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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'?

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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 $P'$. – Christian Blatter Jul 11 '12 at 15:41
@ChristianBlatter, Agreed. Any thoughts about the vertex set of P'? – Dan Moore Jul 11 '12 at 15:47
up vote 2 down vote accepted

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 $\bar P$.

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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 $\mathbb{R}^{n+1}$ - $\mathbb{R}^n$ ? – – Dan Moore Jul 12 '12 at 15:30
@Dan Moore: See my edit. – Christian Blatter Jul 13 '12 at 13:13

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