Lecture 15

Rips’s Short Exact Sequence.

1 → N → G → Q → 1

1 → ⟨x, y⟩ → ⟨x, y, g_{1}, …, g_{s}|R_{1}, …, R_{t}⟩ → ⟨ḡ_{1}, …, ḡ_{s}|\overline{R}_{1}, …, \overline{R}_{t}⟩ → 1 Note. N is finitely presented ⇔ Q is finite. Example. Finitely generated normal subgroup of a free group F_{n} is either

trivial or finite index. (figure)

Theorem 1. (Bieri, Strebel) If X is a finite aspherical 2-complex and N ◁ π_{1}X with N finitely presented then either

0) N is trivial

1) N is free and 1 → N → π_{1}X → (\begin{matrix}virtually \\ free \end{matrix}) → 1 2)N has finite index in π_{1}X

Example. There is a map Q → Out(N) = Aut(N)/In(N)

\xymatrix{G \ar[r] \ar[d] & Aut(N) \ar[d] \\ G/N \ar[r] & Aut(N)/In(N)}

The map Q → Out(N) is injective in our case. With a bit of a extra work (modify construction) arrange Q ≃ Out(N). So every finitely presented group ≅ Out(N) for some finitely generated N.

Example. N = G ⇔ Q is trivial.

Fact. There is no algorithm to detect if a group determined by finite presentation is nontrivial. Thus “generation problem” for G is undecidable (in general).

F_{2} ⋆ F_{2} has undecidable generation problem, membership problem.

The Construction. Consider presentation

⟨\begin{matrix}q_{1}, …, q_{s} \\ x, y \end{matrix}| \begin{matrix}x_{q_{i}} = X_{i+} & x_{q_{i}^{−1}} = X_{i−} \\ y_{q_{i}} = Y_{i+} & y_{q_{i}^{−1}} = Y_{i−} \end{matrix} ⟩ ,

where X_{i+}, X_{i−}, Y_{i+}, Y_{i−} are words in x, y. If X_{i±}, Y_{i±} are chosen randomly of length < 1000000 then we obtain C′(\frac{1}{6}) presentation, and subgroup ⟨x, y⟩ is normal but nontrivial. (C′(\frac{1}{6}) ⇒ It’s Dehn’s presentation ⇒ x is nontrivial, since too short.)

⟨\begin{matrix}q_{1}, …, q_{s} \\ x, y \end{matrix}| \begin{matrix}\begin{matrix}x_{q_{i}} = X_{i+} & x_{q_{i}^{−1}} = X_{i−} \\ y_{q_{i}} = Y_{i+} & y_{q_{i}^{−1}} = Y_{i−} \end{matrix}\} & make N ; normal \\ R_{i} = W_{i} & make G/N ≅ Q \end{matrix}⟩ ,

where W_{i} are words in x, y, with lengths > 10(max{R_{i}}).

For example,

X_{i+} = xyxy^{2} ⋯ xy^{1000M}
X_{i−} = xy^{1000M+1}x ⋯ xy^{2000M}
Y_{i+} = xy^{2000M+1}x ⋯ xy^{3000M}
…

this way pieces are tiny.

Fact. (Guba) Let X and Y be reduced cyclically reduced cyclically distinct words in a_{i}^{±1} which are not proper powers. Suppose U has a subword both of X^{n} and Y^{n}. Then |U| ≤ |X| + |Y |.

Theorem 2. Let ⟨a_{1}, …, a_{n}|R_{1}, …, R_{m}⟩ (R_{i} are not proper powers) be a presentation with R_{i} ≁ R_{j}^{±1}. Then ⟨a_{1}, …, a_{n}|R_{1}^{d_{1}}, …, R_{m}^{d_{m}}⟩ is C′(\frac{1}{6}) provided |R_{i}^{d_{i}}| > 6(max_{j}{|R_{i}|}) for each i.

The Conjugacy Problem. Given ⟨a_{1}, … a_{s}|R_{1}, …, R_{t}⟩, words U, V in a_{i}^{±1}. Is U ∼_{G} V. This problem is (in general) hard, is not a geometric problem. In particular, exists G with finite index subguoup H such that CP (G) solvable, CP (H) not (CP (H) solvable, CP (G) not). This is opposite to word problem, for which G_{1} quasiisometric to G_{2} implies W P (G_{1}) decidable ⇔ W P (G_{2}) decidable.

Geometric Interpretation. Annular diagram A is a finite 2-complex with specific planar embedding such that π_{1}A ≅ ℤ.

(figure: examples)

Annular diagram has 2 boundary paths.

(figure)

Annular diagram in X is just a combinatorial map A → X.

A → X is reduced if it contains no cancellable pairs. (Obtained by removing cancelable pairs.) Area (A) is number of its 2-cells.

Theorem 3. Paths P_{1} and P_{2} are homotopic to each other (P_{1} and P_{2} are conjugated in π_{1}X) iff exists an annular diagram in X with boundary paths P_{1}, P_{2}.

Proof. Suppose A exists, then P_{1} ∼ P_{2}. Indeed, P_{1} ∼ P_{2} in A, this projects to a homotopy in X.

Conversely, if P_{1} ∼ P_{2} in X then we construct A by e.g. P_{1} = QP_{2}Q^{−1} for some path Q, so exists disc diagram D → X with ∂_{p}D = P_{1}QP_{2}^{−1}Q^{−1}. Now form A by identifying Q = Q.

(figure: identifying)

When X is small-cancellation then a minimal area diagram is also small cancellation. But C′(\frac{1}{6}) diagrams are very limited because of Combinatorial Gauss–Bonet Theorem.

Greendlinger’s Lemma.

Theorem 4. Any C(6) annular diagram must contain an i-shell or spur with i ≤ 4. Any C(4) − T (4) annular diagram must contain an i-shell or spur with i ≤ 3.

Proof. (Similar argument to Greendlinger’s Lemma.)

\begin{aligned}2πχ(A) & = & ∑ κ(v) + ∑ κ(f) \\ 0 & = & ∑ κ(v) + ∑ κ(f) \\ 0 & ≤ & π#_{S} + π#_{0} + π#_{1} + \frac{2π}{3} #_{2} + \frac{π}{3} #_{3} + 0#_{4} + 0#_{C} \end{aligned}

(where #_{C} stands for number of connector - see figure)

(figure: connector)

If A is a C′(\frac{1}{8}) annular disc diagram, then either:

• A contains a spur, 0-shell, 1-shell, 2-shell, or 3-shell (in which case A can be simplified by shortening R_{1} or R_{2}), or

• A has a very simple structure:

(figure: list of simple possibilities)

3

This solves the conjugacy problem.