The proof that the embedding of B(m,n) into Hm,n and in general any recursively presented group G into is undistorted also uses bands and annuli, and the structure of diagrams over the infinite presentation of H described in the previous section.
Here we present the main points of the proof that the embeddings in Theorems 10, 14 and 17 are quasi-isometric.
For simplicity consider the case of the group H=Hm,n from Theorem 17. The general case of is similar. By definition of quasi-isometric embeddings, we have to find a constant c>0 such that for any element represented by a geodesic (in B(m,n)) word in the alphabet and for any word Z in the generators of H, that represents the same element, we have .
In order to achieve this goal, consider the minimal diagram over the infinite presentation of H considered in the previous section, with boundary label UZ-1. Then the boundary of has the form p-1 p' where , . We need to show that for some constant c. It suffices to make a correspondence between b-edges of p and edges of p', such that any edge of p'corresponds to at most c-1 edges of p.
First of all notice that we can assume that no B-cell in has a common edge with p. Indeed, if such a B-cell exists, we can cut it off reducing the type of the diagram and replacing the path p with a not shorter path p1 (recall that U was a geodesic word representing g).
Therefore for every edge e on p, there is a maximal b-band in , starting at e. It can end neither on p nor on the boundary of a Gb-cell (both cases are ruled out in the same manner as it was done in the previous section: we can do a type reducing surgery again).
If ends on the path p', we associate the terminal edge of with e. Another possibility is that terminates on the boundary of some maximal a-band (at the cell labeled by a relator ). A lemma similar to Lemma 2 shows that at most 2 maximal b-bands starting on p can end on the boundary of the same a-band. This means that we can consider the set of a-bands where these b-bands end.
The most pleasant (for us) among these a-bands are those which start or end on p' (they cannot end on p because p does not have A-edges). Other a-bands can terminate either on contours of r-bands or on disks.
We need to consider two cases. In the first case the number of those r-bands is large (proportional to the number of a-bands in ). Since there are no r-annuli in (see the previous section), each of these r-bands starts and ends on p', and we obtain a desired inequality .
In the second case we have to assume that the number of r-bands where a-bands terminate is ``small". Since by a variation of Lemma 2 the number of a-bands terminating on the same r-band is bounded by a constant, in this case most a-bands in terminate on discs.
In this situation we use the so called ovals and their shadows (see [31] and [7]).
An oval is a simple closed path h in the disk graph of the diagram
.
It divides the plane into two regions. One of them, denoted by
O(h), must possess the following property. For every disk
on h,
the number n1 of maximal k-bands going from
into O(h) and the
number n2 of the maximal k-bands going from
into the exterior of
O(h) satisfy the following inequalities:
The hyperbolicity of the disk graph and the high degrees of its interior vertices allow us to draw an oval h passing via any interior edge of the disc graph of .
One of the main properties of ovals is that if an r-band starts outside the subdiagram O(h) bounded by the oval and then intersects the oval, it cannot leave O(h). Indeed otherwise the hyperbolicity of the disk graph would imply the existence of either a (k,r)-annulus (which is impossible, see the previous section) or an r-band intersecting too many maximal k-bands starting on the same disk (again it is impossible because of the Moving r-bands construction from the previous section).
Thus any maximal r-band crossing an oval h, must intersect its shadow, i.e. the boundary subpath of the diagram lying in O(h). This allows us to prove that the shadow of any oval h is sufficiently long compared to the perimeter of any disk crossed by h. We can also choose h in such a way that the shadow of h is inside p' (because pdoes not contain k-edges). Therefore the number of a-bands ending on the contour of a disc does not exceed the length of the shadow of any oval passing through the disc. If the bands in end on different discs ,..., then the hyperbolicity of the disc graph allows us to find ovals passing through these discs which have disjoint shadows, all inside p'. Thus the length of p' cannot be smaller than the number of a-bands in , which in turn, as we know, cannot be much smaller than the length of p.
The proof of the result that the shadow of an oval h is sufficiently long compared to the perimeter of a disc in h consists of two cases. In the first case the number of maximal r-band in O(h) is sufficiently large (greater than, say, of the number of all a-edges between successive k-edges of ). This case is clear since all the r-bands must terminate on the shadow.
The second case is complementary to the first one. Since the number of the r-bands is small, the number of the a-bands going from into O(p)and terminating on r-bands, is small too (Lemma 2 works again). Therefore a majority of them terminates either in the shadow of (this is the best alternative for us) or on some disks inside of O(h).
This situation can be analyzed by induction: as before we can draw ovals passing through respectively, whose shadows are disjoint and are inside the shadow of h.