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


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

\section{\hfill\DTMdate{2019-02-25}}
\begin{definition}
A knot $K$ in $S^3$ is a smooth (PL - smooth) embedding of a circle $S^1$ in $S^3$:
\begin{align*}
\varphi:  S^1 \hookrightarrow  S^3
\end{align*}   
\end{definition}
\noindent
Usually we think about a knot as an image of an embedding: $K = \varphi(S^1)$.
\begin{example}
\begin{itemize}
\item 
Knots:
\includegraphics[width=0.08\textwidth]{unknot.png},
\includegraphics[width=0.08\textwidth]{trefoil.png}.
\item 
Not knots: 
\includegraphics[width=0.12\textwidth]{not_injective_knot.png}
(it is not an injection),
\includegraphics[width=0.08\textwidth]{not_smooth_knot.png}
(it is not smooth).
\end{itemize}
\end{example}
\begin{definition}
%\hfill\\
Two knots $K_0 = \varphi_0(S^1)$, $K_1 = \varphi_1(S^1)$ are equivalent if the embeddings $\varphi_0$ and $\varphi_1$ are isotopic, that is there exists a continues function 
\begin{align*}
&\Phi: S^1 \times  [0, 1] \hookrightarrow S^3 \\
&\Phi(x, t) = \Phi_t(x)
\end{align*}
 such that $\Phi_t$ is an embedding for any $t \in [0,1]$, $\Phi_0 = \varphi_0$ and
$\Phi_1 = \varphi_1$.
\end{definition}

\begin{theorem}
Two knots $K_0$ and $K_1$ are isotopic if and only if they are ambient isotopic, i.e. there exists a family of self-diffeomorphisms $\Psi = \{\psi_t: t \in [0, 1]\}$ such that:
\begin{align*}
&\psi(t) = \psi_t \text{ is continius on $t\in [0,1]$}\\
&\psi_t: S^3 \hookrightarrow S^3,\\
& \psi_0 = id ,\\
& \psi_1(K_0) = K_1.
\end{align*}
\end{theorem}
\begin{definition}
A knot is trivial (unknot) if it is equivalent to an embedding $\varphi(t) = (\cos t, \sin t, 0)$, where $t \in [0, 2 \pi] $ is a parametrisation of $S^1$.
\end{definition}
\begin{definition}
A link with k - components is a (smooth) embedding of $\overbrace{S^1 \sqcup \ldots \sqcup S^1}^k$ in $S^3$
\end{definition}
\begin{example}
Links:
\begin{itemize}
\item
a trivial link with $3$ components:
\includegraphics[width=0.2\textwidth]{3unknots.png},
\item
a hopf link: \includegraphics[width=0.13\textwidth]{Hopf.png},
\item
a Whitehead link:
\includegraphics[width=0.13\textwidth]{WhiteheadLink.png},
\item
Borromean link:
\includegraphics[width=0.1\textwidth]{BorromeanRings.png},
\end{itemize}
\end{example}
%
%
%
\begin{definition}
A link diagram is a picture over projection of a link is $S^3$($\mathbb{R}^3$) such that:
\begin{enumerate}[label={(\arabic*)}]
\item 
${D_{\pi}}_{\big|L}$ is non degenerate: \includegraphics[width=0.05\textwidth]{LinkDiagram1.png},
\item the double points are not degenerate: \includegraphics[width=0.05\textwidth]{LinkDiagram2.png},
\item there are no triple point:  \includegraphics[width=0.05\textwidth]{LinkDiagram3.png}.
\end{enumerate}
\end{definition}
There are under- and overcrossings  (tunnels and bridges) on a link diagrams with an obvious meaning.\\
Every link admits a link diagram.
%\begin{comment}

\section*{Reidemeister moves}
A Reidemeister move is one of the three types of operation on a link diagram as shown in Figure~\ref{fig: reidemeister}.
%
The first Reidemeister move inserts or removes a coil. 
%
The second Reidemeister move slides a strand and inserts or removes two crossings of opposite sign. 
%
The third Reidemeister move slides a strand over or under  a crossing.
\begin{figure}[H]
\centering
\includegraphics[width=0.7\textwidth]{moves.png}
\caption{\label{fig: reidemeister}Reidemeister moves (adapted from Adams).}
\end{figure}
\begin{theorem} [Reidemeister’s Theorem]
Two diagrams of the same link can be
deformed into each other by a finite sequence of Reidemeister moves (and isotopy of the plane).
\end{theorem}

\section{}
\begin{example}
\begin{align*}
&F: \mathbb{C}^2 \rightarrow \mathbb{C}  \text{ a polynomial} \\ 
&F(0) = 0
\end{align*}
Fact (Milnor Singular Points of Complex Hypersurfaces):
\end{example}
%\end{comment}

An oriented knot is called negative amphichiral if the mirror image $m(K)$ if $K$ is equivalent the reverse knot of $K$. \\
\begin{example}[Problem]
Prove that if $K$ is negative amphichiral, then $K \# K$ in 
$\mathbf{C}$
\end{example}

