \\
\overbar { g} & = \sum { g_ i t^ { -i} } \\
\overbar { g} (\xi ) & = \sum g_ i \xi ^ i \quad \xi \in S^ 1\\
\overbar { g} (\xi ) & =\overbar { g(\xi )}
\end { align*}
Suppose $ g = ( t - \xi ) ^ { \alpha } g ^ { \prime } $ . Then $ ( t - \xi ) ^ { k - \alpha } $ goes to $ 0 $ in $ \quot { \Lambda } { p ^ k \Lambda } $ .
\begin { theorem}
Every sesquilinear non-degenerate pairing
\begin { align*}
\quot { \Lambda } { p^ k} \times \quot { \Lambda } { p}
\longleftrightarrow \frac { h} { p^ k}
\end { align*}
is isomorphic either to the pairing wit $ h = 1 $ or to the paring with $ h = - 1 $ depending on sign of $ h ( \xi ) $ (which is a real number).
\end { theorem}
\begin { proof}
There are two steps of the proof:
\begin { enumerate}
\item
Reduce to the case when $ h $ has a constant sign on $ S ^ 1 $ .
\item
Prove in the case, when $ h $ has a constant sign on $ S ^ 1 $ .
\end { enumerate}
\begin { lemma}
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If $ P $ is a symmetric polynomial such that $ P ( \eta ) \geq 0 $ for all $ \eta \in S ^ 1 $ , then $ P $ can be written as a product $ P = g \overbar { g } $ for some polynomial $ g $ .
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\end { lemma}
\begin { proof} [Sketch of proof]
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Induction over $ \deg P $ .\\
Let $ \zeta \notin S ^ 1 $ be a root of $ P $ , $ P \in \mathbb { R } [ t, t ^ { - 1 } ] $ . Assume $ \zeta \notin \mathbb { R } $ . We know that polynomial $ P $ is divisible by
$ ( t - \zeta ) $ , $ ( t - \overbar { \zeta } ) $ , $ ( t ^ { - 1 } - \zeta ) $ and $ ( t ^ { - 1 } - \overbar { \zeta } ) $ .
Therefore:
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\begin { align*}
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& P^ { \prime } = \frac { P} { (t - \zeta )(t - \overbar { \zeta } )(t^ { -1} - \zeta )(t^ { -1} - \overbar { \zeta } )} \\
& P^ { \prime } = g^ { \prime } \overbar { g}
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\end { align*}
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We set $ g = g ^ { \prime } ( t - \zeta ) ( t - \overbar { \zeta } ) $ and
$ P = g \overbar { g } $ . Suppose $ \zeta \in S ^ 1 $ . Then $ ( t - \zeta ) ^ 2 \mid P $ (at least - otherwise it would change sign). Therefore:
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\begin { align*}
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& P^ { \prime } = \frac { P} { (t - \zeta )^ 2(t^ { -1} - \zeta )^ 2} \\
& g = (t - \zeta )(t^ { -1} - \zeta ) g^ { \prime } \quad \text { etc.}
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\end { align*}
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The map $ ( 1 , 1 ) \mapsto \frac { h } { p ^ k } = \frac { g \overbar { g } h } { p ^ k } $ is isometric whenever $ g $ is coprime with $ P $ .
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\end { proof}
\begin { lemma} \label { L:coprime polynomials}
Suppose $ A $ and $ B $ are two symmetric polynomials that are coprime and that $ \forall z \in S ^ 1 $ either $ A ( z ) > 0 $ or $ B ( z ) > 0 $ . Then there exist
symmetric polynomials $ P $ , $ Q $ such that
$ P ( z ) , Q ( z ) > 0 $ for $ z \in S ^ 1 $ and $ PA + QB \equiv 1 $ .
\end { lemma}
\begin { proof} [Idea of proof]
For any $ z $ find an interval $ ( a _ z, b _ z ) $ such that if $ P ( z ) \in ( a _ z, b _ z ) $ and $ P ( z ) A ( z ) + Q ( z ) B ( z ) = 1 $ , then $ Q ( z ) > 0 $ , $ x ( z ) = \frac { az + bz } { i } $ is a continues function on $ S ^ 1 $ approximating $ z $ by a polynomial .
