# Tuesday, March 22

## Farey Graphs

:::{.remark}
Build a graph on the hyperbolic plane in the Poincare disc model:

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![](figures/2022-03-22_11-20-24.png)

Here every midpoint corresponds to adding numerators and denominators respectively.

Associate slopes:

- $0/1 \leadsto 1\alpha + 0 \beta$
- $1/0 \leadsto 0\alpha + 1 \beta$
- $1/1 \leadsto 1\alpha + 1 \beta$

Any pair of these is a $\ZZ\dash$basis for $H^1(T^2; \ZZ)\cong \ZZ\cartpower{2}$.
Use $\SL_2(\ZZ) \embeds \PSL_2(\CC)$ to realize any change of basis as an isometry of $\mfh$.
This makes the interior/exterior of any tile isometric to the full upper/lower half-disc.
:::

:::{.remark}
Basic moves: bypasses


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![](figures/2022-03-22_11-29-28.png)

The first case corresponds to slopes $r\in (-\infty , -1)$ and the second to $r\in (-1, -1/2)$.
Idea: the resulting dividing set is locally constant in perturbations of $r$, provided one doesn't cross the endpoints of the curve for the bypass move.
This produces a continued fraction defined inductively by $r_0 = \floor{-{p\over q}}$, writing $-{p\over q} = r_0 -{1\over {p'/q'}} = - {q'\over p'}$ with $-p/q < -p'/q' < -1$ and thus $0 < -{p\over q} - r_0 < 1$, so set $r_1 = \floor{-{p'\over q'}}$.
This yields 
\[
-{p\over q} = r_0 - {1\over r_1 - {1\over r_2 - \cdots}} = [r_0, r_1,\cdots, r_m]
,\]
which terminates in finitely many steps since $p/q$ is rational.
Note that $r_i \leq -1 \implies \floor{r_i}\leq -2$.
:::

:::{.proposition title="?"}
If $r=-p/q = [r_0, \cdots, r_k]$ in a continued fraction expansion and $s=a/b$ is the first point connected to $p/q$ while moving counterclockwise from $0/1$ on the Farey graph, then $-a/b = [r_0, \cdots, r_{k}+1]$.
:::

:::{.remark}
This gives the minimal graph path from $p/q$ back to $0/1$ by jumping the maximal distance along the circle to $a/b$.
Noting that $[r_1, \cdots, r_{k-1}, -1] = [r_1, \cdots, r_{k-1} + 1]$ which is a shorter continued fraction.
:::

:::{.example title="?"}
Let $p/q = 53/17$, then

- $r_0 = -4 = -68/17$
- $r_1 = -2 = -30/15$
- $r_2 = -2 = -26/13$
- $r_3 = -2 = \cdots$


So this yields $[-4, -2, \cdots_7, -2, -3]$.
:::

:::{.remark}
Idea: decompose $p/q = [r_0, \cdots, r_k]$ surgery into integer surgeries on a link with $k$ components.
:::