some new text

lectures_on_knot_theory.tex

@@ -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