diff --git a/lec_03_06.tex b/lec_03_06.tex
index fefe472..669340b 100644
--- a/lec_03_06.tex
+++ b/lec_03_06.tex
@@ -6,6 +6,16 @@ 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. 
 \\
+???????????????
+\\
+\begin{eqnarray*}
+\chi (D^2) = 1 \\
+\chi (\Delta) = 1 - u\\
+\gamma = 0 \in \pi_1(B^4 \setminus S)
+\end{eqnarray*}
+
+??????????????
+\\
 \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}
diff --git a/lec_04_03.tex b/lec_04_03.tex
index 1519178..a76580f 100644
--- a/lec_04_03.tex
+++ b/lec_04_03.tex
@@ -41,16 +41,19 @@ Let us consider a long exact sequence of cohomology of a pair $(S^3, S^3 \setmin
 &  \mathbb{Z} \ar[u,isomorphic] &\\
  \end{tikzcd}
 \end{center}
+The tubular neighbourhood of the knot is homomorphic to 
+$D^2 \times S^1$. 
+So its boundary 
+$\partial N \cong \  S^1 \times S^1$ and therefore:
+$H^1(N, \partial N) \cong \ \mathbb{Z} \oplus \mathbb{Z}$. By excision theorem we have:
 \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}
+H^* (S^3, S^3 \setminus N) &\cong H^* (N, \partial N).
 \end{align*}
+Therefore:
 \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}
+H^	1 (S^3\setminus N) &\cong H^1(S^3\setminus K) \cong \mathbb{Z}.
 \end{align*}
+Let us consider the following diagram: 
 \begin{equation*}
 \begin{tikzcd}[row sep=huge]
 H^1(S^3 \setminus K) \arrow[r,] \arrow[d,"\widetilde{\Theta}"] & 
@@ -62,9 +65,11 @@ H^1(N \setminus K) \arrow[d,"\Theta"] \\
 \noindent
 $\Sigma = \widetilde{\Theta}^{-1}(X)$ is a surface, such that $\partial \Sigma = K$, so it is a Seifert surface. 
 %
-% 
+% picture for excision theorem
 % Thom isomorphism, 
 \end{proof}
+%$S$ - equivalence $\Sigma$\\
+%simple closed curves $\alpha_1, ... \alpha_n \in H_1(\Sigma, \mathbb{Z})$ basis for $H_1$
 \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:
@@ -256,13 +261,13 @@ Suppose $K \subset S^3$ and $\pi_1(S^3 \setminus K)$ is infinite cyclic ($\mathb
 \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 ${\lambda \in \ker (\pi_1(S^1 \times S^1) \longrightarrow \pi_1(M)}$.  
+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. \\
+%$g_3$
+If $g_3(\Sigma) = 0$, 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)).
diff --git a/lec_06_05.tex b/lec_06_05.tex
index e98986c..8af4b6a 100644
--- a/lec_06_05.tex
+++ b/lec_06_05.tex
@@ -38,21 +38,6 @@ $a_j^- = \sum_k v_{kj} t^{-1} \alpha_j$.
 \\
 \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 $S_k$. 
 \[
@@ -60,19 +45,26 @@ Let $c, d \in H(\widetilde{X})$ (see Figure \ref{fig:covering_pairing}), $\Delta
 \] 
 \\
 ?????????????\\
-\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}
-
+There is at least one paper where the structure of (Alexander module?) is calculated from a specific knot (?minimal number of generators?) 
+\\
+C. Kearton, S. M. J. Wilson
+\\
+\begin{fact}
+Let $A$ be a matrix over principal ideal domain $R$.  Than there exist matrices $C$, $D$ and $E$ such that $A = CDE$, 
+\[D = \begin{bmatrix}
+d_1 & 0 & \cdots & \cdots & 0 \\
+0 & d_2 & 0 & \cdots & 0 \\
+\sdots &  & \ddots &  & \sdots &  \\
+0 & \cdots & 0 &  d_{n-1} & 0\\
+0 & \cdots & \cdots & 0 & d_n
+\end{bmatrix},\] 
+where $d_{i + 1} | d_i$, and matrices 
+$C$ and $E$ are invertible over $R$.\\
+$D$ is called a Smith normal form of the matrix $A$.
+\end{fact}
 
