Lecture 13

Example. ⟨a, b|abab^{2}ab^{3} ⋯ ab^{100}, baba^{2}ba^{3} ⋯ ba^{100}⟩. Symbols ab^{n}a, ba^{n}b occur uniquely for n > 1. Thus largest piece is smaller than b^{99}ab^{100}, a^{99}ba^{100}, so has length < 200. But each relator has length 5150. Therefore X satisfies C′(\frac{200}{5150}) ⇒ C′(\frac{1}{25}) ⇒ C(26) ⇒ C(7)

Example. ⟨a, b|W^{n}⟩ satisfies C′(\frac{1}{n}), C(n + 1), as any piece has length < |W |. (A longer piece would imply W = V^{m} for some m > 1.)

Lemma 1. A positive presentation satisfies T (4). (All relators are positive words in generators, e.g. ⟨a, b, c|abac, bbaabccb⟩.)

More generally, if X is a 2-complex and X^{1} is a directed graph, and all attaching maps are directed cycles, then X satisfies T (4) condition.

Proof. There can’t be any valence 3 vertices.

⟨a, b|a^{n}bab^{2}ab^{3} ⋯ ab^{100}⟩, n → ∞ satisfies C(p) with p constant, C′(α) with α → 1. So in a sense C(p) condition is far more general than C′(α).

Example. ⟨a, b|R_{1}, …, R_{m}⟩. Fix m. Suppose each R_{i} has length < n. Then presentation satisfies C′(α), α → 0 as n → ∞.

Thus C′(\frac{1}{6}) presentations are quite common for presentations with few long relators. T (4) is rare.

Lemma 2. T (5) ⇒ all pieces have length 1.

Proof. Suppose R_{1}, R_{2} share piece of length ≥ 2. Then there is a degree 2 branch cover which fails T (5) condition.

(figure: degree 2 branch cover)

So, for example, T (5), C(10) ⇒ C′(\frac{1}{10}).

T (5) condition is rather rare.

Lemma 3. For q ≥ 5 X is T (q) iff each link(v) of each 0-cell has girth ≥ q. (More generally, we can require “all pieces have length 1” instead of “X is T (q ≥ 5)”

Theorem 1. If X is compact and finite 2-complex and either of {C(3)−T (7), C(4)− T (5), C(5) − T (4), C(7) − T (3)} then π_{1}X has a linear isoperimetric function, thus word-hyperbolic.

Proof. Any reduced disc diagram D → X satisfies a (p, q) condition after ignoring valence 2 vertices in its interior, and subdividing the boundary a bounded amount. (As X satisfies negative weight test: \frac{1}{3} + \frac{1}{7} > \frac{1}{2}, \frac{1}{4} + \frac{1}{5} < \frac{1}{2}). Thus Area (D) ≤ k|∂_{p}D| for some constant k = k(X).

(figure: subdividing)

Amount of new vertices is bounded by |∂_{p}D| · p ≤ 6|∂_{p}D|. Subdividing didn’t change area, so

Area (D) = Area (D′) ≤ k|∂_{p}D′| ≤ 6|∂_{p}D|.

Theorem 2. X is either of {C(3) − T (6), C(4) − T (4), C(6) − T (3)} and compact then X has at most quadric isoperimetric function.

Proof. Analogous.

Dehn Presentation.

Definition 1. G = ⟨a_{1}, …, a_{m}|R_{1}, …, R_{n}⟩ is a Dehn presentation if for each word P with P =_{G} 1_{G} there either 1) P has a backtrack, or

2) P has a subpath Q such that • QS = some cyclic permutation of R_{i}^{±1}

• |Q| ≥ |S| i.e. |Q| ≥ \frac{1}{2}|R_{i}|.

(figures: cases 1 and 2)

Theorem 3. If ⟨a_{1}, …, a_{m}|R_{1}, …, R_{n}⟩ is a Denh presentation then its group G has linear isoperimetric function.

Proof. Proved by induction that Area (P) ≤ |P |. Let P ba a word P =_{g} 1_{G}. Base case: 0 = Area (1) , 0 = |1|. By hypothesis either

1) P = Aee^{−1}B, where e = a_{i}^{±1}, or

2) P = AQB , where Q ⊂ ∂R_{i}, |Q| > |R_{i}|.

Then P′ = AB or P′ = AS^{−1}B respectively also represents 1_{G}. |P′| < |P |, therefore Area (P′) ≤ |P′| by induction. Let D′ be a reduced disc diagram for P′. Then

Area (P) ≤ Area (P′) + 1 ≤ |P′| + 1 ≤ |P |,

because to obtain a diagram D for P from D′, we add a spur in case 1) and add a copy of R_{i} along S in case 2).

(figures: adding spur/cell)

Definition 2. i-shell is a 2-cell R in disc diagram D whose boundary such that |∂_{p}R = QS where Q is a subpath of ∂_{p}D and S is an internal path (i.e. no external 1-cells in S) which is concatenation of exactly i pieces.

A spur is a 1-cell ending at valence 1 cell.

(figures: i-shells, spur)

Theorem 4. (Greendlinger’s Lemma) Any nontrivial C(6) − T (3) disc diagram has at least two i-shells or spurs (i ≤ 3).

Any nontrivial C(4)−T (4) disc diagram has at least two i-shells or spurs (i ≤ 2). (“Nontrivial” means it’s not a single point or loop.)

Corollary 1. If X is a C′(\frac{1}{4}) − T (4) or C′(\frac{1}{6}) − T (3) complex, then it is a Dehn complex (presentation).

Proof of Corollary. Let P be a nullhomotopic path in X. Let D be a minimal area disc diagram D → X. By Greendlinger’s Lemma, if D is nontrivial, it has 2 i-shells or spurs, so at least one i-shell or spur away from a base point.

1) Spur has a backtrack.

2) i-shell with interior subpath S and exterior subpath Q. (figure: i-shell)

C′(\frac{1}{6}) ⇒ i ≤ 3, then |S| ≤ |P_{1}| + |P_{2}| + |P_{3}| < \frac{1}{6}|R| + \frac{1}{6}|R| + \frac{1}{6}|R| =

\frac{1}{2}|R|, thus |S| < |Q|, C′(\frac{1}{4}) ⇒ i ≤ 2, then |S| ≤ |P_{1}| + |P_{2}| < \frac{1}{4}|R| + \frac{1}{4}|R| = \frac{1}{2}|R|, thus

|S| < |Q|.