Hyperbolic knot complements and volume — Examples and Counterexamples

The action preserves $\mathrm{\mid }z{\mathrm{\mid }}^2+\mathrm{\mid }w{\mathrm{\mid }}^2$, so it acts smoothly on $S^3$. The orbit of $(2^{-1\mathrm{/}2},2^{-1\mathrm{/}2})$ is exactly $K$. A group element stabilizes this point only if $3\theta$ and $2\theta$ are both multiples of $2\pi$; coprimality of $2$ and $3$ makes the stabilizer trivial. The regions $\mathrm{\mid }z\mathrm{\mid }\le \mathrm{\mid }w\mathrm{\mid }$ and $\mathrm{\mid }w\mathrm{\mid }\le \mathrm{\mid }z\mathrm{\mid }$ are solid tori, parametrized respectively by $(z,w)=(z,\sqrt{1-\mathrm{\mid }z{\mathrm{\mid }}^2}\eta )$ and $(z,w)=(\sqrt{1-\mathrm{\mid }w{\mathrm{\mid }}^2}\eta ,w)$ with the disk coordinate bounded by $2^{-1\mathrm{/}2}$ and $\eta \in S^1$. The $w$ coordinate is the core direction of the first solid torus and winds twice along $K$; the $z$ coordinate is the core direction of the second and winds three times. This proves the claimed $(3,2)$ torus-knot type. On the axis $z=0$ the stabilizer has order $2$, and on the axis $w=0$ it has order $3$.

On $zw\mathrm{\ne }0$, write $\alpha =\arg z$ and $\beta =\arg w$ modulo $2\pi$. The integer coordinate change $\theta =\alpha -\beta$, $\varphi =2\alpha -3\beta$ has determinant $-1$ and is invertible on the phase torus. The circle action changes $\theta$ by its parameter and fixes $\varphi$, while $r=\mathrm{\mid }z\mathrm{\mid }\in (0,1)$ is fixed; thus the quotient of this open set is the cylinder with coordinates $(r,\varphi )$, and the inverse image of any disk $D$ around $q(K)$ in it is explicitly $S_{\theta }^1\times D$. At $r=0$ and $r=1$ the phase circle collapses to one point at each end, so the full orbit space is a sphere; the two axis orbits have stabilizer orders $2$ and $3$. The complement of $D$ is a disk with those two cone points. Its orbits give the fibers of $E_K$, and [F2] identifies the axis fibers as exceptional of orders $2$ and $3$.

Choose a basepoint $x$ on a regular fiber, let $h$ be that fiber loop, and let $\gamma$ be any loop in $E_K$ based at $x$. The map $F:[0,1{]}^2\to E_K$ defined by $F(s,t)=g_{2\pi s}(\gamma (t))$ is well-defined because the tube is invariant. Its boundary represents $h\gamma h^{-1}{\gamma }^{-1}$: the two vertical sides are $\gamma$ and ${\gamma }^{-1}$ since $g_0=g_{2\pi }=\mathrm{id}$, and the horizontal sides are the fiber loop and its reverse. The square contracts its boundary, so $h$ commutes with $[\gamma ]$. Since $\gamma$ was arbitrary, the regular fiber class is central in ${\pi }_1(E_K)$.

By [F3], a Seifert-fibered exterior does not satisfy the required non-Seifert-fibered hypothesis. Step 2.1 establishes that $E_K$ is Seifert-fibered, so this example refutes the statement that every knot exterior is non-Seifert-fibered and also fails that hypothesis of the hyperbolization contract.

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.