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Maria Marchwicka 2019-06-21 23:24:32 -05:00
parent 145832c5a4
commit 00d5474731
15 changed files with 1785 additions and 110 deletions
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8
lec_2.tex
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@ -211,7 +211,7 @@ S^3 = \partial D^4 = \partial (D^2 \times D^2) = (D^2 \times S^1) \cup (S^1 \ti
\]
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 \vert, \vert z_2\vert) = 1.}
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*}
@ -219,7 +219,7 @@ 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
$S_1 \cap S_2 = \{(z_1, z_2): \vert z_1 \vert = \vert z_2 \vert = 1 \} \cong S^1 \times S^1$
$S_1 \cap S_2 = \{(z_1, z_2): \vert z_1 \vert = \vert z_2 \vert = 1 \} \cong S^1 \times S^1$.
\begin{figure}[h]
\centering{
\def\svgwidth{\linewidth}
@ -256,7 +256,9 @@ 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 ${\lambda \in \ker (\pi_1(S^1 \times S^1) \longrightarrow \pi_1(M)}$. 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
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 $\partial D^2$.
\\???? $g_3$?\\

16
lec_3.tex
View File

@ -92,7 +92,7 @@ TO WRITE REFERENCE!!!!!!!!!!!
\end{proof}
\noindent
Consequences:
\begin{enumerate}
\begin{enumerate}[label={(\arabic*)}]
\item
the Alexander polynomial is the characteristic polynomial of $h$:
\[
@ -139,10 +139,16 @@ $g_3(K_1 \# K_2) = g_3(K_1) + g_3(K_2)$.
A knot (link) is called alternating if it admits an alternating diagram.
\end{definition}
\begin{example}
Figure eight knot is an alternating knot. \hfill\\
\includegraphics[width=0.5\textwidth]{figure8.png}
\end{example}
\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}

72
lec_5.tex
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@ -5,6 +5,8 @@ then $\sigma_K(t)
is zero except possibly of finitely many points and $\sigma_K(-1) = \sign(S + S^T) \neq 0$.
\end{theorem}
\begin{proof}
\noindent
We will use the following lemma.
\begin{lemma}
\label{lem:metabolic}
If $V$ is a Hermitian matrix ($\bar{V} = V^T$), $V$ is of size $2n \times 2n$ and
@ -34,7 +36,6 @@ Let $t \in S^1 \setminus \{1\}$. Then:
\end{align*}
As $\det (S + S^T) \neq 0$, so $S - \bar{t}S^T \neq 0$.
\end{proof}
?????????????????s\\
\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}
@ -69,9 +70,76 @@ If $K$ bounds a genus $g$ surface $X \in B^4$ and $S$ is a Seifert form then ${S
B & C
\end{pmatrix}$, where $0$ is $(n - g) \times (n - g)$ submatrix.
\end{lemma}
??????????????????????\\
\begin{align*}
\dim H_1(Z) = 2 n\\
\dim H_1 (Y) = 2 n + 2 g\\
\dim (\ker (H_1, Y) \longrightarrow H_1(\Omega)) = n + g\\
Y = X \sum \Sigma
\end{align*}
\noindent
If $\alpha, \beta \in \ker(H_1(\Sigma \longrightarrow H_1(\Omega))$, then ${\Lk(\alpha, \beta^+) = 0}$.
\begin{corollary}
If $t$ is nota ???? of $\det $ ????
then $\vert \sigma_K(t) \vert \leq 2g$.\\
\end{corollary}
\noindent
If there exists cobordism of genus $g$ between $K$ and $K^\prime$ like shown in Figure \ref{fig:genus_2_bordism}, then $K \# -K^\prime$ bounds a surface of genus $g$ in $B^4$.
\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
Remark: $3$ - genus is additive under taking connected sum, but $4$ - genus is not.
\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 $g_(K \# K^\prime) = 0$.
\\?????????????????????\\
\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.
\end{itemize}
\end{example}
\begin{proposition}
$g_4 (T(p, q) \# -T(r, s))$ is in general hopelessly unknown.
\\???????????????\\
essentially $\sup \vert \sigma_K(t) \vert \leq 2 g_n(K)$
\end{proposition}
\begin{definition}
A knot $K$ is called topologically slice if $K$ bounds a topological locally flat disc in $B^4$ (it has tubular neighbourhood).
\end{definition}
\begin{theorem}[Freedman, '82]
If $\Delta_K(t) \geq 1$, then $K$ is topologically slice, but not necessarily smoothly slice.
\end{theorem}
\begin{theorem}[Powell, 2015]
If $K$ is genus g
\\(top. loc.?????????)\\
cobordant to $K^\prime$,
then $\vert \sigma_K(t) - \sigma_{K^\prime}(t) \vert \leq 2 g$. \\
If $g_4^{\mytop}(K) \geq $ ?????ess $\sup \vert \sigma_K(t) \vert$ and ?????????\\
$\dim \ker (H_1 (Y) \longrightarrow H_1(\Omega)) = \frac{1}{2} \dim H_1(Y)$.
\end{theorem}
???????????????
\[
H^1 (B^4 \setminus Y, \mathbb{Z}) = [B^4 \setminus Y, S^1]
\]
\noindent
Remark: unless $p=2$ or $p = 3 \wedge q = 4$:
\[
g_4^\top (T(p, q)) < q_4(T(p, q))
\]
%??????????????????????
\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: $A^T = A: (a, b) = \bar{(a, b)}$.

72
lectures_on_knot_theory.tex
View File

@ -85,6 +85,8 @@
\DeclareMathOperator{\Hom}{Hom}
\DeclareMathOperator{\rank}{rank}
\DeclareMathOperator{\ord}{ord}
\DeclareMathOperator{\mytop}{top}
\DeclareMathOperator{\Gl}{GL}
\DeclareMathOperator{\Sl}{SL}
\DeclareMathOperator{\Lk}{lk}
@ -148,28 +150,84 @@ 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 $[A], [B] \in H_2(X, \mathbb{Z})$. \\
$A \cdot B$ doesn't depend of choice of $A$ and $B$ in their homology classes:
\[
[A], [B] \in H_2(X, \mathbb{Z}).
\]
\end{proposition}
\noindent
\\
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:
\[
= [\omega]
\int_M \omega = [\omega]([M]), \quad [\omega] \in H^m_\Omega(M), \ [M] \in H_m(M).
\]
\end{example}
\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 $\alpha$.
$T_X \alpha + T_X \beta = T_Z \Sigma$
}\label{fig:torus_alpha_beta}
\end{figure}
\begin{theorem}
Any non-degenerate form
\[
A : \mathbb{Z}^n \times \mathbb{Z}^n \longrightarrow \mathbb{Z}
\]
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 defined 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 ??
\end{proposition}
\section{\hfill\DTMdate{2019-04-15}}
In other words:\\
\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 $\mid \det A\mid = \# H_1(Y, \mathbb{Z})$.\\
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}
@ -426,7 +484,7 @@ Therefore:
&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 $(t - \zeta)^2 \mid P$ (at least - otherwise it would change sign). Therefore:
$P = g \overbar{g}$. 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.}