DESDE EL CONCEPTO DE VALENCIA Y LA TETRAVALENCIA DEL CARBONO

The content of this paragraph is the proof that the filling ratios satisfy a comparison axiom and an extension axiom. A numerical invariant for sim- plicial complexes is said to fulfill a comparison axiom if the existence of a ‘degree one’ map V → W implies an inequality between the respective val- ues of this invariant. It has to be specified which maps are of ‘degree one’ (for instance (n,1)-monotone maps as in Definition 2.9), and often there are further assumptions on the maps, like surjectivity on fundamental groups or on some homology groups. (Other examples can be found in chapters 2 and 4.)

An extension axiom is satisfied if attaching cells with dimension less than the topdimension does not change the value of the considered invariant. Again, further examples are provided by chapters 2 and 4.

The proof of Theorems 3.1 and 3.13 actually uses only these two axioms and no other properties of the filling ratios. It will be given in section 3.2.

Lemma 3.9(Comparison axiom for FR and FV). Iff :V →W is an (n,1)- monotone map between connected finite (_K-orientable) simplicial complexes of dimension n such that f∗ :Hn(V;K)Hn(W;K) is surjective, then

F_K(V)≥F_K(W)

for both F = FR and F = FV.

Proof. Let g2 be a Riemannian metric on W. Choose a Riemannian metric

g1 on V and setgt1 :=f ∗_g

2+t2g1. This is again a Riemannian metric on V. One may choose t >0 so small that

Vol(V, g₁t)≤Vol(W, g2) +ε

for a given ε > 0. Denote the corresponding Kuratowski embeddings by ιt₁

and ι2.

Since f : (V, gt

1) → (W, g2) is nonexpanding, there is a nonexpanding map F : L∞(V) → L∞(W) that extends ι2 ◦f by the universal property of L∞(W). (Think of V ⊂ L∞(V) via ιt

1.) Thus, there is a commutative diagram V_{ _} f // ιt 1 W_{ _} ι2 L∞(V) F //L∞(W) which gives for everyr >0

Hn(V;K) f∗ _//// (ιt 1)∗ Hn(W,K) (ι2)∗ Hn(Ur(ιt1V);K) F∗ // Hn(Ur(ι2W);K) Therefore FillRad_K(V, gt

1) ≥ FillRadK(W, g2), and this gives the desired in-

equality FR_K(V)≥FR_K(W).

Let z ∈ Cn(V;K) represent [V]K. Then f∗z represents f∗[V]K = ±[W]K

(in the case_K=_Qlook at the local degree and use thatfis (n,1)-monotone), and one finds

FillVol_K((ι2)∗(f∗z)) = FillVol_K(F∗((ιt₁)∗z))≤FillVol_K((ιt₁)∗z). Hence, FillVol_K(W, g2)≤FillVol_K(V, g₁t) and FV_K(W)≤FV_K(V).

The proof of the extension axiom is more complicated. We will frequently use the following fact, which is a direct consequence of the universal property (Lemma 3.3).

Corollary 3.10. If i : (V, dg) ,→ L∞(S) is an isometric embedding with S

any set, then

FillRad_K(i) = FillRad_K(V, g) and

FillVol_K(i) = FillVol_K(V, g).

We first investigate the extension axiom for the filling radius.

Proposition 3.11 (Extension axiom for FR). Let V0 be an extension of

V, that means V0 is obtained from V by attaching finitely many cells of dimension less than n := dimV. Then

FR_K(V0) = FR_K(V).

Proof. Since the inclusion i : V ,→ V0 is (n,1)-monotone and induces an isomorphism

i∗ :Hn(V;K)

∼ =

−→Hn(V0;K),

the comparison axiom implies FR_K(V)≥FR_K(V0).

To prove the converse inequality, it suffices by induction to attach one

k-cell at a time (with k < n). Leth:Sk−1 _{→V be the (simplicial) attaching} map, and letg be a Riemannian metric on V.

Consider all R > 0 such that h : (Sk−1_{, g}

R) → (V, g) is nonexpanding,

where gR denotes the round metric with radius R on Sk−1. (In the case k = 1 choose R >0 such that 23πR≥dg(h(−1), h(1)).) Define Riemannian

metricsg0_R onV0 by thinking of V0 as

V ∪h (Sk−1×[−1,0]) ∪ (Sk−1×[0,6]) ∪ S+k with S₊k a k-dimensional hemisphere and taking

g, (−sh∗g+ (1 +s)gR)⊕πRds2, gR⊕πRds2, gR

on the respective parts. Here, the lastgR denotes the round metric of radius Ron the k-dimensional hemisphere. (Fork= 1 use 5πRds2 onS0×[−1,0].)