\section{\hfill\DTMdate{2019-03-04}}
\begin{proof}("joke")\\
Let $K \in S^3$ be a knot and $N$ be its tubular neighbourhood. 
\begin{align*}
H^1(S^3 \setminus N ) \cong H^1(S^3 \setminus K)
\end{align*}
For a pair $(S^3, S^3 \setminus N)$ we have:
\begin{align*}
H^0(S^3)
\end{align*}
\end{proof}
\section{\hfill\DTMdate{2019-03-18}}
\begin{definition}
A knot $K$ is called (smoothly) slice if $K$ is smoothly concordant to an unknot. \\
A knot $K$ is smoothly slice if and only if $K$ bounds a smoothly embedded disk in $B^4$.
\end{definition}
\begin{definition}
Two knots $K$ and $K^{\prime}$ are called (smoothly) concordant if there exists an annulus $A$ that is smoothly embedded in $S^3 \times [0, 1]$ such that
$\partial A = K^{\prime} \times \{1\} \; \sqcup \; K \times \{0\} $.
\end{definition}
Let $m(K)$ denote a mirror image of a knot $K$. 
\begin{fact}
For any $K$, $K \# m(K)$ is slice.
\end{fact}
\begin{fact}
Concordance is an equivalence relation.
\end{fact}
\begin{fact}
If $K_1 \sim {K_1}^{\prime}$ and $K_2 \sim {K_2}^{\prime}$, then
$K_1 \# K_2 \sim {K_1}^{\prime} \# {K_2}^{\prime}$.
\end{fact}
\begin{fact}
$K \# m(K) \sim $ the unknot.
\end{fact}
\noindent
Let $\mathscr{C}$ denote all equivalent classes for knots. $\mathscr{C}$ is a group under taking connected sums, with neutral element (the class defined by) an unknot and inverse element (a class defined by) a mirror image.\\
The figure eight knot is a torsion element in $\mathscr{C}$ ($2K \sim $ the unknot).\\
\begin{example}[Problem]
Are there in concordance group torsion elements that are not $2$ torsion elements? (open)
\end{example}
\noindent
Remark: $K \sim K^{\prime} \Leftrightarrow K \# -K^{\prime}$ is slice.

%
%
\section{\hfill\DTMdate{2019-04-08}}
%
%
$X$ 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*}
H_2(X, \mathbb{Z}) \xrightarrow{\text{Poincaré duality}} H^2(X, \mathbb{Z} ) \xrightarrow{\text{evaluation}}\Hom(H_2(X, \mathbb{Z}), \mathbb{Z})
\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 $X$. 


\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*}
In particular $\mid \det A\mid = \# H_1(Y, \mathbb{Z}$.\\
That means - what is happening on boundary is a measure of degeneracy.  
\\
\vspace{1cm}
\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}
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 
$A = V \times V^T$, where $n = \rank V$.
%\input{ink_diag}
\begin{figure}[H]
\fontsize{40}{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}
\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\\
&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 

\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: $H_2(M, \mathbb{Z})
\times  H_2(M, \mathbb{Z})
\longrightarrow 
\mathbb{Z}$ can be degenerate in the sense that
\begin{align*}
H_2(M, \mathbb{Z}) \longrightarrow \Hom (H_2(M, \mathbb{Z}), \mathbb{Z})\\
(a, b) \mapsto \mathbb{Z}\\
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,  \\
$\widetilde{X} \xrightarrow{\enspace \rho \enspace} X$ - 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*}
&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) &\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. 
\begin{align*}
&\xi \in S^1 \setminus \{ \pm 1\} 
\quad
p_{\xi} = 
(t - \xi)(1 - \xi^{-1}) t^{-1}\\
&\xi \in \mathbb{R} \setminus \{ \pm 1\}
\quad
q_{\xi} = (t - \xi)(1 - \xi^{-1}) t^{-1}\\
&\xi \notin \mathbb{R} \cup S^1 \quad
q_{\xi} = (t - \xi)(t - \overbar{\xi})(1 - \xi^{-1})(1 - \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}}
( \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
\begin{align*}
(t - \zeta) \mid p,\\
(t - \overbar{\zeta}) \mid p,\\
(t^{-1} - \zeta) \mid p,\\
(t^{-1} - \overbar{\zeta}) \mid p,\\
\end{align*}
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}\\
\text{we set } g = g^{\prime}(t - \zeta)(t - \overbar{\zeta}\\
p = g \overbar{g}
\end{align*}
Suppose $\zeta \in S^1$. Then $(t - \zeta)^2 \mid p$ (at least - otherwise it would change sign). 
\begin{align*}
p^{\prime} &= \frac{p}{(t - \zeta)^2(t^{-1} - \zeta)^2}\\
g &= (t - \zeta)(t^{-1} - \zeta) g^{\prime} \quad \text{etc.}\\
(1, 1) \mapsto \frac{h}{p^k} = \frac{g\overbar{g}h}{p^k}  \quad \text{ isometry whenever $g$ is coprime with $p$.}
\end{align*}
\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{align*}
(1, 1) \mapsto \frac{h}{p^k} \mapsto \frac{g\overbar{g}h}{p^k}\\
g\overbar{g} h + p^k\omega = 1
\end{align*}
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, Conway-Borodzik-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 $n$.  
\end{definition}

field of fractions


\section{balagan}

\end{document}