algebraic-topologydifferential-topologygeneral-topologygeometry
Let $p$ and $ q$ be a relatively prime integers.
I want to know how to prove that a Hopf link with framing $-p$ and $-q$ is a surgery link for a lens space $L(p,q)$.
The lens space is first a result of rational surgery with coefficient $ p/q$. I especially want a good explanation of transition from rational surgery to an integer surgery expression.
Thank you in advance.
Maybe not what you want but this may help:
Surgery on the Hopf link is the same as gluing two solid tori to the two sides of $S^1 \times S^1 \times [0,1]$ by the gluing maps determined by the framings. Let $T_1,T_2$ be solid tori where $T_1$ is glued to $S^1 \times S^1 \times \{0\}$ with framing $p$, and $T_2$ is glued to $S^1 \times S^1 \times \{1\}$ with framing $q$.
The diagonal circle $x \times x \times \{.5\} : x \in S^1$ is a meridian plus a longitude of $S^1 \times S^1 \times \{.5\}$. It is homotopic to $p$ longitudes of $T_1$ and $q$ longitudes of $T_2$ by the surgery framings.
So if we just contract the $S^1 \times S^1 \times [0,1]$ to $S^1 \times S^1 \times \{.5\}$, then we have $T_1$ glued to $T_2$ such that $p$ longitudes of $T_1$ is homotopic to $q$ longitudes of $T_2$, so it is the $p,q$ lens space.
In the first picture, I visualize $S^{2n-3}$ as $S^0$ (I really doubt the legitmacy of this, since $S^0$ is not connected and other $S^n$'s are connected), visualize $S^{2n-1}$ as $S^2$, $B_j^{2n-2}$ and $B_{j+1}^{2n-2}$ as arcs in $S^{2}$ connecting "north pole" and "south pole" along great ciecle, and region between them is $B_j^{2n-1}$ in the second figure.
In the third picture, I visualize $S^{2n-3}$ as $S^1$, $B_j^{2n-2}$ and $B_{j+1}^{2n-2}$ as upper and lower disk $D^{2}_+$ and $D^{2}_-$.
Arcs in red, green, yellow and purple are identified respectively.
As far as I understand right now, the unit circle $C$ in the $n$-th $\mathbb C$ factor is vertical to $S^{2n-3}$, so arcs along graet circles in $S^{2n-1}$ from points $e^{2\pi ij/m}$ $(j=1,\cdots, m)$ to points in $S^{2n-3}$ yield $(2n-2)$-dimensional balls $B_{j}^{2n-1}$, which form an equal division of $S^{2n-1}$.
$\rho^r$ is a generator of $\mathbb Z_m$ and takes each $B^{2n−1}_j$ and $B^{2n−2}_j$ to the next one, so we can obtain $L$ as quotient of only one $B^{2n−1}_j$ by identifying its two faces $B^{2n−2}_j$ and $B^{2n−1}_{j+1}$ via $ρ^r$.
Since $\rho$ acts on different coordinates respectively, we can factorize $\rho^{r}$ as $\rho_{2} \circ \rho_{1}$.
$\rho_{1}$ is $\rho^{r}$ restrcting on last $\mathbb C_{n}$ factor, producing a reflection from $B_{j}^{2n-2}$ to $B_{j+1}^{2n-1}$ fixing $S^{2n-3}$.
It's indeed a reflection since the unit circle $C$ in the $n$-th $\mathbb C$ factor is vertical to $S^{2n-3}$, and $B_{j}^{2n-2}$ is formed by arcs along great circles in $S^{2n-1}$.
$\rho_{2}$ is $\rho^{r}$ restrcting on $S^{2n-3}$, producing a rotation on $S^{2n-3}$.
Best Answer
I see from your other questions you are reading Saveliev's book on three manifolds, where this is - IMO - explained extremely well. Can you let us know where you are having trouble?
You can also think about things this way: I can write $S^3$ as a union of two solid tori, $T_1$ and $T_2$. If I let $K_1\subset T_1$ and $K_2\subset T_2$ be central fibers, and $B\in GL(2,\mathbb{Z})$ be the map from $\partial T_1$ to $\partial T_2$ which interchanges meridians and longitudes, then after the identification via $B$, $K_1$ and $K_2$ will be linked, as in your Hopf link.
Now I can do the surgery along this Hopf link by doing it in $T_1$ and $T_2$ separately. In particular, give $K_1$ a $-p$ framing, and $K_2$ a $-q$ framing. After the surgeries, I get two spaces $S_1$ and $S_2$. But both of these are still just solid tori; in fact, if $A_i\in GL(2,\mathbb{Z})$ is the gluing map used for the surgery along $K_i$, then the map $A_i^{-1}:\ \partial T_i\rightarrow \partial S_i$ extends to a homeomorphism from $T_i$ to $S_i$.
The gist of all this is that I can still think about the space I get from gluing $S_1$ to $S_2$ as just gluing two plain, ordinary solid tori. That is, I can get the same space by gluing $T_1$ to $T_2$ via the composition $A_1^{-1}BA_2$. You can check (by multiplying these matrices out) this gives the lens space $L(pq-1,q)$.
The general idea is pretty much the same; note that I never really needed that $K_2$ was an unknot. Suppose $K_2$ is some knot in $S^3$ with some framing. I can consider this knot as lying in $T_2$, and so the space I get via surgery is the same as doing the surgery in $T_2$, then gluing this to $T_1$, where no surgery has taken place. If the resulting space "post-op" is a lens space, this surgery (in $T_2$) can be represented (as before) by some gluing matrix $A_2\in GL(2,\mathbb{Z})$ (i.e., I can get the same effect via gluing two plain, ordinary solid tori via $A_2:\ T_1\rightarrow T_2$.)
If I want to add a new unknot to the mix, I think of it as being the central fiber $K_1\subset T_1$ (because it will become linked in $S^3$), and then just as before, surgery along $K_1$ replaces the gluing map $A_2$ by $A_1BA_2$.
This is pretty much the idea in Saveliev's book, but he doesn't spell out everything as much. His proof that you can use framings coming from the continued fraction expansion of $p/q$ is very easy (just matrix multiplication).
Let me know if you need me to clarify anything, or (even better) where in Saveliev's approach you have gotten stuck.