Euclidean simplices with the same facet-normal Gram matrix are similar facet to facet

(Complements of one normal span.) Fix $k\in I$ and put $W_k:=\mathrm{span}\{u_j:j\mathrm{\ne }k\}$. If $v\in W_k^{\perp }$, then $⟨u_j,v⟩=0$ for all $j\mathrm{\ne }k$. For $\phi \in \sigma$ and $\lambda \in \mathbb{R}$, the point $\phi +\lambda v$ satisfies those other inequalities for every $\lambda$; the $k$-th slack $⟨u_k,\phi +\lambda v⟩-c_k$ is affine in $\lambda$ and is nonnegative at $0$, so if $⟨u_k,v⟩\mathrm{\ne }0$ its admissible values contain an unbounded half-line, while if $⟨u_k,v⟩=0$ they are all of $\mathbb{R}$. This contradicts boundedness of $\sigma$, so $W_k^{\perp }=\{0\}$ [F5]. By [F6], every $x\in V$ is $w+z$ with $w\in W_k$ and $z\in W_k^{\perp }=\{0\}$; hence $W_k=V$. Thus every family obtained by omitting one $u_k$ spans $V$, and in particular all the $u_i$ span $V$. The same argument applied to ${\sigma }^{\mathrm{\prime }}$ shows that the $u_i^{\mathrm{\prime }}$ span $V^{\mathrm{\prime }}$.

(Vertices opposite facets.) Write $\sigma =\mathrm{conv}\{v_0,\dots ,v_n\}$ with affinely independent vertices [F2]. Every face $F$ of $\sigma$ is the convex hull of the vertices it contains: if $z={\sum }_k{\lambda }_k{v}_k\in F$ is a convex combination, a single positive coefficient gives $z=v_k\in F$; otherwise split off any term with $0<{\lambda }_k<1$ as $z={\lambda }_k{v}_k+(1-{\lambda }_k)y$. The face property [F4] puts $v_k,y$ in $F$, and induction on the number of positive coefficients puts every vertex used in $z$ in $F$. Thus $z\in \mathrm{conv}(F\cap \{v_0,\dots ,v_n\})$; the reverse inclusion follows from convexity [F3]. For the facet $F_i=\sigma \cap H_i$, let $S_i:=F_i\cap \{v_0,\dots ,v_n\}$. Then $F_i=\mathrm{conv}(S_i)$ and $S_i$ is nonempty; its points are affinely independent, so its affine span has dimension $\mathrm{\mid }{S}_i\mathrm{\mid }-1$ (subtracting one point gives $\mathrm{\mid }{S}_i\mathrm{\mid }-1$ linearly independent vectors spanning the direction space). Since this affine span is $H_i$, of dimension $n-1$, $\mathrm{\mid }{S}_i\mathrm{\mid }=n$. Hence there is a unique vertex $w_i$ outside $F_i$, and $⟨u_i,w_i⟩>c_i$. If $j\mathrm{\ne }i$ and $w_i\mathrm{\notin }{S}_j$, then $S_j\subseteq S_i$ and both have size $n$, so $S_j=S_i$ and $F_j=F_i$, contradicting that the indexed facets are distinct. Thus $w_i\in S_j$, so $⟨u_j,w_i⟩=c_j$ for every $j\mathrm{\ne }i$. Applying the same argument to ${\sigma }^{\mathrm{\prime }}$ gives opposite vertices $w_i^{\mathrm{\prime }}$ with $⟨u_j^{\mathrm{\prime }},w_i^{\mathrm{\prime }}⟩=c_j^{\mathrm{\prime }}$ for $j\mathrm{\ne }i$ and $⟨u_i^{\mathrm{\prime }},w_i^{\mathrm{\prime }}⟩>c_i^{\mathrm{\prime }}$.

(A nonzero relation valid for both normal families.) Fix $k\in I$. By step 1.1, $u_k={\sum }_{j\mathrm{\ne }k}{a}_j{u}_j$ for some real $a_j$; put $p_k:=1$ and $p_j:=-a_j$ for $j\mathrm{\ne }k$. Then $p\in {\mathbb{R}}^I$ is nonzero and ${\sum }_i{p}_i{u}_i=0$, so $p$ is a relation among the normals. The same vector is a relation among the primed normals: $\parallel {\sum }_i{p}_i{u}_i^{\mathrm{\prime }}{\parallel }^2={\sum }_{i,j}{p}_i{p}_j⟨u_i^{\mathrm{\prime }},u_j^{\mathrm{\prime }}⟩={\sum }_{i,j}{p}_i{p}_j⟨u_i,u_j⟩=\parallel {\sum }_i{p}_i{u}_i{\parallel }^2=0$, hence ${\sum }_i{p}_i{u}_i^{\mathrm{\prime }}=0$ by [F1].

