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