Invariants And Pictures: Low-dimensional Topology And Combinatorial Group Theory. Vassily Olegovich Manturov. Читать онлайн. Newlib. NEWLIB.NET

Автор: Vassily Olegovich Manturov
Издательство: Ingram
Серия: Series On Knots And Everything
Жанр произведения: Математика
Год издания: 0
isbn: 9789811220135
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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 figure-edges e1, e2 are called immediately close in Δ if either e1 = e2 or e1 and figure (or figure 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, ei+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 . . . un−1en, figure, the paths ui, vj consist of 0-edges, ei, fi are figure-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.

      Now we can formulate the Greendlinger theorem.

      Then there exists an exterior arc p ofsome figure-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), figure. 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 figure-cell of the diagram by definition is some relation from the set figure (or its cyclic permutation). Then, due to Greendlinger Theorem 1.3 there is an figure-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), figure. Let WF 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 Rfigure 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 figure* obtained from R by symmetrisation satisfies the condition figure. 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 figure have a common subword p of length figure.

      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, bG [a, b]2 = 1 but [a1000, b1000]1000 ≠ 1.

      Now let us prove Theorem