\\ ??????????????????????????\\
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\begin { flalign*}
(1, 1) \mapsto \frac { h} { p^ k} \mapsto \frac { g\overbar { g} h} { p^ k} & \\
g\overbar { g} h + p^ k\omega = 1&
\end { flalign*}
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Apply Lemma \ref { L:coprime polynomials} for $ A = h $ , $ B = p ^ { 2 k } $ . Then, if the assumptions are satisfied,
\begin { align*}
Ph + Qp^ { 2k} = 1\\
p>0 \Rightarrow p = g \overbar { g} \\
p = (t - \xi )(t - \overbar { \xi } )t^ { -1} \\
\text { so } p \geq 0 \text { on } S^ 1\\
p(t) = 0 \Leftrightarrow
t = \xi or t = \overbar { \xi } \\
h(\xi ) > 0\\
h(\overbar { \xi } )>0\\
g\overbar { g} h + Qp^ { 2k} = 1\\
g\overbar { g} h \equiv 1 \mod { p^ { 2k} } \\
g\overbar { g} \equiv 1 \mod { p^ k}
\end { align*}
???????????????????????????????\\
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If $ P $ has no roots on $ S ^ 1 $ then $ B ( z ) > 0 $ for all $ z $ , so the assumptions of Lemma \ref { L:coprime polynomials} are satisfied no matter what $ A $ is.
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\end { proof}
?????????????????\\
\begin { align*}
(\quot { \Lambda } { p_ { \xi } ^ k} \times
\quot { \Lambda } { p_ { \xi } ^ k} ) & \longrightarrow
\frac { \epsilon } { p_ { \xi } ^ k} , \quad \xi \in S^ 1 \setminus \{ \pm 1\} \\
(\quot { \Lambda } { q_ { \xi } ^ k} \times
\quot { \Lambda } { q_ { \xi } ^ k} ) & \longrightarrow
\frac { 1} { q_ { \xi } ^ k} , \quad \xi \notin S^ 1\\
\end { align*}
??????????????????? 1 ?? epsilon?\\
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\begin { theorem} (Matumoto, Borodzik-Conway-Politarczyk)
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Let $ K $ be a knot,
\begin { align*}
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& H_ 1(\widetilde { X} , \Lambda ) \times
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H_ 1(\widetilde { X} , \Lambda )
= \bigoplus _ { \substack { k, \xi , \epsilon \\ \xi in S^ 1} }
(\quot { \Lambda } { p_ { \xi } ^ k} , \epsilon )^ { n_ k, \xi , \epsilon } \oplus \bigoplus _ { k, \eta }
(\quot { \Lambda } { p_ { \xi } ^ k} )^ { m_ k}
\end { align*}
\begin { align*}
\text { Let } \delta _ { \sigma } (\xi ) = \lim _ { \varepsilon \rightarrow 0^ { +} }
\sigma (e^ { 2\pi i \varepsilon } \xi )
- \sigma (e^ { -2\pi i \varepsilon } \xi ),\\
\text { then }
\sigma _ j(\xi ) = \sigma (\xi ) - \frac { 1} { 2} \lim _ { \varepsilon \rightarrow 0}
\sigma (e^ { 2\pi i \varepsilon } \xi )
+ \sigma (e^ { -2 \pi i \varepsilon } \xi )
\end { align*}
The jump at $ \xi $ is equal to
$ 2 \sum \limits _ { k _ i \text { odd } } \epsilon _ i $ . The peak of the signature function is equal to $ \sum \limits _ { k _ i \text { even } } \epsilon _ i $ .
%$(\eta_{k, \xi_l^{+}} -\eta_{k, \xi_l^{-}}$
\end { theorem}
\end { proof}
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\section { \hfill \DTMdate { 2019-05-27} }
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....
\begin { definition}
A square hermitian matrix $ A $ of size $ n $ .
\end { definition}
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field of fractions
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\section { \hfill \DTMdate { 2019-06-03} }
\begin { theorem}
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Let $ K $ be a knot and $ u ( K ) $ its unknotting number. Let $ g _ 4 $ be a minimal four genus of a smooth surface $ S $ in $ B ^ 4 $ such that $ \partial S = K $ . Then:
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\[
u(K) \geq g_ 4(K)
\]
\begin { proof}
Recall that if $ u ( K ) = u $ then $ K $ bounds a disk $ \Delta $ with $ u $ ordinary double points.
\\
\noindent
Remove from $ \Delta $ the two self intersecting and glue the Seifert surface for the Hopf link. The reality surface $ S $ has Euler characteristic $ \chi ( S ) = 1 - 2 u $ . Therefore $ g _ 4 ( S ) = u $ .
\end { proof}
???????????????????\\
\begin { example}
The knot $ 8 _ { 20 } $ is slice: $ \sigma \equiv 0 $ almost everywhere but $ \sigma ( e ^ { \frac { 2 \pi i } { 6 } } ) = + 1 $ .
