GSAC TALK: THE WORD PROBLEM. 1. The 8-8 Tessellation. Consider the following infinite tessellation of the open unit disc. Actually, we want to think

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1 GSAC TALK: THE WORD PROBLEM 1. The 8-8 Tessellation Consider the following infinite tessellation of the open unit disc. Actually, we want to think of the disc as a model for R 2 and we can do that using the homeomorphism:f(rcis(θ)) = arctan( π r)cis(θ). You might object by saying that the tiles aren t polygonal, because the lines 2 aren t straight but we ll overlook this at this point. After recovering from the initial shock, you might come to realize that this is actually a pretty wonderful tiling. It is symmetric with respect to rotation by 45, actually there s a lot more symmetry involved: the picture is symmetric with respect to inversion by any of the half circles meeting the boundary. More importantly, it is also regular: all shapes are octagons, and the valence of each vertex is 8. In many ways this resembles the tessellation of R 2 by squares, which we are all comfortable with. However, this 8-8 tessellation has a curious property: Take any edge loop L, and mark the (bounded) region it bounds by M, then there s a tile of M which is sticking out, i.e. there s a tile D of M, which shares more than 1 D = 4 consecutive edges with L. Actually, 2 for each disc we can find many tiles which share 6 out of 8 boundary edges with the loop L. Try it out!! Does this also hold for the square tiling? Date: March

2 This fact has a neat proof. You argue that any such disc region M contains a vertex of valence 1 or at least 2 vertices of valence 2. Notice that these vertices must be lie the boundary of M, because if v is internal, its valence is 8. Now you apply this fact not to M but to its dual (the green diagram), which is also a disc with an 8-8 tiling. If the dual contains a valence 1 vertex then M contains a tile with 7 edges on L. If the dual contains a valence two vertex then M contains a tile sharing 6 edges with the boundary, but we must be careful since they might not be consecutive, and so the tile D might not be sticking out. But this has an easy fix: Such a tile corresponds to a separating vertex. The dual has an extremal disc - one that has only one separating vertex (otherwise, it would be an annulus). The extremal disc has at least two valence 2 vertices, the one that is non-separating corresponds to a tile which is sticking out. So it is enough to prove: Lemma 1.1. Any disc M with an 8-8 tiling has a valence 1 vertex or at least two valence 2 vertices. Proof : The main topological ingredient is that the Euler characteristic of a disc χ(m) is 1. Explicitly: Let ɛ = the set of edges in M, V be the set of vertices in M, and Λ the set of 2

3 tiles in M. Denote: V = V, E = ɛ, and F = Λ. Then 1 = V E + F We can assume F 1 and that the disc is non-degenerate, so that every edge lies in one of the tiles. (Why can we assume this?) Now count the number of pairs (e, D) such that e ɛ, D Λ and e is an edge of the tile D: on the one hand every tile has 8 edges so it appears in 8 pairs hence the number of pairs is 8F. On the other hand, each interior edge is counted twice, and each boundary edge once, so the number of pairs is: 2E E. Hence: 2E E = 8F Now count the number of pairs (v, e) such that e ɛ, v V and v is a vertex in the edge e: on the one hand every edge is counted twice so this number is 2E. On the other hand, each vertex is counted d(v) times (where d(v) is the valence of v) so the number of pairs is: d(v). Hence: 2E = d(v) Also notice that E = V 3

4 . Now we just have to combine the above formulas: 4 = 4V 4E+4F = 4V 2E 2E+4F = 4V (d(v)) E 8F +4F = (4 d(v)) E 4F (4 d(v)) V = (4 d(v)) V + (4 d(v)) = (3 d(v))+ (4 8) int int (3 d(v)). So either there s a vertex with valence 1 or at least four vertices of valence Presentations and the word problem Let F 2 =< a, b > denote the set of words in a, ā, b, b where we allow deletion (or insertion) of aā, āa, b b, bb: for instance abāa bā is equivalent to the empty word. This set is a group with the operation of concatenation, and the empty word as the identity. A word in < a, b > is reduced if is has no instances of aā, āa, b b, bb as subwords. It is called the free group on 2 letters. Analogously, one can define the free group on n letters, F n. G =< a, b, c ab = c 2 > then G is the set of words in a, b, c and their inverses where we allow deletion or insertion of ab c 2 or c 2 bā into a word. For instance: aca b = c 2 bca b. This sort of description of a group: < generators relations > is called a presentation. If the set of generators and relations is finite we say that this is a finite presentation. As you might suspect from this example, it might be quite tricky to tell whether two elements of a group presented in this way are in fact one and the same. Indeed, it was shown 4

5 by Novikov (1955) and (independently) by Boone (1959) that there are groups with a finite presentation where there is no algorithm to decide if a given word is trivial (empty) or not. In general we formulate the word problem: Given a presentation of the group G =< x 1,..., x n r 1 = 1, r 2 = 1,..., r m = 1 > is there an algorithm to decide if a given word w in the generators and their inverses is the identity? For example, in Z 2 =< a, b ab = ba >=< a, b abā b = 1 >=< a, b [a, b] = 1 > there is an easy algorithm. 3. Dehn s Algorithm Consider the following presentation: < a, b, c, d aba 1 b 1 = cdc 1 d 1 >. The relator r is aba 1 b 1 dcd 1 c 1, with length 8. We will show that if a word w is equivalent to the identity, then it contains a subword of r with 6 or more letters (the subword may start at the end of the relator and continue at the beginning of it, for example d 1 c 1 aba 1 b 1 ). Thus, here s an algorithm: given a word we scan it for a piece of the relator of length 6, if we find one, we replace it with its complement in the relator, thus decreasing the length of the word. If we can t find such a piece and we end up with something other than the empty word - our word was not equivalent to the identity. 5

6 4. Surface groups Our relator has length 8. Label the edges of the octagon with a, b, c, d according to the relation. If we identify edges with identical labels we get a surface - the genus two torus. Therefore, any word in a, b, c, d represents a loop on this surface. Notice that the word aba 1 b 1 dcd 1 c 1 bounds a disc. Fact 4.1. A word in a, b, c, d is the identity in G if and only if the corresponding path bounds a disc in the surface. We say that G is the fundamental group of the surface. The surface is covered by R 2. This means that we have an action of G on R 2 whose quotient is S. The fundamental domain of this action is an octagon. If we take this octagon and move it around by the action, we get the tessellation we analyzed at the beginning of this lecture. After choosing a vertex to be our base point we can label the edges with a, b, c, d and their inverses. The label of each tile reads r the relation. Each vertex v is labelled by a word corresponding to a path from the base point to v. The graph with these labels is called the Cayley graph of G. Fact 4.2. An edge loop in the surface lifts to an edge path in R 2. An edge loop bounds a disc on S if and only if it lifts to a loop in R 2. 6

7 Now, if a word w is trivial, then it is represented by an edge loop in our tessellation. So it has a tile D which is sticking out. So when we read the label of w, those edges shared with D will read a piece of the relation of length 6. 7