change of names

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}
+
+
+

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}

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}
+

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}

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 $p<q$ and $p,q>1$. \\
+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}

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 $<x, y > = 0$.
+\end{lemma}
+\begin{proof}
+\begin{align*}
+x \in U_{p_1} 
+\end{align*}
+\end{proof}
+\begin{align*}
+H_1(Y, \mathbb{Z}) \cong \quot{\mathbb{Z}^n}{A\mathbb{Z}}\\
+A \longrightarrow BAC^T \quad \text{Smith normal form}
+\end{align*}
+???????????????????????\\
+In general 
+
+%no lecture at 29.04

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}
+
+
+

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}

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}

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}

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