Markoff numbers are integers that appear a Markoff triple
which are solutions of a Diophantine equation
the so-called Markoff cubic
The largest integer in a triple determines the two other numbers.
If
There is a natural map (we'll see why shortly)
Definition: Let
Natural map:
Theorem The shortest representative for a non trivial homology class is always a multiple of a closed simple geodesic.
comes from Bass-Serre tree of
arc = Poincaré geodesic joining
comes from Bass-Serre tree of
H. Cohn Approach to Markoff’s Minimal Forms Through Modular Functions (1955)
modular torus =
representation
Definition character map
Theorem: (Cohn and many others) The semi-algebraic set:
identified with the Teichmueller space of the punctured torus.
group of the automorphisms is induced by the action of the mapping class group
the permutation
the (Vieta) involution
Proof:
(trace)
(Ptolemy)
The multiplicity of any number in the complementary regions to the tree is at most 6.
If
then there is a unique triple
Let
If one of the Ford circles is
the midpoint of the arc is at Euclidean height
One of the Ford circles is
the midpoint of the arc is at Euclidean height
blue arc
Normalisation, fundamental triple
Every Markoff number
Lemma C Let
The "number of ways" of writing
is equal to the number of arcs satisfying:
joins
integers
arc meets the orbit
The "number of ways" of writing
Lemma C' Let
The "number of ways" of writing
is equal to the number of arcs
Every Markoff number
(Ptolemy)
In fact the Ptolemy relation factorises over
yields a recursion for a set of Gaussian integers
such that each Markoff number
Fibonacci numbers
x, y = 1 + 0J, 1 + 1J
fib = []
for k in range(10):
z = y*y.conjugate() + 1J
u = x*z/ (x*x.conjugate())
fib.append(f'{u}')
x, y = y, u
' '.join(fib)
Back to Button exactly 6 simple arcs of
every Markoff number
If
then there is a unique triple
Theorem F1: Let
has a solution over
if
Theorem F2: Let
has a solution over
iff
Theorem F3: Let
has a solution over
if
An immersed ideal triangles with sides of
Acting on
Acting on
"Geometric" proof: Group acting on
Count fixed points
identity
Acting on
or on the arcs of
of
Let
The "number of ways" of writing
with
subgroup of automorphisms
fixing the cusp labeled
First the automorphism
The automorphism
fixes two and exactly two arcs in
Can then apply Burnside Lemma to prove Theorem F2
Can other elementary results for quadratic forms?
using immersed "equilateral" ideal triangles.
(Elsholtz)
using arcs
Baragar ? m, 3m - 2, 3m + 2 prime
More detailed analysis of the spectrum of
Orthotree, orthoshapes and ortho-integral surfaces
Nhat Minh Doan
#
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- slides : google **greg mcshane github**
- click on **serfest**
- if there is a bug in my slides blame [this guy](https://github.com/yhatt)
#
## infinity of Markoff triples: $z=1$
$\begin{pmatrix} 3 & -1 \\ 1 & 0 \end{pmatrix}$
is an automorph of
$$x^2 + y^2 - 3x y.$$
So $( v_n,v_{n+1},1)$ is a Markoff triple where
$\begin{pmatrix} x \\ y \end{pmatrix}= \begin{pmatrix}v_{n+1} \\ v_n \end{pmatrix} = \begin{pmatrix} 3 & -1 \\ 1 & 0 \end{pmatrix}^n \begin{pmatrix}1 \\ 1 \end{pmatrix}$
#
### Odd index Pell numbers are Markoff numbers
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$0, 1, 2, 5, 12, 29, 70, 169, 408, 985, 2378, 5741, 13860,\ldots$
$(1,5,2), (5,29,2),(29,169,2)\ldots$
https://oeis.org/A000057
and satisfy divisibility relations
$F_{2n} = (F_{n+1} + F_{n-1})F_n \Rightarrow F_n | F_{2n}$
$\begin{pmatrix}
F_{n+1} & F_{n} \\
F_{n} & F_{n-1}
\end{pmatrix}
= \begin{pmatrix}
1 & 1\\
1 & 0
\end{pmatrix}^n \Rightarrow
\begin{pmatrix}
F_{2n+1} & F_{2n} \\
F_{2n} & F_{2n-1}
\end{pmatrix}=
\begin{pmatrix}
F_{n+1} & F_{n} \\
F_{n} & F_{n-1}
\end{pmatrix}^2$
* [ Bugeaud, Reutenauer, Siksek](https://core.ac.uk/download/pdf/82088222.pdf)
* Conclusion too hard!!!
