261 lines
9.9 KiB
TeX
261 lines
9.9 KiB
TeX
% I don't have this first fragent in my notes
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Let $M$ be compact, oriented, connected four-dimensional manifold. If ${H_1(M, \mathbb{Z}) = 0}$ then there exists a
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bilinear form - the intersection form on $M$:
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\begin{center}
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\begin{tikzcd}
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[
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column sep=tiny,
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row sep=small,
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ar symbol/.style = {draw=none,"\textstyle#1" description,sloped},
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isomorphic/.style = {ar symbol={\cong}},
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]
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H_2(M, \mathbb{Z})&
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\times & H_2(M, \mathbb{Z})
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\longrightarrow &
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\mathbb{Z}
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\\
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\ar[u,isomorphic] \mathbb{Z}^n && &\\
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\end{tikzcd}
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\end{center}
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\noindent
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Let us consider a specific case: $M$ has a boundary $Y = \partial M$.
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Betti number $b_1(Y) = 0$, $H_1(Y, \mathbb{Z})$ is finite.
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Then the intersection form can be degenerated in the sense that:
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\begin{align*}
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H_2(M, \mathbb{Z})
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\times H_2(M, \mathbb{Z})
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&\longrightarrow
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\mathbb{Z} \quad&
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H_2(M, \mathbb{Z}) &\longrightarrow \Hom (H_2(M, \mathbb{Z}), \mathbb{Z})\\
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(a, b) &\mapsto \mathbb{Z} \quad&
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a &\mapsto (a, \_) \in H_2(M, \mathbb{Z})
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\end{align*}
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has coker precisely $H_1(Y, \mathbb{Z})$.
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\\???????????????\\
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% Here my notes begin:
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Let $K \subset S^3$ be a knot, $X = S^3 \setminus K$ a knot complement and
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$\widetilde{X} \xrightarrow{\enspace \rho \enspace} X$ an infinite cyclic cover (universal abelian cover).
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%By Hurewicz theorem we know that:
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%\begin{align*}
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%\pi_1(X) \longrightarrow \quot{\pi_1(X)}{[\pi_1(X), \pi_1(X)]} = H_1(X, \mathbb{Z} ) \cong \mathbb{Z}
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%\end{align*}
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\noindent
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$C_{*}(\widetilde{X})$ has a structure of a $\mathbb{Z}[t, t^{-1}] \cong \mathbb{Z}[\mathbb{Z}]$ module. \\
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Let $H_1(\widetilde{X}, \mathbb{Z}[t, t^{-1}])$ be the Alexander module of the knot $K$ with an intersection form:
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\begin{align*}
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H_1(\widetilde{X}, \mathbb{Z}[t, t^{-1}]) \times
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H_1(\widetilde{X}, \mathbb{Z}[t, t^{-1}]) \longrightarrow \quot{\mathbb{Q}}{\mathbb{Z}[t, t^{-1}]}
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\end{align*}
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\begin{lemma}
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\begin{align*}
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&H_1(\widetilde{X}, \mathbb{Z}[t, t^{-1}]) \cong
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\quot{\mathbb{Z}{[t, t^{-1}]}^n}{(tV - V^T)\mathbb{Z}[t, t^{-1}]^n}\;, \\
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&\text{where $V$ is a Seifert matrix.}
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\end{align*}
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\end{lemma}
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\begin{lemma}
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\begin{align*}
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H_1(\widetilde{X}, \mathbb{Z}[t, t^{-1}]) \times
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H_1(\widetilde{X}, \mathbb{Z}[t, t^{-1}]) &\longrightarrow \quot{\mathbb{Q}}{\mathbb{Z}[t, t^{-1}]}\\
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(\alpha, \beta) \quad &\mapsto \alpha^{-1}{(t -1)(tV - V^T)}^{-1}\beta
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\end{align*}
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\end{lemma}
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\noindent
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Note that $\mathbb{Z}[t, t^{-1}]$ is not PID.
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Therefore we don't have primary decomposition of this module.