Then the induced distance functions of g and g_R0 coincide on V:

dg0_R|V ≡dg,

i. e. the inclusioni: (V, dg),→(V0, dg_R0 ) is isometric as a map of metric spaces.

Hence by Corollary 3.10, we have

and therefore

FillRad_K(V0, g0_R)≤FillRad_K(V, g)

since for any r > 0 the r-neighborhood of ιg0_RV0 is larger than the one of ιg_R0 V.

Furthermore, note that the inclusion ((Sk−1_×[1,6])∪Sk

+, dg0

R)⊂(V

0_{, d}

g0 R)

is isometric with respect to the induced path metrics. We will write V_R0 :=

ιg0 RV

0_{, V}

R :=ιg0

RV and so on.

Next, we restrict our attention to radiiR >FillRad_K(V, g). Note that the

r-neighborhood of ((Sk−1×[1,6])∪S₊k)Rdoes not meet ther-neighborhood of VR for anyr < R. Thus, we may think ofUr(((Sk−1×[1,6])∪S+k)R) as some

kind of tubular neighborhood and try to retract it to its core. This core isk- dimensional and plays therefore no role for n-dimensional homology. Hence, if Hn(V;K) vanishes inUr(VR0), then also inUr(VR). We now concretize this

idea.

Choose one of the radii R with R > FillRad_K(V, g) and with R > 1 as reference radius and call it R0. Since the Kuratowski embedding ι0 := ιg0

R0

is not differentiable on the attached k-cell, we need to choose a smooth approximation to get an actual tubular neighborhood. Therefore, let

ι : (Sk−1×[3 +δ/πR0,6])∪S+k ,→L ∞

(V0) be a smooth embedding such that

d(ι(x), ι0(x))< δ

for allx∈(Sk−1×[3 +δ/πR0,6])∪S+k. Using the Kuratowski embeddingι0 onV0\((Sk−1_×[3,6])∪Sk

+) and linear interpolation onSk

−1_{×[3,3 +δ/πR} 0], this defines a Lipschitz map ι : (V0, g_R0 ₀) → L∞(V0) which is 3δ-close to

ι0. Denote by K := dil(ι) its Lipschitz constant, and think of ι as a map

V_R0

0 →L

∞_(V0_).

With respect to y∈V_R0₀ \((Sk−1_×[2,6])∪Sk

+)R0 the dilation dil(ι, y) of

ι is at most 1 +δ. This holds because

d(ι(y), ι(y0)) d(y, y0₎ ≤ d(y, y0) +d(y0, ι(y0)) d(y, y0₎ ≤1 + 3δ d(y, y0₎ ≤1 +δ for every y0 ∈ ((Sk−1 _×[3,6])∪Sk

+)R0 and because ι is isometric on V

0 R0 \ ((Sk−1_×[3,6])∪Sk +)R0. Let F : L∞(V0) → L∞(V0) be an extension of ι : V_R0₀ → L∞(V0) as in Lemma 3.3. Then F(Ur(VR00\((S k−1_× [2,6])∪S₊k)R0))⊂Ur(1+δ)(V 0 R0\((S k−1_× [2,6])∪S₊k)R0)

by the fact that dil(F, y) = dil(ι, y)≤1 +δ for ally ∈V_R0₀\((Sk−1_×[2,6])∪

S₊k)R0.

Denote by E :=ι((Sk−1_{×[3 +δ/πR}

0,6])∪S+k) the smooth part ofι(V 0_). Let ν(E) →E be its normal bundle, and let τ :N ,→L∞(V0) be a tubular neighborhood for the trivial spray (i. e. the exponential map is given by vector addition) where N ⊂ν(E) is open, fiberwise starshaped with respect to the zero section (thus it allows a deformation retraction to the zero section), and fiberwise bounded by FillRad_K(V, g)/2. Furthermore, assume that

τ(N|ι((Sk−1×[31₂,6])∪S₊k))⊂Ur0(((S

k−1_×

[2,6])∪S₊k)R0)

with r0 := FillRadK(V

0_{, g}0

R0). By compactness there is anε >0 such that

F(Uε(((Sk−1×[4,6])∪S₊k)R0))⊂τ(N|ι((S

k−1 _×

[31₂,6])∪S₊k)).

