Читать книгу Invariants And Pictures: Low-dimensional Topology And Combinatorial Group Theory - Vassily Olegovich Manturov - Страница 16

An important problem of combinatorial group theory is the word problem: the question whether for a given word W in a group G holds the equality W =1 (or, more generally, whether two given words are equal in a given group). Usually the difficult question is to construct a group where a certain set of relations holds but a given word is nontrivial. In other words, to prove that a set of relations does not yield W = 1 (note that there may exist groups where both the relations and W = 1 hold due to the presence of additional relations). Small cancellation theory proves to be a very powerful and useful instrument in that situation. In particular, an important role in solution of that kind of problems plays the Greendlinger theorem which we will formulate in this section. We will always assume that the presentation (1.1) is symmetrised.

The Greendlinger theorem deals with the length of the common part of cells’ boundary. Sometimes two cells are separated by 0-cells. Naturally, we should ignore those 0-cells. To formulate that accurately we need some preliminary definitions.

Let Δ be a reduced diagram over a symmetrised presentation (1.1). Two -edges e₁, e₂ are called immediately close in Δ if either e₁ = e₂ or e₁ and (or and e₂) belong to the contour of some 0-cell of the diagram Δ. Furthermore, two edges e and f are called close if there exists a sequence e = e₁, e₂, . . . , el = f such that for every i = 1, . . . , l − 1 the edges ei, e_i+1 are immediately close.

Now for two cells Π₁, Π₂ a subpath p₁ of the contour of the cell Π₁ is a boundary arc between Π₁ and Π₂ if there exists a subpath p₂ of the contour of the cell Π₂ such that p₁ = e₁u₁e₂ . . . u_n−1en, , the paths ui, vj consist of 0-edges, ei, fi are -edges such that for each i = 1, . . . , n the edge fi is close to the edge ei. In the same way a boundary arc between a cell and the contour of the diagram Δ is defined.

Informally we can explain this notion in the following way. Intuitively, a boundary arc between two cells is the common part of the boundaries of those cells. The boundary arc defined above becomes exactly that if we collapse all 0-cells between the cells Π₁ and Π₂.

Now consider a maximal boundary arc, that is a boundary arc which does not lie in a longer boundary arc. It is called interior if it is a boundary arc between two cells, and exterior if it is a boundary arc between a cell and the contour ∂Δ.

Remark 1.2. It is easy to see that the small cancellation conditions C′(λ) and C(p) have a natural geometric interpretation in terms of boundary arcs. Namely, an interior arc of a cell Π of a reduced diagram of a group satisfying the C′(λ) condition has length smaller than λ|∂Π|. Likewise, the C(p) condition means that the boundary of every cell of the corresponding diagram consists of at least p arcs.

Now we can formulate the Greendlinger theorem.

Theorem 1.3. Let Δ be a reduced disc diagram over a presentation of a group G satisfying the small cancellation condition C′(λ) for some and let Δ have at least one -cell. Further, let the label φ(q) of the contour q = ∂Δ be cyclically irreducible and such that the cyclic word φ(q) does not contain any proper subwords equal to 1 in the group G.

Then there exists an exterior arc p ofsome -cell Π such that

Remark 1.3. If we formulate the Greendlinger theorem for unoriented diagrams, the theorem still holds.

Before proving this theorem, let us interpret it in terms of group presentation and the word problem. Consider a group G with a presentation (1.1) satisfying the small cancellation condition C′(λ), . Due to van Kampen lemma 1.1 (and its strengthening, Theorem 1.1) for a word W = 1 in the group G there exists a diagram with boundary label W. Boundary label of every -cell of the diagram by definition is some relation from the set (or its cyclic permutation). Then, due to Greendlinger Theorem 1.3 there is an -cell such that at least half of its boundary “can be found” in the boundary of the diagram. Thus we obtain the following corollary (sometimes it is also called the Greendlinger theorem):

Theorem 1.4 (Greendlinger [Greendlinger, 1961]).

Let G be a group with a presentation (1.1) satisfying the small cancellation condition C′(λ), . Let W ∈ F be a nontrivial freely reduced word such that W = 1 in the group G. Then there exist a subword V of W and a relation R ∈ such that V is also a subword of R and

This theorem is very useful in solving the word problem.

