Lecture 11

Last time: Let D be a disc diagram, (4, 4)-comlex. Girth ≥ 4, each 2-cell is square. Then Area (D) ≤ |∂_{P} D|^{2}

Observe. Let X be (4, 4)-complex, its first cubical subdivision is a (4, 4)- complex whose 2-cells are squares.

(figure: cubical subdivision) “old” links did not change. New ones have girth at least 4. So cubical subdivision preserves “(4,4)”.

X′ = subdivide X. Isoperimetric function are ∼ the same. Therefore we have the following theorem.

Theorem 1. Let X be finite (4, 4)-complex. Then its isoperimetric function f(n) is ≤ kn^{2}.

Theorem 2. Let X be finite (3, 6)-complex or (3, 6)-complex. Then its isoperimetric function f(n) is ≤ kn^{2}.

Theorem 3. Let D be a (3, 6) disc diagram whose 2-cells are triangles. Then Area (D) ≤ |∂_{P} D|^{2}.

Proof. As usual, consider non-singular disk diagram D. We consider the dual curves in D modeled on

(figure: types of odd/even dual curves)

Lemma 1. No dual curve is a circle. (figure)

Lemma 2. No dual curve self-intersects. (figure)

Lemma 3. No two dual curves intersect in a bigon. (figure)

Proof. Barycentrycally subdivide D to form D′, so dual curves occupy 1skeleth of D′. Apply combinatorial Gauss-Bonet theorem to E, conclude E has ≥ 3 corners.

(figure: E inside D)

2πχ(E) = ∑_{f} κ(f) + ∑_{v∈∂E} κ(v) + ∑_{v∈IntE} κ(v)
2π ≤  0  + \frac{5π}{6} (\begin{matrix}number \\ of \\ corners \end{matrix}) +  0

Note. D′ satisfies non-positive weight test in an obvious way.

(figure: contributions of acute corners)

Only acute corners give positive contribution. Number of acute corners ≥ \frac{12}{5}. Thus lemmas 1, 2, 3.

Back to proving theorem. Bound number of 2-cells by number·(dual curves)^{2}.

(figure: dual curves corresponding to triangle)

{triangles} ⟶^{h} {dual curves}^{2}

h is injective by lemma 2.

{dual curves} → {cells in∂_{P} D}

We are done once {dual curves} ≤ k|∂_{P} D|.

(figure: how dual curves behave when hit the boundary)

When curve hits the boundary vertex, the 6 middle ones stop, the rest are continued symmetrically. So {dual curves} ≤ 14|∂_{P} D|.

We proved that for D a triangular (3, 6) disk diagram, Area (D) ≤ |∂_{P} D|^{2}.

Theorem 4. Let X be (3, 6)-complex or (6, 3)-complex. Then the isoperimetric function is bounded by a quadratic function f(n) ≤ kn^{2}.

Proof. First triangle subdivision of (6, 3)-complex is a (3, 6)-complex. Moreover, any (3, 6)-complex can be subdivided into a triangular (3, 6)-complex, meaning all 2-cells are triangles. (Just divide each n-gon into (n − 2) triangles without new 0-cells.) But a triangular (3, 6)-complex has f(n) ≤ n^{2}, because this holds for any triangular (3, 6)-diagram. That completes the proof.

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Construction of “thick” (3, 6)-complexes. (“Thick” means each 1-cell has at least three 2-cells attached locally.)

Branched covering spaces. (figures)

X = ⟨a|aaa^{−1}⟩, Y = X \ D_{ϵ}(p).

π_{1}Y is free, so Y has many finite covering spaces. Choose a finite cover Ŷ → Y. Then add in the points p to form a branch cover X̂ → X. If you choose Ŷ appropriately (large enough), then X is (3, 6)-complex, X̂ is always thick.

(figures: link of the 0-cell, Z in Y)

Y = X \ p

Deformation retraction Y → Z yields immersion link(p) → Z.

Covering spaces of Z ⟷ covering spaces of Y ⟷ branch covering spaces of X Deformation retraction Y → Z ↕ Deformation retraction Ŷ → Ẑ induces immersion link(p̂) → Ẑ.

Thus to make (3, 6)-complex X̂ choose Ẑ to have girth ≥ 12 (any graph has a finite cover with a given girth).