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Maria Marchwicka 2019-06-24 21:58:03 -05:00
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6
images/torus_alpha_beta.pdf_tex
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@ -50,10 +50,10 @@
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2
lec_2.tex
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@ -277,5 +277,5 @@ H_1(M) \cong \mathbb{Z} \Longrightarrow \lambda \in \ker ( \pi_1(S^1 \times S^1)
\label{fig:meridian_and_longitude} \label{fig:meridian_and_longitude}
\end{figure} \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}). 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 $H_1(S^3\setminus K, \mathbb{X})$. 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 $H_1(M)$. $[\mu]$ 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} \end{proof}

10
lec_3.tex
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@ -1,4 +1,4 @@
\subsection{Algebraic knot} \subsection{Algebraic knots}
\noindent \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 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 $2$. The dimension of sphere is $3$ and $F^{-1}(0)$ has codimension $2$.
@ -31,7 +31,7 @@ algebraic link (in older books on knot theory there is another notion of algebra
\begin{example} \begin{example}
Let $p$ and $q$ be coprime numbers such that $p<q$ and $p,q>1$. \\ 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. 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 $S^3 = \{ (z, w) \in \mathbb{C} : \max( \vert z \vert, \vert w \vert ) = \varepsilon$. Consider $S^3 = \{ (z, w) \in \mathbb{C} : \max( \vert z \vert, \vert w \vert )\} = \varepsilon$.
The intersection The intersection
$F^{-1}(0) \cap S^3$ is a torus $T(p, q)$. $F^{-1}(0) \cap S^3$ is a torus $T(p, q)$.
\\??????????????????? \\???????????????????
@ -128,9 +128,9 @@ $g_3(K_1 \# K_2) = g_3(K_1) + g_3(K_2)$.
\fontsize{12}{10}\selectfont \fontsize{12}{10}\selectfont
\centering{ \centering{
\def\svgwidth{\linewidth} \def\svgwidth{\linewidth}
\resizebox{1\textwidth}{!}{\input{images/satellite.pdf_tex}} \resizebox{1\textwidth}{!}{\input{images/satellite.pdf_tex}}}
} \caption{Whitehead double satellite knot.\\
\caption{Whitehead double satellite knot. Its pattern knot embedded non-trivially in an unknotted solid torus $T$ (e.i. $K \not\subset S^3\subset T$) and pattern in a companion knot.} 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} \label{fig:sattelite}
\end{figure} \end{figure}
\noindent \noindent

2
lec_4.tex
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@ -65,7 +65,7 @@ Remark: $K \sim K^{\prime} \Leftrightarrow K \# -K^{\prime}$ is slice.
\def\svgwidth{\linewidth} \def\svgwidth{\linewidth}
\resizebox{0.5\textwidth}{!}{\input{images/ball_4_pushed_seifert.pdf_tex}} \resizebox{0.5\textwidth}{!}{\input{images/ball_4_pushed_seifert.pdf_tex}}
} }
\caption{$Y = F \cup \Sigma$ is a smooth close surface.} \caption{$Y = F \cup \Sigma$ is a smooth closed surface.}
\label{fig:closed_surface} \label{fig:closed_surface}
\end{figure} \end{figure}
\noindent \noindent

132
lec_5.tex
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@ -1,15 +1,14 @@
\subsection{Slice knots and metabolic form}
\begin{theorem} \begin{theorem}
\label{the:sign_slice}
If $K$ is slice, If $K$ is slice,
then $\sigma_K(t) then $\sigma_K(t)
= \sign ( (1 - t)S +(1 - \bar{t})S^T)$ = \sign ( (1 - t)S +(1 - \bar{t})S^T)$
is zero except possibly of finitely many points and $\sigma_K(-1) = \sign(S + S^T) \neq 0$. is zero except possibly of finitely many points and $\sigma_K(-1) = \sign(S + S^T) \neq 0$.
