Spherical Simplex Metrics, Angular Links, and Cones — Examples

The quadratic form. Since $C$ has diagonal $1$ and off-diagonal $c$, expanding gives $x^{T}Cx={\sum }_i{x}_i^2+2c{\sum }_{i<j}{x}_i{x}_j=(1-c)\parallel x{\parallel }^2+c({\sum }_i{x}_i{)}^2$; with $c=\frac{1}{2}$ both coefficients are positive, so $x^{T}Cx>0$ for $x\mathrm{\ne }0$ and $C$ is positive definite.

The Gram realisation. By step 1.1 and [F1] the Cholesky factorisation $C=L{L}^T$ exists with $L$ invertible, and [F3] gives that its rows $u_0,\dots ,u_3$ are unit vectors with $u_i\cdot u_j=c_{ij}$; thus $u_i\cdot u_j=\frac{1}{2}$ for $i\mathrm{\ne }j$, the pairwise angular distances are $\arccos \frac{1}{2}$, and every point of $\mathrm{\Sigma }(C)$ has unique barycentric ray coordinates.

The hemisphere functional. Let $w:=\frac{2}{5}(u_0+u_1+u_2+u_3)$ and $\phi (x):=⟨w,x⟩$. Since every row of $C$ sums to $1+3c=\frac{5}{2}$, for each fixed $j$ one has $⟨w,u_j⟩=\frac{2}{5}{\sum }_i⟨u_i,u_j⟩=\frac{2}{5}\cdot \frac{5}{2}=1$; hence $\phi (u_i)=1$ for all $i$, so by [F3] $\phi ({\sum }_i{\lambda }_i{u}_i)={\sum }_i{\lambda }_i$ and $\phi \equiv 1$ on $\mathrm{\Delta }(C)$. Therefore $\mathrm{\Sigma }(C)\subseteq \{x:\phi (x)>0\}$, the open hemisphere.

The vertex link. By [F3] and the Schur formula with $c=\frac{1}{2}$, the link of $u_0$ has Gram matrix $C^{\mathrm{l}\mathrm{k}}$ whose diagonal entries are $1$ and whose off-diagonal entries are $\frac{c-c^2}{1-c^2}=\frac{1\mathrm{/}2-1\mathrm{/}4}{3\mathrm{/}4}=\frac{1}{3}$. Its quadratic form is $y^T{C}^{\mathrm{l}\mathrm{k}}y=\frac{2}{3}\parallel y{\parallel }^2+\frac{1}{3}({\sum }_i{y}_i{)}^2$, which is positive for $y\mathrm{\ne }0$; so the link is the spherical triangle whose three angular distances are $\arccos \frac{1}{3}$.

The face link by projection. Let $G$ be the Gram matrix of $u_0,u_1$, so $G=\left(\begin{array}{cc}1& 1\mathrm{/}2\\ 1\mathrm{/}2& 1\end{array}\right)$ and $G^{-1}=\frac{4}{3}\left(\begin{array}{cc}1& -1\mathrm{/}2\\ -1\mathrm{/}2& 1\end{array}\right)$; the orthogonal projection onto $\mathrm{span}(u_0,u_1)$ is $Px={\sum }_{i,j}{g}^{ij}⟨x,u_j⟩u_i$ by [F2], so $P{u}_2=\frac{1}{3}(u_0+u_1)=P{u}_3$ and $\mathrm{\mid }P{u}_2{\mathrm{\mid }}^2=\frac{1}{3}$; hence $\mathrm{\mid }\mathrm{proj}{u}_2{\mathrm{\mid }}^2=\mathrm{\mid }\mathrm{proj}{u}_3{\mathrm{\mid }}^2=1-\frac{1}{3}=\frac{2}{3}$ and $⟨\mathrm{proj}{u}_2,\mathrm{proj}{u}_3⟩=⟨u_2,u_3⟩-⟨P{u}_2,P{u}_3⟩=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}$, so the projected cosine is $\frac{1\mathrm{/}6}{2\mathrm{/}3}=\frac{1}{4}$ and the link of the face is an arc of angular length $\arccos \frac{1}{4}$.