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We can simplify this problem by replacing $\mathbb{Z}$ by $\mathbb{R}$. We lose some date by doing this transition, but we can
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\begin{align*}
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\xi \in S^1 \setminus \{ \pm 1\}
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&\quad
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p_{\xi} =
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(t - \xi)(t - \xi^{-1}) t^{-1}
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\\
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\xi \in \mathbb{R} \setminus \{ \pm 1\}
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&\quad
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q_{\xi} = (t - \xi)(t - \xi^{-1}) t^{-1}
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\\
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\xi \notin \mathbb{R} \cup S^1
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&\quad
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q_{\xi} = (t - \xi)(t - \overbar{\xi})(t - \xi^{-1})
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(t - \overbar{\xi}^{-1}) t^{-2}
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\end{align*}
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Let $\Lambda = \mathbb{R}[t, t^{-1}]$. Then:
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\begin{align*}
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H_1(\widetilde{X}, \Lambda) \cong \bigoplus_{\substack{\xi \in S^1 \setminus \{\pm 1 \}\\ k\geq 0}}
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( \quot{\Lambda}{p_{\xi}^k })^{n_k, \xi}
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\oplus
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\bigoplus_{\substack{\xi \notin S^1 \\ l\geq 0}}
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(\quot{\Lambda}{q_{\xi}^l})^{n_l, \xi}&
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\end{align*}
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We can make this composition orthogonal with respect to the Blanchfield paring.
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\vspace{0.5cm}\\
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Historical remark:
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\begin{itemize}
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\item John Milnor, \textit{On isometries of inner product spaces}, 1969,
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\item Walter Neumann, \textit{Invariants of plane curve singularities}
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%in: Knots, braids and singulari- ties (Plans-sur-Bex, 1982), 223–232, Monogr. Enseign. Math., 31, Enseignement Math., Geneva
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, 1983,
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\item András Némethi, \textit{The real Seifert form and the spectral pairs of isolated hypersurfaceenumerate singularities}, 1995,
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%Compositio Mathematica, Volume 98 (1995) no. 1, p. 23-41
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\item Maciej Borodzik, Stefan Friedl
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\textit{The unknotting number and classical invariants II}, 2014.
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\end{itemize}
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\vspace{0.5cm}
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Let $p = p_{\xi}$, $k\geq 0$.
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\begin{align*}
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\quot{\Lambda}{p^k \Lambda} \times
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\quot{\Lambda}{p^k \Lambda} &\longrightarrow \quot{\mathbb{Q}(t)}{\Lambda}\\
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(1, 1) &\mapsto \kappa\\
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\text{Now: } (p^k \cdot 1, 1) &\mapsto 0\\
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p^k \kappa = 0 &\in \quot{\mathbb{Q}(t)}{\Lambda}\\
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\text{therfore } p^k \kappa &\in \Lambda\\
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\text{we have } (1, 1) &\mapsto \frac{h}{p^k}\\
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\end{align*}
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$h$ is not uniquely defined: $h \rightarrow h + g p^k$ doesn't affect paring. \\
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Let $h = p^k \kappa$.
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\begin{example}
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\begin{align*}
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\phi_0 ((1, 1))=\frac{+1}{p}\\
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\phi_1 ((1, 1)) = \frac{-1}{p}
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\end{align*}
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$\phi_0$ and $\phi_1$ are not isomorphic.