(The linear isometry.) Define $T:V\to V^{\mathrm{\prime }}$ on a combination of the $u_i$ by $T\left({\sum }_i{a}_i{u}_i\right):={\sum }_i{a}_i{u}_i^{\mathrm{\prime }}$. This is well defined: if ${\sum }_i{a}_i{u}_i=0$ then $\parallel {\sum }_i{a}_i{u}_i^{\mathrm{\prime }}{\parallel }^2={\sum }_{i,j}{a}_i{a}_j⟨u_i^{\mathrm{\prime }},u_j^{\mathrm{\prime }}⟩={\sum }_{i,j}{a}_i{a}_j⟨u_i,u_j⟩=\parallel {\sum }_i{a}_i{u}_i{\parallel }^2=0$, so ${\sum }_i{a}_i{u}_i^{\mathrm{\prime }}=0$ by [F1]. By construction $T$ is linear in the sense of [F10], satisfies $T(u_i)=u_i^{\mathrm{\prime }}$, and preserves inner products: $⟨T\left({\sum }_i{a}_i{u}_i\right),T\left({\sum }_j{b}_j{u}_j\right)⟩={\sum }_{i,j}{a}_i{b}_j⟨u_i^{\mathrm{\prime }},u_j^{\mathrm{\prime }}⟩={\sum }_{i,j}{a}_i{b}_j⟨u_i,u_j⟩=⟨{\sum }_i{a}_i{u}_i,{\sum }_j{b}_j{u}_j⟩$; so $T$ is a linear isometry [F1, F7, F10]. Since the $u_i^{\mathrm{\prime }}$ span $V^{\mathrm{\prime }}$ [step 1.1], the image of $T$ is $V^{\mathrm{\prime }}$, and $T$ is bijective.

(Sign consistency and the two offset sums.) Put $R:={\sum }_i{p}_i{c}_i$ and $R^{\mathrm{\prime }}:={\sum }_i{p}_i{c}_i^{\mathrm{\prime }}$. For each $i$, using $⟨u_j,w_i⟩=c_j$ for $j\mathrm{\ne }i$ from step 1.2 and ${\sum }_k{p}_k{u}_k=0$ from step 2.1, $0=⟨{\sum }_k{p}_k{u}_k,w_i⟩=p_i⟨u_i,w_i⟩+{\sum }_{k\mathrm{\ne }i}{p}_k{c}_k=p_i(⟨u_i,w_i⟩-c_i)+R$. Since $⟨u_i,w_i⟩-c_i>0$ [step 1.2], the number $R$ is nonzero and $p_i$ has the sign of $-R$, for every $i$: all coefficients $p_i$ are nonzero and of one sign. Applying the same computation to the primed simplex, using the opposite vertices from step 1.2 and the primed relation from step 2.1, gives $p_i(⟨u_i^{\mathrm{\prime }},w_i^{\mathrm{\prime }}⟩-c_i^{\mathrm{\prime }})=-R^{\mathrm{\prime }}$ for every $i$, so $R^{\mathrm{\prime }}\mathrm{\ne }0$ and $p_i$ has the sign of $-R^{\mathrm{\prime }}$. Thus $R^{\mathrm{\prime }}$ and $R$ have the same sign, and $t:=R^{\mathrm{\prime }}\mathrm{/}R>0$.

(The translation solving the offset equations.) Put $\mathrm{\Delta }:=(c_i^{\mathrm{\prime }}-t{c}_i{)}_{i\in I}\in {\mathbb{R}}^I$. By step 3.1, ${\sum }_i{p}_i{\mathrm{\Delta }}_i=R^{\mathrm{\prime }}-tR=0$, i.e. $\mathrm{\Delta }\perp p$ [F5]. Let $A:V^{\mathrm{\prime }}\to {\mathbb{R}}^I$ be the linear map $A(y):=(⟨u_i^{\mathrm{\prime }},y⟩{)}_{i\in I}$. Its kernel is trivial: if $A(y)=0$ then $⟨u_i^{\mathrm{\prime }},y⟩=0$ for all $i$, so $y\perp \mathrm{span}\{u_i^{\mathrm{\prime }}\}=V^{\mathrm{\prime }}$ by step 1.1 and $y=0$ by [F1]. By [F8] (applied to the coordinate basis of ${\mathbb{R}}^I$, of $\mathrm{\mid }I\mathrm{\mid }=n+1$ elements) $\dim \mathrm{im}A=\dim V^{\mathrm{\prime }}=n$; the kernel of the nonzero functional ${\mathbb{R}}^I\to \mathbb{R}$, $x\mapsto {\sum }_i{p}_i{x}_i$, is $p^{\perp }$. It is nonzero because $p\mathrm{\ne }0$ [step 2.1], so its image is the one-dimensional space $\mathbb{R}$; [F8] therefore gives $\dim p^{\perp }=(n+1)-1=n$. Moreover $\mathrm{im}A\subseteq p^{\perp }$, because for $x=A(y)$ one has ${\sum }_i{p}_i{x}_i=⟨{\sum }_i{p}_i{u}_i^{\mathrm{\prime }},y⟩=0$ by step 2.1. As $\mathrm{im}A$ and $p^{\perp }$ are subspaces of $p^{\perp }$ of the same dimension $n$, [F9] gives $\mathrm{im}A=p^{\perp }$; since $\mathrm{\Delta }\perp p$, there is $y_0\in V^{\mathrm{\prime }}$ with $⟨u_i^{\mathrm{\prime }},y_0⟩=c_i^{\mathrm{\prime }}-t{c}_i$ for every $i\in I$.