The two-step Schur value. Iterating the Schur formula of [F3] over the vertices $u_0,u_1$ means applying it first to the Gram matrix of the link of $u_0$, whose off-diagonal entries are $\frac{1}{3}$ by step 3.2, and then to a $2\times 2$ block; the resulting off-diagonal entry is $\frac{1\mathrm{/}3-(1\mathrm{/}3{)}^2}{1-(1\mathrm{/}3{)}^2}=\frac{2\mathrm{/}9}{8\mathrm{/}9}=\frac{1}{4}$, the same value as the projected cosine of step 3.3.

Order independence and conclusion. By [F3] the link of the face spanned by $u_0,u_1$ equals the Gram matrix of the normalised orthogonal projections of the remaining vertices onto $\mathrm{span}(u_0,u_1{)}^{\perp }$, so the projection computation of step 3.3 and the iterated Schur computation of step 4.1 are two descriptions of the same matrix; they agree at the value $\frac{1}{4}$, the face link is an arc of angular length $\arccos \frac{1}{4}$, and the positivity check is the one of step 3.2 applied to this $2\times 2$ block, whose determinant $1-\frac{1}{16}$ is positive.

Interior angle. Put $a=\pi \mathrm{/}(2m)$. At $v_0=(1,0)$, the addition and double-angle formulas give $v_1-v_0=2\sin a(-\sin a,\cos a)$ and $v_{2m-1}-v_0=2\sin a(-\sin a,-\cos a)$. Since $\sin a>0$, the unit incident edge directions are $u_{+}=(-\sin a,\cos a)$ and $u_{-}=(-\sin a,-\cos a)$; their inner product is ${\sin }^{2}a-{\cos }^{2}a=-\cos (2a)=\cos (\pi -\pi \mathrm{/}m)$. Their angular distance is therefore $\pi -\pi \mathrm{/}m$ by [F2]. The directions bound the inward wedge at the vertex, so this is the interior angle. Rotation by $j\pi \mathrm{/}m$ preserves inner products and carries the configuration to $v_j$, giving the same angle at every vertex.

Inward normals. At $v_0=(1,0)$ the edge $[v_0,v_1]$ has midpoint direction $(\cos \frac{\pi }{2m},\sin \frac{\pi }{2m})$ and the edge $[v_{2m-1},v_0]$ has midpoint direction $(\cos \frac{\pi }{2m},-\sin \frac{\pi }{2m})$; the outward unit normals are these midpoint directions, and the inward unit normals are their negatives $n_{+}=(-\cos \frac{\pi }{2m},-\sin \frac{\pi }{2m})$ and $n_{-}=(-\cos \frac{\pi }{2m},\sin \frac{\pi }{2m})$. Then $⟨n_{+},n_{-}⟩={\cos }^2\frac{\pi }{2m}-{\sin }^2\frac{\pi }{2m}=\cos \frac{\pi }{m}$ by [F3], so $d_{\mathrm{a}\mathrm{n}\mathrm{g}}(n_{+},n_{-})=\arccos (\cos \frac{\pi }{m})=\frac{\pi }{m}$, since $0<\frac{\pi }{m}\le \frac{\pi }{2}$ and $\cos$ is injective on $[0,\pi ]$.

The mirror lines. With $c:=\cos (\pi \mathrm{/}m)$ and $v_s:=c{e}_s+e_t$, $v_t:=e_s+c{e}_t$, evaluating the bilinear form [F4] gives $B(v_s,e_s)=c-c=0$ and $B(v_t,e_t)=-c+c=0$, so $\mathbb{R}{v}_s\subseteq H_s$ and $\mathbb{R}{v}_t\subseteq H_t$; since $B{\mathrm{\mid }}_P$ is positive definite by [F5] and $v_s,v_t\mathrm{\ne }0$, each of the subspaces $\{v\in P:B(v,e_s)=0\}$ and $\{v\in P:B(v,e_t)=0\}$ is a line, so these inclusions are equalities. Moreover $B(v_s,v_s)=c^2-2c^2+1=1-c^2$, $B(v_t,v_t)=1-c^2$ and $B(v_s,v_t)=2c-c^3-c=c(1-c^2)$, so the ratio of the Statement is $c(1-c^2)\mathrm{/}(1-c^2)=c$, the denominator being positive because $c=\cos (\pi \mathrm{/}m)<1$ for $m\ge 2$.