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\end{example}
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\begin{proof}
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Let $\Phi:
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\quot{\Lambda}{p^k \Lambda} \longrightarrow
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\quot{\Lambda}{p^k \Lambda}$
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be an isomorphism. \\
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Let: $\Phi(1) = g \in \lambda$
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\begin{align*}
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\quot{\Lambda}{p^k \Lambda}
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\xrightarrow{\enspace \Phi \enspace}&
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\quot{\Lambda}{p^k \Lambda}\\
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\phi_0((1, 1)) = \frac{1}{p^k} \qquad&\qquad
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\phi_1((g, g)) = \frac{1}{p^k} \quad \text{($\Phi$ is an isometry).}
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\end{align*}
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Suppose for the paring $\phi_1((g, g))=\frac{1}{p^k}$ we have $\phi_1((1, 1)) = \frac{-1}{p^k}$. Then:
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\begin{align*}
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\frac{-g\overbar{g}}{p^k} = \frac{1}{p^k} &\in \quot{\mathbb{Q}(t)}{\Lambda}\\
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\frac{-g\overbar{g}}{p^k} - \frac{1}{p^k} &\in \Lambda \\
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-g\overbar{g} &\equiv 1\pmod{p} \text{ in } \Lambda\\
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-g\overbar{g} - 1 &= p^k \omega \text{ for some } \omega \in \Lambda\\
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\text{evalueting at $\xi$: }\\
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\overbrace{-g(\xi)g(\xi^{-1})}^{>0} - 1 = 0 \quad \contradiction
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\end{align*}
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\end{proof}
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????????????????????\\
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\begin{align*}
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g &= \sum{g_i t^i}\\
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\overbar{g} &= \sum{g_i t^{-i}}\\
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\overbar{g}(\xi) &= \sum g_i \xi^i \quad \xi \in S^1\\
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\overbar{g}(\xi) &=\overbar{g(\xi)}
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\end{align*}
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Suppose $g = (t - \xi)^{\alpha} g^{\prime}$. Then $(t - \xi)^{k - \alpha}$ goes to $0$ in $\quot{\Lambda}{p^k \Lambda}$.
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\begin{theorem}
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Every sesquilinear non-degenerate pairing
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\begin{align*}
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\quot{\Lambda}{p^k} \times \quot{\Lambda}{p}
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\longrightarrow \frac{h}{p^k}
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\end{align*}
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is isomorphic either to the pairing wit $h=1$ or to the paring with $h=-1$ depending on sign of $h(\xi)$ (which is a real number).
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\end{theorem}
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\begin{proof}
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There are two steps of the proof:
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\begin{enumerate}
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\item
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Reduce to the case when $h$ has a constant sign on $S^1$.
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\item
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Prove in the case, when $h$ has a constant sign on $S^1$.
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\end{enumerate}
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\begin{lemma}
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If $P$ is a symmetric polynomial such that $P(\eta)\geq 0$ for all $\eta \in S^1$, then $P$ can be written as a product $P = g \overbar{g}$ for some polynomial $g$.
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\end{lemma}
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\begin{proof}[Sketch of proof]:
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Induction over $\deg P$.\\
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Let $\zeta \notin S^1$ be a root of $P$, $P \in \mathbb{R}[t, t^{-1}]$. Assume $\zeta \notin \mathbb{R}$. We know that polynomial $P$ is divisible by
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$(t - \zeta)$, $(t - \overbar{\zeta})$, $(t^{-1} - \zeta)$ and $(t^{-1} - \overbar{\zeta})$.
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Therefore:
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\begin{align*}
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&P^{\prime} = \frac{P}{(t - \zeta)(t - \overbar{\zeta})(t^{-1} - \zeta)(t^{-1} - \overbar{\zeta})}\\
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&P^{\prime} = g^{\prime}\overbar{g}
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\end{align*}
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We set $g = g^{\prime}(t - \zeta)(t - \overbar{\zeta})$ and
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$P = g \overbar{g}$. Suppose $\zeta \in S^1$. Then $(t - \zeta)^2 \vert P$ (at least - otherwise it would change sign). Therefore:
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\begin{align*}
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&P^{\prime} = \frac{P}{(t - \zeta)^2(t^{-1} - \zeta)^2}\\
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&g = (t - \zeta)(t^{-1} - \zeta) g^{\prime} \quad \text{etc.}
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\end{align*}
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The map $(1, 1) \mapsto \frac{h}{p^k} = \frac{g\overbar{g}h}{p^k}$ is isometric whenever $g$ is coprime with $P$.
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\end{proof}
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\begin{lemma}\label{L:coprime polynomials}
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Suppose $A$ and $B$ are two symmetric polynomials that are coprime and that $\forall z \in S^1$ either $A(z) > 0$ or $B(z) > 0$. Then there exist
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symmetric polynomials $P$, $Q$ such that
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$P(z), Q(z) > 0$ for $z \in S^1$ and $PA + QB \equiv 1$.
