we need to write it down. Formally:
$Area(w) = \min { n | \prod_{i=1}^n u_i r+i^{\pm 1} u_i'}$

I don't understand the geometric content of this definition. I asked
on mathoverflow .
#### § Isopermetric function for a group

$f: \mathbb N \rightarrow \mathbb N$ is a Dehn function or isoperimetric function
if the area of the word is upper bounded by $f(|word|)$. In some sense, the
length of the word is the perimeter to the area, and this gives us a form
of the isoperimetric inequality. Formally, $f$ is a Dehn function if for all
words $w \in F(S)$ such that $[[w]] = e$, we have $A(w) \leq f(|w|)$. depending
on the growth of $f$, we say that $G$ has linear, quadratic, exponential etc.
Dehn function.
#### § Geometric content of Area

We define a map to be aninite, planar, oriented, connected and simply connected
simplicial2-complex (!). A map $D$ is a diagram over an alphabet $S$ iff every
edge $e \in D$ has a label $lbl(e) \in S$ such that $lbl(e^{-1}) = (lbl(e))^{-1}$.
Hang on: what does it mean to invert an edge? I presume it
means to go backwards along an edge. So we assume the graph is directed, and we
have edges in both directions.
A Van Kampen diagram over a group $G = \langle S | R \rangle$ is a diagram $D$
over $S$ such that for all faces of $D$, the label of the boundary of $f$
is labelled by some $r^{\pm} : r \in R$. The area of such a diagram
is the number of faces.
#### § Hyperbolic iff Linear Isopermetric Inequality is satisfied

A finitely presented group is hyperbolic if and of if its cayley grah
satisfies the linear isoperimetric inequality.
#### § Deciding if elements are conjugate to each other

- If we can answer the question of whether two elements are conjugate to each other (does there exist a $g$ such that $ghg' =?= k$), we can solve that an element is equal to the identity:
- Pick $k = e$. Then if we have $ghg' = k = e$, then $gh = g$ hence $h = e$.

- If we can check that an element is equal to the identity, we can check for equality of elements. two elements $k, l$ are equal iff $kl' = e$.

- So solving conjugacy automatically allows us to check of equality.

#### § Proof that conjugacy is solvable for hyperbolic groups

Consider a solution to the problem of finding an $x$ such that $xgx^{-1} = h$.
We claim that due to the hyperbolicity of the space, such an $x$ cannot be
"too long".
#### § Key intuition for hyperbolicity allows us to control word length

- Suppose we are interested to find a $g$ such that $ghg^{-1} = e$
- We can think of this as a case where $h$ is the base edge of a triangle, $e$ is the opposite vertex, and $g, g^{-1}$ are the other two sides:

```
e
/ \
g g^{-1}
/ \
+---h---+
```

- If the space is euclidea, then $g$ and $g^{-1}$ can be as long as they want to be while $h$ stays the same. The angles $gh, hg^{-1}$ will become larger, and the angle $gg^{-1}$ will become smaller as the length $g$increases.
- In a hyperbolic space, because the angle sum is less than 180, if we move $e$ too far away, the $h$ will "bend" to maintain the angle sum to be less than 180. But this means that we have distorted the $h$! There is a bound to how long we can make $g$ before we start distorting $h$. This bound on $g$ is exponential in the length of $h$.
- Alternatively, in a CAT-0 space, the base of the triangle cannot be too far away from the centroid. So we can't have $e$ be moved too far away, because if $g$gets too large.

#### § References