 \begin{definition}
-The $\mathbb{Z}[t, t^{-1}]$ module $H_1(\widetilde{X})$ is called the Alexander module of knot $K$.
+The $\mathbb{Z}[t, t^{-1}]$ module $H_1(\widetilde{X})$ is called the Alexander module of a knot $K$.
 \end{definition}
 \noindent
 Let $R$ be a PID, $M$ a finitely generated $R$ module. Let us consider
@@ -92,53 +84,36 @@ For knots the order of the Alexander module is the Alexander polynomial.
 \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}
+\\
+??????????????????\\
+General picture : $K$, $X$ knot complement... 
+\begin{eqnarray*}
+H_1( X, \mathbb{Z}) = \mathbb{Z} \\
+H_1(\widetilde{X}, \mathbb{Z}[t, t^{-1}]) \\
+\pi_1(X)
+\end{eqnarray*}
+\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
-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}
+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
-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
-
+$\Sigma_?(K) \rightarrow S^3$ ?????\\
+$H_1(\Sigma_?(K), \mathbb{Z}) = h$\\
+$H \times H \longrightarrow \quot{\mathbb{Q}}{\mathbb{Z}}$\\
+...\\
 \begin{figure}[h]
 \fontsize{10}{10}\selectfont
 \centering{
 \def\svgwidth{\linewidth}
-\resizebox{0.5\textwidth}{!}{\input{images/ball_4_alpha_beta.pdf_tex}}
+\resizebox{1\textwidth}{!}{\input{images/covering_pairing.pdf_tex}}
+\caption{$c, d \in H_1(\widetilde{X})$.}
+\label{fig:covering_pairing}
 }
-%\caption{Sketch for Fact %%\label{fig:concordance_m}
 \end{figure}
 
-\end{document}
-
+\subsection*{Blanchfield pairing}
diff --git a/lec_08_04.pdf b/lec_08_04.pdf
new file mode 100644
index 0000000..01a1369
Binary files /dev/null and b/lec_08_04.pdf differ
diff --git a/lec_08_04.tex b/lec_08_04.tex
index 6b588f1..e432602 100644
--- a/lec_08_04.tex
+++ b/lec_08_04.tex
@@ -4,7 +4,7 @@ $H_2$ 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*}
-
+\noindent
 Intersection form: 
 $H_2(X, \mathbb{Z}) \times 
 H_2(X, \mathbb{Z}) \longrightarrow \mathbb{Z}$ is symmetric and non singular. 
@@ -110,9 +110,9 @@ If $CUC^T = W$, then for
 $\binom{a}{b} = C^{-1} \binom{1}{0}$ we have:
 \[
 \binom{a}{b} W 
-\binom{a}{b} = \binom{1}{0} U \binom{1}{0} = 1.
+\binom{a}{b} = \binom{1}{0} U \binom{1}{0} = 1 \notin 2 \mathbb{Z}.
 \]
-
+% if we switch to \mathbb{Q} it will be possible? 
 \begin{theorem}[Whitehead]
 Any non-degenerate form 
 \[
diff --git a/lec_10_06.tex b/lec_10_06.tex
new file mode 100644
index 0000000..9541a40
--- /dev/null
+++ b/lec_10_06.tex
@@ -0,0 +1,48 @@
+
+
+
+
+\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}
diff --git a/lec_11_03.tex b/lec_11_03.tex
index 41e6075..4b35f79 100644
--- a/lec_11_03.tex
+++ b/lec_11_03.tex
@@ -1,6 +1,6 @@
 \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. 
+Suppose $F: \mathbb{C}^2 \rightarrow \mathbb{C}$ is a polynomial and $F(0) = 0$. Let take a 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 $S^3$. 
diff --git a/lec_15_04.pdf b/lec_15_04.pdf
new file mode 100644
index 0000000..d2799cb
Binary files /dev/null and b/lec_15_04.pdf differ
diff --git a/lec_15_04.tex b/lec_15_04.tex
index 40710ee..8928830 100644
--- a/lec_15_04.tex
+++ b/lec_15_04.tex
@@ -52,9 +52,11 @@ $(a, b) \mapsto aA^{-1}b^T$
 The intersection form on a four-manifold determines the linking on the boundary. \\
 
