How to see transformations on polytopes?

Question

I have a polytope in six dimension with extreme points

$(1,0,0,0,0,0)$
$(0,1,0,0,0,0)$
$(0,0,1,0,0,0)$
$(1,1,0,1,0,0)$
$(1,0,1,0,1,0)$
$(0,1,1,0,0,1)$
$(1,1,1,1,1,1)$
$(0,0,0,0,0,0)$

Each of the above vector is interpreted as $(p_1,p_2,p_3,p_{12},p_{13},p_{23})$ where each $p_i$ is 0 or 1 and $p_{ij}$ is boolean AND of $p_i$ and $p_j$ that is $p_i$ ^ $p_j$. I apply the following transformation $\sigma_i$ defined as $$p_j=p_j, \; \; \; j \ne i$$ $$p_i=1-p_i $$ $$p_{ij} = p_j-p_{ij} $$ $$p_{jk}=p_{jk}, \; \; \; j,k \ne i$$ I can see that $\sigma_i$ ( $1 \le i \le 3$ ) maps all the eight extreme points in a one to one way amongst themselves. The point I am unable to prove is that if each point inside the polytope is mapped to some other point in the polytope itself. Also is there a transformation possible such that the extreme points are mapped amongst themselves but some of the points inside the polytope are mapped outside of the polytope ?

Answer 1

Any map preserving the extreme points of a convex polytope preserves the entire polytope. In particular, let $\operatorname{conv}(A)$ be the convex hull of a set of points $A$ - that is the set of points $p$ writable as convex sums: $$p=x_1a_1+x_2a_2+x_3a_3+\ldots+x_na_n$$ where $a_i\in A$ and $0\leq x_i \leq 1$ and $x_1+x_2+\ldots+ x_n= 1$. Then, for any linear map $f$ and set $A$, the following holds $$\operatorname{conv}(f(A))=f(\operatorname{conv}(A))$$ The proof of this is relatively simple when we note that $$f(x_1a_1+x_2a_2+\ldots +x_na_n)=x_1f(a_1)+x_2f(a_2)+\ldots+x_nf(a_n)$$ which allows us to show that any element of the right hand set is in the left hand set and vice versa by "translating" between the two forms.

In your case, you have a polyhedron with some extreme points $A$. Presumably the set of points in the polyhedron is $P=\operatorname{conv}(A)$. Additionally, you have a linear map for which $f(A)=A$. Then the above reads: $$\operatorname{conv}(f(A))=f(\operatorname{conv}(A))$$ and if we just simplify from the given relations we get $$\operatorname{conv}(A)=f(P)$$ $$P=f(P)$$ as desired.