Worked case: the figure-eight complement

The two tetrahedra have disjoint interiors in the face-glued triangulation, and their common faces have three-dimensional volume zero. Additivity of the Riemannian volume integral and [F2] therefore give total volume $2v_3=6\mathrm{\Lambda }(\pi \mathrm{/}3)$.

For $N=\mathrm{10,000}$ and $Q={10}^{30}$, let $a_k=\lfloor Q\mathrm{/}(3k+1{)}^2\rfloor$ and $b_k=\lfloor Q\mathrm{/}(3k+2{)}^2\rfloor$. Exact integer division gives $781302412526189990283977435641\le {\sum }_{k=0}^N(a_k-b_k-1)\le Q{\sum }_{k=0}^N((3k+1{)}^{-2}-(3k+2{)}^{-2})\le {\sum }_{k=0}^N(a_k-b_k+1)=781302412526189990283977455643$. Also $r=1732050807568877293527446341505<Q\sqrt{3}<r+1=1732050807568877293527446341506$, as verified by squaring both integer endpoints against $3Q^2$. The truncated volume is $V_N=\frac{3\sqrt{3}}{2}{\sum }_{k=0}^N((3k+1{)}^{-2}-(3k+2{)}^{-2})$, because [F3] gives $\mathrm{\Lambda }(\pi \mathrm{/}3)=\frac{\sqrt{3}}{4}$ times that series. The preceding integer bounds and $r<Q\sqrt{3}<r+1$ give $\frac{3r(781302412526189990283977435641)}{2Q^2}\le V_N\le \frac{3(r+1)(781302412526189990283977455643)}{2Q^2}$. Exact cross-multiplication places $V_N$ between $2.02988321185$ and $2.02988321187$. Its omitted positive tail is less than $1.445\times {10}^{-9}$ because $6\sqrt{3}\mathrm{/}(72N^2)=\sqrt{3}\mathrm{/}(12N^2)<1.445\times {10}^{-9}$. Thus the complete interval lies strictly between $2.029875$ and $2.029885$, giving the stated five-decimal rounding.

Steps 1.1–2.1 compute the volume of the explicit complete gluing metric. For any other complete finite-volume hyperbolic metric on the same exterior interior, the identity fundamental-group isomorphism satisfies [F4], which supplies a volume-preserving isometry. Therefore the computed value is independent of the chosen complete hyperbolic metric and is the asserted complete hyperbolic volume of the figure-eight complement.