(The similarity and the facet correspondence.) For $\phi \in V$ and $i\in I$, using $T(u_i)=u_i^{\mathrm{\prime }}$ and inner-product preservation of step 2.2 and the point $y_0$ of step 4.1, $⟨u_i^{\mathrm{\prime }},y_0+tT(\phi )⟩=(c_i^{\mathrm{\prime }}-t{c}_i)+t⟨u_i,\phi ⟩$. Hence $y_0+tT(\phi )\in {\sigma }^{\mathrm{\prime }}$ if and only if $⟨u_i,\phi ⟩\ge c_i$ for all $i$, that is, if and only if $\phi \in \sigma$; so $f(\phi ):=y_0+tT(\phi )$ maps $\sigma$ onto ${\sigma }^{\mathrm{\prime }}$, and $⟨u_i^{\mathrm{\prime }},f(\phi )⟩=c_i^{\mathrm{\prime }}$ if and only if $⟨u_i,\phi ⟩=c_i$, so $f$ carries the facet $F_i$ of $\sigma$ onto the facet $F_i^{\mathrm{\prime }}$ of ${\sigma }^{\mathrm{\prime }}$. Since $t>0$ and $T$ is a linear isometry, $f$ is a similarity.

(Conjugating the facet reflections.) For $i\in I$ define $s_i:V\to V$ by $s_i(\phi ):=\phi +2(c_i-⟨u_i,\phi ⟩)u_i$. Its linear part $r_i(\phi ):=\phi -2⟨u_i,\phi ⟩u_i$ preserves the norm: using $\mathrm{\parallel }{u}_i\mathrm{\parallel }=1$ and bilinearity, $\mathrm{\parallel }{r}_i(\phi ){\mathrm{\parallel }}^2=\mathrm{\parallel }\phi {\mathrm{\parallel }}^2-4⟨u_i,\phi {⟩}^2+4⟨u_i,\phi {⟩}^2\mathrm{\parallel }{u}_i{\mathrm{\parallel }}^2=\mathrm{\parallel }\phi {\mathrm{\parallel }}^2$. It fixes $u_i^{\perp }$ and sends $u_i$ to $-u_i$, so it is the reflection across $u_i^{\perp }$ and is a linear isometry [F11]. Since $s_i=r_i+2c_i{u}_i$, it preserves distances; direct substitution gives $s_i^2=\mathrm{i}\mathrm{d}$, and $s_i$ fixes $H_i$ pointwise. Thus $s_i$ is the affine reflection in $H_i$ and is an isometry [F12]. Let $s_i^{\mathrm{\prime }}$ be the corresponding reflection of $E^{\mathrm{\prime }}$ defined by the primed data. Then for every $\psi \in V^{\mathrm{\prime }}$, writing $f^{-1}(\psi )=T^{-1}((\psi -y_0)\mathrm{/}t)$ and using $⟨u_i,f^{-1}(\psi )⟩=(⟨u_i^{\mathrm{\prime }},\psi ⟩-c_i^{\mathrm{\prime }}+t{c}_i)\mathrm{/}t$, $f(s_i(f^{-1}(\psi )))=\psi +2t(c_i-⟨u_i,f^{-1}(\psi )⟩)u_i^{\mathrm{\prime }}=\psi +2(c_i^{\mathrm{\prime }}-⟨u_i^{\mathrm{\prime }},\psi ⟩)u_i^{\mathrm{\prime }}=s_i^{\mathrm{\prime }}(\psi )$. Thus $f\circ s_i\circ f^{-1}=s_i^{\mathrm{\prime }}$ for every $i$, and the map $C_f:g\mapsto f\circ g\circ f^{-1}$ carries the group generated by the $s_i$ onto the group generated by the $s_i^{\mathrm{\prime }}$. It preserves composition since $C_f(g\circ h)=C_f(g)\circ C_f(h)$, and its inverse is $h\mapsto f^{-1}\circ h\circ f$; hence it is an isomorphism of the two reflection groups.