 \noindent
+\begin{fact}
 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 
-$A = V \times V^T$, $n = \rank V$.
+\[H_1(\Sigma(K), \mathbb{Z}) \cong  \quot{\mathbb{Z}^n}{A\mathbb{Z}}\ \  ,\] where 
+$A = V \times V^T$ and $n = \rank V$.
+\end{fact}
 %\input{ink_diag}
 \begin{figure}[h]
 \fontsize{20}{10}\selectfont
diff --git a/lec_18_03.tex b/lec_18_03.tex
index f2873d7..24d3aa9 100644
--- a/lec_18_03.tex
+++ b/lec_18_03.tex
@@ -70,9 +70,10 @@ Remark: $K \sim K^{\prime} \Leftrightarrow K \# -K^{\prime}$ is slice.
 \end{figure}
 \noindent
 \\
-Pontryagin-Thom construction tells us that there exists a compact oriented three - manifold $\Omega \subset B^4$ such that $\partial \Omega = Y$.
+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 $\partial B = \beta$. 
+\[
+\alpha, \beta \in \ker (H_1(\Sigma, \mathbb{Z}) \longrightarrow H_1(\Omega, \mathbb{Z})).\] 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 
@@ -151,19 +152,43 @@ Let $V =
 \begin{pmatrix}
           0 & A\\
           B & C
-\end{pmatrix}$
+\end{pmatrix}$. Then 
 \begin{align*}
+tV - V^T =  
+\begin{pmatrix}
+          0 & tA\\
+          tB & tC
+\end{pmatrix}
+-
+\begin{pmatrix}
+          0 & B^T\\
+          A^T & C^T
+\end{pmatrix}
+=
+\begin{pmatrix}
+          0 & tA - B^T\\
+          tB - A^T & tC - C^T
+\end{pmatrix}
+\\
 \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 $\Delta_K(t) = f(t) \cdot f(t^{-1})$.
+If $K$ is a slice knot then there exists $f \in \mathbb{Z}[t, t^{-1}]$ 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 $\sigma(K) \equiv 0$.
+If $K$ is slice, then the signature $\sigma(K) \equiv 0$:
+\[V + V^T = 
+\begin{pmatrix}
+          0 & A + B^T\\
+          B + A^T & C + C^T
+\end{pmatrix}
+\Rightarrow \sigma = 0 
+.\]
 \end{fact}
 
 
diff --git a/lec_20_05.pdf b/lec_20_05.pdf
new file mode 100644
index 0000000..eeb7447
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diff --git a/lec_20_05.tex b/lec_20_05.tex
index cf6649d..020911b 100644
--- a/lec_20_05.tex
+++ b/lec_20_05.tex
@@ -1,3 +1,4 @@
+% I don't have this first fragent in my notes 
 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 $M$: 
 
@@ -32,6 +33,7 @@ a &\mapsto (a, \_) \in  H_2(M, \mathbb{Z})
 \end{align*}
 has coker precisely $H_1(Y, \mathbb{Z})$. 
 \\???????????????\\
+% Here my notes begin: 
 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). 
 
