From 42c73cbb864daca988ae4fc687f298c6869e6337 Mon Sep 17 00:00:00 2001 From: Maria Marchwicka Date: Wed, 3 Jul 2019 00:06:01 -0500 Subject: [PATCH] change of names --- lec_03_06.tex | 63 +++++++++++ lec_04_03.tex | 281 ++++++++++++++++++++++++++++++++++++++++++++++++++ lec_06_05.tex | 144 ++++++++++++++++++++++++++ lec_08_04.tex | 150 +++++++++++++++++++++++++++ lec_11_03.tex | 176 +++++++++++++++++++++++++++++++ lec_15_04.tex | 135 ++++++++++++++++++++++++ lec_18_03.tex | 170 ++++++++++++++++++++++++++++++ lec_20_05.tex | 252 ++++++++++++++++++++++++++++++++++++++++++++ lec_25_02.tex | 280 +++++++++++++++++++++++++++++++++++++++++++++++++ lec_25_03.tex | 234 +++++++++++++++++++++++++++++++++++++++++ lec_27_05.tex | 8 ++ lec_7.tex | 0 lec_8.tex | 0 lec_9.tex | 0 14 files changed, 1893 insertions(+) create mode 100644 lec_03_06.tex create mode 100644 lec_04_03.tex create mode 100644 lec_06_05.tex create mode 100644 lec_08_04.tex create mode 100644 lec_11_03.tex create mode 100644 lec_15_04.tex create mode 100644 lec_18_03.tex create mode 100644 lec_20_05.tex create mode 100644 lec_25_02.tex create mode 100644 lec_25_03.tex create mode 100644 lec_27_05.tex delete mode 100644 lec_7.tex delete mode 100644 lec_8.tex delete mode 100644 lec_9.tex diff --git a/lec_03_06.tex b/lec_03_06.tex new file mode 100644 index 0000000..fefe472 --- /dev/null +++ b/lec_03_06.tex @@ -0,0 +1,63 @@ +\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} + + + diff --git a/lec_04_03.tex b/lec_04_03.tex new file mode 100644 index 0000000..1519178 --- /dev/null +++ b/lec_04_03.tex @@ -0,0 +1,281 @@ +\subsection{Existence of Seifert surface - second proof} +%\begin{theorem} +%For any knot $K \subset S^3$ there exists a connected, compact and orientable surface $\Sigma(K)$ such that $\partial \Sigma(K) = K$ +%\end{theorem} +\begin{proof}(Theorem \ref{theo:Seifert})\\ +Let $K \in S^3$ be a knot and $N = \nu(K)$ be its tubular neighbourhood. Because $K$ and $N$ are homotopy equivalent, we get: +\begin{align*} +H^1(S^3 \setminus N ) \cong H^1(S^3 \setminus K). +\end{align*} +Let us consider a long exact sequence of cohomology of a pair $(S^3, S^3 \setminus N)$ with integer coefficients: + +\begin{center} +\begin{tikzcd} +[ + column sep=0cm, fill=none, + row sep=small, + ar symbol/.style =% + {draw=none,"\textstyle#1" description,sloped}, + isomorphic/.style = {ar symbol={\cong}}, +] +&\mathbb{Z} +\\ + +& H^0(S^3) \ar[u,isomorphic] \to +&H^0(S^3 \setminus N) \to +\\ +\to H^1(S^3, S^3 \setminus N) \to + & H^1(S^3) \to + & H^1(S^3\setminus N) \to + \\ +& 0 \ar[u,isomorphic]& + \\ + \to H^2(S^3, S^3 \setminus N) \to + & H^2(S^3) \ar[u,isomorphic] \to + & H^2(S^3\setminus N) \to + \\ +\to H^3(S^3, S^3\setminus N)\to +& H^3(S) \to +& 0 +\\ +& \mathbb{Z} \ar[u,isomorphic] &\\ + \end{tikzcd} +\end{center} +\begin{align*} +N \cong & D^2 \times S^1\\ +\partial N \cong & S^1 \times S^1\\ +H^1(N, \partial N) \cong & \mathbb{Z} \oplus \mathbb{Z} +\end{align*} +\begin{align*} +H^* (S^3, S^3 \setminus N) &\cong H^* (N, \partial N)\\ +\\ +H^ 1 (S^3\setminus N) &\cong H^1(S^3\setminus K) \cong \mathbb{Z} +\end{align*} +\begin{equation*} +\begin{tikzcd}[row sep=huge] +H^1(S^3 \setminus K) \arrow[r,] \arrow[d,"\widetilde{\Theta}"] & +H^1(N \setminus K) \arrow[d,"\Theta"] \\ +{[S^3 \setminus K, S^1]} \arrow[r,]& +{[N \setminus K, S^1]} +\end{tikzcd} +\end{equation*} +\noindent +$\Sigma = \widetilde{\Theta}^{-1}(X)$ is a surface, such that $\partial \Sigma = K$, so it is a Seifert surface. +% +% +% Thom isomorphism, +\end{proof} +\subsection{Alexander polynomial} +\begin{definition} +Let $S$ be a Seifert matrix for a knot $K$. The Alexander polynomial $\Delta_K(t)$ is a Laurent polynomial: +\[ +\Delta_K(t) := \det (tS - S^T) \in +\mathbb{Z}[t, t^{-1}] \cong \mathbb{Z}[\mathbb{Z}] +\] +\end{definition} + +\begin{theorem} +$\Delta_K(t)$ is well defined up to multiplication by $\pm t^k$, for $k \in \mathbb{Z}$. +\end{theorem} +\begin{proof} +We need to show that $\Delta_K(t)$ doesn't depend on $S$-equivalence relation. +\begin{enumerate}[label={(\arabic*)}] +\item Suppose $S\prime = CSC^T$, $C \in \Gl(n, \mathbb{Z})$ (matrices invertible over $\mathbb{Z}$). Then $\det C = 1$ and: +\begin{align*} +&\det(tS\prime - S\prime^T) = +\det(tCSC^T - (CSC^T)^T) =\\ +&\det(tCSC^T - CS^TC^T) = +\det C(tS - S^T)C^T = +\det(tS - S^T) +\end{align*} +\item +Let \\ +$ A := t +\begin{pmatrix} + \begin{array}{c|c} + S & + \begin{matrix} + \ast & 0 \\ + \sdots & \sdots\\ + \ast & 0 + \end{matrix} \\ + \hline + \begin{matrix} + \ast & \dots & \ast\\ + 0 & \dots & 0 + \end{matrix} + & + \begin{matrix} + 0 & 0\\ + 1 & 0 + \end{matrix} + \end{array} +\end{pmatrix} +- +\begin{pmatrix} + \begin{array}{c|c} + S^T & + \begin{matrix} + \ast & 0 \\ + \sdots & \sdots\\ + \ast & 0 + \end{matrix} \\ + \hline + \begin{matrix} + \ast & \dots & \ast\\ + 0 & \dots & 0 + \end{matrix} + & + \begin{matrix} + 0 & 1\\ + 0 & 0 + \end{matrix} + \end{array} +\end{pmatrix} += +\begin{pmatrix} + \begin{array}{c|c} + tS - S^T & + \begin{matrix} + \ast & 0 \\ + \sdots & \sdots\\ + \ast & 0 + \end{matrix} \\ + \hline + \begin{matrix} + \ast & \dots & \ast\\ + 0 & \dots & 0 + \end{matrix} + & + \begin{matrix} + 0 & -1\\ + t & 0 + \end{matrix} + \end{array} +\end{pmatrix} +$ +\\ +\\ +Using the Laplace expansion we get $\det A = \pm t \det(tS - S^T)$. +\end{enumerate} +\end{proof} +% +% +% +\begin{example} +If $K$ is a trefoil then we can take +$S = \begin{pmatrix} +-1 & -1 \\ +0 & -1 +\end{pmatrix}$. Then +\[ +\Delta_K(t) = \det +\begin{pmatrix} +-t + 1 & -t\\ +1 & -t +1 +\end{pmatrix} += (t -1)^2 + t = t^2 - t +1 \ne 1 +\Rightarrow \text{trefoil is not trivial.