diff --git a/images/11n34.png b/images/11n34.png new file mode 100644 index 0000000..123e67d Binary files /dev/null and b/images/11n34.png differ diff --git a/images/11n34.svg b/images/11n34.svg new file mode 100644 index 0000000..51a3751 --- /dev/null +++ b/images/11n34.svg @@ -0,0 +1,135 @@ + + + + + + + + + + + + + + + + + + image/svg+xml + + + + + + + + + diff --git a/images/11n34.svg.2019_06_03_08_08_16.0.svg b/images/11n34.svg.2019_06_03_08_08_16.0.svg new file mode 100644 index 0000000..8567cb2 --- /dev/null +++ b/images/11n34.svg.2019_06_03_08_08_16.0.svg @@ -0,0 +1,125 @@ + + + + + + + + + + + + + + + + + image/svg+xml + + + + + + + + + diff --git a/images/11n34_v2.svg b/images/11n34_v2.svg new file mode 100644 index 0000000..6eb4990 --- /dev/null +++ b/images/11n34_v2.svg @@ -0,0 +1,134 @@ + + + + + + + + + + + + + + + + + + image/svg+xml + + + + + + + + + diff --git a/lectures_on_knot_theory.tex b/lectures_on_knot_theory.tex index 99b6955..773992e 100644 --- a/lectures_on_knot_theory.tex +++ b/lectures_on_knot_theory.tex @@ -550,8 +550,12 @@ If $K$ is a knot, then $n$ is necessarily even, and so $\Delta_K(t^{-1}) = t^{-n \end{align*} \end{lemma} \begin{proof} - +If $\Sigma$ is a genus $g$ - Seifert surface for $K$ then $H_1(\Sigma) = \mathbb{Z}^{2g}$, so $S$ is an $2g \times 2g$ matrix. Therefore $\det (tS - S^T)$ is a polynomial of degree at most $2g$. \end{proof} +\begin{example} +There are not trivial knots with Alexander polynomial equal $1$, for example: +$\Delta_{11n34} \equiv 1$. +\end{example} %removing one disk from surface doesn't change $H_1$ (only $H_2$) \section{} \begin{example}