Kenneth Kuttler

A.5. THE SUBDIVISION OPERATION 469

It is a point of [v0, · · · ,vn] because it is a convex combination of the vk. It is called thebarycenter because it is at the very center. Then for c an n simplex and b its barycenter, say∂̂c ≡ ∑

mi=1 aici,ai an integer ±1 and ci is of the form [v0, · · · ,vi−1,vi+1, · · · ,vn] an n− 1

simplex, then Ŝ (c)≡Cb

(Ŝ(

∂̂c))

where b is the barycenter of c. We extend Ŝ to makeit a homomorphism on the free group of geometric simplices.

Here is how the subdivision operator and boundary interact. Here is how the subdivisionoperator and boundary interact

Theorem A.5.4 ∂̂Ŝ (c) = Ŝ(

∂̂c).

Proof: This is clearly true for n = 0. In case n = 1, say c = [v0,v1]. Then the right isclearly [v1]− [v0] and the left is

∂̂

(CbŜ

(∂̂ [v0,v1]

))= ∂̂

(CbŜ ([v1]− [v0])

)= ∂̂ (Cb ([v1]− [v0]))

= ∂̂ ([b,v1]− [b,v0]) = [v1]− [v0]

Assume then that Ŝ is a homomorphism defined up to n− 1 for which the conclusionholds. Let c be a geometric n simplex. Then ∂̂Ŝ (c)≡ ∂̂CbŜ

(∂̂c). Say c = [v0, · · · ,vn]

so ∂̂c = ∑ni=0 (−1)i [v0, · · · ,vi−1,vi+1, · · · ,vn] , and so

Ŝ ∂̂c =n

∑i=0

(−1)i [bi,v0, · · · ,vi−1,vi+1, · · · ,vn]

where bi is the barycenter of some boundary simplex. Then when Cb is done to this lastexpression, the individual terms in the resulting sum are of the form

(−1)i [b,bi,v0, · · · ,vi−1,vi+1, · · · ,vn] .

When ∂̂ is done to sum of these, the terms not having b reduce to

∑i=0

(−1)i [bi,v0, · · · ,vi−1,vi+1, · · · ,vn]

which yields Ŝ(

∂̂c)

. For those which do have b, it reduces to Cb

(∂̂Ŝ

(∂̂c))

which, by

induction is the same as Cb

(Ŝ(

∂̂ ∂̂c))

= 0. ■

Note that Ŝ(

∂̂c)

is composed of n− 1 simplices which are contained in ∂̂c if c iscomposed of n simplices. Thus n−1 simplices on the “interior” disappeared.

What is accomplished by this subdivision applied to a simplex S? It yields a chainwhose union is S for which each simplex in the chain is small. If we do this subdivisionoperation enough, we can make the resulting simplices as small as desired. Here is why: Ifyou have an n simplex [x0, · · · ,xn] , its diameter is the maximum of |xk−xl | for all k ̸= l.Consider

∣∣b−x j∣∣ . It equals∣∣∣∣∣ n

∑i=0

1n+1

(xi−x j)

∣∣∣∣∣=∣∣∣∣∣∑i ̸= j

1n+1

(xi−x j)

∣∣∣∣∣≤ nn+1

diam(S) .

A.5. THE SUBDIVISION OPERATION 469It is a point of [vo,-+: ,Un| because it is a convex combination of the v,. It is called thebarycenter because it is at the very center. Then for c ann simplex and b its barycenter, sayOc = VE | aici, a; an integer +1 and ¢; is of the form [vo0,-++ ,Vi-1,Vit1,°** ,Un| ann—1—simplex, then SF (c) =Cp (7 (c)) where b is the barycenter of c. We extend F to makeit a homomorphism on the free group of geometric simplices.Here is how the subdivision operator and boundary interact. Here is how the subdivisionoperator and boundary interactTheorem A.5.4 OF (c) = F (de) .Proof: This is clearly true for n = 0. In case n = 1, say c = [vo, v1]. Then the right isclearly [v1] — [vo] and the left is3 (CoA (Ilvo.vi])) = 9 (Co (vi) [vo})) = 9 (Co ((e1] — feo)))= 9((b,v1] ~[b,v0]) = [v1] — volAssume then that -Y is a homomorphism defined up to n— 1 for which the conclusionholds. Let c be a geometric n simplex. Then 0.7 (c) = 0Cp.7 (dc) . Say c = [v0,°°+ , Un]s0 0c = ” 9 (—1)' wo, ,Vi-1,Vit1;°°* ; Un], and so—~A n .SOC = Y (=1)'[bi, v0, »Vji-1,Vi+1,°°° Un0Lwhere 0; is the barycenter of some boundary simplex. Then when Cy is done to this lastexpression, the individual terms in the resulting sum are of the form(—1)'[b,B;,¥,-* »Vi-1,Vi+1,°°° ,Un .When a is done to sum of these, the terms not having b reduce tonY (=1)' [bi v0," »Uj-1, Vi41,°°° Un]i=0which yields F | (dc). For those which do have b, it reduces to Cp (OF (dc) J which, byinduction is the same as Cy (7 (aac) ) =0. iNote that .% (dc is composed of n— 1 simplices which are contained in dc if c iscomposed of n simplices. Thus n — | simplices on the “interior” disappeared.What is accomplished by this subdivision applied to a simplex S? It yields a chainwhose union is S for which each simplex in the chain is small. If we do this subdivisionoperation enough, we can make the resulting simplices as small as desired. Here is why: Ifyou have an n simplex [%,--- , Z|, its diameter is the maximum of |a;, — a;| for all k #1.Consider |b — x || . It equals” i diam (S).