@@ -62,7 +64,7 @@ H_1(\widetilde{X}, \mathbb{Z}[t, t^{-1}]) &\longrightarrow \quot{\mathbb{Q}}{\ma
 \end{align*}
 \end{fact}
 \noindent
-Note that $\mathbb{Z}$ is not PID. 
+Note that $\mathbb{Z}[t, t^{-1}]$ 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*}
@@ -156,7 +158,7 @@ Suppose $g = (t - \xi)^{\alpha} g^{\prime}$. Then $(t - \xi)^{k - \alpha}$ goes
 Every sesquilinear non-degenerate pairing
 \begin{align*}
 \quot{\Lambda}{p^k} \times \quot{\Lambda}{p}
-\longleftrightarrow \frac{h}{p^k}
+\longrightarrow \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}
@@ -171,7 +173,7 @@ Prove in the case, when $h$ has a constant sign on $S^1$.
 \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]
+\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 $(t^{-1} - \overbar{\zeta})$.
@@ -219,25 +221,23 @@ If $P$ has no roots on $S^1$ then $B(z) > 0$ for all $z$, so the assumptions of
 \end{proof}
 ?????????????????\\
 \begin{align*}
-(\quot{\Lambda}{p_{\xi}^k} \times
-\quot{\Lambda}{p_{\xi}^k}) &\longrightarrow
+\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
+\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)
+\begin{theorem}[Matumoto, Borodzik-Conway-Politarczyk]
 Let $K$ be a knot, 
 \begin{align*}
-&H_1(\widetilde{X}, \Lambda) \times
+H_1(\widetilde{X}, \Lambda) \times
 H_1(\widetilde{X}, \Lambda) 
-= \bigoplus_{\substack{k, \xi, \epsilon\\ \xi in S^1}}
+= \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^{+}} 
+(\quot{\Lambda}{p_{\xi}^k})^{m_k} \text{ and} \\
+ \delta_{\sigma}(\xi) = \lim_{\varepsilon \rightarrow 0^{+}} 
 \sigma(e^{2\pi i \varepsilon} \xi)
 - \sigma(e^{-2\pi i \varepsilon} \xi),\\
 \text{then } 
@@ -246,7 +246,15 @@ H_1(\widetilde{X}, \Lambda)
 + \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 $\sum\limits_{k_i \text{even}} \epsilon_i$.
-%$(\eta_{k, \xi_l^{+}} -\eta_{k, \xi_l^{-}}$
+$2\sum\limits_{k_i \odd} \epsilon_i$.\\
+The peak of the signature function is equal to ${\sum\limits_{k_i \even}}{\epsilon_i}$.
+\\
+?????????????????
+\\
+$(\eta_{k, \xi_l^{+}} -\eta_{k, \xi_l^{-}}$
+% Livingston Pacific Jurnal of M. 2012
 \end{theorem}
 \end{proof}
+
+
+
diff --git a/lec_25_02.pdf b/lec_25_02.pdf
new file mode 100644
index 0000000..5210c48
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diff --git a/lec_25_02.tex b/lec_25_02.tex
index 5a94fae..7586384 100644
--- a/lec_25_02.tex
+++ b/lec_25_02.tex
@@ -25,7 +25,7 @@ Not knots:
 %\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: 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
@@ -35,7 +35,7 @@ $\Phi_1 = \varphi_1$.
 \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) = \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.
@@ -45,7 +45,7 @@ Two knots $K_0$ and $K_1$ are isotopic if and only if they are ambient isotopic,
 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 $S^3$
+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:
@@ -54,12 +54,12 @@ Links:
 a trivial link with $3$ components:
 \includegraphics[width=0.2\textwidth]{3unknots.png},
 \item
-a hopf link: \includegraphics[width=0.13\textwidth]{Hopf.png},
+a Hopf link: \includegraphics[width=0.13\textwidth]{Hopf.png},
 \item
 a Whitehead link:
 \includegraphics[width=0.13\textwidth]{WhiteheadLink.png},
 \item
-Borromean link:
+a Borromean link:
 \includegraphics[width=0.1\textwidth]{BorromeanRings.png}.
 \end{itemize}
 \end{example}
@@ -76,10 +76,13 @@ $D_{\pi |_L}$ is non degenerate: \includegraphics[width=0.05\textwidth]{LinkDiag
 \end{enumerate}
 \end{definition}
 \noindent
-There are under- and overcrossings  (tunnels and bridges) on a link diagrams with an obvious meaning.\\
+There are under- and overcrossings  (tunnels and bridges) on a link diagrams with an obvious meaning.
+\begin{fact}
 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).\\
+\end{fact}
+\noindent
+
+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.
 
