\classheader{2012-12-05}

\subsection*{Applications to Knots}

\begin{proposition}
The trefoil knot is knotted.
\end{proposition}
\begin{proof}
We know that $\pi_1(\S^3-\text{unknot})\cong\Z$, but it is not too hard to show that  $\Gamma=\pi_1(\S^3-\text{trefoil})=\langle a,b\mid aba=bab\rangle$. We claim that $\Gamma$ is not isomorphic to $\Z$. To see this, note that there is a surjection $f:\Gamma\to S_3$ (a non-abelian group), defined by $a\mapsto (12)$ and $b\mapsto (23)$, because these elements of $S_3$ satisfy the relation $aba=bab$.
\end{proof}

Let $T(p,q)$ be the $(p,q)$-torus knot, i.e. the knot obtained by embedding the torus $\T^2$ in $\R^3$ in the standard way and taking the curve on the torus that wraps around $p$ times one way and $q$ times the other way.

\begin{proposition}
$T(p,q)$ is equivalent to $T(m,n)$ $\iff$ $m=p$ and $n=q$, or $m=q$ and $n=p$.
\end{proposition}
\begin{proof}
It turns out that $\Gamma=\pi_1(\S^3-T(p,q))=\langle a,b\mid a^pb^q\rangle$. Thus, $a^p$ commutes with both $a$ and $b$, so that $a^p\in Z(\Gamma)$, hence $\langle a^p\rangle\triangleleft \Gamma$. It is easy to see that $\Gamma/\langle a^p\rangle\cong \Z/p\Z\free\Z/q\Z$, hence $(\Gamma/\langle a^p\rangle)^{\ab}\cong\Z/p\Z\times\Z/q\Z$, which allows you to distinguish a lot of things.
\end{proof}

\begin{proposition}
If $p:Y\to X$ is a cover with $p(y)=x$, then $p_*:\pi_1(Y,y)\to\pi_1(X,x)$ is an injection, and $p_*(\pi_1(Y,y))$ consists of those $[\gamma]$ such that any lift of $\gamma$ to $Y$ is a loop.
\end{proposition}
\begin{proof}
The map sends $\gamma$ to $\alpha=p\circ \gamma$, and if $\alpha$ is trivial in $\pi_1(X,x)$ there is some homotopy $\alpha_t$ from $\alpha$ to a constant map.  Lifting this homotopy to $Y$ gives a homotopy from $\gamma$ to a constant map.
\end{proof}

\begin{proposition}
Let $p:Y\to X$ be a covering map, with $X,Y$ connected. Then $\# p^{-1}(X)=[\pi^{-1}(X):p_*(\pi_1(Y))]$.
\end{proposition}
\begin{proof}
Define $\varphi:\pi_1(X)/p_*(\pi_1(Y))\to\{p^{-1}(x)\}$ by sending the coset of $[\gamma]$ to $\gamma(1)$. Show that this map is bijective.
\end{proof}

Note that $\pi_1(X)$ acts on the set $p^{-1}(x)$ via $[\gamma]\cdot y\mapsto \widetilde{\gamma}(1)$, where $\widetilde{\gamma}$ is the lift of $\gamma$ starting at $y$. For example, $\pi_1(\S^1)$ acts on $p^{-1}(1)=\Z\subset\R$ by translations.

The stabilizer $\mathrm{Stab}(y)$ is conjugate to $p_*(\pi_1(Y))$.


\begin{definition}
We say that two coverings $p_1:Y_1\to X$ and $p_2:Y_2\to X$ are isomorphic when there is a homeomorphism $f:Y_1\to Y_2$ such that $p_2=p_1\circ f$, i.e. $f$ takes the set $p_1^{-1}(x)$ to $p_2^{-1}(x)$ for any $x\in X$. For any covering $p:Y\to X$, the group of deck transformations of $p$ is just $\Aut(p:Y\to X)$.
\end{definition}
\begin{example}
The group of deck transformations of the covering $\R\to\S^1$ is just $\Z$, consisting of the translations of $\R$ by integers.
\end{example}

As we mentioned before, there is a correspondence
\[\{\text{subgroups }H\subseteq \pi_1(X,x)\}\longleftrightarrow\{\text{coverings }p:Y\to X\text{ up to isomorphism}\}.\]
In particular, the trivial subgroup $\{e\}$ corresponds to a simply-connected cover $\widetilde{X}$ of $X$, called the universal cover of $X$. You can look in Hatcher for details, but essentially $\widetilde{X}$ consists of certain equivalence classes of paths in $X$.

Looking at universal covers for graphs is good practice for the general case.


\begin{definition}
A covering map $p:Y\to X$ is called a regular cover (a.k.a. normal cover) when for all $x\in X$ and $y_1,y_2\in p^{-1}(x)$, there is a deck transformation $h\in\Aut(p:Y\to X)$ such that $h(y_1)=y_2$. 
\end{definition}

It turns out that $p:Y\to X$ is regular if and only if $p_*(\pi_1(Y))\triangleleft\pi_1(X)$, i.e. for any $h\in p_*(\pi_1(Y))$ and $g\in\pi_1(X)$, we have $ghg^{-1}\in p_*(\pi_1(Y))$. We can see this because $ghg^{-1}(g(y))=gh(y)=g(y)$, since $h\in \pi_1(Y)$, so that $ghg^{-1}\in\mathrm{Stab}(g(y))\cong \pi_1(Y,g(y))$. This is a special case of a very important general fact:
\[\mathrm{Fix}(ghg^{-1})=g\mathrm{Fix}(h).\]
The exam will cover only the things we've done in class, but you should be sure to learn the lifting criterion on your own.
\begin{center}
\begin{tikzcd}
& Y\ar{d}{p} \\ Z\ar[dashed]{ur}{?} \ar{r}[swap]{f} & X
\end{tikzcd} can lift $\iff$ $f_*(\pi_1(Z))\subseteq p_*(\pi_1(Y))$
\end{center}