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\end{lemma}
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\begin{proof}[Idea of proof]
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For any $z$ find an interval $(a_z, b_z)$ such that if $P(z) \in (a_z, b_z)$ and $P(z)A(z) + Q(z)B(z) = 1$, then $Q(z) > 0$, $x(z) = \frac{az + bz}{i}$ is a continues function on $S^1$ approximating $z$ by a polynomial .
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\\??????????????????????????\\
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\begin{flalign*}
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(1, 1) \mapsto \frac{h}{p^k} \mapsto \frac{g\overbar{g}h}{p^k}&\\
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g\overbar{g} h + p^k\omega = 1&
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\end{flalign*}
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Apply Lemma \ref{L:coprime polynomials} for $A=h$, $B=p^{2k}$. Then, if the assumptions are satisfied,
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\begin{align*}
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Ph + Qp^{2k} = 1\\
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p>0 \Rightarrow p = g \overbar{g}\\
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p = (t - \xi)(t - \overbar{\xi})t^{-1}\\
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\text{so } p \geq 0 \text{ on } S^1\\
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p(t) = 0 \Leftrightarrow
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t = \xi or t = \overbar{\xi}\\
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h(\xi) > 0\\
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h(\overbar{\xi})>0\\
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g\overbar{g}h + Qp^{2k} = 1\\
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g\overbar{g}h \equiv 1 \mod{p^{2k}}\\
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g\overbar{g} \equiv 1 \mod{p^k}
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\end{align*}
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???????????????????????????????\\
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If $P$ has no roots on $S^1$ then $B(z) > 0$ for all $z$, so the assumptions of Lemma \ref{L:coprime polynomials} are satisfied no matter what $A$ is.
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\end{proof}
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?????????????????\\
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\begin{align*}
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\quot{\Lambda}{p_{\xi}^k} \times
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\quot{\Lambda}{p_{\xi}^k} &\longrightarrow
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\frac{\epsilon}{p_{\xi}^k}, \quad \xi \in S^1 \setminus\{\pm 1\}\\
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\quot{\Lambda}{q_{\xi}^k} \times
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\quot{\Lambda}{q_{\xi}^k} &\longrightarrow
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\frac{1}{q_{\xi}^k}, \quad \xi \notin S^1\\
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\end{align*}
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??????????????????? 1 ?? epsilon?\\
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\begin{theorem}[Matumoto, Borodzik-Conway-Politarczyk]
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Let $K$ be a knot,
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\begin{align*}
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H_1(\widetilde{X}, \Lambda) \times
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H_1(\widetilde{X}, \Lambda)
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= \bigoplus_{\substack{k, \xi, \epsilon\\ \xi \in S^1}}
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(\quot{\Lambda}{p_{\xi}^k}, \epsilon)^{n_k, \xi, \epsilon} \oplus \bigoplus_{k, \eta}
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(\quot{\Lambda}{p_{\xi}^k})^{m_k} \text{ and} \\
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\delta_{\sigma}(\xi) = \lim_{\varepsilon \rightarrow 0^{+}}
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\sigma(e^{2\pi i \varepsilon} \xi)
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- \sigma(e^{-2\pi i \varepsilon} \xi),\\
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\text{then }
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\sigma_j(\xi) = \sigma(\xi) - \frac{1}{2} \lim_{\varepsilon \rightarrow 0}
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\sigma(e^{2\pi i \varepsilon}\xi)
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+ \sigma(e^{-2 \pi i \varepsilon}\xi)
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\end{align*}
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The jump at $\xi$ is equal to
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$2\sum\limits_{k_i \odd} \epsilon_i$.\\
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The peak of the signature function is equal to ${\sum\limits_{k_i \even}}{\epsilon_i}$.
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\\
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?????????????????
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\\
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$(\eta_{k, \xi_l^{+}} -\eta_{k, \xi_l^{-}}$
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% Livingston Pacific Jurnal of M. 2012
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\end{theorem}
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\end{proof}
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