§ Hyperbolic groups have solvable word problem

I have never seen an elementary account of this in a 'trust these facts, now here is why hyperbolic groups have a solvable word problem'. I am writing such an account for myself. It's an account for building intuition, so no proofs will be provided except for the final theorem. All facts will be backed by intuition. Since most of it is geometric, it's easy to convey intuition.

§ Graphs of groups, quasi-isometry.

• NOTE: I will consistently denote the inverse of $g$ by $g^{-1}$.

We can convert any group into a graph by using the cayley graph of the group. We characterize hyperbolic space as a space where we can build 'thin triangles'. We also think of hyperbolic space as where geodesics from a given point diverge (in terms of angle) exponentially fast.
The choice of generators for the cayley graph gives different graphs. We can assign a unique geometric object by considering cayley graphs upto quasi isometry. The cayley graph of a group with respect to different generating sets are quasi-isometric. We can now try to study properties that are invariant under quasi-isometry, since these are somehow 'represented faithfully by the geometry'.

§ Hyperbolicity enters the picture

We now say that a graph is hyperbolic if the cayley graph of the group is hyperbolic. We can show that hyperbolicity is preserved by quasi-isometry. So this property does not depend on the generating set.

§ Isoperimetric Inequality

If we have a finitely presented group $G = \langle S | R \rangle$, and $w$ is a word in the free group $Free(S)$, if $[[w]] = 1$, we will have

This follows almost by definition. Since we have quotiented by $r$ we can have elements of $r$ in between $u u'$. We will need to have a $u'$ since there's nothing else to cancel off the $u$.

§ Area of a word

Let $G = \langle S | R \rangle$. Let $w$ be a word in $S$ such that $[[w]] = e$. The area of the word $w$ is the minimal number of such $u r u'$ components we need to write it down. Formally:

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

• Notes On Hyperbolic and Automatic Groups: Michael Batty
• The geometry of the word problem by Martin R Bridson
• My math.stackexchange question asking about why delta thin triangles help to solve conjugacy