\end { example}
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%ref Structure in the classical knot concordance group
%Tim D. Cochran, Kent E. Orr, Peter Teichner
%Journal-ref: Comment. Math. Helv. 79 (2004) 105-123
\subsection * { Surgery}
%Rolfsen, geometric group theory, Diffeomorpphism of a torus, Mapping class group
Recall that $ H _ 1 ( S ^ 1 \times S ^ 1 , \mathbb { Z } ) = \mathbb { Z } ^ 3 $ . As generators for $ H _ 1 $ we can set $ { \alpha = [ S ^ 1 \times \pt ] } $ and $ { \beta = [ \pt \times S ^ 1 ] } $ . Suppose $ { \phi : S ^ 1 \times S ^ 1 \longrightarrow S ^ 1 \times S ^ 1 } $ is a diffeomorphism.
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Consider an induced map on homology group:
\begin { align*}
H_ 1(S^ 1 \times S^ 1, \mathbb { Z} ) \ni \phi _ * (\alpha ) & = p\alpha + q \beta , \quad p, q \in \mathbb { Z} ,\\
\phi _ *(\beta ) & = r \alpha + s \beta , \quad r, s \in \mathbb { Z} , \\
\phi _ * & =
\begin { pmatrix}
p & q\\
r & s
\end { pmatrix}
\end { align*}
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As $ \phi _ * $ is diffeomorphis, it must be invertible over $ \mathbb { Z } $ . Then for a direction preserving diffeomorphism we have $ \det \phi _ * = 1 $ . Therefore $ \phi _ * \in \Sl ( 2 , \mathbb { Z } ) $ .
\end { theorem}
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\vspace { 10cm}
\begin { theorem}
Every such a matrix can be realized as a torus.
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\end { theorem}
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\begin { proof}
\begin { enumerate} [label={ (\Roman * )} ]
\item
Geometric reason
\begin { align*}
\phi _ t:
S^ 1 \times S^ 1 & \longrightarrow S^ 1 \times S^ 1 \\
S^ 1 \times \pt & \longrightarrow \pt \times S^ 1 \\
\pt \times S^ 1 & \longrightarrow S^ 1 \times \pt \\
(x, y) & \mapsto (-y, x)
\end { align*}
\item
\end { enumerate}
\end { proof}
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\section { balagan}
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\noindent
\noindent
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\section { \hfill \DTMdate { 2019-05-06} }
\begin { definition}
Let $ X $ be a knot complement.
Then $ H _ 1 ( X, \mathbb { Z } ) \cong \mathbb { Z } $ and there exists an epimorphism
$ \pi _ 1 ( X ) \overset { \phi } \twoheadrightarrow \mathbb { Z } $ .\\
The infinite cyclic cover of a knot complement $ X $ is the cover associated with the epimorphism $ \phi $ .
\[
\widetilde { X} \longtwoheadrightarrow X
\]
\end { definition}
%Rolfsen, bachalor thesis of Kamila
\begin { figure} [h]
\fontsize { 10} { 10} \selectfont
\centering {
\def \svgwidth { \linewidth }
\resizebox { 1\textwidth } { !} { \input { images/covering.pdf_ tex} }
\caption { Infinite cyclic cover of a knot complement.}
\label { fig:covering}
}
\end { figure}
\begin { figure} [h]
\fontsize { 10} { 10} \selectfont
\centering {
\def \svgwidth { \linewidth }
\resizebox { 0.8\textwidth } { !} { \input { images/knot_ complement.pdf_ tex} }
\caption { A knot complement.}
\label { fig:complement}
}
\end { figure}
\noindent
Formal sums $ \sum \phi _ i ( t ) a _ i + \sum \phi _ j ( t ) \alpha _ j $ \\
finitely generated as a $ \mathbb { Z } [ t, t ^ { - 1 } ] $ module.
\\
Let $ v _ { ij } = \Lk ( a _ i, a _ j ^ + ) $ . Then
$ V = \{ v _ ij \} _ { i, j = 1 } ^ n $ is the Seifert matrix associated to the surface $ \Sigma $ and the basis $ a _ 1 , \dots , a _ n $ . Therefore $ a _ k ^ + = \sum _ { j } v _ { jk } \alpha _ j $ . Then
$ \Lk ( a _ i, a _ k ^ + ) = \Lk ( a _ k ^ + , a _ i ) = \sum _ j v _ { jk } \Lk ( \alpha _ j, a _ i ) = v _ { ik } $ .
We also notice that $ \Lk ( a _ i, a _ j ^ - ) = \Lk ( a _ i ^ + , a _ j ) = v _ { ij } $ and
$ a _ j ^ - = \sum _ k v _ { kj } t ^ { - 1 } \alpha _ j $ .
\\
\noindent
The homology of $ \widetilde { X } $ is generated by $ a _ 1 , \dots , a _ n $ and relations.
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\begin { definition}
The Nakanishi index of a knot is the minimal number of generators of $ H _ 1 ( \widetilde { X } ) $ .