\end{theorem} \end{theorem}
\begin{proof}
\noindent
We will use the following lemma.
\begin{lemma} \begin{lemma}
\label{lem:metabolic} \label{lem:metabolic}
If $V$ is a Hermitian matrix ($\bar{V} = V^T$), $V$ is of size $2n \times 2n$ and If $V$ is a Hermitian matrix ($\bar{V} = V^T$), $V$ is of size $2n \times 2n$,
$ $
V = \begin{pmatrix} V = \begin{pmatrix}
0 & A \\ 0 & A \\
@ -25,11 +24,16 @@ $\begin{pmatrix}
\end{pmatrix}$ with half-dimensional null-space. \end{pmatrix}$ with half-dimensional null-space.
\end{definition} \end{definition}
\noindent \noindent
In other words: non-degenerate metabolic hermitian form has vanishing signature.\\ Theorem \ref{the:sign_slice} can be also express as follow:
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}. \\ non-degenerate metabolic hermitian form has vanishing signature.
Let $t \in S^1 \setminus \{1\}$. Then: \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*} \begin{align*}
&\det((1 - t) S + (1 - \bar{t}) S^T) = \det((1 - t) S + (1 - \bar{t}) S^T) =&
\det((1 - t) S + (t\bar{t} - \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)) =
\det((1 -t)(S - \bar{t} S^T)). \det((1 -t)(S - \bar{t} S^T)).
@ -37,71 +41,118 @@ Let $t \in S^1 \setminus \{1\}$. Then:
As $\det (S + S^T) \neq 0$, so $S - \bar{t}S^T \neq 0$. As $\det (S + S^T) \neq 0$, so $S - \bar{t}S^T \neq 0$.
\end{proof} \end{proof}
\begin{corollary} \begin{corollary}
If $K \sim K^\prime$ then for all but finitely many $t \in S^1 \setminus \{1\}: \sigma_K(t) = \sigma_{K^\prime}(t)$. 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} \end{corollary}
\begin{proof} \begin{proof}
If $ K \sim K^\prime$ then $K \# K^\prime$ is slice. If $ K \sim K^\prime$ then $K \# K^\prime$ is slice.
\[ \[
\sigma_{-K^\prime}(t) = -\sigma_{K^\prime}(t) \sigma_{-K^\prime}(t) = -\sigma_{K^\prime}(t)
\] \]
\\??????????????\\ The signature gives a homomorphism from the concordance group to $\mathbb{Z}$.
The signature give a homomorphism from the concordance group to $\mathbb{Z}$.\\
??????????????????\\
Remark: if $t \in S^1$ is not algebraic over $\mathbb{Z}$, then $\sigma_K(t) \neq 0$ Remark: if $t \in S^1$ is not algebraic over $\mathbb{Z}$, then $\sigma_K(t) \neq 0$
(we can is the argument that $\mathscr{C} \longrightarrow \mathbb{Z}$ as well). (we can use the argument that $\mathscr{C} \longrightarrow \mathbb{Z}$ as well).
\end{proof} \end{proof}
\subsection{Four genus}
\begin{figure}[h] \begin{figure}[h]
\fontsize{20}{10}\selectfont \fontsize{20}{10}\selectfont
\centering{ \centering{
\def\svgwidth{\linewidth} \def\svgwidth{\linewidth}
\resizebox{0.5\textwidth}{!}{\input{images/genus_2_bordism.pdf_tex}} \resizebox{0.7\textwidth}{!}{\input{images/genus_2_bordism.pdf_tex}}
} }
\caption{$K$ and $K^\prime$ are connected by a genus $g$ surface of genus.}\label{fig:genus_2_bordism} \caption{$K$ and $K^\prime$ are connected by a genus $g$ surface.}\label{fig:genus_2_bordism}
\end{figure} \end{figure}
???????????????????????\\
\begin{proposition}[Kawauchi inequality] \begin{proposition}[Kawauchi inequality]
If there exists a genus $g$ surface as in Figure \ref{fig:genus_2_bordism} 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 $\vert \sigma_K(t) - \sigma_{K^\prime}(t) \vert \leq 2 g$. 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} \end{proposition}
% Kawauchi Chapter 12 ??? % Kawauchi Chapter 12 ???
