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Jul 9, 2002, 5:41:30 PM7/9/02

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Let S be the n-dimensional simplex, i.e. the set of (n+1)-tuples of

nonnegative real numbers (x_0,...,x_n) such that x_0+...+x_n=1. Let F

be the map sending an (n+1)-tuple (x_0,...,x_n) of nonnegative real

numbers, not all zero, to F(x_0,...,x_n) = (x_0/x,...,x_n/x), where

x=x_0+...+x_n, so that F(x_0,...,x_n) is in S.

nonnegative real numbers (x_0,...,x_n) such that x_0+...+x_n=1. Let F

be the map sending an (n+1)-tuple (x_0,...,x_n) of nonnegative real

numbers, not all zero, to F(x_0,...,x_n) = (x_0/x,...,x_n/x), where

x=x_0+...+x_n, so that F(x_0,...,x_n) is in S.

If (k_0,...,k_n) is an (n+1)-tuple of natural numbers, let

S(k_0,...,k_n) be the image by F of the cube (k_0+I)*...*(k_n+I)

(where I is the segment [0,1]) consisting of (n+1)-tuples of real

numbers of the form (k_0+t_0,...,k_n+t_n) where each t_i is between 0

and 1 (we possibly need to remove the vertex (0,...,0) from the cube

for the map F to be defined on it, namely when k_0=...=k_n=0; this is

unimportant). This defines a polytope S(k_0,...,k_n) in S.

Now fix k (a natural number) and consider all polytopes S(k_0,...,k_n)

with k_0+...+k_n=k. (There are binomial(k+n,n) such.) It is not too

hard to show that the S(k_0,...,k_n) form a division of S (in the

sense that their union is all of S and the intersection of two of them

is, at most, a face of dimension <n).

My question is: is it true that all S(k_0,...,k_n) (still with

k_0+...+k_n=k a constant) are of the same measure? Here, the measure

on S is the usual one: it is the unique probability measure on S which

is proportional to Lebesgue measure on the hyperplane of equation

x_0+...+x_n=1.

(Background: this division of the simplex corresponds to a classical

electoral system. Namely, if there are k seats in an assembly, n+1

political parties numbered 0 through n and each one receaves a share

x_i of the votes, with x_0+...+x_n=1, we want to distribute the seats

proportionally to the number of votes, as far as that can be arranged.

Then assign to the party i k_i seats in the assembly where k_0,...,k_n

are the (essentially unique) integers such that k_0+...+k_n=k and

(x_0,...,x_n) belongs to S(k_0,...,k_n).)

--

David A. Madore

(david....@ens.fr,

http://www.eleves.ens.fr:8080/home/madore/ )

Jul 12, 2002, 6:58:45 PM7/12/02

to

Yes, all the projections have the same volume. Here is a proof (written

for n=2, I think it will be easier to read -for another n, replace

obviously cubes by hypercubes, triangles by simplexes and so on). s will

denote the sum of the coordinates of a point (what you called x).

for n=2, I think it will be easier to read -for another n, replace

obviously cubes by hypercubes, triangles by simplexes and so on). s will

denote the sum of the coordinates of a point (what you called x).

First note that there is no real reason in the question (except the

background) to restrict oneself to cubes with integer vertices. Admit

vertices for your cube anywhere in (R+)^3 ; and note A(a,b,c) the area

of the projection of the cube of summits (a + epsilon_0, b + epsilon_1,

c + epsilon_2) (epsilon_i being 0 or 1).

A(a,b,c) can be computed in the following way : first note that F(cube)

is divided as the union of the three F(face) where face is a face of the

cube adjacent to (a,b,c). Now watch one of these faces, say the

horizontal one (fix the third variable x_2 to c) ; this can be in turn

be divided as the union of 2! triangles, one for each ordering of the

two variables x_0 and x_1 : the first triangle has summits (a,b,c)

(a+1,b,c) and (a+1,b+1,c) while the second triangle has summits (a,b,c)

(a,b+1,c) and (a+1,b+1,c). Watch the first of these triangles ; write

down the formulas for the image of each summit under F, note that the

area of the resulting triangle is proportional to the determinant of

these three points (by a factor \sqrt{3}\over 3!, with no guarantee),

write down this determinant, you'll see that this area is a rational

function of (a,b,c) whose denominator is s(s+1)(s+2). Now sum up all of

these on the (2+1)2! triangles ; now we know that A(a,b,c) is a rational

fraction with denominator s(s+1)(s+2).

Since we know by advance that the first factor is superfluous, we can

wonder whether we can chase it. You can do that thinking that the

projection of the cube is also the projection of the dark side of the

cube. So you can also prove that A(a,b,c) can also be written with

denominator (s+1)(s+2)(s+3), hence can be written with denominator

(s+1)(s+2) (but disappointingly this trick is not necessary to finish

the proof).

Now move the cube to infinity along the half-line leaving from zero

through its base vertex. Intuitively, the further you go, the more

linear F becomes ; or seen otherwise, this motion has the same result

than leaving the cube motionless while the pole of the perspective leave

to infinity on the other side, that is the perspective tends to a linear

projection (leaving haze around what happens to the image hyperplane) ;

alternately, you can also projectify by allowing the side of the cube to

move, and leave the base vertex of the cube and the pole motionless

while the side tends to zero. This informal argument is easy to shape ;

just compute s[F(a + epsilon_0, b + epsilon_1, c + epsilon_2) -

F(a,b,c)] when (a,b,c) tends to infinity along a half-line : you'll

check the limit position of this point exists (and is the linear

projection of (epsilon_0,epsilon_1,epsilon_2) on x_0+x_1+x_2 = 0 along

the direction of (a,b,c)). Henceforth s^2A(a,b,c) has a limit when

(a,b,c) tends to infinity along a fixed direction. Due to what we

already know about the shape of A's denominator, this proves the

numerator is at most of degree one -and even that it is constant if we

used the dark side argument.

Anyway, A(a,b,c) is obviously symmetric ; the degree of the numerator is

small enough to allow to conclude that A(a,b,c) is a function of s

alone. Since s is the number of seats in a Parliament, its value is

fixed and the images of all your cubes have the same area.

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