1, 2, 4 lec

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Maria Marchwicka 2019-06-17 10:08:38 -05:00
parent a27119a907
commit 58e57a644a
3 changed files with 1086 additions and 3 deletions

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@ -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{

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@ -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 $p<q$ and $p,q>1$.
\\
$F^{-1}(0) \cap S^3$ 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}