% Borodzik 2010 Morse theory for plane algebraic curves
\begin{lemma} \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} 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 $\begin{pmatrix}
0 & A\\ 0 & A\\
B & C B & C
\end{pmatrix}$, where $0$ is $(n - g) \times (n - g)$ submatrix. \end{pmatrix}$, where $0$ is $(n - g) \times (n - g)$ submatrix.
\end{lemma} \end{lemma}
??????????????????????\\
\begin{align*} \begin{proof}
\dim H_1(Z) = 2 n\\ \begin{figure}[h]
\dim H_1 (Y) = 2 n + 2 g\\ \fontsize{20}{10}\selectfont
\dim (\ker (H_1, Y) \longrightarrow H_1(\Omega)) = n + g\\ \centering{
Y = X \sum \Sigma \def\svgwidth{\linewidth}
\end{align*} \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 \noindent
If $\alpha, \beta \in \ker(H_1(\Sigma \longrightarrow H_1(\Omega))$, then ${\Lk(\alpha, \beta^+) = 0}$. 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} \begin{corollary}
If $t$ is nota ???? of $\det $ ???? If $t$ is not a root of
then $\vert \sigma_K(t) \vert \leq 2g$.\\ $\det S S^T - $ \\
????????????????\\
then
$\vert \sigma_K(t) \vert \leq 2g$.
\end{corollary} \end{corollary}
\noindent \begin{fact}
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 $B^4$. 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} \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 $\partial \Sigma = K$. 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} \end{definition}
\noindent \noindent
Remark: $3$ - genus is additive under taking connected sum, but $4$ - genus is not. 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{example}
\begin{itemize} \begin{itemize}
\item Let $K = T(2, 3)$. $\sigma(K) = -2$, therefore $T(2, 3)$ isn't a slice knot. \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 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 \item
?????????????\\ the equality:
The equality:
\[ \[
g_4(T(p, q) ) = \frac{1}{2} (p - 1) (g -1) g_4(T(p, q) ) = \frac{1}{2} (p - 1) (g -1)
\] \]
was conjecture in the '70 and proved by P. Kronheimer and T. Mrówka. was conjecture in the '70 and proved by P. Kronheimer and T. Mrówka (1994).
% OZSVATH-SZABO AND RASMUSSEN
\end{itemize} \end{itemize}
\end{example} \end{example}
\begin{proposition} \begin{proposition}
@ -143,3 +194,14 @@ $(1 - t) S + (1 - \bar{t}^{-1}) S^T$ is Witt equivalence to $(1 - t) S^\prime +
%??????????????????????????? %???????????????????????????
\noindent \noindent
A form is meant as hermitian with respect to this involution: $A^T = A: (a, b) = \bar{(a, b)}$. A form is meant as hermitian with respect to this involution: $A^T = A: (a, b) = \bar{(a, b)}$.
\\
????????????????????????????
\\
\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}

147
lec_6.tex
View File

@ -0,0 +1,147 @@
$X$ 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
\\
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:
\[
\int_M \omega = [\omega]([M]), \quad [\omega] \in H^m_\Omega(M), \ [M] \in H_m(M).
\]
\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}
?????????????????????????\\
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 $CVC^T = W$, then for
$\binom{a}{b} = C^{-1} \binom{1}{0}$ we have $\binom{a}{b} $ \\
????????????????\\
$\omega \binom{a}{b} = \binom{1}{0} u \binom{1}{0} = 1$.
