Lecture 9

Definition 1. An immersion Y → X is a local injection. A near-immersion Y → X is a map such that its restriction to Y \ Y^{0} is an immersion.

Note. A combinatorial map φ : Y → X is an immersion iff φ : link(v) → link(φ(v)) is an embedding.

It is a near-immersion iff φ : link(v) → link(φ(v)) is an immersion. (Also, it is the same being embedding of links of 1-cells.)

(figure: six slices pie mapping onto three slices pie)

Given a map of a 1-cell, we can continue it on the corresponding 2-cell by coning

Y ⇝ ^{Y × [0, 1]/} Y × {1} = c

(figures: cones)

Lemma 1. Suppose Y → X is a near-immersion, suppose also X is an angles 2-complex which satisfies the [negative] weight test. Then Y satisfies the same condition.

Proof. Pull back the angles on corners of X to corners of Y. Now the map φ : Y → X is an angle preserving map between angled 2-complexes. κ(f) ≤ 0 for each 2-cell of Y.

Let σ be immersed closed path in in link(y), y ∈ Y^{0}. Then φ∘σ immersed closed path in in link(φ(y)), φ(y) ∈ X^{0}, and |σ|_{∠(Y)} = |φ ∘ σ|_{∠(X)} ≥ 2π.

Corollary 1. Let X satisfy [negative] weight test. If S → X is a near-immersion of a close surface then χ(S) ≤ 0 [χ(S) < 0], so there are no near-immersion S^{2} → X [T^{2} → X].

or, in other words,

Corollary 2. X is aspherical, i.e. π_{i}X = 1 for i ≥ 2, i.e. universal cover of X is contractible [ℤ^{2} ↛ π_{1}X, i.e. non-hyperbolic].

Definition 2. X is diagrammatically reducible (DR) if there is no near-immersion S^{2} → X.

Conjecture. (Whitehead) Let X, a 2-complex, be aspherical. Let A be a subcomplex. Then A is aspherical.

Theorem 1. If X is DR then X is aspherical.

Idea of proof. Let S^{2} →^{f} X. Simplitial approximation theorem: allows to choose a subdivision of S^{2}, X and a new cellular map g : S^{2} → X such that f ≃ g (homotopic). You can snush/ignore cells that get squashed. Left with h : S^{2} → X combinatorial, h ≃ g ≃ f. But h is not yet near-immersion.

Let Y be a surface. Suppose Y → X is a near-immersion.

(figure: removing a “cancellable pair”)

(figure: dealing with deviant cases)

Idea. Suppose X satisfies the negative weight test. Suppose ℤ^{2} ↪ π_{1}X. There exists a map T^{2} → X, π_{1}T^{2} → π_{1}X, a ↦ p_{a}, b ↦ p_{b}, closed paths in X, ℤ^{2} = ⟨p_{a}, p_{b}⟩.

(figure: a cell corresponding to a, b in disc diagram) It can’t be singular, as p_{a}, p_{b} generate ℤ^{2}.

Futzing around obtain a near-immersion T^{2} → X. Pull back angles. Apply combinatorial Gauss-Bonet theorem: 0 = 2πχ(T^{2}) = ∑ κ(v) + ∑ κ(f) < 0.

Isoperimetric Function.

Theorem 2. Let X be a 2-complex. Let P → X be a nullhomotopic path. Then there exists a disc diagram D → X for P such that the map D → X is a nearimmersion.

Proof. There exists some disc diagram D_{i} → X for P. If D_{i} → X is not a near-immersion, then D_{i} has a cancelable pair. Removing this cancelable pair yeilds a lower area disc diagram D_{i+1} → X with boundary path P. Eventually reach a near-immersion D → X.

(However, disc diagram has not necessarily the minimal square:

figure: a cube — area 1 and area 5 diagrams.)

Theorem 3. Let X be a compact angled 2-complex satisfying negative weight test. Then X has a linear isoperimetric function.

Proof. There exists a constant M = M(X) such that for any near-immersion D → X, Area (D) ≤ M|∂_{P} D|.

Let c_{1} be such that κ(f) ≤ −c_{1} < 0 for all f in X.

Let c_{2} be such that |σ|_{∠} > 2π + c_{2} for any σ.

Let D → X be a nonsingular dias disc diagram (singularity just increase length of boundary without increasing area). Pull back angles to D.

2πχ(D) =  ∑_{f} κ(f)  + ∑_{v∈∂D} κ(v) + ∑_{v∈IntD} κ(v)
2π ≤ Area (D) · (−c_{1}) +  c_{2}|∂D|  +  0
c_{1}Area (D)  ≤  c_{2}|∂D|
Area (D)  ≤  \frac{c_{2}}{c_{1}} |∂D|.

Done.