The affine ${\tilde{A}}_2$ nerve: every edge exists, the Gram determinant vanishes, and the perimeter is exactly $2\pi$

Clause (i): for every two-element subset $T=\{s_i,s_j\}$ the cosine matrix is $C_T=\left(\begin{array}{cc}1& -\frac{1}{2}\\ -\frac{1}{2}& 1\end{array}\right)$ with leading principal minors $1>0$ and $1-\frac{1}{4}=\frac{3}{4}>0$, hence positive definite by [F4]; by the finiteness criterion [F3] the group $W_T$ is finite. Its restricted presentation is $⟨s_i,s_j\mid s_i^2=s_j^2=(s_i{s}_j{)}^3=1⟩$ by [F1]. Put $r=s_i{s}_j$; then $s_{i}r{s}_i=r^{-1}$, so every word is $r^k$ or $r^k{s}_i$ with $0\le k<3$. The homomorphism sending $s_i$ to $(12)$ and $s_j$ to $(23)$ in $\mathrm{Sym}(\{1,2,3\})$ has six distinct images of these words, so all six forms are distinct: $W_T$ is the dihedral group of order $6$ (Coxeter matrices, the presented Coxeter group, reduced words, length, and standard parabolic subgroups, The symmetric group $\mathrm{Sym}(X)$: the bijections of a set $X$ under composition, $\mathrm{Sym}(X)$ is a group under composition, and it is non-abelian whenever $X$ has at least three distinct elements). Thus $T$ is spherical, all three edges of $L$ exist and each has length $\pi -\pi \mathrm{/}3=\frac{2\pi }{3}$ [F1].

Clause (i), link: each row of $C_S$ sums to zero, so $(1,1,1)$ is a nonzero kernel vector and $C_S$ is not positive definite. Thus the vertex link has two vertices and no edge by [F1]; its two components are points, so their truncated angular distance is $\pi$. Algebraically the unnormalized Schur complement of $C_{\{s_i\}}$ is $\left(\begin{array}{cc}3\mathrm{/}4& -3\mathrm{/}4\\ -3\mathrm{/}4& 3\mathrm{/}4\end{array}\right)$; diagonal normalization gives $\left(\begin{array}{cc}1& -1\\ -1& 1\end{array}\right)$. The value $-1$ here corresponds to a nonedge, not to a spherical link-edge cell.

Clause (ii): the matrix $C_S$ has rows with diagonal $1$ and all off-diagonal entries $-\frac{1}{2}$, and $(1,1,1)$ is a nonzero kernel vector since each row sums to $1-1=0$; hence $C_S$ is not invertible, its determinant is $0$ by [F5], and it is not positive definite by [F4]; by the finiteness criterion [F3] the group $W$ is infinite. The vertex set $S$ is pairwise adjacent but not a simplex of $L$ (simplices have positive-definite cosine matrices [F1]), so by the metric flag condition the triple is not filled [F2]; since all three edges exist and no $2$-simplex does, $L$ is exactly the cycle formed by the three edges and their three vertices.

Clause (ii), metric: the cycle $L$ has three edges of length $\frac{2\pi }{3}$ joined at the three vertices; the path metric of that cycle is the circle metric of circumference $2\pi$, because the distance between two points is the minimum of the lengths of their two circular arcs and the total length is $3\cdot \frac{2\pi }{3}=2\pi$; hence $L$ is (isometric to) the round circle $S_{2\pi }^1$ and is an isometrically embedded circle of length $2\pi$ in itself [F6].

Clause (iii): step 3.1 exhibits an isometric copy of $S_{2\pi }^1$ in $L$, so $g(L)\le 2\pi$. If an isometric embedding $f:S_{\mathrm{\ell }}^1\to L$ existed for some $\mathrm{\ell }<2\pi$, the two semicircles between $0$ and $\mathrm{\ell }\mathrm{/}2$ in $S_{\mathrm{\ell }}^1$ would be distinct geodesic segments of length $\mathrm{\ell }\mathrm{/}2<\pi$, and their images would be distinct geodesics between $f(0)$ and $f(\mathrm{\ell }\mathrm{/}2)$. In the circle metric of circumference $2\pi$, the unique shorter arc is the only geodesic at distances $<\pi$, because the other circular arc is strictly longer. This contradiction rules out every embedded circle shorter than $2\pi$, hence $g(L)=2\pi$. The boundary 3-cycle has perimeter exactly $2\pi$, outside the strict comparison tests.