The link at a vertex. At $v_0=(1,0)$ the two facets through $v_0$ are the edges $[v_0,v_1]$ and $[v_{2m-1},v_0]$, with edge directions $u_{+}:=\frac{v_1-v_0}{\mid v_1-v_0\mid }$ and $u_{-}:=\frac{v_{2m-1}-v_0}{\mid v_{2m-1}-v_0\mid }$; the angle between $u_{+}$ and $u_{-}$ is the interior angle $\pi -\pi \mathrm{/}m$ of step 1.1. By [F1] the tangent cone is $\{\xi :⟨\xi ,n_{+}⟩\ge 0,⟨\xi ,n_{-}⟩\ge 0\}$ for the inward normals $n_{\pm }$ of the two edges. Its boundary lines are $\mathbb{R}{u}_{+}$ and $\mathbb{R}{u}_{-}$, since $u_{\pm }$ is the unit direction along the facet whose inward normal is $n_{\pm }$, so the cone is the intersection of the two closed half-planes bounded by these lines that contain $u_{+}$ and $u_{-}$; that intersection is exactly the wedge $\{a{u}_{+}+b{u}_{-}:a,b\ge 0\}$, whose unit section is the arc from $u_{+}$ to $u_{-}$. The angular distance of its endpoints is $\arccos ⟨u_{+},u_{-}⟩=\pi -\pi \mathrm{/}m$.

The mirror angle. The angle $\theta$ of the nonzero vectors $v_s,v_t$ in the positive definite plane $(P,B)$ is the number in $[0,\pi ]$ with $\cos \theta =B(v_s,v_t)\mathrm{/}(\mid v_s{\mid }_B\mid v_t{\mid }_B)$, where $\mid v{\mid }_B=\sqrt{B(v,v)}$; by step 1.3 this cosine is $c\in [0,1]$, so $\theta \in [0,\pi \mathrm{/}2]$ and $\theta =\arccos c=\pi \mathrm{/}m$ by [F2]. The mirror lines $H_s,H_t$ therefore meet at angle $\pi \mathrm{/}m$.

The product. Let $A=r_s{r}_t$. By [F5] $A$ is a $B$-preserving linear map of the positive definite plane $(P,B)$ with $\det A=1$ and $\mathrm{tr}A=2\cos (2\pi \mathrm{/}m)$. Choose a $B$-orthonormal basis $(u,w)$ of $P$ by [F6] and let $M=\left(\begin{array}{cc}p& q\\ r& s\end{array}\right)$ be the matrix of $A$ in it; $B$-preservation and $\det M=1$ give $M^{T}M=I$, so $M^{-1}=\left(\begin{array}{cc}s& -q\\ -r& p\end{array}\right)=M^T=\left(\begin{array}{cc}p& r\\ q& s\end{array}\right)$, whence $s=p$ and $r=-q$; thus $M=\left(\begin{array}{cc}p& q\\ -q& p\end{array}\right)$ with $p^2+q^2=1$ and $2p=\mathrm{tr}M=2\cos (2\pi \mathrm{/}m)$. Hence $p=\cos (2\pi \mathrm{/}m)$ and $q^2=1-p^2={\sin }^2(2\pi \mathrm{/}m)$ by [F6], so $q=\pm \sin (2\pi \mathrm{/}m)$ for $m\ge 3$, where $0<2\pi \mathrm{/}m<\pi$ makes $\sin (2\pi \mathrm{/}m)>0$, and $q=0$ for $m=2$. In every $B$-orthonormal basis, therefore, $A$ acts by the rotation of angle $2\pi \mathrm{/}m$ or of angle $-2\pi \mathrm{/}m$: the product $r_s{r}_t$ is a rotation of $(P,B)$ through $2\pi \mathrm{/}m$ up to the choice of orientation.