} +\] +\end{example} +\begin{fact} +$\Delta_K(t)$ is symmetric. +\end{fact} +\begin{proof} +Let $S$ be an $n \times n$ matrix. +\begin{align*} +&\Delta_K(t^{-1}) = \det (t^{-1}S - S^T) = (-t)^{-n} \det(tS^T - S) = \\ +&(-t)^{-n} \det (tS - S^T) = (-t)^{-n} \Delta_K(t) +\end{align*} +If $K$ is a knot, then $n$ is necessarily even, and so $\Delta_K(t^{-1}) = t^{-n} \Delta_K(t)$. +\end{proof} +\begin{lemma} +\begin{align*} +\frac{1}{2} \deg \Delta_K(t) \leq g_3(K), +\text{ where } deg (a_n t^n + \dots + a_1 t^l )= k - l. +\end{align*} +\end{lemma} +\begin{proof} +If $\Sigma$ is a genus $g$ - Seifert surface for $K$ then $H_1(\Sigma) = \mathbb{Z}^{2g}$, so $S$ is an $2g \times 2g$ matrix. Therefore $\det (tS - S^T)$ is a polynomial of degree at most $2g$. +\end{proof} +\begin{example} +There are not trivial knots with Alexander polynomial equal $1$, for example: +\includegraphics[width=0.3\textwidth]{11n34.png} +$\Delta_{11n34} \equiv 1$. +\end{example} + +\subsection{Decomposition of $3$-sphere} +We know that $3$ - sphere can be obtained by gluing two solid tori: +\[ +S^3 = \partial D^4 = \partial (D^2 \times D^2) = (D^2 \times S^1) \cup (S^1 \times D^2). +\] +So the complement of solid torus in $S^3$ is another solid torus.\\ +Analytically it can be describes as follow. \\ +Take $(z_1, z_2) \in \mathbb{C}$ such that ${\max(\vert z_1 \vert, \vert z_2\vert) = 1.} +$ +Define following sets: +\begin{align*} +S_1 = \{ (z_1, z_2) \in S^3: \vert z_1 \vert = 0\} \cong S^1 \times D^2 ,\\ +S_2 = \{(z_1, z_2) \in S ^3: \vert z_2 \vert = 1 \} \cong D^2 \times S^1. +\end{align*} +The intersection +$S_1 \cap S_2 = \{(z_1, z_2): \vert z_1 \vert = \vert z_2 \vert = 1 \} \cong S^1 \times S^1$. +\begin{figure}[h] +\centering{ +\def\svgwidth{\linewidth} +\resizebox{0.3\textwidth}{!}{\includegraphics[width=0.3\textwidth]{sphere_as_torus.png}} +\caption{The complement of solid torus in $S^3$ is another solid torus.} +\label{fig:sphere_as_tori} +} +\end{figure} + +\subsection{Dehn lemma and sphere theorem} +%removing one disk from surface doesn't change $H_1$ (only $H_2$) +% +% +% +\begin{lemma}[Dehn] +Let $M$ be a $3$-manifold and $D^2 \overset{f} \rightarrow M^3$ be a map of a disk such that $f\big|_{\partial D^2}$ is an embedding. Then there exists an embedding +${D^2 \overset{g}\longhookrightarrow M}$ such that: +\[ +g\big|_{\partial D^2} = f\big|_{\partial D^2.} +\] +\end{lemma} +\noindent +Remark: Dehn lemma doesn't hold for dimension four.\\ +Let $M$ be connected, compact three manifold with boundary. +Suppose $\pi_1(\partial M) \longrightarrow \pi_1(M)$ has non-trivial kernel. Then there exists a map $f: (D^2, \partial D^2) \longrightarrow (M, \partial M)$ such that $f\big|_{\partial D^2}$ is non-trivial loop in $\partial M$. +\begin{theorem}[Sphere theorem] +Suppose $\pi_1(M) \ne 0$. Then there exists an embedding $f: S^2 \hookrightarrow M$ that is homotopy non-trivial. +\end{theorem} +\begin{problem} +Prove that $S^3 \ K$ is Eilenberg–MacLane space of type $K(\pi, 1)$. +\end{problem} +\begin{corollary} +Suppose $K \subset S^3$ and $\pi_1(S^3 \setminus K)$ is infinite cyclic ($\mathbb{Z})$. Then $K$ is trivial. +\end{corollary} +\begin{proof} +Let $N$ be a tubular neighbourhood of a knot $K$ and $M = S^3 \setminus N$ its complement. Then $\partial M = S^1 \times S^1$. Let $f : \pi_1(\partial M ) \longrightarrow \pi_1(M)$. +If $\pi_1(M)$ is infinite cyclic group then the map $f$ is non-trivial. Suppose ${\lambda \in \ker (\pi_1(S^1 \times S^1) \longrightarrow \pi_1(M)}$. +There is a map $g: (D^2, \partial D^2) \longrightarrow (M, \partial M)$ such that $g(\partial D^2) = \lambda$.\\ + By Dehn's lemma there exists an embedding ${h: (D^2, \partial D^2) \longhookrightarrow (M, \partial M)}$ such that +$h\big|_{\partial D^2} = f \big|_{\partial D^2}$ and $h(\partial D^2) = \lambda$. +Let $\Sigma$ be a union of the annulus and the image of $\partial D^2$. +\\???? $g_3$?\\ +If $g(\Sigma) = 0$, then $K$ is trivial. \\ +Now we should proof that: +\[ +H_1(M) \cong \mathbb{Z} \Longrightarrow \lambda \in \ker ( \pi_1(S^1 \times S^1) \longrightarrow \pi_1(M)). +\] +\begin{figure}[h] +\fontsize{40}{10}\selectfont +\centering{ +\def\svgwidth{\linewidth} +\resizebox{0.4\textwidth}{!}{\input{images/torus_lambda.pdf_tex}} +} +\caption{$\mu$ is a meridian and $\lambda$ is a longitude.} +\label{fig:meridian_and_longitude} +\end{figure} +Choose a meridian $\mu$ such that $\Lk (\mu, K) = 1$. Recall the definition of linking number via homology group (Definition \ref{def:lk_via_homo}). +$[\mu]$ represents the generator of $H_1(S^3\setminus K, \mathbb{Z})$. From definition of $\lambda$ we know that $\lambda$ is trivial in $H_1(M)$ ($\Lk(\lambda, K) =0$, therefore $[\lambda]$ was trivial in $pi_1(M)$). If $K$ is non-trivial then $\lambda$ is non-trivial in $\pi_1(M)$, but it is trivial in $H_1(M)$. +\end{proof} diff --git a/lec_06_05.tex b/lec_06_05.tex new file mode 100644 index 0000000..e98986c --- /dev/null +++ b/lec_06_05.tex @@ -0,0 +1,144 @@ +\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} + diff --git a/lec_08_04.tex b/lec_08_04.tex new file mode 100644 index 0000000..6b588f1 --- /dev/null +++ b/lec_08_04.tex @@ -0,0 +1,150 @@ +$X$ 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{Poincar\'e duality}} H^2(X, \mathbb{Z} ) \xrightarrow{\text{evaluation}}\Hom(H_2(X, \mathbb{Z}), \mathbb{Z}) +\end{align*} + +Intersection form: +$H_2(X, \mathbb{Z}) \times +H_2(X, \mathbb{Z}) \longrightarrow \mathbb{Z}$ is symmetric and non singular. +\\ +Let $A$ and $B$ be closed, oriented surfaces in $X$. +\\ +\begin{figure}[h] +\fontsize{20}{10}\selectfont +\centering{ +\def\svgwidth{\linewidth} +\resizebox{0.5\textwidth}{!