**Theorem (Fermat)**
* Button's theorem follows from "unicity" of $c,d$
* unique factorisation $p = (ci+d)(-ci+d)\in \mathbb{Z}[i]$
* Prove his conjectures with one figure?
* snake graph and its perfect matchings
* "lengths" that verify a Ptolemy inequality
$\mathbb{Q}\cup \infty \subset$ circle/projective line
$\mathbb{Q}\cup \infty \subset$ circle/projective line
#
![width:600px](./pozzi.jpg.png)
[source](https://www.mathi.uni-heidelberg.de/~pozzetti/trees/4.pdf)
#
## Visualizing using natural map
$\mathbb{Q}\cup \infty \rightarrow$ Markoff numbers
$p/q \mapsto m_{p/q} = \frac23 \cosh\left(\frac12\ell_{\gamma_{p/q}} \right)= \frac23 \cosh(\| (q,p) \|_s)$
* $SL(2, \mathbb{Z})$ action on $\mathbb{Q}\cup \infty$
* mapping class group action on simple curves on $\mathbb{H}/\Gamma'$
{: style="text-align: left"}
#
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![w:650](./farey_tree.png)
#
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![w:550](./ford_circles.gif)
**Ford circles**
<!-1- # Definitions -1->
**arc** = Poincaré geodesic joining $a/c, b/d \in \mathbb{Q}\cup \infty$
* **$\lambda$-length of arc** $= |ad - bc|$
* $\lambda$-length of arc on $\mathbb{H}/\Gamma'$ is the length of a lift to $\mathbb{H}$
* $\mathrm{SL}(2,\mathbb{Z})$ acts by Mobius transformations on $\mathbb{H}$
* orbit of $F := \{ z, \mathrm{Im}\, z > 1\}$ are the Ford circles
<!-1- - $\begin{pmatrix} a & b \\ c & d \end{pmatrix}.z = \frac{az+b}{cz+d}$ -1->
#
<!-1- _transition: slide -1->
![w:500](./ford_circles_again.png)
* $p/q$ point of tangency with $\mathbb{R}$, diameter = $1/q^2$
* hyperbolic midpoint of the arc joining $F$ to this Ford circle is at euclidean height $1/q$
<!-1- - ie diameter is the square of the inverse of the denominator of $p/q$ -1->
#
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![bg left 100%](./lambda.svg)
- $p/q$ point of tangency with $\mathbb{R}$, diameter = $1/q^2$
- hyperbolic midpoint of the arc joining $F$ to this Ford circle is at euclidean height $1/q$
<!-1- - ie diameter is the square of the inverse of the denominator of $p/q$ -1->
$\mathrm{SL}(2,\mathbb{Z})$ transitive on $\mathbb{P}(\mathbb{Q}^2)$,
* can suppose $a/c= \infty=1/0$ and $b/d = k/(ad - bc)$
* joins Ford circle $F$ tangent at $\infty$ and another of diameter $1/(ad - bc)^2$
* hyperbolic length of portion outside these is $2\log(ad - bc)$
<!-1- - the **midpoint** of this vertical arc is at height $1/|ad - bc|$ -1->
#
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# Example
text_align: top
# In fact....
<p style = "text-align: left">
if m is a Markoff number and
</p>
- $m = p^k$
- or $m = 2p^k$
<p style = "text-align: left">
then m satisfies the uniqueness conjecture
</p>
#
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## Let's begin then...
#
<!-1- _transition: cube -1->
## QED
#
## Recall
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![bg left 100%](./sami.jpg)
- **arc** = Poincaré geodesic joining $a/b, c/d \in \mathbb{Q}\cup \infty$
- **$\lambda$- length of arc** $= |ad - bc|$
- Let $n$ be a positive integer.
- The number of ways of writing $n$ as a sum of squares $n = c^2 + d^2$ with $c,d$ coprime integers
- is equal to the number of integers $0 \leq k < n-1$ coprime to $n$ such that the line $\{ k/n + i t,\, t>0 \}$ contains a point in the $\Gamma$ orbit of $i$.
$\simeq \mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}$
- whose fixed point is $i+1$.