lecture from 20.05

lectures_on_knot_theory.tex

@@ -1,8 +1,8 @@
 \documentclass[12pt, twoside]{article}
 
-\usepackage{comment}
 \usepackage{amssymb}
 \usepackage{amsmath}
+\usepackage{xfrac}
 \usepackage[english]{babel}
 \usepackage{csquotes}
 \usepackage{graphicx}
@@ -10,18 +10,25 @@
 \usepackage{titlesec}
 \usepackage{comment}
 \usepackage{pict2e}
-
+\usepackage{hyperref}
 \usepackage{advdate}
+\usepackage{amsthm}
+\usepackage[useregional]{datetime2}
 
-%... Set the first lecture date
-\ThisYear{2019}
-\ThisMonth{3}
-\ThisDay{5}
-
+\hypersetup{
+    colorlinks,
+    citecolor=black,
+    filecolor=black,
+    linkcolor=black,
+    urlcolor=black
+}
+\usepackage{fontspec}
+\usepackage{mathtools}
+\usepackage{unicode-math}
 
 \graphicspath{ {images/} }
 
-\newtheorem{lemama}{Lemma}
+\newtheorem{lemma}{Lemma}
 \newtheorem{fact}{Fact}
 \newtheorem{example}{Example}
 %\theoremstyle{definition}
@@ -29,43 +36,66 @@
 %\theoremstyle{plain}
 \newtheorem{theorem}{Theorem}
 \newtheorem{proposition}{Proposition}
+\newcommand{\contradiction}{%
+  \ensuremath{{\Rightarrow\mspace{-2mu}\Leftarrow}}%
+}
+
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+%%%% For quotient groups / modding equiv relations
+%%%% Use: \quot{A}{B} --> A/B
+\newcommand*\quot[2]{{^{\textstyle #1}\big/_{\textstyle #2}}}
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+\newcommand{\overbar}[1]{\mkern 1.5mu\overline{\mkern-1.5mu#1\mkern-1.5mu}\mkern 1.5mu}
+\DeclareMathOperator{\Hom}{Hom}
+\AtBeginDocument{\renewcommand{\setminus}{\mathbin{\backslash}}}
 
 \input{knots_macros}
 
-
-\titleformat{\section}{\normalfont \Large \bfseries}
+\titleformat{\section}{\normalfont \large \bfseries}
 {Lecture\ \thesection}{2.3ex plus .2ex}{} 
 \titlespacing{\subsection}{2em}{*1}{*1}
 
 
 \begin{document}
+%\tableofcontents
+%\newpage
 %\input{myNotes}
 
 \section{}
+\begin{flushright}
+\DTMdate{2019-02-25}
+\end{flushright}
 \begin{definition}
-A \textbf{knot} $K$ in $S^3$ is a smooth (PL - smooth) embedding of a circle $S^1$ in $S^3$:
+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}
+!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
+knot and not a knot (not inection), not smooth, 
+\end{example}
 \begin{definition}
-\hfill\\
+%\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
-$\Phi_1 = \varphi_1$
-\\
-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 $\Phi$ such that:
-\begin{align*}
-&\Psi: S^3 \hookrightarrow S^3\\
-& \psi_0 = id\\
-& \psi_1(K_0) = K_1
-\end{align*}
+$\Phi_1 = \varphi_1$.
 \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 $\Psi$ such that:
+\begin{align*}
+&\Psi: 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}
@@ -73,21 +103,36 @@ A knot is trivial (unknot) if it is equivalent to an embedding $\varphi(t) = (\c
 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}
-A trivial link with $3$ components\\
-A hopf link\\
-Whitehead link\\
+Links:
+\begin{itemize}
+\item
+a trivial link with $3$ components:
+\includegraphics[width=0.13\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 $S^3$/$R^3$ such that:
+A link diagram is a picture over projection of a link is $S^3$($\mathbb{R}^3$) such that:
 \begin{enumerate}
-\item is non degenerate 
-\item The double points are not degenerated 
-\item There are no triple point
+\item 
+${D_{\pi}}_{\big|L}$ is non degenerate
+\includegraphics[width=0.02\textwidth]{LinkDiagram1.png},
+\item the double points are not degenerate
+\includegraphics[width=0.02\textwidth]{LinkDiagram2.png},
+\item there are no triple point
+\includegraphics[width=0.03\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}
 
 \subsection{Reidemeister moves}
 A Reidemeister move is one of the three types of operation on a link diagram as shown in Figure~\ref{fig: reidemeister}.
@@ -106,14 +151,328 @@ The third Reidemeister move slides a strand over or under  a crossing.
 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{Z nagrania Kamili}
+
+\section{}
 \begin{example}
 \begin{align*}
-&F: \mathbb{C}^2 \rightarrow \mathbb{C}  \text{a polynomial} \\ 
+&F: \mathbb{C}^2 \rightarrow \mathbb{C}  \text{ a polynomial} \\ 
 &F(0) = 0
 \end{align*}
 Fact (Milnor Singular Points of Complex Hypersurfaces):
 \end{example}
-\section{} 25.03.19 
+\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{}
+\begin{flushright}
+\DTMdate{2019-03-04}
+\end{flushright}
+\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{}
+\begin{flushright}
+\DTMdate{2019-03-18}
+\end{flushright}
+\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{}
+\begin{flushright}
+\DTMdate{2019-04-08}
+\end{flushright}
+$X$ 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{} 
+\begin{flushright}
+\DTMdate{2019-05-20}
+\end{flushright}
+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}
+\end{align*}
+where $V$ is a Seifert matrix.
+\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{}
+\begin{flushright}
+\DTMdate{2019-05-27}
+\end{flushright}
+....
+\begin{definition}
+A square hermitian matrix $A$ of size $n$.  
+\end{definition}
+
+field of fractions
+
+\section{}
+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{align*}
+H_1(Y, \mathbb{Z}) \times
+H_1(Y, \mathbb{Z}) \longrightarrow \quot{\mathbb{Q}}{\mathbb{Z}} \text{ - a linking form}
+\end{align*}
 \end{document}