lectures_on_knot_theory/lectures_on_knot_theory.tex

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\input{knots_macros}
\graphicspath{ {images/} }


\begin{document}
\tableofcontents
%\newpage
%\input{myNotes}

\section{Basic definitions \hfill\DTMdate{2019-02-25}}
%\input{lec_1.tex}

\section{Alexander polynomial \hfill\DTMdate{2019-03-04}}
%\input{lec_2.tex}
%add Hurewicz theorem?


\section{Examples of knot classes
\hfill\DTMdate{2019-03-11}}
%\input{lec_3.tex}

\section{Concordance group \hfill\DTMdate{2019-03-18}}
%\input{lec_4.tex}

\section{Genus $g$ cobordism \hfill\DTMdate{2019-03-25}}
%\input{lec_5.tex}

\section{\hfill\DTMdate{2019-04-08}}
\input{lec_6.tex}

\section{\hfill\DTMdate{2019-04-15}}
???????????????????\\
\begin{theorem}
Suppose that $K \subset S^3$ is a slice knot (i.e. $K$ bound a disk in $B^4$). 
Then if $F$ is a Seifert surface of $K$ and $V$ denotes the associated Seifet matrix, then there exists $P \in \Gl_g(\mathbb{Z})$ such that:
\\??????????????? T ????????
\begin{align}
PVP^{-1} = 
\begin{pmatrix}
0 & A\\
B & C
\end{pmatrix}, \quad A, C, C \in M_{g \times g} (\mathbb{Z})
\end{align}
\end{theorem}
In other words you can find rank $g$ direct summand $\mathcal{Z}$ of $H_1(F)$ \\
????????????\\
such that for any 
$\alpha, \beta \in  \mathcal{L}$ the linking number $\Lk (\alpha, \beta^+) = 0$. 
\begin{definition}
An abstract Seifert matrix (i. e. 
\end{definition}
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*}
In particular $\vert \det A\vert = \# H_1(Y, \mathbb{Z})$.\\
That means - what is happening on boundary is a measure of degeneracy.  

\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_1(Y, \mathbb{Z}) &
  \times \quad H_1(Y, \mathbb{Z})&
  \longrightarrow &
  \quot{\mathbb{Q}}{\mathbb{Z}}
  \text{ - a linking form}
\\
   \quot{\mathbb{Z}^n}{A\mathbb{Z}} \ar[u,isomorphic] & 
   \quot{\mathbb{Z}^n}{A\mathbb{Z}} \ar[u,isomorphic] &\\
\end{tikzcd}
$(a, b) \mapsto aA^{-1}b^T$
\end{center}
?????????????????????????????????\\
\noindent
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 
$A = V \times V^T$, $n = \rank V$.
%\input{ink_diag}
\begin{figure}[h]
\fontsize{20}{10}\selectfont
\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}
\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}
\noindent
Let  $Y = \Sigma(K)$. Then:
\begin{align*}
H_1(Y, \mathbb{Z}) \times H_1(Y, \mathbb{Z}) &\longrightarrow \quot{\mathbb{Q}}{\mathbb{Z}}
\\
(a,b) &\mapsto a A^{-1} b^{T},\qquad
A = V + V^T.
\end{align*}
????????????????????????????
\\
We have a primary decomposition of $H_1(Y, \mathbb{Z}) = U$ (as a group). For any $p \in \mathbb{P}$ we define $U_p$ to be the subgroup of elements annihilated by the same power of $p$. We have $U = \bigoplus_p U_p$. 
\begin{example}
\begin{align*}
\text{If } U &= 
\mathbb{Z}_3 \oplus
\mathbb{Z}_{45} \oplus
\mathbb{Z}_{15} \oplus
\mathbb{Z}_{75} 
\text{ then }\\
U_3 &= 
\mathbb{Z}_3 \oplus
\mathbb{Z}_9 \oplus
\mathbb{Z}_3 \oplus
\mathbb{Z}_3
\text{ and }\\
U_5 &= 
(e) \oplus
\mathbb{Z}_5 \oplus
\mathbb{Z}_5 \oplus
\mathbb{Z}_{25}.
\end{align*}
\end{example}

\begin{lemma}
Suppose $x \in U_{p_1}$, $y \in U_{p_2}$ and $p_1 \neq p_2$. Then $<x, y > = 0$.
\end{lemma}
\begin{proof}
\begin{align*}
x \in U_{p_1} 
\end{align*}
\end{proof}
\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*}
???????????????????????\\
In general 