\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}

117
lectures_on_knot_theory.tex
View File

@ -8,6 +8,7 @@
\usepackage[english]{babel} \usepackage[english]{babel}
\usepackage{caption}
\usepackage{comment} \usepackage{comment}
\usepackage{csquotes} \usepackage{csquotes}
@ -23,6 +24,7 @@
\usepackage{mathtools} \usepackage{mathtools}
\usepackage{pict2e} \usepackage{pict2e}
\usepackage[section]{placeins}
\usepackage[pdf]{pstricks} \usepackage[pdf]{pstricks}
\usepackage{tikz} \usepackage{tikz}
@ -84,9 +86,9 @@
\DeclareMathOperator{\Hom}{Hom} \DeclareMathOperator{\Hom}{Hom}
\DeclareMathOperator{\rank}{rank} \DeclareMathOperator{\rank}{rank}
\DeclareMathOperator{\coker}{coker}
\DeclareMathOperator{\ord}{ord} \DeclareMathOperator{\ord}{ord}
\DeclareMathOperator{\mytop}{top} \DeclareMathOperator{\mytop}{top}
\DeclareMathOperator{\Gl}{GL} \DeclareMathOperator{\Gl}{GL}
\DeclareMathOperator{\Sl}{SL} \DeclareMathOperator{\Sl}{SL}
\DeclareMathOperator{\Lk}{lk} \DeclareMathOperator{\Lk}{lk}
@ -126,80 +128,18 @@
%add Hurewicz theorem? %add Hurewicz theorem?
\section{\hfill\DTMdate{2019-03-11}} \section{Examples of knot classes
\hfill\DTMdate{2019-03-11}}
\input{lec_3.tex} \input{lec_3.tex}
\section{Concordance group \hfill\DTMdate{2019-03-18}} \section{Concordance group \hfill\DTMdate{2019-03-18}}
\input{lec_4.tex} \input{lec_4.tex}
\section{Genus $g$ cobordism \hfill\DTMdate{2019-03-25}}
\section{\hfill\DTMdate{2019-03-25}}
\input{lec_5.tex} \input{lec_5.tex}
\section{\hfill\DTMdate{2019-04-08}} \section{\hfill\DTMdate{2019-04-08}}
% \input{lec_6.tex}
%
$X$ 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:
$H_2(X, \mathbb{Z}) \times
H_2(X, \mathbb{Z}) \longrightarrow \mathbb{Z}$ - 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:
\[
[A], [B] \in H_2(X, \mathbb{Z}).
\]
\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:
\[
\int_M \omega = [\omega]([M]), \quad [\omega] \in H^m_\Omega(M), \ [M] \in H_m(M).
\]
\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 $\alpha$.
$T_X \alpha + T_X \beta = T_Z \Sigma$
}\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 ??
\section{\hfill\DTMdate{2019-04-15}} \section{\hfill\DTMdate{2019-04-15}}
\begin{theorem} \begin{theorem}
@ -341,14 +281,17 @@ a &\mapsto (a, \_) H_2(M, \mathbb{Z})
\end{align*} \end{align*}
has coker precisely $H_1(Y, \mathbb{Z})$. has coker precisely $H_1(Y, \mathbb{Z})$.
\\???????????????\\ \\???????????????\\
Let $K \subset S^3$ be a knot, \\ Let $K \subset S^3$ be a knot, $X = S^3 \setminus K$ a knot complement and
$X = S^3 \setminus K$ - a knot complement, \\ $\widetilde{X} \xrightarrow{\enspace \rho \enspace} X$ an infinite cyclic cover (universal abelian cover). By Hurewicz theorem we know that:
$\widetilde{X} \xrightarrow{\enspace \rho \enspace} X$ - an infinite cyclic cover (universal abelian cover).
