156 CHAPTER 6. FIXED POINT THEOREMS

thenS1∩S2 =

[xk0 , · · · ,xkr

]where

[xk0 , · · · ,xkr

]is in the triangulation and{xk0 , · · · ,xkr

}={z1

0, · · · ,z1m}∩{z2

0, · · · ,z2p}

or else the two simplices do not intersect.

The following proposition is geometrically fairly clear. It will be used without commentwhenever needed in the following argument about triangulations.

Proposition 6.1.3 Say [x1, · · · ,xr] , [x̂1, · · · , x̂r] , [z1, · · · ,zr] are all r−1 simplices and

[x1, · · · ,xr] , [x̂1, · · · , x̂r]⊆ [z1, · · · ,zr]

and [z1, · · · ,zr,b] is an r+1 simplex and

[y1, · · · ,ys] = [x1, · · · ,xr]∩ [x̂1, · · · , x̂r] (6.1)

where{y1, · · · ,ys}= {x1, · · · ,xr}∩{x̂1, · · · , x̂r} (6.2)

Then[x1, · · · ,xr,b]∩ [x̂1, · · · , x̂r,b] = [y1, · · · ,ys,b] (6.3)

Proof: If you have ∑si=1 tiyi + ts+1b in the right side, the ti summing to 1 and nonneg-

ative, then it is obviously in both of the two simplices on the left because of 6.2. Thus[x1, · · · ,xr,b]∩ [x̂1, · · · , x̂r,b]⊇ [y1, · · · ,ys,b].

Now suppose xk = ∑rj=1 tk

jz j, x̂k = ∑rj=1 t̂k

jz j, as usual, the scalars adding to 1 andnonnegative.

Consider something in both of the simplices on the left in 6.3. Is it in the right? Theelement on the left is of the form

r

∑α=1

sαxα + sr+1b=r

∑α=1

ŝα x̂α + ŝr+1b

where the sα , are nonnegative and sum to one, similarly for ŝα . Thus

r

∑j=1

r

∑α=1

sα tαj z j + sr+1b=

r

∑α=1

r

∑j=1

ŝα t̂αj z j + ŝr+1b (6.4)

Now observe that

∑j∑α

sα tαj + sr+1 = ∑

α

∑j

sα tαj + sr+1 = ∑

α

sα + sr+1 = 1.

A similar observation holds for the right side of 6.4. By uniqueness of the coordinates inan r+1 simplex, and assumption that [z1, · · · ,zr,b] is an r+1 simplex, ŝr+1 = sr+1 and so

r

∑α=1

sα

1− sr+1xα =

r

∑α=1

ŝα

1− sr+1x̂α