}{\input{images/intersection_form_A_B.pdf_tex}} +} +\caption{$T_X A + T_X B = T_X X$ +}\label{fig:torus_alpha_beta} +\end{figure} +??????????????????????? +\begin{align*} +x \in A \cap B\\ +T_XA \oplus T_X B = T_X X\\ +\{\epsilon_1, \dots , \epsilon_n \} = A \cap C\\ +A \cdot B = \sum^n_{i=1} \epsilon_i +\end{align*} +\begin{proposition} +Intersection form $A \cdot B$ doesn't depend of choice of $A$ and $B$ in their homology classes: +\[ +[A], [B] \in H_2(X, \mathbb{Z}). +\] +\end{proposition} +\noindent +\\ +\subsection{Fundamental cycle} +If $M$ is an $m$ - dimensional close, connected and orientable manifold, then $H_m(M, \mathbb{Z})$ and the orientation of $M$ determined a cycle $[M] \in H_m(M, \mathbb{Z})$, called the fundamental cycle. +\begin{example} +If $\omega$ is an $m$ - form then: +\[ +\int_M \omega = [\omega]([M]), \quad [\omega] \in H^m_\Omega(M), \ [M] \in H_m(M). +\] +\end{example} +\begin{figure}[h] +\fontsize{20}{10}\selectfont +\centering{ +\def\svgwidth{\linewidth} +\resizebox{0.8\textwidth}{!}{\input{images/torus_alpha_beta.pdf_tex}} +} +\caption{$\beta$ cross $3$ times the disk bounded by $\alpha$. +$T_X \alpha + T_X \beta = T_X \Sigma$ +}\label{fig:torus_alpha_beta} +\end{figure} +\begin{example} +K{\"u}nneth +?????????????????????????\\ +Let $X = S^2 \times S^2$. +We know that: +\begin{align*} +&H_2(S^2, \mathbb{Z}) =\mathbb{Z}\\ +&H_1(S^2, \mathbb{Z}) = 0\\ +&H_0(S^2, \mathbb{Z}) =\mathbb{Z} +\end{align*} +We can construct a long exact sequence for a pair: +\begin{align*} +&H_2(\partial X) \to H_2(X) +\to H_2(X, \partial X) \to \\ +\to &H_1(\partial X) \to H_1(X) \to H_1(X, \partial X) \to +\end{align*} +????????????????????\\ +Simple case $H_1(\partial X)$ \\????????????\\ + is torsion. +$H_2(\partial X)$ is torsion free (by universal coefficient theorem),\\ +???????????????????????\\ +therefore it is $0$. +\\?????????????????????\\ +We know that $b_1(X) = b_2(X)$. Therefore by Poincar\'e duality: +\begin{align*} +b_1(X) = +\dim_{\mathbb{Q}} H_1(X, \mathbb{Q}) +\overset{\mathrm{PD}}{=} +\dim_{\mathbb{Q}} H^2(X, \mathbb{Q}) = +\dim_{\mathbb{Q}} H_2(X, \mathbb{Q}) = b_2(X) +\end{align*} +???????????????????????????????\\ +$H_2(X, \mathbb{Z})$ is torsion free and +$H_2(X_1, \mathbb{Q}) = 0$, therefore $H_2(X, \mathbb{Z}) = 0$. +The map +$H_2(X, \mathbb{Z}) \longrightarrow H_2(X, \partial X, \mathbb{Z})$ is a monomorphism. \\??????????\\ (because it is an isomorphism after tensoring by $\mathbb{Q}$. +\\ +Suppose $\alpha_1, \dots, \alpha_n$ is a basis of $H_2(X, \mathbb{Z})$. +Let $A$ be the intersection matrix in this basis. Then: +\begin{enumerate} +\item +A has integer coefficients, +\item +$\det A \neq 0$, +\item +$\vert \det A \vert = +\vert H_1 (\partial X, \mathbb{Z}) \vert = +\vert \coker H_2(X) \longrightarrow H_2(X, \partial X) \vert$. +\end{enumerate} +\end{example} +??????????????????? +\\ +\\ +If $CUC^T = W$, then for +$\binom{a}{b} = C^{-1} \binom{1}{0}$ we have: +\[ +\binom{a}{b} W +\binom{a}{b} = \binom{1}{0} U \binom{1}{0} = 1. +\] + +\begin{theorem}[Whitehead] +Any non-degenerate form +\[ +A : \mathbb{Z}^4 \times \mathbb{Z}^4 \longrightarrow \mathbb{Z} +\] +can be realized as an intersection form of a simple connected $4$-dimensional manifold. +\end{theorem} +?????????????????????????? +\begin{theorem}[Donaldson, 1982] +If $A$ is an even definite intersection form of a smooth $4$-manifold then it is diagonalizable over $\mathbb{Z}$. +\end{theorem} +?????????????????????????? +?????????????????????????? +?????????????????????????? +?????????????????????????? +\begin{definition} +even define +\end{definition} +Suppose $X$ us $4$ -manifold with a boundary such that $H_1(X) = 0$. + +%$A \cdot B$ gives the pairing as ?? +\begin{proof} +Obviously: +\[H_1(\partial X, \mathbb{Z}) = \coker H_2(X) \longrightarrow H_2(X, \partial X) = \quot{H_2(X, \partial X)}{H_2(X)}. +\] +Let $A$ be an $n \times n$ matrix. $A$ determines a \\ +??????????????/\\ +\begin{align*} +\mathbb{Z}^n \longrightarrow \Hom (\mathbb{Z}^n, \mathbb{Z})\\ +a \mapsto (b \mapsto b^T A a)\\ +\vert \coker A \vert = \vert \det A \vert +\end{align*} +all homomorphisms $b = (b_1, \dots, b_n) $???????\\?????????\\ + +\end{proof} diff --git a/lec_11_03.tex b/lec_11_03.tex new file mode 100644 index 0000000..41e6075 --- /dev/null +++ b/lec_11_03.tex @@ -0,0 +1,176 @@ +\subsection{Algebraic knots} +\noindent +Suppose $F: \mathbb{C}^2 \rightarrow \mathbb{C}$ is a polynomial and $F(0) = 0$. Let take small small sphere $S^3$ around zero. This sphere intersect set of roots of $F$ (zero set of $F$) transversally and by the implicit function theorem the intersection is a manifold. +The dimension of sphere is $3$ and $F^{-1}(0)$ has codimension $2$. +So there is a subspace $L$ - compact one dimensional manifold without boundary. +That means that $L$ is a link in $S^3$. +\begin{figure}[h] +\fontsize{40}{10}\selectfont +\centering{ +\def\svgwidth{\linewidth} +\resizebox{0.2\textwidth}{!}{\input{images/milnor_singular.pdf_tex}} +} +\caption{The intersection of a sphere $S^3$ and zero set of polynomial $F$ is a link $L$.} +\label{fig:milnor_singular} +\end{figure} +%ref: Milnor Singular Points of Complex Hypersurfaces +\begin{theorem} + +$L$ is an unknot if and only if +zero is a smooth point, i.e. +$\bigtriangledown F(0) \neq 0$ (provided $S^3$ has a sufficiently small radius). +\end{theorem} +\noindent +Remark: if $S^3$ is large it can happen that $L$ is unlink, but $F^{-1}(0) \cap B^4$ is "complicated". \\ +%Kyle M. Ormsby +\noindent +In other words: if we take sufficiently small sphere, the link is non-trivial if and only if the point $0$ is singular and the isotopy type of the link doesn't depend on the radius of the sphere. +A link obtained is such a way is called an +algebraic link (in older books on knot theory there is another notion of algebraic link with another meaning). +%ref: Eisenbud, D., Neumann, W. +\begin{example} +Let $p$ and $q$ be coprime numbers such that $p1$. \\ +Zero is an isolated singular point ($\bigtriangledown F(0) = 0$). $F$ is quasi - homogeneous polynomial, so the isotopy class of the link doesn't depend on the choice of a sphere. +Consider $S^3 = \{ (z, w) \in \mathbb{C} : \max( \vert z \vert, \vert w \vert )\} = \varepsilon$. +The intersection +$F^{-1}(0) \cap S^3$ is a torus $T(p, q)$. +\\??????????????????? +$F(z, w) = z^p - w^q$\\ +.