diff --git a/images/3unknots.png b/images/3unknots.png new file mode 100644 index 0000000..aaf791c Binary files /dev/null and b/images/3unknots.png differ diff --git a/images/BorromeanRings.png b/images/BorromeanRings.png new file mode 100644 index 0000000..36e252a Binary files /dev/null and b/images/BorromeanRings.png differ diff --git a/images/Hopf.png b/images/Hopf.png new file mode 100644 index 0000000..96d429f Binary files /dev/null and b/images/Hopf.png differ diff --git a/images/LinkDiagram1.png b/images/LinkDiagram1.png new file mode 100644 index 0000000..d8b750d Binary files /dev/null and b/images/LinkDiagram1.png differ diff --git a/images/LinkDiagram2.png b/images/LinkDiagram2.png new file mode 100644 index 0000000..5a713a9 Binary files /dev/null and b/images/LinkDiagram2.png differ diff --git a/images/LinkDiagram3.png b/images/LinkDiagram3.png new file mode 100644 index 0000000..10db52f Binary files /dev/null and b/images/LinkDiagram3.png differ diff --git a/images/WhiteheadLink.png b/images/WhiteheadLink.png new file mode 100644 index 0000000..eecf317 Binary files /dev/null and b/images/WhiteheadLink.png differ diff --git a/images/unknot_and_trefoil.png b/images/unknot_and_trefoil.png new file mode 100644 index 0000000..0893080 Binary files /dev/null and b/images/unknot_and_trefoil.png differ diff --git a/lectures_on_knot_theory.tex b/lectures_on_knot_theory.tex index 8727e23..c9d37b2 100644 --- a/lectures_on_knot_theory.tex +++ b/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} \ No newline at end of file