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Maria Marchwicka 2019-06-07 15:13:19 +02:00
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@ -582,6 +582,7 @@ A knot $K$ is smoothly slice if and only if $K$ bounds a smoothly embedded disk
Two knots $K$ and $K^{\prime}$ are called (smoothly) concordant if there exists an annulus $A$ that is smoothly embedded in $S^3 \times [0, 1]$ such that
$\partial A = K^{\prime} \times \{1\} \; \sqcup \; K \times \{0\} $.
\end{definition}
\noindent
Let $m(K)$ denote a mirror image of a knot $K$.
\begin{fact}
For any $K$, $K \# m(K)$ is slice.
@ -700,9 +701,7 @@ In general
\section{\hfill\DTMdate{2019-05-20}}
Let $M$ be compact, oriented, connected four-dimensional manifold.\\
??????????????????????????????????\\
If $H_1(M, \mathbb{Z}) = 0$ then there exists a
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}
@ -723,13 +722,13 @@ H_2(M, \mathbb{Z})&
\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: $H_2(M, \mathbb{Z})
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}$ can be degenerate in the sense that
\begin{align*}
\mathbb{Z}\\
H_2(M, \mathbb{Z}) \longrightarrow \Hom (H_2(M, \mathbb{Z}), \mathbb{Z})\\
(a, b) \mapsto \mathbb{Z}\\
a \mapsto (a, \_) H_2(M, \mathbb{Z})
@ -863,30 +862,24 @@ Reduce to the case when $h$ has a constant sign on $S^1$.
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 $g$.
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
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*}
(t - \zeta) \mid p,\\
(t - \overbar{\zeta}) \mid p,\\
(t^{-1} - \zeta) \mid p,\\
(t^{-1} - \overbar{\zeta}) \mid p,\\
&P^{\prime} = \frac{P}{(t - \zeta)(t - \overbar{\zeta})(t^{-1} - \zeta)(t^{-1} - \overbar{\zeta})}\\
&P^{\prime} = g^{\prime}\overbar{g}
\end{align*}
therefore:
We set $g = g^{\prime}(t - \zeta)(t - \overbar{\zeta})$ and
$P = g \overbar{g}$. Suppose $\zeta \in S^1$. Then $(t - \zeta)^2 \mid P$ (at least - otherwise it would change sign). 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}\\
\text{we set } g = g^{\prime}(t - \zeta)(t - \overbar{\zeta}\\
p = g \overbar{g}
\end{align*}
Suppose $\zeta \in S^1$. Then $(t - \zeta)^2 \mid p$ (at least - otherwise it would change sign).
\begin{align*}
p^{\prime} &= \frac{p}{(t - \zeta)^2(t^{-1} - \zeta)^2}\\
g &= (t - \zeta)(t^{-1} - \zeta) g^{\prime} \quad \text{etc.}\\
(1, 1) \mapsto \frac{h}{p^k} = \frac{g\overbar{g}h}{p^k} \quad \text{ isometry whenever $g$ is coprime with $p$.}
&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