@@ -112,8 +115,8 @@ deformed into each other by a finite sequence of Reidemeister moves (and isotopy
 \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
+\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 $\partial \Sigma = L$.\\ 
 
@@ -128,7 +131,7 @@ We smooth all the crossings, so we get a disjoint union of circles on the plane.
 \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 $\partial D_2$. 
+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 $D_2$. Then we glue both components on the boundaries: $\partial D_1$ and $\partial D_2$. 
 
 \begin{figure}[h]
 \begin{center}
@@ -180,7 +183,7 @@ Let $\nu(\beta)$ be a tubular neighbourhood of $\beta$.  The linking number can
 \begin{example}
 \begin{itemize}
 \item
-Hopf link:
+A Hopf link:
 \begin{figure}[h]
 \fontsize{20}{10}\selectfont
 \centering{
@@ -200,16 +203,16 @@ $T(6, 2)$ link:
 \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 $L$ 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 curves $\alpha_1, \dots, \alpha_n$. 
+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 $\alpha_i^+$ 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]
@@ -272,9 +275,9 @@ V \rightarrow
                0 & 0
            \end{matrix}
     \end{array}
-\end{pmatrix}$
+\end{pmatrix},$
 \item
-inverse of (2)
+inverse of (2).
 
 \end{enumerate} 
 \end{theorem}
diff --git a/lec_25_03.tex b/lec_25_03.tex
index 9cdf66b..453b5f4 100644
--- a/lec_25_03.tex
+++ b/lec_25_03.tex
@@ -8,11 +8,11 @@ is zero except possibly of finitely many points and $\sigma_K(-1) = \sign(S + S^
 \end{theorem}
 \begin{lemma}
 \label{lem:metabolic}
-If $V$ is a Hermitian matrix ($\bar{V} = V^T$), $V$ is of size  $2n \times 2n$, 
+If $V$ is a Hermitian matrix ($\overline{V} = V^T$) of size  $2n \times 2n$, 
 $
 V = \begin{pmatrix}
 0 & A \\
-\bar{A}^T & B
+\overline{A^T} & B
 \end{pmatrix}
 $ and $\det V \neq 0$ then $\sigma(V) = 0$. 
 \end{lemma}
@@ -20,7 +20,7 @@ $ and $\det V \neq 0$ then $\sigma(V) = 0$.
 A Hermitian form $V$ is metabolic if $V$ has structure
 $\begin{pmatrix}
 0 & A\\
-\bar{A}^T & B
+\overline{A^T} & B
 \end{pmatrix}$ with half-dimensional null-space. 
 \end{definition}
 \noindent
diff --git a/lec_27_05.tex b/lec_27_05.tex
index 3228e3b..1b7b240 100644
--- a/lec_27_05.tex
+++ b/lec_27_05.tex
@@ -1,8 +1,68 @@
-....
+
+
+???????
+\begin{theorem}
+Such a pairing is isometric to a pairing:
+\[
+\begin{bmatrix}
+1
+\end{bmatrix}
+\times 
+\begin{bmatrix}
+1
+\end{bmatrix}
+\rightarrow
+\frac{\epsilon}{p^k_{\xi}}, 
+\: \epsilon \in {\pm 1}
+\]
+\end{theorem}
+?????????????
+\[ 
+\begin{bmatrix}
+1
+\end{bmatrix} = 1 \in \quot{\Lambda}{p^k_{\xi}  \Lambda }
+\]
+????????
+\begin{theorem}
+The jump of the signature function at $\xi$ is equal to 
+$2 \sum\limits_{k_i \odd} \epsilon_i$. \\
+The peak of the signature function is equal to $\sum\limits_{k_i \even} \epsilon_i$.
+\[
+(\quot{\Lambda}{p^{k_1} \Lambda}, \epsilon_1) \oplus \dots \oplus (\quot{\Lambda}{p^{k_n} \Lambda}, \epsilon_n)
+\]
+%$(\eta_{k, \xi_l^{+}} -\eta_{k, \xi_l^{-}}$
+\end{theorem}
+
 \begin{definition}
-A square hermitian matrix $A$ of size $n$ with coefficients in \\
-the Blanchfield pairing if:
-$H_1(\bar{X}$
+A matrix $A$ is called Hermitian if 
+$\overline{A(t)} = {A(t)}^T$
 \end{definition}
 
