Recall that isoperimetric functions of a finitely presented group measure areas of van Kampen diagrams over the presentation of this group. Figure 1 shows what a van Kampen diagram may look like.
It is a directed planar labeled graph where every edge is labeled by a
generator from X, and the contour of every 2-cell (face) is labeled by a
relator from R. By van Kampen lemma [25] a word
is equal
to 1 in the group G if and only if there exists a van Kampen diagram
over the
presentation of G with boundary label w. The number of cells in is equal to the number of factors in a representation of w as a
product of conjugates of relators from R:
If a van Kampen diagram has minimal number of cells, n, among all diagrams
with the same boundary label w then we say that w has area n. A
function f(m) is called an isoperimetric function of
the presentation
of
the group G if every word of length at most m which is equal to 1 in the
group has area at most f(m). On the set of functions
,
one can define a quasi-order saying that if
We can also define isodiametric functions introduced by Gersten [16]. These functions measure the diameter of a van Kampen diagram with given perimeter2. More precisely, with every word w which is equal to 1 in G, we associate its diameter, that is the smallest diameter of a van Kampen diagram with boundary label w. Then if d(n) is an isodiametric function of , d(n) must exceed the diameter of every word w= 1 (mod G) of length . The equivalence of isodiametric functions is defined as before. Isodiametric functions of different finite presentations of the same group are always equivalent [19].
Both isoperimetric and isodiametric functions reflect the decidability of the word problem in the group. In particular it is well known (see, for example, [26]) that the word problem is decidable if and only if the Dehn function (the smallest isodiametric function) is recursive. Nevertheless, the word problem in a group with huge Dehn function may be easy. For example the word problem in the Baumslag-Solitar group can be solved in quadratic time (since this group is representable by integer matrices) while the Dehn function is exponential [11]. One of our main goals is to show that still there exists a very close connection between the Dehn functions and the computational complexity of the word problem.
If and are finitely presented, one can also define the area function of G in H. This function is defined on the set W of all words in the alphabet X which are equal to 1 in G. It takes every word from W to the area of this word in H.
Other important concepts are the distortion function and the length function. Let be a finitely generated subgroup of a finitely generated group . Then the distortion function takes every natural number n to . In other words, in order to compute dG,H(n) we consider all (finitely many) elements of G whose lengths in H are at most n, for each of these elements we compute its length in the alphabet X, and then take the maximum of these lengths.
The corresponding length function of G inside H is the function which takes every element g of G to |g|Y, the length of g in H.
Two functions are called O-equivalent if for some constant c and every . Different choices of generating sets in G and H lead to O-equivalent length functions and equivalent distortion functions associated with this embedding.
If |g|X= O(|g|Y) for every , or, equivalently, if the distortion function is at most linear we say that G is quasi-isometrically embedded into H. Otherwise we say that G is distorted.
For example, every subgroup of the free group is (obviously) quasi-isometrically embedded, but the (cyclic) center of the 3-dimensional Heisenberg group has quadratic distortion. Indeed cn2=[an,bn] for every n, the length of cn2 in C is n2 and the length of this element in H3 is .
Just as Dehn functions and isodiametric functions reflect the decidability of the word problem, the distortion function reflects the decidability of the membership problem for subgroups: if G is a finitely generated subgroup of a finitely generated group H which has solvable word problem, then the membership in G for elements of H is decidable if and only if the distortion function of G in H is recursive [12].
In this paper, we survey recent results about isoperimetric, isodiametric, length and area functions of groups obtained by the authors in collaboration with J.-C. Birget, V. Guba and E. Rips.
There are several important connections between Dehn functions and length functions. We present two connections here. The first of them plays an important role in several papers by Bridson and others on isoperimetric functions. The second connection is new.
Theorem 2 is new although a remark in [19] hints to a possibility of some connection between distortion of subgroups in and Dehn functions. Here is a proof of this theorem. It uses the well known Mikhailova's trick (see [25]) and a result from Baumslag and Roseblade [5].