\begin{align*} \begin{align*}
\pi_1(X) \longrightarrow \quot{\pi_1(X)}{[\pi_1(X), \pi_1(X)]} = H_1(X, \mathbb{Z} ) \cong \mathbb{Z} \pi_1(X) \longrightarrow \quot{\pi_1(X)}{[\pi_1(X), \pi_1(X)]} = H_1(X, \mathbb{Z} ) \cong \mathbb{Z}
\end{align*} \end{align*}
????????????????????????????????????????????????????????????????????????\\
????????????????????????????????????????????????????????????????????????\\
????????????????????????????????????????????????????????????????????????\\
????????????????????????????????????????????????????????????????????????\\
$C_{*}(\widetilde{X})$ has a structure of a $\mathbb{Z}[t, t^{-1}] \cong \mathbb{Z}[\mathbb{Z}]$ module. \\ $C_{*}(\widetilde{X})$ has a structure of a $\mathbb{Z}[t, t^{-1}] \cong \mathbb{Z}[\mathbb{Z}]$ module. \\
$H_1(\widetilde{X}, \mathbb{Z}[t, t^{-1}])$ - Alexander 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*} \begin{align*}
H_1(\widetilde{X}, \mathbb{Z}[t, t^{-1}]) \times 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}]} H_1(\widetilde{X}, \mathbb{Z}[t, t^{-1}]) \longrightarrow \quot{\mathbb{Q}}{\mathbb{Z}[t, t^{-1}]}
@ -365,27 +308,31 @@ H_1(\widetilde{X}, \mathbb{Z}[t, t^{-1}]) \longrightarrow \quot{\mathbb{Q}}{\mat
\begin{align*} \begin{align*}
H_1(\widetilde{X}, \mathbb{Z}[t, t^{-1}]) \times 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}]}\\ H_1(\widetilde{X}, \mathbb{Z}[t, t^{-1}]) &\longrightarrow \quot{\mathbb{Q}}{\mathbb{Z}[t, t^{-1}]}\\
(\alpha, \beta) &\mapsto \alpha^{-1}(t -1)(tV - V^T)^{-1}\beta (\alpha, \beta) \quad &\mapsto \alpha^{-1}(t -1)(tV - V^T)^{-1}\beta
\end{align*} \end{align*}
\end{fact} \end{fact}
\noindent \noindent
Note that $\mathbb{Z}$ is not PID. Therefore we don't have primer decomposition of this moduli. We can simplify this problem by replacing $\mathbb{Z}$ by $\mathbb{R}$. We lose some date by doing this transition. 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*} \begin{align*}
&\xi \in S^1 \setminus \{ \pm 1\} \xi \in S^1 \setminus \{ \pm 1\}
\quad &\quad
p_{\xi} = p_{\xi} =
(t - \xi)(t - \xi^{-1}) t^{-1} (t - \xi)(t - \xi^{-1}) t^{-1}
\\ \\
&\xi \in \mathbb{R} \setminus \{ \pm 1\} \xi \in \mathbb{R} \setminus \{ \pm 1\}
\quad &\quad
q_{\xi} = (t - \xi)(t - \xi^{-1}) t^{-1} q_{\xi} = (t - \xi)(t - \xi^{-1}) t^{-1}
\\ \\
& \xi \notin \mathbb{R} \cup S^1
\xi \notin \mathbb{R} \cup S^1 \quad &\quad
q_{\xi} = (t - \xi)(t - \overbar{\xi})(t - \xi^{-1})(t - \overbar{\xi}^{-1}) t^{-2}\\ q_{\xi} = (t - \xi)(t - \overbar{\xi})(t - \xi^{-1})
& (t - \overbar{\xi}^{-1}) t^{-2}
\Lambda = \mathbb{R}[t, t^{-1}]\\ \end{align*}
&\text{Then: } H_1(\widetilde{X}, \Lambda) \cong \bigoplus_{\substack{\xi \in S^1 \setminus \{\pm 1 \}\\ k\geq 0}} 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} ( \quot{\Lambda}{p_{\xi}^k })^{n_k, \xi}
\oplus \oplus
\bigoplus_{\substack{\xi \notin S^1 \\ l\geq 0}} \bigoplus_{\substack{\xi \notin S^1 \\ l\geq 0}}
@ -557,7 +504,9 @@ $2 \sum\limits_{k_i \text{ odd}} \epsilon_i$. The peak of the signature function
.... ....
\begin{definition} \begin{definition}
A square hermitian matrix $A$ of size $n$. A square hermitian matrix $A$ of size $n$ with coefficients in \\
the Blanchfield pairing if:
$H_1(\bar{X}$
\end{definition} \end{definition}
field of fractions field of fractions