Example 1.3 (A.A. Klyachko). Consider the relation R = [x, y]² and the word

The question is, whether in every group with the relation R the equality W = 1 holds.

First, it is easy to see that the set of relations _* obtained from R by symmetrisation satisfies the condition . Therefore, due to the Greendlinger theorem every word V such that V = 1 in the group G has a cyclic permutation such that both its irreducible form and some relation have a common subword p of length .

On the other hand, the longest common subwords of the word W = [x¹⁰⁰⁰, y¹⁰⁰⁰]¹⁰⁰⁰ and any of the relations have length 2: those are

Therefore we can state that there exists a group G where for some two elements a, b ∈ G [a, b]² = 1 but [a¹⁰⁰⁰, b¹⁰⁰⁰]¹⁰⁰⁰ ≠ 1.

Now let us prove Theorem 1.3.

Proof. Let G be a group with a presentation (1.1) and Δ be a diagram as in the statement of the theorem. Since Δ is a disc diagram, we can place it on the sphere. We construct the dual graph Φ to the 1-skeleton of the diagram Δ in the following way.

Place a vertex of the new graph Φ into each -cell of the diagram Δ and one vertex O into the outer region on the sphere. To construct the edges of the graph Φ (note that they are not edges of a diagram) we perform the following procedure. For every interior boundary arc between cells Π₁ and Π₂ we chose an arbitrary -edge e₁ and an edge e₂ close to it such that lies in ∂Π₂. Now we connect the vertices of the graph Φ, lying inside the cells Π₁, Π₂ by a smooth curve γ transversally intersecting the interior of the edges e₁ and e₂ once and crossing all 0-cells lying between them. This curve γ is now considered as an edge of the graph Φ. This procedure is performed for every vertex of the graph Φ and every boundary arc (if the arc is exterior, we connect the vertex with the “exterior” vertex O).

The main idea of the proof is to study the dual graph Φ and to prove the necessary inequality by the reasoning of Euler characteristic of sphere.

First, note that among the faces of the graph Φ there are no 1-gons (because the boundary labels of the cells of the diagram Δ are cyclically irreducible) or 2-gons (because in the previous paragraphs we have constructed exactly one edge crossing every maximal boundary arc of Δ). Therefore every face of Φ is at least a triangle. Thus, denoting the number of vertices, edges and faces of the graph Φ by V, E and F respectively, and considering the usual Euler formula V − E + F = 2 we get the following inequality:

Now, suppose that the statement of the theorem does not hold. For each edge of the graph Φ connecting the vertices oi, oj lying in the cells Π_i, Π_j of the diagram Δ we attribute this edge to each of those vertices with the coefficient ; if an edge connects a vertex ok with the vertex O, we attribute it to the vertex ok with the coefficient 1.

Fix an arbitrary vertex o ≠ O. Denote the cell where the vertex o lies by Π and consider the following possibilities.

(1)Let the contour ∂Π consist only of interior arcs. Due to the condition the length of each of those arcs is smaller than (see Remark 1.2). Therefore, their number is not smaller than seven and thus at least edges of the graph Φ is attributed to the vertex o.

(2)Let there be exactly one exterior arc p. Since we suppose that |p| ≤ |∂Π|, the number of interior arcs of the contour of the cell Π is at least four. Therefore we attribute at least edges to the vertex o.

(3)Let the cell Π have two exterior edges p₁, p₂. Note that the end of the arc p₁ cannot be the beginning of the arc p₂ (and vice versa) and they cannot be separated by 0-edges only because otherwise we could cut the diagram Δ with edges f₁, . . . , fl such that φ(fi) = 1 for all i = 1, . . . , l and due to van Kampen lemma obtain two proper subwords φ(q₁), φ(q₂) of the word φ(q) (where q is the contour of the diagram Δ) equal to 1 in the group G. That contradicts the condition of φ(q) being cyclically irreducible.

Therefore, the cell Π has at least two distinct interior boundary arcs and there are at least edges attributed to the vertex o.

(4)If the cell Π has at least three exterior arcs, there are at least three edges of the graph Φ attributed to the vertex o.

Considering those possibilities for all vertices of the graph Φ except the “exterior” vertex O we see that

and that contradicts the inequality (1.3). This contradiction completes the proof.