Moreover, choosing ε <(r0−3δ)/K guarantees that

F(Uε((Sk−1×[2,4])R0))⊂Ur0((S

k−1_×

[2,4])R0).

Claim. There is anR ≥R0 such thatHn(V;K)vanishes for all real numbers r >FillRad_K(V0, g0_R) inside Ur(VR00 \((S k−1_× [2,6])∪S₊k)R0) ∪ Uε(((S k−1_× [2,6])∪S₊k)R0).

Proof of the claim. Choose C ≤ 1 such that CπFillRad_K(V, g) < ε, and choose

R≥max(R0/C,diam(VR00)/Cπ).

The identity on V0 gives a nonexpanding homeomorphism f :V_R0 →V_R0

0.

Let ˜F :L∞(V0)→L∞(V0) be an extension off as in Lemma 3.3. Then, for any FillRad_K(V0, g_R0 )< r≤πFillRad_K(V, g) one finds

˜

F(Ur(((Sk−1×[2,6])∪S+k)R))⊂Uε(((Sk−1×[2,6])∪S+k)R0)

since dil( ˜F , y) = dil(f, y) ≤ C for any y ∈ ((Sk−1 _{× [2,6]) ∪S}k

+)R. This

follows from

d(f(y), f(y0))

d(y, y0₎ ≤C

which holds for all y0 ∈ V_R0 since on ((Sk−1 _{×[1,6]) ∪S}k

+)R the map f is

the contraction by the factor R0/R ≤ C, and for the other y0 note that the numerator is bounded from above by diam(V_R0

0) and the denominator is

bounded from below by πR. Hence ˜F maps Ur(V_R0) to Ur(VR00 \((S k−1_× [2,6])∪S₊k)R0)∪Uε(((S k−1_× [2,6])∪S₊k)R0).

Note that FillRad_K(V0, g_R0 ) ≥ FillRad_K(V0, g0_R0) = r0 by the universal property. Therefore, applying F shows thatHn(V;_K) also vanishes in

Ur(1+δ)(VR00 \((S

k−1_×

[4,6])∪S₊k)R0)∪τ(N|ι((S

k−1 _×

[31₂,6])∪S₊k)) for all δ > 0 and r > FillRad_K(V0, g0_R). Using the tubular neighborhood retraction (this is where the fiberwise bound on N comes in) and a Mayer- Vietoris argument one sees that Hn(V;_K) maps to zero in

Ur(1+δ)((V0\S˚+k)R0).

Since this holds for all δ >0 and r >FillRad_K(V0, g_R0 ) it follows that FillRad_K(V0\S˚₊k, g0_R0)≤FillRad_K(V0, g_R0 ).

The retraction (V0\S˚k

+, g 0

R0) → (V, g) is nonexpanding by the choice of

g_R0

0, therefore

FillRad_K(V, g)≤FillRad_K(V0\S˚₊k, g_R0 ₀)≤FillRad_K(V0, g0_R).

Note that we have actually proved that FillRad_K(V, g) = FillRad_K(V0, g0_R). Since Vol(V, g) = Vol(V0, g0_R), one gets FR_K(V)≤FR_K(V0). This finishes the proof of Proposition 3.11.

To finish this paragraph, we will proof the extension axiom for FV. Note that an extension of a _K-orientable simplicial complex is again_K-orientable and has the same fundamental class.

Lemma 3.12 (Extension axiom for FV). Let V0 be an extension of V. Then

FV_K(V0) = FV_K(V).

Proof. Since the inclusion i : V ,→ V0 is (n,1)-monotone and induces an isomorphism in n-dimensional homology, the inequality FV_K(V)≥FV_K(V0) holds by the comparison axiom.

Let g be a Riemannian metric on V, and extend it overV0 such that the inclusion i:V ,→V0 is isometric as a map of metric spaces (see the proof of Proposition 3.11). Call this Riemannian metricg0. Thenιg0◦i:V ,→L∞(V0)

is an isometric embedding and FillVol_K(V, g) = FillVol_K(ιg0 ◦i) by Corollary

3.10. Choose a Lipschitz chain z∈Cn(V;K) that represents [V]K. Then i∗z

is a Lipschitz chain that represents [V0]_K and

FillVol_K(V, g) = FillVol_K(ιg0_∗(i_∗z)) = FillVol

K(V

0_{, g}0_).

In document La transposición del concepto de aromaticidad en los libros de texto (página 34-38)