Kenneth Kuttler

6.1. SIMPLICES AND TRIANGULATIONS 157

where ∑αsα

1−sr+1= ∑α

ŝα

1−sr+1= 1, which would say that both sides are a single element of

[x1, · · · ,xr]∩ [x̂1, · · · , x̂r] = [y1, · · · ,ys] and this shows both are equal to something of theform ∑

si=1 tiyi,∑i ti = 1, ti ≥ 0. Therefore,

∑α=1

sα

1− sr+1xα =

∑i=1

tiyi,r

∑α=1

sαxα =s

∑i=1

(1− sr+1) tiyi

It follows thatr

∑α=1

sαxα + sr+1b=s

∑i=1

(1− sr+1) tiyi + sr+1b ∈ [y1, · · · ,ys,b]

which proves the other inclusion. ■Next I will explain why any simplex can be triangulated in such a way that all sub-

simplices have diameter less than ε .This is obvious if n ≤ 2. Supposing it to be true for n− 1, is it also so for n? The

barycenter b of a simplex [x0, · · · ,xn] is 11+n ∑ixi. This point is not in the convex hull of

any of the faces, those simplices of the form [x0, · · · , x̂k, · · · ,xn] where the hat indicatesxk has been left out. Thus, placing b in the kth position, [x0, · · · ,b, · · · ,xn] is a n simplexalso. First note that [x0, · · · , x̂k, · · · ,xn] is an n−1 simplex. To be sure [x0, · · · ,b, · · · ,xn]is an n simplex, we need to check that certain vectors are linearly independent. If

0 =k−1

∑j=1

c j (x j−x0)+ak

n+1

∑i=0xi−x0

∑j=k+1

d j (x j−x0)

then does it follow that ak = 0 = c j = d j?

0 =k−1

∑j=1

c j (x j−x0)+ak1

n+1

∑i=0

(xi−x0)

∑j=k+1

d j (x j−x0)

0 =k−1

∑j=1

(c j +

n+1

)(x j−x0)+ak

1n+1

(xk−x0)

∑j=k+1

(d j +

n+1

)(x j−x0)

Thus akn+1 = 0 and each c j +

akn+1 = 0 = d j +

akn+1 so each c j and d j are also 0. Thus, this is

also an n simplex.Actually, a little more is needed. Suppose [y0, · · · ,yn−1] is an n−1 simplex such that

[y0, · · · ,yn−1]⊆ [x0, · · · , x̂k, · · · ,xn] . Why is [y0, · · · ,yn−1,b] an n simplex? We know thevectors

{y j−y0

}n−1k=1 are independent and that y j = ∑i̸=k t j

i xi where ∑i̸=k t ji = 1 with each

being nonnegative. Suppose

n−1

∑j=1

c j(y j−y0

)+ cn (b−y0) = 0 (6.5)

If cn = 0, then by assumption, each c j = 0. The proof goes by assuming cn ̸= 0 and derivinga contradiction. Assume then that cn ̸= 0. Then you can divide by it and obtain modified

6.1. SIMPLICES AND TRIANGULATIONS 157where Y' ny =Ya ar = 1, which would say that both sides are a single element of[a1,---,@,-][@1,---,@,] = [y,,---,y,] and this shows both are equal to something of theform )F_) tiy;, Uti = 1,t; => 0. Therefore,r Ss r ad | Ga = ViVi Y, Sata = Yi (1 —srpi)tiy;i=l a=1 i=]=! 1—Sp41gIt follows thatY sata t5p.1b= y( 1 = sp41) try; +5410 € [yy,--+ .Ys55]a=1 i=]which proves the other inclusion.Next I will explain why any simplex can be triangulated in such a way that all sub-simplices have diameter less than €.This is obvious if n < 2. Supposing it to be true for n — 1, is it also so for n? Thebarycenter b of a simplex [a,--- ,&p] is ih y; x;. This point is not in the convex hull ofany of the faces, those simplices of the form [x,--- ,&x,+-: ,@,| where the hat indicatesax, has been left out. Thus, placing 6 in the k’” position, [a,--- ,b,--+ ,@n] is an simplexalso. First note that [ao,--- ,@,,--+ ,@n] is ann — 1 simplex. To be sure [x,--- ,b,--- , Xn]is ann simplex, we need to check that certain vectors are linearly independent. If0= Dole — xo j-eyta(, mi be on) + y dj(J=k+1then does it follow that a, =0 =c; =d;?0= Dole — 20) + Okk=1ak0 = _L( A \(e )+a Sy (we x0)+ ) (« +— Jee a)Jj esjok+l n+1Thus “4 = 0 and each cj +also an} in simplex.Actually, a little more is needed. Suppose [Yyo,--- , Y,—1] is ann — 1 simplex such that[Yo.'*+ >Yn—1] © [®0,°++ ,Be,-++ » Ln]. Why is [yo,---, Y,_1,6] an simplex? We know theah =0=d;+ “4 so each c; and d; are also 0. Thus, this isvectors {yj - yo}, are independent and that y ; = Vix t! x; where ))j-4x t! = | with eachbeing nonnegative. SupposeYalu — Yo) +n (b— Yo) =0 (6.5)If c, =0, then by assumption, each c; = 0. The proof goes by assuming c, 4 0 and derivinga contradiction. Assume then that c, 4 0. Then you can divide by it and obtain modified