\end { definition}
%see Maciej page
\noindent
Remark about notation: sometimes one writes $ H _ 1 ( X; \mathbb { Z } [ t, t ^ { - 1 } ] ) $ (what is also notation for twisted homology) instead of $ H _ 1 ( \widetilde { X } ) $ .
\\
?????????????????????
\\
\noindent
$ \Sigma _ ? ( K ) \rightarrow S ^ 3 $ ?????\\
$ H _ 1 ( \Sigma _ ? ( K ) , \mathbb { Z } ) = h $ \\
$ H \times H \longrightarrow \quot { \mathbb { Q } } { \mathbb { Z } } $ \\
...\\
Let now $ H = H _ 1 ( \widetilde { X } ) $ . Can we define a paring? \\
Let $ c, d \in H ( \widetilde { X } ) $ (see Figure \ref { fig:covering_ pairing} ), $ \Delta $ an Alexander polynomial. We know that $ \Delta c = 0 \in H _ 1 ( \widetilde { X } ) $ (Alexander polynomial annihilates all possible elements). Let consider a surface $ F $ such that $ \partial F = c $ . Now consider intersection points $ F \cdot d $ . This points can exist in any $ N _ k $ or $ S _ k $ .
\[
\frac { 1} { \Delta } \sum _ { j\in \mathbb { Z} t^ { -j} } (F \cdot t^ j d) \in \quot { \mathbb { Q} [t, t^ { -1} ]} { \mathbb { Z} [t, t^ { -1} ]}
\]
\\
?????????????\\
\begin { figure} [h]
\fontsize { 10} { 10} \selectfont
\centering {
\def \svgwidth { \linewidth }
\resizebox { 1\textwidth } { !} { \input { images/covering_ pairing.pdf_ tex} }
\caption { $ c, d \in H _ 1 ( \widetilde { X } ) $ .}
\label { fig:covering_ pairing}
}
\end { figure}
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\begin { definition}
The $ \mathbb { Z } [ t, t ^ { - 1 } ] $ module $ H _ 1 ( \widetilde { X } ) $ is called the Alexander module of knot $ K $ .
\end { definition}
\noindent
Let $ R $ be a PID, $ M $ a finitely generated $ R $ module. Let us consider
\[
R^ k \overset { A} \longrightarrow R^ n \longtwoheadrightarrow M,
\]
where $ A $ is a $ k \times n $ matrix, assume $ k \ge n $ . The order of $ M $ is the $ \gcd $ of all determinants of the $ n \times n $ minors of $ A $ . If $ k = n $ then $ \ord M = \det A $ .
\begin { theorem}
Order of $ M $ doesn't depend on $ A $ .
\end { theorem}
\noindent
For knots the order of the Alexander module is the Alexander polynomial.
\begin { theorem}
\[
\forall x \in M: (\ord M) x = 0.
\]
\end { theorem}
\noindent
$ M $ is well defined up to a unit in $ R $ .
\subsection * { Blanchfield pairing}
\section { balagan}
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\begin { fact} [Milnor Singular Points of Complex Hypersurfaces]
\end { fact}
%\end{comment}
\noindent
An oriented knot is called negative amphichiral if the mirror image $ m ( K ) $ of $ K $ is equivalent the reverse knot of $ K $ : $ K ^ r $ . \\
\begin { problem}
Prove that if $ K $ is negative amphichiral, then $ K \# K = 0 $ in
$ \mathscr { C } $ .
%
%\\
%Hint: $ -K = m(K)^r = (K^r)^r = K$
\end { problem}
\begin { example}
Figure 8 knot is negative amphichiral.
\end { example}
%
%
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\begin { theorem}
Let $ H _ p $ be a $ p $ - torsion part of $ H $ . There exists an orthogonal decomposition of $ H _ p $ :
\[
H_ p = H_ { p, 1} \oplus \dots \oplus H_ { p, r_ p} .
\]
$ H _ { p, i } $ is a cyclic module:
\[
H_ { p, i} = \quot { \mathbb { Z} [t, t^ { -1} ]} { p^ { k_ i} \mathbb { Z} [t, t^ { -1} ]}
\]
\end { theorem}
\noindent
The proof is the same as over $ \mathbb { Z } $ .
\noindent
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%Add NotePrintSaveCiteYour opinionEmailShare
%Saveliev, Nikolai
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%Lectures on the Topology of 3-Manifolds
%An Introduction to the Casson Invariant
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\begin { figure} [h]
\fontsize { 10} { 10} \selectfont
\centering {
\def \svgwidth { \linewidth }
\resizebox { 0.5\textwidth } { !} { \input { images/ball_ 4_ alpha_ beta.pdf_ tex} }
}
%\caption{Sketch for Fact %%\label{fig:concordance_m}
\end { figure}
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\end { document}
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