I am working through some of Rolfsen's "Knots and Links" and I have needed to go back and take a more careful look at the first few sections where he carefully discusses curves on solid tori. Let $V$ be a solid tori. A simple (i.e. injective) closed curve that is essential (i.e. not homotopic to a point) in $\partial V$ is called a meridian if it bounds a disc in $V$.
1) Why is such a curve $J$ bounding a disc in $V$ imply that $J$ is the image of $\{ 1\} \times S^1$ under a homeomorphism $S^1 \times D^2 \to V$?
2) Why are all meridians in $V$ ambiently isotopic (thus permitting to talk about "the" meridian)?
My next questions regard the equivalence of various definitions of a longitude in a solid torus. A simple closed curve $K$ is called a longitude if it is the image of $S^1 \times \{1\}$ under some homeomorphism $S^1 \times D^2 \to V$. If $K$ is a longitude of $V$ then we see that $K$ represents a generator of $H_1(V) = \pi_1(V)$ and $K$ represents some meridian of $V$ transversely at a single point. However, I do not understand the converses to these statements:
3) If a simple closed curve in $\partial V$ represents a generator of $H_1(V)$ why is a longitude of $V$?
4) If $K$ intersects some meridian of $V$ (transversely) in a single point, then why does it follow that $K$ is a longitude?
Finally, I understand that a automorphism of $\partial V$ that extends to an automorphism of $V$ must map a meridian to a meridian but:
5) Why is it that if an automorphism of $\partial V$ that maps a meridian on $V$ to a meridian of $V$ extends to an automorphism of $V$? Is there an analogous result giving a criterion for when automorphisms of boundaries of arbitrary genus handlebodies extend to automorphisms over the whole handlebody?
Thanks – sorry if this is too many questions packed into one post but I figured that they are all interrelated.
Best Answer
There are a couple of theorems from algebraic/geometric topology that you need to address these questions. The first theorem is that the set of isotopy classes of nontrivial simple closed curves on $\partial V$ corresponds bijectively with the set of ordered pairs $(m,n)$ of relatively prime integers, where the isotopy class of $c$ corresponds to $(m,n)$ if and only if $c$ is homologous to the cycle $m \cdot \mu + n \cdot \lambda$, if and only if (letting $p$ be a common base point of $\mu$ and $\lambda$) the curve $c$ is freely homotopic to the concatenated closed curve $\mu^m \lambda^n$.
Since $\mu$ is trivial in $\pi_1(V)$ whereas $\lambda$ generates $\pi_1(V)$, it follows that $c$ is homotopically trivial in $V$ if and only if $(m,n) = (\pm 1,0)$. Thus all homotopically trivial curves in $V$ are in a single isotopy class modulo change of orientation, which answers 2). Question 1) follows because an ambient isotopy in $\partial V$ between two meridian curves in $\partial V$ easily extends to an ambient isotopy of $V$ (why? that's your question 5) which I'll address below).
To address 3), a curve $c$ corresponding to $(m,n)$ represents a generator of $H_1(V)$ if and only if $n = \pm 1$. So now all one has to do is to explicitly construct a longitude representing $(m,\pm 1)$. To do this, take a longitude representing $(0,\pm 1)$, and take its image under the $m^{\text{th}}$ power of a Dehn twist on $V$ around a meridian disc.
To address 4), we need an additional theorem: given $c,d$ corresponding to pairs $(m,n)$ and $(p,q)$, the algebraic intersection number of $c,d$ is equal to the determinant $mq-np$. If $c,d$ intersect transversely in a single point then $mq-np=\pm 1$. If in addition $c$ is a meridian and so $(m,n)=(\pm 1,0)$, then $p=\pm 1$, and so $d$ is a longitude.
Finally, to turn to question 5), suppose $c,c'$ are isotopic simple closed curves in $\partial V$. First let me show that there exists an ambient isotopy of $V$ that takes $c$ to $c'$. Let $h : \partial V \times [0,1] \to \partial V$ be the ambient isotopy between $c,c'$, where $h_0(x)=h(x,0)=x$ and $h_1(x)=h(x,t)$ takes $c$ to $c'$. Pick a collar neighborhood $N(\partial V) \subset V$ and a homeomorphism $f : N(\partial V) \approx \partial V \times [0,1]$. By exploiting the fact that the domain of $h$ equals the range of $f$, you can combine $f$ and $h$ into an isotopy of $V$ between $c$ and $c'$ which is stationary in $h(V)-N(\partial V)$. You can think of this as saying that any isotopy of $\partial V$ can be "absorbed" into $V$. I'll leave further details to you.
So now you ask, why can any automorphism of $\partial V$ taking $c$ to $c'$ be extended to an automorphism of $V$ taking $c$ to $c'$? From what I've said above, your question reduces to the special case that $c=c'$. Now we need a third theorem: the automorphism group of $\partial V$, or more precisely the mapping class group which is the group of self-homeomorphisms modulo isotopy, is isomorphic to $GL_2(\mathbb{Z})$, where an automorphism of $\partial V$ corresponds to a matrix $\begin{pmatrix}m & n \\ p & q\end{pmatrix}$ if and only if the meridian maps to a $(m,n)$ curve and the longitude maps to a $(p,q)$ curve. Thus an automorphism fixes the meridian if and only if it corresponds to a matrix of the form $\begin{pmatrix}1 & 0 \\ p & \pm 1\end{pmatrix}$, so all you have to do is to construct specific automorphisms of $V$ whose restriction to $\partial V$ corresponds to each of those matrices. When the lower right entry is $+1$ then these are simply the Dehn twist powers around a meridian disc. Otherwise, when it is $-1$, then compose the Dehn twist powers with a reflection.