From 2606bba01555f1106c5275c6680a53d7b15a560c Mon Sep 17 00:00:00 2001 From: Maria Marchwicka Date: Mon, 24 Jun 2019 21:58:03 -0500 Subject: [PATCH] small progress --- images/genus_2_bordism.pdf | Bin 18661 -> 15072 bytes images/genus_2_bordism.svg | 31 +- images/genus_bordism_proof.pdf | Bin 0 -> 31051 bytes images/genus_bordism_proof.pdf_tex | 66 + images/genus_bordism_proof.svg | 1276 +++++++++++++++++++ images/genus_bordism_zeros.pdf | Bin 0 -> 3836 bytes images/genus_bordism_zeros.pdf_tex | 60 + images/genus_bordism_zeros.svg | 1684 ++++++++++++++++++++++++++ images/intersection_form_A_B.pdf | Bin 0 -> 3903 bytes images/intersection_form_A_B.pdf_tex | 55 + images/intersection_form_A_B.svg | 100 ++ images/milnor_singular.pdf | Bin 4117 -> 4111 bytes images/milnor_singular.pdf_tex | 16 +- images/milnor_singular.svg | 47 +- images/torus_alpha_beta.pdf_tex | 6 +- lec_2.tex | 2 +- lec_3.tex | 10 +- lec_4.tex | 2 +- lec_5.tex | 134 +- lec_6.tex | 147 +++ lectures_on_knot_theory.tex | 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b/images/torus_alpha_beta.pdf_tex @@ -50,10 +50,10 @@ \makeatother% \begin{picture}(1,0.41568239)% \put(0.40129099,4.95648293){\color[rgb]{0,0,0}\makebox(0,0)[lt]{\begin{minipage}{0.37778386\unitlength}\raggedright \end{minipage}}}% - \put(0.63089219,0.39833514){\color[rgb]{0,0,0}\makebox(0,0)[lt]{\begin{minipage}{0.38670889\unitlength}\raggedright ${\alpha \cdot \beta = - \beta \cdot \alpha}$\\  \end{minipage}}}% + \put(0.63089219,0.39833514){\color[rgb]{0,0,0}\makebox(0,0)[lt]{\begin{minipage}{0.38670889\unitlength}\raggedright ${\alpha \cdot \beta = - \beta \cdot \alpha}$\\ \end{minipage}}}% \put(0,0){\includegraphics[width=\unitlength,page=1]{torus_alpha_beta.pdf}}% - \put(0.94608607,0.30060216){\color[rgb]{0,0,0}\makebox(0,0)[lt]{\begin{minipage}{0.38670889\unitlength}\raggedright $\beta$\\  \end{minipage}}}% - \put(0.44999409,0.1665571){\color[rgb]{0,0,0}\makebox(0,0)[lt]{\begin{minipage}{0.38670889\unitlength}\raggedright $\alpha$\\  \end{minipage}}}% + \put(0.94608607,0.30060216){\color[rgb]{0,0,0}\makebox(0,0)[lt]{\begin{minipage}{0.38670889\unitlength}\raggedright $\beta$\\ \end{minipage}}}% + \put(0.44999409,0.1665571){\color[rgb]{0,0,0}\makebox(0,0)[lt]{\begin{minipage}{0.38670889\unitlength}\raggedright $\alpha$\\ \end{minipage}}}% \put(0,0){\includegraphics[width=\unitlength,page=2]{torus_alpha_beta.pdf}}% \end{picture}% \endgroup% diff --git a/lec_2.tex b/lec_2.tex index 94dbb4a..1519178 100644 --- a/lec_2.tex +++ b/lec_2.tex @@ -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} \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}). -$[\mu]$ 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} diff --git a/lec_3.tex b/lec_3.tex index 2aef0ee..ac2c94b 100644 --- a/lec_3.tex +++ b/lec_3.tex @@ -1,4 +1,4 @@ -\subsection{Algebraic knot} +\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. 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} Let $p$ and $q$ be coprime numbers such that $p1$. \\ 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 $F^{-1}(0) \cap S^3$ 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 \centering{ \def\svgwidth{\linewidth} -\resizebox{1\textwidth}{!}{\input{images/satellite.pdf_tex}} -} -\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.} +\resizebox{1\textwidth}{!}{\input{images/satellite.pdf_tex}}} +\caption{Whitehead double satellite 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} \end{figure} \noindent diff --git a/lec_4.tex b/lec_4.tex index 3a0a316..f2873d7 100644 --- a/lec_4.tex +++ b/lec_4.tex @@ -65,7 +65,7 @@ Remark: $K \sim K^{\prime} \Leftrightarrow K \# -K^{\prime}$ is slice. \def\svgwidth{\linewidth} \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} \end{figure} \noindent diff --git a/lec_5.tex b/lec_5.tex index 56128a2..db2b191 100644 --- a/lec_5.tex +++ b/lec_5.tex @@ -1,15 +1,14 @@ +\subsection{Slice knots and metabolic form} \begin{theorem} +\label{the:sign_slice} 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} -\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 +If $V$ is a Hermitian matrix ($\bar{V} = V^T$), $V$ is of size $2n \times 2n$, $ V = \begin{pmatrix} 0 & A \\ @@ -25,11 +24,16 @@ $\begin{pmatrix} \end{pmatrix}$ with half-dimensional null-space. \end{definition} \noindent -In other words: non-degenerate metabolic hermitian form has vanishing signature.\\ -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: +Theorem \ref{the:sign_slice} can be also express as follow: +non-degenerate metabolic hermitian form has vanishing signature. +\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*} -&\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 - \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 $S - \bar{t}S^T \neq 0$. \end{proof} \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} \begin{proof} If $ K \sim K^\prime$ then $K \# K^\prime$ is slice. \[ \sigma_{-K^\prime}(t) = -\sigma_{K^\prime}(t) \] -\\??????????????