\\ +$F^{-1}(0) = \{t = t^q, w = t^p\}.$ For unknot $t = \max (\vert t\vert ^p, \vert t \vert^q) = \varepsilon$. +\end{example} +as a corollary we see that $K_T^{n, }$ ???? \\ +is not slice unless $m=0$. \\ +$t = re^{i \Theta}, \Theta \in [0, 2\pi], r = \varepsilon^{\frac{i}{p}}$ + +\begin{figure}[h] +\fontsize{40}{10}\selectfont +\centering{ +\def\svgwidth{\linewidth} +\resizebox{0.2\textwidth}{!}{\input{images/polynomial_and_surface.pdf_tex}} +} +\caption{Sa.} +\label{fig:polynomial_and_surface} +\end{figure} +\begin{theorem} +Suppose $L$ is an algebraic link. $L = F^{-1}(0) \cap S^3$. Let +\begin{align*} +&\varphi : S^3 \setminus L \longrightarrow S^1 \\ +&\varphi(z, w) =\frac{F(z, w)}{\vert F(z, w) \vert}\in S^1, \quad (z, w) \notin F^{-1}(0). +\end{align*} +The map $\varphi$ is a locally trivial fibration. +\end{theorem} +???????\\ +$ rh D \varphi \equiv 1$ +\begin{definition} +A map $\Pi : E \longrightarrow B$ is locally trivial fibration with fiber $F$ if for any $b \in B$, there is a neighbourhood $U \subset B$ such that $\Pi^{-1}(U) \cong U \times $ \\ +????????????\\ $\Gamma$ ?????????????\\ +FIGURES\\ +!!!!!!!!!!!!!!!!!!!!!!!!!!\\ +\end{definition} + +\begin{theorem} +The map $j: \mathscr{C} \longrightarrow \mathbb{Z}^{\infty}$ is a surjection that maps ${K_n}$ to a linear independent set. Moreover $\mathscr{C} \cong \mathbb{Z}$ +\end{theorem} +... +\\ +In general $h$ is defined only up to homotopy, but this means that +\[ +h_* : H_1 (F, \mathbb{Z}) \longrightarrow H_1 (F, \mathbb{Z}) +\] +is well defined \\ +???????????\\ map. +\begin{theorem} +\label{thm:F_as_S} +Suppose $S$ is a Seifert matrix associated with $F$ then $h = S^{-1}S^T$. +\end{theorem} +\begin{proof} +TO WRITE REFERENCE!!!!!!!!!!! +%see Arnold Varchenko vol II +%Picard - Lefschetz formula +%Nemeth (Real Seifert forms +\end{proof} +\noindent +Consequences: +\begin{enumerate}[label={(\arabic*)}] +\item +the Alexander polynomial is the characteristic polynomial of $h$: +\[ +\Delta_L (t) = \det (h - t I d) +\] +In particular $\Delta_L $ is monic (i.e. the top coefficient is $\pm 1$), +???????????????? +\item +S is invertible, +\item +$F$ minimize the genus (i.e. $F$ is minimal genus Seifert surface). +\\??????????????????\\ +\end{enumerate} +% +\begin{definition} +A link $L$ is fibered if there exists a map ${\phi: S^3\setminus L \longrightarrow S^1}$ which is locally trivial fibration. +\end{definition} +\noindent +If $L$ is fibered then Theorem \ref{thm:F_as_S} holds and all its consequences. +\begin{problem} +If $K_1$ and $K_2$ are fibered knots, then also $K_1 \# K_2$ is fibered. +\end{problem} +\noindent +?????????????????????\\ +\begin{problem} +Prove that connected sum is well defined:\\ +$\Delta_{K_1 \# K_2} = +\Delta_{K_1} + \Delta_{K_2}$ and +$g_3(K_1 \# K_2) = g_3(K_1) + g_3(K_2)$. + +\end{problem} +\begin{figure}[h] +\fontsize{12}{10}\selectfont +\centering{ +\def\svgwidth{\linewidth} +\resizebox{1\textwidth}{!}{\input{images/satellite.pdf_tex}}} +\caption{Example for a satellite knot: a Whitehead double of a trefoil.\\ +The pattern knot embedded non-trivially in an unknotted solid torus $T$ (e.i. $K \not\subset S^3\subset T$) on the left and the pattern in a companion knot - trefoil - on the right.} +\label{fig:sattelite} +\end{figure} +\noindent +\subsection{Alternating knot} +\begin{definition} +A knot (link) is called alternating if it admits an alternating diagram. +\end{definition} + +\begin{figure}[h] +\fontsize{12}{10}\selectfont +\centering{ +\def\svgwidth{\linewidth} +\includegraphics[width=0.3\textwidth]{figure8.png} +} +\caption{Example: figure eight knot is an alternating knot.} +\label{fig:fig8} +\end{figure} + +\begin{definition} +A reducible crossing in a knot diagram is a crossing for which we can find a circle such that its intersection with a knot diagram is exactly that crossing. A knot diagram without reducible crossing is called reduced. +\end{definition} +\begin{fact} +Any reduced alternating diagram has minimal number of crossings. +\end{fact} +\begin{definition} +The writhe of the diagram is the difference between the number of positive and negative crossings. +\end{definition} +\begin{fact}[Tait] +Any two diagrams of the same alternating knot have the same writhe. +\end{fact} +\begin{fact} +An alternating knot has Alexander polynomial of the form: +$ +a_1t^{n_1} + a_2t^{n_2} + \dots + a_s t^{n_s} +$, where $n_1 < n_2 < \dots < n_s$ and $a_ia_{i+1} < 0$. +\end{fact} +\begin{problem}[open] +What is the minimal $\alpha \in \mathbb{R}$ such that if $z$ is a root of the Alexander polynomial of an alternating knot, then $\Re(z) > \alpha$.\\ +Remark: alternating knots have very simple knot homologies. +\end{problem} +\begin{proposition} +If $T_{p, q}$ is a torus knot, $p < q$, then it is alternating if and only if $p=2$. +\end{proposition} \ No newline at end of file diff --git a/lec_15_04.tex b/lec_15_04.tex new file mode 100644 index 0000000..40710ee --- /dev/null +++ b/lec_15_04.tex @@ -0,0 +1,135 @@ +???????????????????\\ +\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 $ = 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 diff --git a/lec_18_03.tex b/lec_18_03.tex new file mode 100644 index 0000000..f2873d7 --- /dev/null +++ b/lec_18_03.tex @@ -0,0 +1,170 @@ +\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} + +\begin{figure}[h] +\fontsize{20}{10}\selectfont +\centering{ +\def\svgwidth{\linewidth} +\resizebox{0.8\textwidth}{!