diff --git a/images/torus_alpha_beta.svg b/images/torus_alpha_beta.svg new file mode 100644 index 0000000..2bed741 --- /dev/null +++ b/images/torus_alpha_beta.svg @@ -0,0 +1,1039 @@ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + image/svg+xml + + + + + + + $\Sigma$ genus $3$  + + + + + + + + diff --git a/lec_2.tex b/lec_2.tex index 51ea39a..af21196 100644 --- a/lec_2.tex +++ b/lec_2.tex @@ -208,7 +208,7 @@ $\Delta_{11n34} \equiv 1$. 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(\mid z_1 \mid, \mid z_2\mid) = 1 +Take $(z_1, z_2) \in \mathbb{C}$ such that $\max(\mid z_1 \mid, \mid z_2\mid) = 1 $. Define following sets: $S_1 = \{ (z_1, z_2) \in S^3: \mid z_1 \mid = 0\} \cong S^1 \times D^2 $ and $S_2 = \{(z_1, z_2) \in S ^3: \mid z_2 \mid = 1 \} \cong D^2 \times S^1$. The intersection $S_1 \cap S_2 = \{(z_1, z_2): \mid z_1 \mid = \mid z_2 \mid = 1 \} \cong S^1 \times S^1$ \begin{figure}[h] \centering{ diff --git a/lectures_on_knot_theory.tex b/lectures_on_knot_theory.tex index ac1ba7b..10b4152 100644 --- a/lectures_on_knot_theory.tex +++ b/lectures_on_knot_theory.tex @@ -89,6 +89,8 @@ \DeclareMathOperator{\Sl}{SL} \DeclareMathOperator{\Lk}{lk} \DeclareMathOperator{\pt}{\{pt\}} +\DeclareMathOperator{\sign}{sign} + \titleformat{\subsection}{% \normalfont \fontsize{12}{15}\bfseries}{% @@ -141,9 +143,26 @@ \end{figure} ???????????? \\ +$L$ is a link in $S^3$ \\ +\\?????????????????\\ +$L$ is an unknot if and only if $F(0) \neq 0$ (provided $S^3$ has a sufficiently small radius. +\\ \noindent +Remark: if $S^3$ is large it can happen that $L$ is unlink, but $F^{-1} \cap B^4$ is "complicated". \\ +????????????\\ +\noindent +\begin{example} +Let $p$ and $q$ be coprime numbers such that $p1$. +\\ +$F^{-1}(0) \cap S^3$ is a solid torus $T(p, q)$. \\ +$F(z, w) = z^p - w^q$\\ +Consider $S^3 = \{ (z, w) \in \mathbb{C} : \max( \mid z \mid, \mid w \mid ) = \varepsilon$.\\ +$F^{-1}(0) = \{t = t^q, w = t^p\}.$ For unknot $t = \max (\mid t\mid ^p, \mid t \mid^q) = \varepsilon$. +\end{example} as a corollary we see that $K_T^{n, }$ ???? \\ -is not slice unless $m=0$. +is not slice unless $m=0$. \\ +$t = re^{i \Theta}, \Theta \in [0, 2\pi], r = \varepsilon^{\frac{i}{p}}$ ?????????????????????????\\ +Suppose $L$ is a diagonal link. $L = F^{-1}(0) \cap S^3$. \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} @@ -178,6 +197,23 @@ A link $L$ is fibered if there exists a map ${\phi: S^3\setminus L \longleftarro \section{\hfill\DTMdate{2019-03-25}} +\begin{theorem} +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{proof} +\begin{lemma} +If $V$ is a Hermitian matrix ($\bar{V} = V^T$), $V$ is of size $2n \times 2n$ and +$ +V = \begin{pmatrix} +0 & A \\ +\bar{A}^T & B +\end{pmatrix} +$ +\end{lemma} +\end{proof} \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} @@ -198,7 +234,15 @@ H_2(X, \mathbb{Z}) \longrightarrow \mathbb{Z}$ - symmetric, non singular. \\ Let $A$ and $B$ be closed, oriented surfaces in $X$. \begin{proposition} -$A \cdot B$ doesn't depend of choice of $A$ and $B$ in their homology classes. +$A \cdot B$ doesn't depend of choice of $A$ and $B$ in their homology classes $[A], [B] \in H_2(X, \mathbb{Z})$. \\ +\noindent +If $M$ is an $m$ - dimensional close, connected and orientable manifold, then $H_m(M, \mathbb{Z})$ and the orientation if $M$ determined a cycle $[M] \in H_m(M, \mathbb{Z})$, called the fundamental cycle. +\begin{example} +If $\omega$ is an $m$ - form then: +\[ + = [\omega] +\] +\end{example} %$A \cdot B$ gives the pairing as ?? \end{proposition}