%no lecture at 29.04

\section{\hfill\DTMdate{2019-05-20}}

Let $M$ be compact, oriented, connected four-dimensional manifold. If ${H_1(M, \mathbb{Z}) = 0}$ then there exists a 
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 $Y = \partial M$. 
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})
\times  H_2(M, \mathbb{Z})
&\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})
\end{align*}
has coker precisely $H_1(Y, \mathbb{Z})$. 
\\???????????????\\
Let $K \subset S^3$  be a knot, $X = S^3 \setminus K$  a knot complement and
$\widetilde{X} \xrightarrow{\enspace \rho \enspace} X$  an infinite cyclic cover (universal abelian cover). By Hurewicz theorem we know that:
\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. \\
Let $H_1(\widetilde{X}, \mathbb{Z}[t, t^{-1}])$ be the Alexander module of the knot $K$ with an intersection form: 
\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*}
&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.}
\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) \quad &\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 primary decomposition of this module.  
We can simplify this problem by replacing $\mathbb{Z}$ by $\mathbb{R}$. We lose some date by doing this transition, but we can 
\begin{align*}
\xi \in S^1 \setminus \{ \pm 1\} 
&\quad
p_{\xi} = 
(t - \xi)(t - \xi^{-1}) t^{-1}
\\
\xi \in \mathbb{R} \setminus \{ \pm 1\}
&\quad
q_{\xi} = (t - \xi)(t - \xi^{-1}) t^{-1}
\\
\xi \notin \mathbb{R} \cup S^1 
&\quad
q_{\xi} = (t - \xi)(t - \overbar{\xi})(t - \xi^{-1})
(t - \overbar{\xi}^{-1}) t^{-2}
\end{align*}
Let $\Lambda = \mathbb{R}[t, t^{-1}]$. Then: 
\begin{align*}
H_1(\widetilde{X}, \Lambda) \cong  \bigoplus_{\substack{\xi \in S^1 \setminus \{\pm 1 \}\\ k\geq 0}}
( \quot{\Lambda}{p_{\xi}^k })^{n_k, \xi}
\oplus
\bigoplus_{\substack{\xi \notin S^1 \\ l\geq 0}}
(\quot{\Lambda}{q_{\xi}^l})^{n_l, \xi}&
\end{align*}
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}
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 $g$. 
\end{lemma}
\begin{proof}[Sketch of proof]
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:
\begin{align*}
&P^{\prime} = \frac{P}{(t - \zeta)(t - \overbar{\zeta})(t^{-1} - \zeta)(t^{-1} - \overbar{\zeta})}\\
&P^{\prime} = g^{\prime}\overbar{g}
\end{align*}
We set  $g = g^{\prime}(t - \zeta)(t - \overbar{\zeta})$ and
$P = g \overbar{g}$. Suppose $\zeta \in S^1$. Then $(t - \zeta)^2 \vert P$ (at least - otherwise it would change sign). Therefore:
\begin{align*}
&P^{\prime} = \frac{P}{(t - \zeta)^2(t^{-1} - \zeta)^2}\\
&g = (t - \zeta)(t^{-1} - \zeta) g^{\prime} \quad \text{etc.}
\end{align*}
The map $(1, 1) \mapsto \frac{h}{p^k} = \frac{g\overbar{g}h}{p^k}$ is isometric whenever $g$ is coprime with $P$.
\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 .
\\??????????????????????????\\
\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*}
Apply Lemma \ref{L:coprime polynomials} for $A=h$, $B=p^{2k}$. 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*}
???????????????????????????????\\
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. 
\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?\\
\begin{theorem}(Matumoto, Borodzik-Conway-Politarczyk)
Let $K$ be a knot, 
\begin{align*}
&H_1(\widetilde{X}, \Lambda) \times
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}
\section{\hfill\DTMdate{2019-05-27}}

....
\begin{definition}
A square hermitian matrix $A$ of size $n$ with coefficients in \\
the Blanchfield pairing if:
$H_1(\bar{X}$
\end{definition}

field of fractions

\section{Surgery \hfill\DTMdate{2019-06-03}}
\begin{theorem}
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:
\[
u(K) \geq g_4(K)
\]
\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 disks and glue the Seifert surface for the Hopf link. The reality surface $S$ has Euler characteristic $\chi(S) = 1 - 2u$. Therefore $g_4(S) = u$.
\end{proof}
%Tim D. Cochran and Peter Teichner
\begin{example}
The knot $8_{20}$ is slice: $\sigma \equiv 0$ almost everywhere but $\sigma(e^{\frac{ 2\pi i}{6}}) = + 1$.
\end{example}
%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 $H_1(S^1 \times S^1, \mathbb{Z}) = \mathbb{Z}^2$. 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. 
Consider an induced map on the 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*} 
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}
\vspace{10cm}
\begin{theorem}
Every such a matrix can be realized as a torus. 
\end{theorem}
\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}
\begin{figure}[h]
\fontsize{20}{10}\selectfont
\centering{
\def\svgwidth{\linewidth}
\resizebox{0.5\textwidth}{!}{\input{images/dehn_twist.pdf_tex}}
\caption{Dehn twist.}
\label{fig:dehn_twist}
}
\end{figure}




\section{balagan}

\noindent
\noindent

\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. 
\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}


\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}

\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}
%
%
\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
%Add NotePrintSaveCiteYour opinionEmailShare
%Saveliev, Nikolai

%Lectures on the Topology of 3-Manifolds
%An Introduction to the Casson Invariant

\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}

\end{document}