}{\input{images/concordance.pdf_tex}} +} +\end{figure} +\begin{definition} +A knot $K$ is called (smoothly) slice if $K$ is smoothly concordant to an unknot. \\ +Put differently: a knot $K$ is smoothly slice if and only if $K$ bounds a smoothly embedded disk in $B^4$. +\end{definition} + + + +\noindent +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}\label{fact:concordance_connected} +If $K_1 \sim {K_1}^{\prime}$ and $K_2 \sim {K_2}^{\prime}$, then +$K_1 \# K_2 \sim {K_1}^{\prime} \# {K_2}^{\prime}$. + +\begin{figure}[h] +\fontsize{10}{10}\selectfont +\centering{ +\def\svgwidth{\linewidth} +\resizebox{1\textwidth}{!}{\input{images/concordance_sum.pdf_tex}} +} +\caption{Sketch for Fact \ref{fact:concordance_connected}.} +\label{fig:concordance_sum} +\end{figure} + +\end{fact} +\begin{fact} +$K \# m(K) \sim $ the unknot. +\end{fact} +\noindent +\begin{theorem} +Let $\mathscr{C}$ denote a set of all equivalent classes for knots and $[0]$ denote class of all knots concordant to a trivial knot. +$\mathscr{C}$ is a group under taking connected sums. The neutral element in the group is $[0]$ and the inverse element of an element $[K]\in \mathscr{C}$ is $-[K] = [mK]$. +\end{theorem} +\begin{fact} +The figure eight knot is a torsion element in $\mathscr{C}$ ($2K \sim $ the unknot). +\end{fact} +\begin{problem}[open] +Are there in concordance group torsion elements that are not $2$ torsion elements? +\end{problem} +\noindent +Remark: $K \sim K^{\prime} \Leftrightarrow K \# -K^{\prime}$ is slice. +\\ +\begin{figure}[h] +\fontsize{20}{10}\selectfont +\centering{ +\def\svgwidth{\linewidth} +\resizebox{0.5\textwidth}{!}{\input{images/ball_4_pushed_seifert.pdf_tex}} +} +\caption{$Y = F \cup \Sigma$ is a smooth closed surface.} +\label{fig:closed_surface} +\end{figure} +\noindent +\\ +Pontryagin-Thom construction tells us that there exists a compact oriented three - manifold $\Omega \subset B^4$ such that $\partial \Omega = Y$. +Suppose $\Sigma$ is a Seifert surface and $V$ a Seifert form defined on $\Sigma$: ${(\alpha, \beta) \mapsto \Lk(\alpha, \beta^+)}$. Suppose $\alpha, \beta \in H_1(\Sigma, \mathbb{Z})$, i.e. there are cycles and +$\alpha, \beta \in \ker (H_1(\Sigma, \mathbb{Z}) \longrightarrow H_1(\Omega, \mathbb{Z}))$. Then there are two cycles $A, B \in \Omega$ such that $\partial A = \alpha$ and $\partial B = \beta$. +Let $B^+$ be a push off of $B$ in the positive normal direction such that +$\partial B^+ = \beta^+$. +Then +$\Lk(\alpha, \beta^+) = A \cdot B^+$. But $A$ and $B$ are disjoint, so $\Lk(\alpha, \beta^+) = 0$. Then the Seifert form is zero. +\\ +\noindent +Let us consider following maps: +\[ +\Sigma \overset{\phi} \longhookrightarrow Y \overset{\psi} \longhookrightarrow \Omega. +\] +Let $\phi_*$ and $\psi_*$ be induced maps on the homology group. If an element $\gamma \in \ker (H_1(\Sigma, \mathbb{Z}) \longrightarrow H_1(\Omega, \mathbb{Z}))$, then $\gamma \in \ker \phi_*$ or $\gamma \in \ker \psi_*$. +% +% +% +\begin{proposition} +\[ +\dim \ker (H_1(Y, \mathbb{Z}) \longrightarrow H_1(\Omega, \mathbb{Z})) = \frac{1}{2} b_1(Y), +\] +where $b_1$ is first Betti number. +\end{proposition} +\begin{proof} +Consider the following long exact sequence for a pair $(\Omega, Y)$: +\begin{align*} +& 0 \to H_3(\Omega) \to H_3(\Omega, Y) \to +\\ +\to & H_2(Y) \to H_2(\Omega) \to H_2(\Omega, Y) \to \\ +\to & H_1(Y) \to H_1(\Omega) \to H_1(\Omega, Y) \to \\ +\to & H_0(Y) \to H_0(\Omega) \to 0 +\end{align*} +By Poincar\'e duality we know that: +\begin{align*} +H_3(\Omega, Y) &\cong H^0(\Omega),\\ +H_2(Y) &\cong H^0(Y),\\ +H_2(\Omega) &\cong H^1(\Omega, Y),\\ +H_1(\Omega, Y) &\cong H^1(\Omega). +\end{align*} +Therefore $\dim_{\mathbb{Q}} \quot{H_1(Y)}{V} += \dim_{\mathbb{Q}} V +$.\\ +\noindent +Suppose $g(K) = 0$ ($K$ is slice). Then $H_1(\Sigma, \mathbb{Z}) \cong H_1(Y, \mathbb{Z})$. Let $g_{\Sigma}$ be the genus of $\Sigma$, $\dim H_1(Y, \mathbb{Z}) = 2g_{\Sigma}$. Then the Seifert form $V$ on a $K$ +has a subspace of dimension $g_{\Sigma}$ on which it is zero: + +\begin{align*} +\newcommand\coolover[2]% + {\mathrlap{\smash{\overbrace{\phantom{% + \begin{matrix} #2 \end{matrix}}}^{\mbox{$#1$}}}}#2} +\newcommand\coolunder[2]{\mathrlap{\smash{\underbrace{\phantom{% + \begin{matrix} #2 \end{matrix}}}_{\mbox{$#1$}}}}#2} +\newcommand\coolleftbrace[2]{% + #1\left\{\vphantom{\begin{matrix} #2 \end{matrix}}\right.} +\newcommand\coolrightbrace[2]{% + \left.\vphantom{\begin{matrix} #1 \end{matrix}}\right\}#2} + \vphantom{% phantom stuff for correct box dimensions + \begin{matrix} + \overbrace{XYZ}^{\mbox{$R$}}\\ \\ \\ \\ \\ \\ + \underbrace{pqr}_{\mbox{$S$}} + \end{matrix}}% + V = +\begin{matrix}% matrix for left braces + \coolleftbrace{g_{\Sigma}}{ \\ \\ \\} + \\ \\ \\ \\ +\end{matrix}% +\begin{pmatrix} + \coolover{g_{\Sigma}}{0 & \dots & 0 } & * & \dots & *\\ + \sdots & & \sdots & \sdots & & \sdots \\ + 0 & \dots & 0 & * & \dots & *\\ + * & \dots & * & * & \dots & *\\ + \sdots & & \sdots & \sdots & & \sdots \\ + * & \dots & * & * & \dots & * + \end{pmatrix}_{2g_{\Sigma} \times 2g_{\Sigma}} +\end{align*} +\end{proof} +\noindent +Let $V = +\begin{pmatrix} + 0 & A\\ + B & C +\end{pmatrix}$ +\begin{align*} +\det (tV - V^T) = \det (tA - B^T) - \det(tB - A^T) +\end{align*} +\begin{corollary} +\label{cor:slice_alex} +If $K$ is a slice knot then there exists $f \in \mathbb{Z}[t^{\pm 1}]$ such that $\Delta_K(t) = f(t) \cdot f(t^{-1})$. +\end{corollary} +\begin{example} +Figure eight knot is not slice. +\end{example} +\begin{fact} +If $K$ is slice, then the signature $\sigma(K) \equiv 0$. +\end{fact} + + + diff --git a/lec_20_05.tex b/lec_20_05.tex new file mode 100644 index 0000000..cf6649d --- /dev/null +++ b/lec_20_05.tex @@ -0,0 +1,252 @@ +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, \_) \in 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*} +\noindent +$C_{*}(\widetilde{X})$ 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} diff --git a/lec_25_02.tex b/lec_25_02.tex new file mode 100644 index 0000000..5a94fae --- /dev/null +++ b/lec_25_02.tex @@ -0,0 +1,280 @@ +\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} (unknot), +\includegraphics[width=0.08\textwidth]{trefoil.png} (trefoil). +\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 $D_{\pi}$ is a picture over projection $\pi$ of a link $L$ in $\mathbb{R}^3$($S^3$) to $\mathbb{R}^2$ ($S^2$) such that: +\begin{enumerate}[label={(\arabic*)}] +\item +$D_{\pi |_L}$ is non degenerate: \includegraphics[width=0.05\textwidth]{LinkDiagram1.png}, +\item the double points are not degenerate: \includegraphics[width=0.03\textwidth]{LinkDiagram2.png}, +\item there are no triple point: \includegraphics[width=0.05\textwidth]{LinkDiagram3.png}. +\end{enumerate} +\end{definition} +\noindent +There are under- and overcrossings (tunnels and bridges) on a link diagrams with an obvious meaning.\\ +Every link admits a link diagram. +\\ +Let $D$ be a diagram of an oriented link (to each component of a link we add an arrow in the diagram).\\ +We can distinguish two types of crossings: right-handed +$\left(\PICorientpluscross\right)$, called a positive crossing, and left-handed $\left(\PICorientminuscross\right)$, called a negative crossing. + +\subsection{Reidemeister moves} +A Reidemeister move is one of the three types of operation on a link diagram as shown below: +\begin{enumerate}[label=\Roman*] +\item\hfill\\ +\includegraphics[width=0.6\textwidth]{rm1.png}, +\item\hfill\\\includegraphics[width=0.6\textwidth]{rm2.png}, +\item\hfill\\\includegraphics[width=0.4\textwidth]{rm3.png}. +\end{enumerate} + +\begin{theorem} [Reidemeister, 1927 ] +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} +% +% +% +%The number of Reidemeister Moves Needed for Unknotting +%Joel Hass, Jeffrey C. Lagarias +%(Submitted on 2 Jul 1998) +% Piotr Sumata, praca magisterska +% proof - transversality theorem (Thom) + +%Singularities of Differentiable Maps +%Authors: Arnold, V.I., Varchenko, Alexander, Gusein-Zade, S.M. + +\subsection{Seifert surface} +\noindent +Let $D$ be an oriented diagram of a link $L$. We change the diagram by smoothing each crossing: +\begin{align*} +\PICorientpluscross \mapsto \PICorientLRsplit\\ +\PICorientminuscross \mapsto \PICorientLRsplit +\end{align*} +We smooth all the crossings, so we get a disjoint union of circles on the plane. Each circle bounds a disks in $\mathbb{R}^3$ (we choose disks that don't intersect). For each smoothed crossing we add a twisted band: right-handed for a positive and left-handed for a negative one. We get an orientable surface $\Sigma$ such that $\partial \Sigma = L$.\\ + +\begin{figure}[h] +\fontsize{15}{10}\selectfont +\centering{ +\def\svgwidth{\linewidth} +\resizebox{0.8\textwidth}{!}{\input{images/seifert_alg.pdf_tex}} +\caption{Constructing a Seifert surface.} +\label{fig:SeifertAlg} +} +\end{figure} + +\noindent +Note: the obtained surface isn't unique and in general doesn't need to be connected, but by taking connected sum of all components we can easily get a connected surface (i.e. we take two disconnected components and cut a disk in each of them: $D_1$ and $D_2$; now we glue both components on the boundaries: $\partial D_1$ and $\partial D_2$. + +\begin{figure}[h] +\begin{center} +\includegraphics[width=0.6\textwidth]{seifert_connect.png} +\end{center} +\caption{Connecting two surfaces.} +\label{fig:SeifertConnect} +\end{figure} + +\begin{theorem}[Seifert] +\label{theo:Seifert} +Every link in $S^3$ bounds a surface $\Sigma$ that is compact, connected and orientable. Such a surface is called a Seifert surface. +\end{theorem} +% +\begin{figure}[h] +\fontsize{12}{10}\selectfont +\centering{ +\def\svgwidth{\linewidth} +\resizebox{1\textwidth}{!}{\input{images/torus_1_2_3.pdf_tex}} +\caption{Genus of an orientable surface.} +\label{fig:genera} +} +\end{figure} +% +% +\begin{definition} +The three genus $g_3(K)$ ($g(K)$) of a knot $K$ is the minimal genus of a Seifert surface $\Sigma$ for $K$. +\end{definition} + +\begin{corollary} +A knot $K$ is trivial if and only $g_3(K) = 0$. +\end{corollary} + +\noindent +Remark: there are knots that admit non isotopic Seifert surfaces of minimal genus (András Juhász, 2008). + +\begin{definition} +Suppose $\alpha$ and $\beta$ are two simple closed curves in $\mathbb{R}^3$. +On a diagram $L$ consider all crossings between $\alpha$ and $\beta$. Let $N_+$ be the number of positive crossings, $N_-$ - negative. Then the linking number: $\Lk(\alpha, \beta) = \frac{1}{2}(N_+ - N_-)$. +\end{definition} +\begin{definition} +\label{def:lk_via_homo} +Let $\alpha$ and $\beta$ be two disjoint simple cross curves in $S^3$. +Let $\nu(\beta)$ be a tubular neighbourhood of $\beta$. The linking number can be interpreted via first homology group, where $\Lk(\alpha, \beta)$ is equal to evaluation of $\alpha$ as element of first homology group of the complement of $\beta$: +\[ +\alpha \in H_1(S^3 \setminus \nu(\beta), \mathbb{Z}) \cong \mathbb{Z}.\] +\end{definition} + +\begin{example} +\begin{itemize} +\item +Hopf link: +\begin{figure}[h] +\fontsize{20}{10}\selectfont +\centering{ +\def\svgwidth{\linewidth} +\resizebox{0.4\textwidth}{!}{\input{images/linking_hopf.pdf_tex}}, +} +\end{figure} +\item +$T(6, 2)$ link: +\begin{figure}[h] +\fontsize{20}{10}\selectfont +\centering{ +\def\svgwidth{\linewidth} +\resizebox{0.4\textwidth}{!