It is proved in [5] that every finitely generated subgroup E of is the equalizer (in [5] it is called the free corner pullback) of two homomorphisms and of two finitely generated subgroups E', E'' of F2onto a finitely presented group G, that is . Since every finitely generated subgroup of F2 is quasi-isometrically embedded in F2, is quasi-isometrically embedded into . So it is enough to show that the distortion function of the equalizer of two homomorphisms and of two free groups onto a finitely presented group G in is equivalent to the Dehn function of G.
Let d(k) be the Dehn function of G. Let , (we assume that these generating sets are closed under taking inverses). As a generating set for we take the set of all pairs (xi,1), (1,yj). Let E be the equalizer of and in H.
Let be generators of the kernel of (as a normal subgroup of Fn). Without loss of generality we assume that the set is closed under cyclic shifts and inverses.
For every i=1,...,m pick one word
such that
.
For every j=1,...,n pick one word
sj such that
.
Then the equalizer E is
generated by the pairs
(xi,ti), (sj, yj), (1,rk), i=1,...,m,
j=1,...,n,
.
Indeed, if
,
that is
and
u=xi1xi2...xip then
We shall prove that the distortion function of E in H is equivalent to d(k). In order to do that we need the following general statement.
Proof. If has an internal edge (i.e. an edge which belongs to the contours of two cells) then it has an internal edge f one of whose vertices belongs to the boundary. Let us cut along f leaving the second vertex of f untouched. We can repeat this operation until we get a diagram which does not have internal edges. It is easy to see that the boundary label of is equal to w in the free group. The number of edges of which do not belong to contours of cells (let us call them edges of type 1) is the same as the number of such edges in and the number of edges which belong to contours of cells in (edges of type 2) is at most twice the number of such edges of (we cut each edge from a contour of a cell at most once, after the cut we get two external edges instead of one internal edge).
Suppose that a cell in has more than one edge which has a common vertex with but does not belong to the contour of . Take any point O on . Let p be the boundary path of starting at O and let q be the boundary path of starting at O. Consider the path qq-1p. The subpath q-1p bounds a subdiagram of containing all cells but . Replace the path q in qq-1pby a loop q' with the same label starting at O and lying inside the cell . Let the region inside q' be a new cell . Then the path q'q-1p bounds a diagram whose boundary label is freely equal to w. Notice that has exactly one edge having a common vertex with and not belonging to the contour of . Thus this operation reduces the number of cells which have more than one edge which has a common vertex with the cell but does not belong to the contour of it.
After a number of such transformations we shall have a diagram which has the form of a tree T with cells hanging like leaves (each has exactly one common vertex with the tree).
The number of edges of type 1 in cannot be bigger than the number of all edges in , so it cannot be more than two times bigger than the total number of edges in .
The boundary label of is freely equal to w, and it has the form u1ri1u2ri2...udridud+1 where d is the number of cells in , u1u2...ud+1 is the label of a tree, so u1u2...ud+1=1 in the free group. The sum of lengths of ui is at most four times the number of edges in because the word u1u2...ud+1 is written on the tree T, and when we travel along the tree, we pass through each edge twice.
The lemma is proved.
Let us consider the distortion function of E in H.
Without loss of generality we can assume that d(k) is the Dehn function of the presentation
(recall that Dehn functions of different finite presentations of the same group are equivalent).
Let (u,v) be any element in E whose length in H, |u|+|v|, is .
Then as before
Notice that the length of the word adoes not exceed
Let
be the minimal area van Kampen diagram over the presentation
By Lemma 1,
This implies that the distortion function of E in H does not exceed a function equivalent to the Dehn function d.
To prove that the Dehn function ddoes not exceed a function equivalent to the distortion function of E in H, it is enough, for every number , to take a word , from the kernel of , , of area d(p). Then it is easy to see that any representation of (1,a) as a product of generators of E must contain at least d(p) factors of the form (1,rk) (because it corresponds to a representation of a in the form u1ri1u2ri2...ridud+1 where u1...ud+1=1 in the free group).