Looking for a (nonlinear) map from $n$-dimensional cube to an $n$-dimensional simplex

Question

I am looking for a (nonlinear) map from $n$-dimensional cube to an $n$-dimensional simplex; to make it simple, assume the following figure which is showing a sample transformation for the case when $n=2$. (sorry for my bad drawing). The two circles in the rectangle, are set of points, which are mapped to the rectangle. Note that, points by this transformation maintain their relative position to each other, as shown in the figure.

http://tinypic.com?ref=30bcxzp

Any suggestion how to find such transformation?

Answer 1

If all you want is a homeomorphism this is pretty simple. I'll do the 2-dimensional case because it's easiest and the leg work for extending to higher dimensions can be annoying if not done carefully.

Let $X=\{(x,y)\in\mathbb{R}^2\:|\:x\in[0,1], y\in[0,1]\}$, let $X'=\{(x,y)\in\mathbb{R}^2\:|\:x\in[0,\frac{1}{2}], y\in[0,\frac{1}{2}]\}$, let $Y=\{(x,y)\in X\:|x+y\leq 1\}\:$. We want a homeomorphism $h\colon X\rightarrow Y$.

Let $g\colon X\rightarrow X'$ be given by $g(x,y)=(\frac{1}{2}x,\frac{1}{2}y)$. Let $f\colon X'\rightarrow Y$ be given by $$f(x,y)=\left\{\begin{array}{lr}(2x-y,y),&\mbox{ if }x\geq y\\ (x,2y-x),&\mbox{ if }x\leq y\end{array}\right.$$ and so both $g$ and $f$ are homeomorphisms and hence their composition $(f\circ g)=h\colon X\rightarrow Y$ is a homeomorphism.

Answer 2

For a point $P=(x_1,...,x_n)$ in the unit cube $0 \le x_i \le 1$, define $$s=x_1+\cdots +x_n, \ \ \ m=\max(x_1,...,x_n).$$ Then map $P$ to the point $$f(P)=(x_1 \cdot \frac{m}{s},...,x_n \cdot \frac{m}{s}).$$ This maps the unit cube onto the simplex $x_i \ge 0, \ \sum x_i \le 1.$

This map works by contracing each ray from the origin to $P$ by the right amount so that it ends up inside or on the boundary of the simplex. A few examples in 3-d:

$F(P)=F((1/3,1/3,1/3))=(1/9,1/9/1/9).$ Here $F(P)$ is strictly inside the simplex. If one follows the ray from the origin through $P$ one will hit the remote corner $P'=(1,1,1)$ of the cube, for which $f(P')=(1/3,1/3,1/3),$ a point exactly on the boundary of the simplex.

Or look at $F(P)=F((1/2,1/3,1/4))=(3/13,2/13,3/26)$ inside the simplex, while if we rescale $P$ until its largest coordinate becomes $1$ to obtain $P'=(1,2/3,1/2)$ (so that $P'$ lies on the boundary of the unit cube) we find that $f(P')=(6/13,4/13,3/13)$, again a point exactly on the boundary since its coordinates sum to 1.