}{\input{images/linking_torus_6_2.pdf_tex}}. +} +\end{figure} +\end{itemize} +\end{example} +\begin{fact} +\[ +g_3(\Sigma) = \frac{1}{2} b_1 (\Sigma) = +\frac{1}{2} \dim_{\mathbb{R}}H_1(\Sigma, \mathbb{R}), +\] +where $b_1$ is first Betti number of $\Sigma$. +\end{fact} + +\subsection{Seifert matrix} +Let $L$ be a link and $\Sigma$ be an oriented Seifert surface for $L$. Choose a basis for $H_1(\Sigma, \mathbb{Z})$ consisting of simple closed $\alpha_1, \dots, \alpha_n$. +Let $\alpha_1^+, \dots \alpha_n^+$ be copies of $\alpha_i$ lifted up off the surface (push up along a vector field normal to $\Sigma$). Note that elements $\alpha_i$ are contained in the Seifert surface while all $\alpha_i^+$ are don't intersect the surface. +Let $\Lk(\alpha_i, \alpha_j^+) = \{a_{ij}\}$. Then the matrix $S = \{a_{ij}\}_{i, j =1}^n$ is called a Seifert matrix for $L$. Note that by choosing a different basis we get a different matrix. + +\begin{figure}[h] +\fontsize{20}{10}\selectfont +\centering{ +\def\svgwidth{\linewidth} +\resizebox{0.8\textwidth}{!}{\input{images/seifert_matrix.pdf_tex}} +} +\end{figure} + +\begin{theorem} +The Seifert matrices $S_1$ and $S_2$ for the same link $L$ are S-equivalent, that is, $S_2$ can be obtained from $S_1$ by a sequence of following moves: +\begin{enumerate}[label={(\arabic*)}] + +\item +$V \rightarrow AVA^T$, where $A$ is a matrix with integer coefficients, + +\item + +$V \rightarrow +\begin{pmatrix} + \begin{array}{c|c} + V & + \begin{matrix} + \ast & 0 \\ + \sdots & \sdots\\ + \ast & 0 + \end{matrix} \\ + \hline + \begin{matrix} + \ast & \dots & \ast\\ + 0 & \dots & 0 + \end{matrix} + & + \begin{matrix} + 0 & 0\\ + 1 & 0 + \end{matrix} + \end{array} +\end{pmatrix} \quad$ +or +$\quad +V \rightarrow +\begin{pmatrix} + \begin{array}{c|c} + V & + \begin{matrix} + \ast & 0 \\ + \sdots & \sdots\\ + \ast & 0 + \end{matrix} \\ + \hline + \begin{matrix} + \ast & \dots & \ast\\ + 0 & \dots & 0 + \end{matrix} + & + \begin{matrix} + 0 & 1\\ + 0 & 0 + \end{matrix} + \end{array} +\end{pmatrix}$ +\item +inverse of (2) + +\end{enumerate} +\end{theorem} diff --git a/lec_25_03.tex b/lec_25_03.tex new file mode 100644 index 0000000..9cdf66b --- /dev/null +++ b/lec_25_03.tex @@ -0,0 +1,234 @@ +\subsection{Slice knots and metabolic form} +\begin{theorem} +\label{the:sign_slice} +If $K$ is slice, +then $\sigma_K(t) + = \sign ( (1 - t)S +(1 - \bar{t})S^T)$ +is zero except possibly of finitely many points and $\sigma_K(-1) = \sign(S + S^T) \neq 0$. +\end{theorem} +\begin{lemma} +\label{lem:metabolic} +If $V$ is a Hermitian matrix ($\bar{V} = V^T$), $V$ is of size $2n \times 2n$, +$ +V = \begin{pmatrix} +0 & A \\ +\bar{A}^T & B +\end{pmatrix} +$ and $\det V \neq 0$ then $\sigma(V) = 0$. +\end{lemma} +\begin{definition} +A Hermitian form $V$ is metabolic if $V$ has structure +$\begin{pmatrix} +0 & A\\ +\bar{A}^T & B +\end{pmatrix}$ with half-dimensional null-space. +\end{definition} +\noindent +Theorem \ref{the:sign_slice} can be also express as follow: +non-degenerate metabolic hermitian form has vanishing signature. +\begin{proof} +\noindent +We note that $\det(S + S^T) \neq 0$. Hence $\det ( (1 - t) S + (1 - \bar{t})S^T)$ is not identically zero on $S^1$, so it is non-zero except possibly at finitely many points. We apply the Lemma \ref{lem:metabolic}. +\\ +Let $t \in S^1 \setminus \{1\}$. +Then: +\begin{align*} +\det((1 - t) S + (1 - \bar{t}) S^T) =& +\det((1 - t) S + (t\bar{t} - \bar{t}) S^T) =\\ +&\det((1 - t) (S - \bar{t} - S^T)) = +\det((1 -t)(S - \bar{t} S^T)). +\end{align*} +As $\det (S + S^T) \neq 0$, so $S - \bar{t}S^T \neq 0$. +\end{proof} +\begin{corollary} +If $K \sim K^\prime$ then for all but finitely many $t \in S^1 \setminus \{1\}: \sigma_K(t) = -\sigma_{K^\prime}(t)$. +\end{corollary} +\begin{proof} +If $ K \sim K^\prime$ then $K \# K^\prime$ is slice. +\[ +\sigma_{-K^\prime}(t) = -\sigma_{K^\prime}(t) +\] +The signature gives a homomorphism from the concordance group to $\mathbb{Z}$. +Remark: if $t \in S^1$ is not algebraic over $\mathbb{Z}$, then $\sigma_K(t) \neq 0$ +(we can use the argument that $\mathscr{C} \longrightarrow \mathbb{Z}$ as well). +\end{proof} +\subsection{Four genus} +\begin{figure}[h] +\fontsize{20}{10}\selectfont +\centering{ +\def\svgwidth{\linewidth} +\resizebox{0.7\textwidth}{!}{\input{images/genus_2_bordism.pdf_tex}} +} +\caption{$K$ and $K^\prime$ are connected by a genus $g$ surface.}\label{fig:genus_2_bordism} +\end{figure} + +\begin{proposition}[Kawauchi inequality] +If there exists a genus $g$ surface as in Figure \ref{fig:genus_2_bordism} +then for almost all +$t \in S^1 \setminus \{1\}$ we have +$\vert +\sigma_K(t) - \sigma_{K^\prime}(t) +\vert \leq 2 g$. +\end{proposition} +% Kawauchi Chapter 12 ??? +% Borodzik 2010 Morse theory for plane algebraic curves +\begin{lemma} +If $K$ bounds a genus $g$ surface $X \in B^4$ and $S$ is a Seifert form then ${S \in M_{2n \times 2n}}$ has a block structure $\begin{pmatrix} +0 & A\\ +B & C +\end{pmatrix}$, where $0$ is $(n - g) \times (n - g)$ submatrix. +\end{lemma} + +\begin{proof} +\begin{figure}[h] +\fontsize{20}{10}\selectfont +\centering{ +\def\svgwidth{\linewidth} +\resizebox{0.5\textwidth}{!}{\input{images/genus_bordism_zeros.pdf_tex}} +} +\caption{There exists a $3$ - manifold $\Omega$ such that $\partial \Omega = X \cup \Sigma$.