-field of fractions
\ No newline at end of file
+\begin{theorem}[Borodzik-Friedl 2015, Borodzik-Conway-Politarczyk 2018]
+A square Hermitian matrix $A(t)$ of size $n$ with coefficients in $\mathbb{Z}[t, t^{-1}]$ 
+(or $\mathbb{R}[t, t^{-1}]$ ) represents 
+the Blanchfield pairing if:
+\begin{eqnarray*}
+H_1(\bar{X}, \Lambda) = \quot{\Lambda^n }{A\Lambda^n },\\
+(x, y) \mapsto {\overline{x}}^T A^{-1} y \in \quot{\Omega}{\Lambda}\\
+H_1(\widetilde{X}, \Lambda) \times 
+H_1(\widetilde{X}, \Lambda) \longrightarrow
+\quot{\Omega}{\Lambda},
+\end{eqnarray*}
+where $\Lambda = \mathbb{Z}[t, t^{-1}]$ or $\mathbb{R}[t, t^{-1}]$, $\Omega = \mathbb{Q}(t)$ or $\mathbb{R}(t)$
+\end{theorem}
+????????\\field of fractions ??????
+\begin{eqnarray*}
+H_1(\Sigma(K), \mathbb{Z}) = \quot{\mathbb{Z}^n}{(V + V^T) \mathbb{Z}^n}\\
+H_1(\Sigma(K), \mathbb{Z}) 
+\times
+H_1(\Sigma(K), \mathbb{Z}) 
+\longrightarrow
+= \quot{\mathbb{Q}}{\mathbb{Z}}\\
+(a, b) \mapsto a{(V + V^T)}^{-1} b
+\end{eqnarray*}
+???????????????????\\
+\begin{eqnarray*}
+y \mapsto y + Az \\
+\overline{x^T} A^{-1}(y + Az) = \overline{x^T} A^{-1} + \overline{x^T} \mathbb{1} z
+\end{eqnarray*}
diff --git a/lectures_on_knot_theory.pdf b/lectures_on_knot_theory.pdf
index f36a8dd..040fc66 100644
Binary files a/lectures_on_knot_theory.pdf and b/lectures_on_knot_theory.pdf differ
diff --git a/lectures_on_knot_theory.tex b/lectures_on_knot_theory.tex
index 57ff132..be651af 100644
--- a/lectures_on_knot_theory.tex
+++ b/lectures_on_knot_theory.tex
@@ -1,3 +1,4 @@
+% use XeLaTex to compile
 \documentclass[12pt, twoside]{article}
 
 
@@ -95,6 +96,10 @@
 \DeclareMathOperator{\Lk}{lk}
 \DeclareMathOperator{\pt}{\{pt\}}
 \DeclareMathOperator{\sign}{sign}
+\DeclareMathOperator{\odd}{odd}
+\DeclareMathOperator{\even}{even}
+
+
 
