2
$\begingroup$

Say P is an n-dimensional polytope in $\mathbb{R}^n$ with vertex set V, and c is a point in $\mathbb{R}^{n+1}$ which is not in $\mathbb{R}^n$. Let W = { c - v | v $\in$ V}, and P' = conv(V $\cup$ W).

Symmetry considerations show that P' is centrally symmetric about c, because for a pair of antipodal vertices v and c - v, they're either both vertices of P' or neither are.

a) Is V $\cup$ W the vertex set of P'?

b) Does the preceding information (plus the combinatorial type of P) determine the combinatorial type of P'?

$\endgroup$
2
  • $\begingroup$ Things are easier to visualize when you assume $c=0$ and $P$ in the hyperplane $x_{n+1}=-1$. It's pretty obvious then that translating $P$ in its plane will not change the combinatorial structure of $P'$. $\endgroup$ Jul 11 '12 at 15:41
  • $\begingroup$ @ChristianBlatter, Agreed. Any thoughts about the vertex set of P'? $\endgroup$
    – Dan Moore
    Jul 11 '12 at 15:47
2
$\begingroup$

For a) we assume $c=0\in{\mathbb R}^{n+1}$ and $P$ in the hyperplane $x_{n+1}=-1$. Let $V=\{(v_i,-1)|1\leq i\leq N\}$ be the vertex set of $P$; then $W=\{(-v_i,1)|1\leq i\leq N\}$ is the vertex set of a polytope $\hat P$ symmetrically congruent to $P$ in the hyperplane $x_{n+1}=1$.

Any point $x\in \bar P:={\rm conv}(V\cup W)$ can be written in the form $$x=\sum_{i=1}^N \lambda_i (v_i,-1) +\sum_{i=1}^N \lambda'_i (- v_i,1)$$ with all $\lambda_i$, $\lambda'_i\geq0$ and $\sum_{i=1}^N (\lambda_i+\lambda'_i)=1$.

If $\lambda':=\sum_{i=1}^N\lambda'_i=0$ then $\lambda:=\sum_{i=1}^N\lambda_i=1$ and $$x=\sum_{i=1}^N\lambda_i (v_i,-1)\in P\ .$$ By assumption on $P$ a point $x\in P$ is not extremal in $P$ (whence in $\bar P$), unless $x$ is a point of $V$. Similarly: If $\lambda=0$ then $x\in \hat P$, and $x$ is not extremal in $\bar P$ unless $x\in W$. Finally, if $\lambda$ and $\lambda'$ are both $>0$ then $x$ is an inner point of a segment $[y,y']$ with $y\in P$,$\ y'\in\hat P$, and is not extremal in $\bar P$ either. Altogether we have shown that the vertex set $E$ of $\bar P$ is contained in $V\cup W$.

Conversely: Consider a point $v\in V$ and assume that it is an inner point of a segment $[a,b]$ with $a$, $b\in \bar P$. One has $$a=\sum_{i=1}^N \lambda_i (v_i,-1) +\sum_{i=1}^N \lambda'_i (- v_i,1)\ ,\qquad b=\sum_{i=1}^N \mu_i (v_i,-1) +\sum_{i=1}^N \mu'_i (- v_i,1)\ .$$ If a single $\lambda_i'$ or $\mu_i'$ would be $>0$ the last coordinate of $v$ could not be $-1$. Therefore $\lambda'=\mu'=0$ and $\lambda>0$, $\ \mu>0$; in particular $a$, $b\in P$. It follows that the point $v\in V$ would not be an extremal point of $P$, contrary to assumption on $P$. Argue similarly for a point $v'\in W$.

That is to say that the answer to your question a) is Yes.

For b) we assume $P$ in the hyperplane $\langle e_1,\ldots, e_n\rangle$ and choose $e_{n+1}:=c$. Then changing $c$ to $c':=\sum_{i=1}^n \alpha_i e_i + \sigma e_{n+1}$, $\ \sigma\ne0$, and constructing $\bar P$ anew amounts to an affine map $T:\ {\mathbb R}^{n+1}\to {\mathbb R}^{n+1}$ with $$T(e_i)=e_i \quad(1\leq i\leq n),\qquad T(e_{n+1})=c'\ .$$ From the way $\bar P$ is constructed it is obvious that $T(P)$ has the same combinatorial structure as $\bar P$.

$\endgroup$
2
  • $\begingroup$ excellent answer to part (a). The argument for part (b) should support rotations of P as well as translations (each rotation inducing an opposite rotation of the antipodal polytope), because in the original formulation, c can move around a lot. So, perhaps, going back to the original formulation for part (b), the argument could use the linear transformation that fixes P and moves c to an arbitrary point c' in $\mathbb{R}^{n+1}$ - $\mathbb{R}^n$ ? – $\endgroup$
    – Dan Moore
    Jul 12 '12 at 15:30
  • $\begingroup$ @Dan Moore: See my edit. $\endgroup$ Jul 13 '12 at 13:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.