\\ -The signature give a homomorphism from the concordance group to $\mathbb{Z}$.\\ -??????????????????\\ +The signature gives a homomorphism from the concordance group to $\mathbb{Z}$. Remark: if $t \in S^1$ is not algebraic over $\mathbb{Z}$, then $\sigma_K(t) \neq 0$ -(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} +\subsection{Four genus} \begin{figure}[h] \fontsize{20}{10}\selectfont \centering{ \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} -???????????????????????\\ + \begin{proposition}[Kawauchi inequality] 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} % Kawauchi Chapter 12 ??? +% Borodzik 2010 Morse theory for plane algebraic curves \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 $\begin{pmatrix} 0 & A\\ B & C \end{pmatrix}$, 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*} + +\begin{proof} +\begin{figure}[h] +\fontsize{20}{10}\selectfont +\centering{ +\def\svgwidth{\linewidth} +\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 -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} -If $t$ is nota ???? of $\det $ ???? -then $\vert \sigma_K(t) \vert \leq 2g$.\\ +If $t$ is not a root of +$\det S S^T - $ \\ +????????????????\\ +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{fact} +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} 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. +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{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 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 -?????????????\\ -The equality: +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. +was conjecture in the '70 and proved by P. Kronheimer and T. Mrówka (1994). +% OZSVATH-SZABO AND RASMUSSEN \end{itemize} \end{example} \begin{proposition} @@ -142,4 +193,15 @@ 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: $A^T = A: (a, b) = \bar{(a, b)}$. \ No newline at end of file +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} diff --git a/lec_6.tex b/lec_6.tex index e69de29..e803a2a 100644 --- a/lec_6.tex +++ b/lec_6.tex @@ -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} diff --git a/lectures_on_knot_theory.tex b/lectures_on_knot_theory.tex index 97e49d5..13561f7 100644 --- a/lectures_on_knot_theory.tex +++ b/lectures_on_knot_theory.tex @@ -8,6 +8,7 @@ \usepackage[english]{babel} +\usepackage{caption} \usepackage{comment} \usepackage{csquotes} @@ -23,6 +24,7 @@ \usepackage{mathtools} \usepackage{pict2e} +\usepackage[section]{placeins} \usepackage[pdf]{pstricks} \usepackage{tikz} @@ -84,9 +86,9 @@ \DeclareMathOperator{\Hom}{Hom} \DeclareMathOperator{\rank}{rank} +\DeclareMathOperator{\coker}{coker} \DeclareMathOperator{\ord}{ord} \DeclareMathOperator{\mytop}{top} - \DeclareMathOperator{\Gl}{GL} \DeclareMathOperator{\Sl}{SL} \DeclareMathOperator{\Lk}{lk} @@ -126,80 +128,18 @@ %add Hurewicz theorem? -\section{\hfill\DTMdate{2019-03-11}} +\section{Examples of knot classes +\hfill\DTMdate{2019-03-11}} \input{lec_3.tex} \section{Concordance group \hfill\DTMdate{2019-03-18}} \input{lec_4.tex} - -\section{\hfill\DTMdate{2019-03-25}} +\section{Genus $g$ cobordism \hfill\DTMdate{2019-03-25}} \input{lec_5.tex} \section{\hfill\DTMdate{2019-04-08}} -% -% -$X$ 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 ?? - +\input{lec_6.tex} \section{\hfill\DTMdate{2019-04-15}} \begin{theorem} @@ -341,14 +281,17 @@ 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, \\ -$\widetilde{X} \xrightarrow{\enspace \rho \enspace} X$ - an infinite cyclic cover (universal abelian cover). +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. \\ -$H_1(\widetilde{X}, \mathbb{Z}[t, t^{-1}])$ - 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*} 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}]} @@ -365,27 +308,31 @@ H_1(\widetilde{X}, \mathbb{Z}[t, t^{-1}]) \longrightarrow \quot{\mathbb{Q}}{\mat \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) &\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{fact} \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*} -&\xi \in S^1 \setminus \{ \pm 1\} -\quad +\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 +\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}\\ -& -\Lambda = \mathbb{R}[t, t^{-1}]\\ -&\text{Then: } H_1(\widetilde{X}, \Lambda) \cong \bigoplus_{\substack{\xi \in S^1 \setminus \{\pm 1 \}\\ k\geq 0}} +\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}} @@ -557,7 +504,9 @@ $2 \sum\limits_{k_i \text{ odd}} \epsilon_i$. The peak of the signature function .... \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} field of fractions