 
 \titleformat{\subsection}{%
@@ -122,642 +127,45 @@
 %\input{myNotes}
 
 \section{Basic definitions \hfill\DTMdate{2019-02-25}}
-%\input{lec_1.tex}
+\input{lec_25_02.tex}
 
 \section{Alexander polynomial \hfill\DTMdate{2019-03-04}}
-%\input{lec_2.tex}
+\input{lec_04_03.tex}
 %add Hurewicz theorem?
 
 
 \section{Examples of knot classes
 \hfill\DTMdate{2019-03-11}}
-%\input{lec_3.tex}
+\input{lec_11_03.tex}
 
 \section{Concordance group \hfill\DTMdate{2019-03-18}}
-%\input{lec_4.tex}
+\input{lec_18_03.tex}
 
 \section{Genus $g$ cobordism \hfill\DTMdate{2019-03-25}}
-%\input{lec_5.tex}
+\input{lec_25_03.tex}
 
 \section{\hfill\DTMdate{2019-04-08}}
-\input{lec_6.tex}
+\input{lec_08_04.tex}
 
-\section{\hfill\DTMdate{2019-04-15}}
-???????????????????\\
-\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
-
-\section{\hfill\DTMdate{2019-05-20}}
-
-Let $M$ be compact, oriented, connected four-dimensional manifold. If ${H_1(M, \mathbb{Z}) = 0}$ then there exists a 
-bilinear form - the intersection form on $M$: 
-
-\begin{center}
-\begin{tikzcd}
-[
-column sep=tiny,
-row sep=small,
-ar symbol/.style = {draw=none,"\textstyle#1" description,sloped},
-isomorphic/.style = {ar symbol={\cong}},
-]
-H_2(M, \mathbb{Z})&
-\times & H_2(M, \mathbb{Z})
-\longrightarrow &
-\mathbb{Z}
-\\
-\ar[u,isomorphic] \mathbb{Z}^n  && &\\
-\end{tikzcd}
-\end{center}
-\noindent
-Let us consider a specific case: $M$ has a boundary $Y = \partial M$. 
-Betti number $b_1(Y) = 0$, $H_1(Y, \mathbb{Z})$ is  finite.
-Then 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, \_) 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*}
-????????????????????????????????????????????????????????????????????????\\
-????????????????????????????????????????????????????????????????????????\\
-????????????????????????????????????????????????????????????????????????\\
-????????????????????????????????????????????????????????????????????????\\
-$C_{*}(\widetilde{X})$ 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}
-\section{\hfill\DTMdate{2019-05-27}}
-
-....
-\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
-
-\section{Surgery \hfill\DTMdate{2019-06-03}}
-\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}
-
-
-
-
-\section{balagan}
-
-\noindent
-\noindent
+\section{Linking form\hfill\DTMdate{2019-04-15}}
+\input{lec_15_04.tex}
 
 \section{\hfill\DTMdate{2019-05-06}}
+\input{lec_06_05.tex}
 
-\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
-$\Sigma_?(K) \rightarrow S^3$ ?????\\
-$H_1(\Sigma_?(K), \mathbb{Z}) = h$\\
-$H \times H \longrightarrow \quot{\mathbb{Q}}{\mathbb{Z}}$\\
-...\\
+\section{\hfill\DTMdate{2019-05-20}}
+\input{lec_20_05.tex}
 
-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}
+\section{\hfill\DTMdate{2019-05-27}}
+\input{lec_27_05.tex}
 
+\section{Surgery \hfill\DTMdate{2019-06-03}}
+\input{lec_03_06.tex}
 
-\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}
+\section{Surgery\hfill\DTMdate{2019-06-03}}
+\input{lec_10_06.tex}
 
 \end{document}
 
+
+