Lecture 14

Theorem 1. (Greendlinger’s Lemma) Let D → X be a C(6) [C(4) − T (4)] disc diagram. Then either

1) D is trivial

2) D is a single 2-cell

3) [C(6)] D contain a “count” of 6 i-shells/spurs, where spurs, 0-shells, 1-shells count as 3, 2-shells count as 2, 3-shells count as 1.

3) [C(4) − T (4)] D contain a “count” of 4 i-shells/spurs, where spurs, 0-shells, 1-shells count as 2, 2-shells count as 1.

(figure: examples) Proof of [C(6)]. Assume D is not a single 2-cell or 0-cell. Ignore all valence 2

0-cells. Assign angles as follows:

(figure) totally internal angles \frac{2π}{3} , (figure) half internal angles \frac{π}{2} , (figure) totally external angles 0. Now D is an angled 2-complex.

1) Each internal 0-cell has curvature ≤ 0.

2) Each external 0-cell has curvature ≤ 0 except endpoint of a spur has curvature = π.

Indeed, if v ∈ ∂D but valence(v) ≠ 1, then the link of v is a forest

with at least 3 vertices. Therefore either link(v) has at least 2 components, so χ(link(v)) ≥ 2, or link(v) is connected and ∑ ∠ ≥ π (figures of 3 possibilities).

So 2-cells must make positive curvature. Each internal 2-cell has non-positive curvature (sum of angles ≥ \frac{2π}{3} ·6). Each 2-cell with ≥ 2 boundary arcs has curvature ≤ 0, since has ≥ 4 angles of value \frac{π}{2} .

Exercise. Check this. (figures: possible cases) Remaining 2-cells have exactly 1 boundary edge. Curvature of a 2-cell is κ(f) =

∑ ∠ − (|f| − 2)π. Sum of angles of an i-shell (i ≥ 1) is ∑ ∠ = \frac{π}{2} + \frac{π}{2} + (i − 1)\frac{2π}{3} . Then κ(f) = π − \frac{π}{3} (i − 1)

κ(1−shell) = π κ(2−shell) = \frac{2π}{3} κ(3−shell) = \frac{π}{3} κ(4−shell) = 0 (By the way: you may have as many 4-shells as you like.)

2πχ(D) = ∑ κ(v) + ∑ κ(f)

\begin{aligned}2π & ≤ & 0 & + & π · #_{S} & + & π · #_{0} & + & π · #_{1} & + & \frac{2π}{3} · #_{2} & + & \frac{π}{3} · #_{3} \\ 6 & ≤ & 0 & + & 3 · #_{S} & + & 3 · #_{0} & + & 3 · #_{1} & + & 2 · #_{2} & + & 1 · #_{3} \end{aligned}

Proof sketch for C(4) − T (4) case. Similar argument except use \frac{π}{2} angles on internal and half-internal corners, 0 angles on totally external corners (ends up κ(f) = (3 − i)\frac{π}{2} ). □

Why no lemma for C(3) − T (6)? (figure: hexagon) Zillion 2-shells, no 1-shells, no 0-shells. Although, there are positive features. (figure: rhombus).

Exercise. Give example of infinitely mane (6, 3) complexes.

Rips’s Short Exact Sequence. A construction producing interesting subgroups of C′(\frac{1}{6}) groups.

Let Q be a finitely presented group ⟨ḡ_{1}, …, ḡ_{s}|\overline{R}_{1}, …, \overline{R}_{t}⟩. There exist a short exact sequence

1 → N → G → Q → 1

s.t. G is a finitely presented C′(\frac{1}{6}) groups, N is finitely generated.

1 → ⟨x, y⟩ → ⟨x, y, g_{1}, …, g_{s}|R_{1}, …, R_{t}⟩ → ⟨ḡ_{1}, …, ḡ_{s}|\overline{R}_{1}, …, \overline{R}_{t}⟩ → 1

Rips’s construction can be used to produce C′(\frac{1}{6}) group G with interesting subgroups from a group Q with interesting properties.

Example. If Q has undecidable word problem then G has undecidable membership problem in N. Induced from a word \overline{w} in generators of Q, (\overline{w} = 1) ⇔ (w ∈ N).

Example. If has a finitely generated subgroup \overline{H} that is finitely generated but not finitely presented then its preimage H in G is finitely generated but not finitely presented.

Indeed, if \overline{H} = ⟨\overline{w}_{1}, …, \overline{w}_{n}⟩ then H = ⟨w_{1}, …, w_{n}, x, y⟩. If H is finitely presented then H = ⟨h_{1}, …, h_{k}|U_{1}, …, U_{m}⟩, so \overline{H} = ⟨h_{1}, …, h_{k}|U_{1}, …, U_{m}, X, Y ⟩.