}\label{fig:omega_in_B_4} +\end{figure} +\noindent +Let $K$ be a knot and $\Sigma$ its Seifert surface as in Figure \ref{fig:omega_in_B_4}. +There exists a $3$ - submanifold +$\Omega$ such that +$\partial \Omega = Y = X \cup \Sigma$ +(by Thom-Pontryagin construction). +If $\alpha, \beta \in \ker (H_1(\Sigma) \longrightarrow H_1(\Omega))$, +then ${\Lk(\alpha, \beta^+) = 0}$. Now we have to determine the size of the kernel. We know that +${\dim H_1(\Sigma) = 2 n}$. When we glue $\Sigma$ (genus $n$) and $X$ (genus $g$) along a circle we get a surface of genus $n + g$. Therefore $\dim H_1 (Y) = 2 n + 2 g$. Then: +\[ +\dim (\ker (H_1(Y) \longrightarrow H_1(\Omega)) = n + g. +\] +So we have $H_1(W)$ of dimension + $2 n + 2 g$ +- the image of $H_1(Y)$ +with a subspace +corresponding to the image of $H_1(\Sigma)$ with dimension $2 n$ and a subspace corresponding to the kernel +of $H_1(Y) \longrightarrow H_1(\Omega)$ of size $n + g$. +We consider minimal possible intersection of this subspaces that corresponds to the kernel of the composition $H_1(\Sigma) \longrightarrow H_1(Y) \longrightarrow H_1(\Omega)$. As the first map is injective, elements of the kernel of the composition have to be in the kernel of the second map. +So we can calculate: +\[ +\dim \ker (H_1(\Sigma) \longrightarrow H_1(\Omega)) = 2 n + n + g -2 n - 2 g = n - g. +\] +\end{proof} +\begin{corollary} +If $t$ is not a root of +$\det (tS - S^T) $, then +$\vert \sigma_K(t) \vert \leq 2g$. +\end{corollary} +\begin{fact} +If there exists cobordism of genus $g$ between $K$ and $K^\prime$ like shown in Figure \ref{fig:proof_for_bound_disk}, then $K \# -K^\prime$ bounds a surface of genus $g$ in $B^4$. +\end{fact} +\begin{figure}[H] +\fontsize{20}{10}\selectfont +\centering{ +\def\svgwidth{\linewidth} +\resizebox{0.7\textwidth}{!}{\input{images/genus_bordism_proof.pdf_tex}} +} +\caption{If $K$ and $K^\prime$ are connected by a genus $g$ surface, then $K \# -K^\prime$ bounds a genus $g$ surface.}\label{fig:proof_for_bound_disk} +\end{figure} + +\begin{definition} +The (smooth) four genus $g_4(K)$ is the minimal genus of the surface $\Sigma \in B^4$ such that $\Sigma$ is compact, orientable and $\partial \Sigma = K$. +\end{definition} +\noindent +Remarks: +\begin{enumerate}[label={(\arabic*)}] +\item +$3$ - genus is additive under taking connected sum, but $4$ - genus is not, +\item +for any knot $K$ we have $g_4(K) \leq g_3(K)$. +\end{enumerate} +\begin{example} +\begin{itemize} +\item Let $K = T(2, 3)$. $\sigma(K) = -2$, therefore $T(2, 3)$ isn't a slice knot. +\item Let $K$ be a trefoil and $K^\prime$ a mirror of a trefoil. $g_4(K^\prime) = 1$, but $g_4(K \# K^\prime) = 0$, so we see that $4$-genus isn't additive, +\item +the equality: +\[ +g_4(T(p, q) ) = \frac{1}{2} (p - 1) (g -1) +\] +was conjecture in the '70 and proved by P. Kronheimer and T. Mrówka (1994). +% OZSVATH-SZABO AND RASMUSSEN +\end{itemize} +\end{example} +\begin{proposition} +$g_4 (T(p, q) \# -T(r, s))$ is in general hopelessly unknown. +\end{proposition} +\begin{proposition} +Supremum of the signature function of the knot is bounded almost everywhere by two times $4$ - genus: +\[ +\ess \sup \vert \sigma_K(t) \vert \leq 2 g_4(K). +\] +\end{proposition} +\subsection{Topological genus} +\begin{definition} +A knot $K$ is called topologically slice if $K$ bounds a topological locally flat disc in $B^4$ (i.e. the disk has tubular neighbourhood). +\end{definition} +\begin{theorem}[Freedman, '82] +If $\Delta_K(t) = 1$, then $K$ is topologically slice (but not necessarily smoothly slice). +\end{theorem} +\begin{theorem}[Powell, 2015] +If $K$ is genus $g$ +(topologically flat) +cobordant to $K^\prime$, +then +\[ +\vert \sigma_K(t) - \sigma_{K^\prime}(t) \vert \leq 2 g +\] +if $g_4^{\mytop}(K) \geq \ess \sup \vert \sigma_K(t) \vert$. +\end{theorem} +\noindent +The proof for smooth category was based on following equality: +\[ +\dim \ker (H_1 (Y) \longrightarrow H_1(\Omega)) = \frac{1}{2} \dim H_1(Y). +\] +For this equality we assumed that there exists a $3$ - dimensional manifold $\Omega$ (as shown in Figure \ref{fig:omega_in_B_4}) which was guaranteed by Pontryagin-Thom Construction.\\ +Pontryagin-Thom Construction relays on taking $\Omega$ as preimage of regular value: +\[ +H^1 (B^4 \setminus Y, \mathbb{Z}) = [B^4 \setminus Y, S^1], +\] +what relies on Sard's theorem, that the set of regular values has positive measure. But Sard's theorem doesn't work for topologically locally flat category. So there was a gap in the proof for topological locally flat category - the existence of $\Omega$.\\ +\noindent +Remark: unless $p=2$ or $p = 3 \wedge q = 4$: +\[ +g_4^{\mytop} (T(p, q)) < q_4(T(p, q)). +\] +% Wilczyński '93 +%Feller 2014 +%Baoder 2017 +%Lemark +\\ +\noindent +From the category of cobordant knots (or topologically cobordant knots) there exists a map to $\mathbb{Z}$ given by signature function. To any element $K$ we can associate a form +\[ +(1 - t)S + (1 - \bar{t})S^T) \in W(\mathbb{Z}[t, t^{-1}]). +\] This association is not well define because id depends on the choice of Seifert form. However, different choices lead ever to congruent forms ($S \mapsto CSC^T$) or induced the change on the form by adding or subtracting a hyperbolic element. +\begin{definition} +The Witt group $W$ of $\mathbb{Z}[t, t^{-1}]$ elements are classes of non-degenerate +forms over $\mathbb{Z}[t, t^{-1}]$ under the equivalence relation $V \sim W$ if $V \oplus - W$ is metabolic. +\end{definition} +\noindent +If $S$ differs from $S^\prime$ by a row extension, then +$(1 - t) S + (1 - \bar{t}^{-1}) S^T$ is Witt equivalence to $(1 - t) S^\prime + (1 - t^{-1})S^T$. +\\ +\noindent +A form is meant as hermitian with respect to this involution: $A^T = A: (a, b) = \bar{(a, b)}$. +\\ +$ +W(\mathbb{Z}_p) = \mathbb{Z}_2 \oplus +\mathbb{Z}_2$ or +$\mathbb{Z}_4$ +\\ +??????????????????????? +\\ +$\sum a_gt^j \longrightarrow \sum a_g t^{-1}$\\ +\begin{theorem}[Levine '68] +\[ +W(\mathbb{Z}[t^{\pm 1}]) +\longrightarrow \mathbb{Z}_2^\infty \oplus +\mathbb{Z}_4^\infty \oplus +\mathbb{Z} +\] +\end{theorem} diff --git a/lec_27_05.tex b/lec_27_05.tex new file mode 100644 index 0000000..3228e3b --- /dev/null +++ b/lec_27_05.tex @@ -0,0 +1,8 @@ +.... +\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 \ No newline at end of file diff --git a/lec_7.tex b/lec_7.tex deleted file mode 100644 index e69de29..0000000 diff --git a/lec_8.tex b/lec_8.tex deleted file mode 100644 index e69de29..0000000 diff --git a/lec_9.tex b/lec_9.tex deleted file mode 100644 index e69de29..0000000