Past exam of the mathematics course of the University of Cambridge 2003 iii Paper 5 3 iii Solution Created 2026-10-03 Updated 2026-10-14
The required quasi-isometry is the identity on the group vertices. For each choose a fixed -word representing it and map an -edge to that word path. The bounded substitutions in Q2(ii) prove the distortion inequalities.
For the area comparison, choose a reverse substitution sending each to an -word. Let bound the lengths of . Every word for a defining relator is null over the first group presentation. Since there are finitely many such words, their areas have a common bound . Similarly every generator correction has area bounded by a constant ; include inverses in this bound.
Let be null over with length at most . Its substituted word has length at most and a diagram of area at most . Replace each -edge by its -path. Each original relator face can then be filled over using at most faces, so has area at most . Attach the correction strips changing its letters back to those of , at cost at most . ThereforeThe reverse substitutions give the corresponding inequality for . Enlarging the constants to the single constant in the paper's comparison proves . This is Dehn functions under a change of finite presentation.
Two qualifications are necessary for the printed continuation. The invariant is the optimal isoperimetric function, namely the Dehn function, not arbitrary upper bounds. For example, the relator-free group presentation of has zero area on all null words, so both and are isoperimetric upper bounds for that same group presentation; they are not equivalent under .
Also, the comparison supplies no recursive bound unless one was available beforehand. Its actual consequence, by part (ii), isThe unconditional assertion that every finitely presented group has a solvable word problem is false. Finitely presented counterexamples are established in the primary paper The Word Problem. Thus the missing solvability premise cannot be inferred from group presentation equivalence alone.