The link Gram matrix. By step 2.1 the vertex link is the arc with endpoints $u_{+},u_{-}$; its Gram matrix as a spherical $1$-simplex is the matrix with diagonal entries $1$ and off-diagonal entries $\cos (\pi -\pi \mathrm{/}m)$, and $\cos (\pi -\pi \mathrm{/}m)=-\cos (\pi \mathrm{/}m)=-c=B(e_s,e_t)$ by [F3] and [F4], so it equals the Gram matrix of $B{\mathrm{\mid }}_P$ displayed in the Statement. It is positive definite since $1-c^2>0$ and the diagonal entries are $1$. Hence for $m\ge 3$ the link edge angular distance $\pi -\pi \mathrm{/}m\in (\pi \mathrm{/}2,\pi )$ and the mirror angle $\pi \mathrm{/}m\in (0,\pi \mathrm{/}2]$ are distinct and supplementary, and they are equal to $\pi \mathrm{/}2$ only in the square case $m=2$; in either case they must not be interchanged.

The nerve and its path distance. Fix distinct $s,t\in S$. Send $s$ to the involution $a(u)=-u$, $t$ to $b(u)=2-u$, and every other generator to the identity in $\mathrm{Sym}(\mathbb{R})$. By [F9] this extends to a homomorphism from $W$. The product $ab$ is the translation $u\mapsto u-2$, whose $k$-th power sends $0$ to $-2k$; these images are distinct for distinct nonnegative integers $k$. Hence $W_{\{s,t\}}$ is infinite. Every $W_T$ with $s,t\in T$ contains it, and is infinite; a singleton generates at most the two elements $1,s$, so the nerve consists exactly of the empty simplex and the $n$ singletons. Every cell $\mathrm{\Sigma }(C_i)$ of its spherical realization $L$ is a point by [F1]. Any step of a chain in $L$ therefore has equal endpoints, and no chain joins distinct points. Thus [F2] gives $d_{\mathrm{p}\mathrm{a}\mathrm{t}\mathrm{h}}(x,x)=0$ and $d_{\mathrm{p}\mathrm{a}\mathrm{t}\mathrm{h}}(x,y)=+\mathrm{\infty }$ for $x\mathrm{\ne }y$.

The truncated metric. By [F3] and the convention $\min \{\pi ,\mathrm{\infty }\}=\pi$, one has $d_{\pi }(x,y)=\min \{\pi ,d_{\mathrm{p}\mathrm{a}\mathrm{t}\mathrm{h}}(x,y)\}=\pi$ for $x\mathrm{\ne }y$, while $d_{\pi }(x,x)=0$ by step 1.1. By [F4] the function $d_{\pi }$ is a metric of diameter at most $\pi$, and since $n\ge 2$ there is a pair $x\mathrm{\ne }y$ with $d_{\pi }(x,y)=\pi$, so the diameter is exactly $\pi$. Comparing values, $d_{\pi }$ agrees with $d_{\mathrm{p}\mathrm{a}\mathrm{t}\mathrm{h}}$ exactly for $x=y$ and nowhere off the diagonal, and by [F5] a metric takes real values while $+\mathrm{\infty }$ is not a real number, so $d_{\mathrm{p}\mathrm{a}\mathrm{t}\mathrm{h}}$ itself is not a metric on $L$ and the truncation is what produces one.

The cone distances. For $r,s>0$ the cone formula of [F3] and step 2.1 give, for $x=y$, $d_C((r,x),(s,x){)}^2=r^2+s^2-2rs\cos 0=(r-s{)}^2$, so that $d_C=\mid r-s\mid$, and, for $x\mathrm{\ne }y$, $d_C((r,x),(s,y){)}^2=r^2+s^2-2rs\cos \pi =(r+s{)}^2$ with $\cos \pi =-1$ by [F7], so that $d_C=r+s$; also $d_C(o,(r,x))=r$. Restricting to one ray, the map $t\mapsto (t,x)$ with $0\mapsto o$ therefore satisfies $d_C=\mid t-t^{\mathrm{\prime }}\mid$ on $[0,\mathrm{\infty })$, an isometry onto the ray $R_x:=\{o\}\cup \{(t,x):t>0\}$, and for $x\mathrm{\ne }y$ all pairs on different rays satisfy $d_C=t+t^{\mathrm{\prime }}$.