On the volume conjecture for polyhedra

Abstract

We formulate a generalization of the volume conjecture for planar graphs. Denoting by $\langle \Gamma , c \rangle ^{\mathrm {U}}$ the Kauffman bracket of the graph $\Gamma$ whose edges are decorated by real “colors” c, the conjecture states that, under suitable conditions, certain evaluations of $\langle \Gamma ,\lfloor kc \rfloor \rangle ^{\mathrm {U}}$ grow exponentially as $k\rightarrow \infty$ and the growth rate is the volume of a truncated hyperbolic hyperideal polyhedron whose one-skeleton is $\Gamma$ (up to a local modification around all the vertices) and with dihedral angles given by c. We provide evidence for it, by deriving a system of recursions for the Kauffman brackets of planar graphs, generalizing the Gordon–Schulten recursion for the quantum 6j-symbols. Assuming that $\langle \Gamma ,\lfloor kc \rfloor \rangle ^{\mathrm {U}}$ does grow exponentially these recursions provide differential equations for the growth rate, which are indeed satisfied by the volume (the Schläfli equation); moreover, any small perturbation of the volume function that is still a solution to these equations, is a perturbation by an additive constant. In the appendix we also provide a proof outlined elsewhere of the conjecture for an infinite family of planar graphs including the tetrahedra.

Appendix 1: Proof of Theorem 1.4

1.1 Volumes of hyperideal hyperbolic tetrahedra

In Murakami and Yano [17] found a formula for the hyperbolic volume of a hyperbolic compact tetrahedron and the formula was later shown by Ushijima to hold also for truncated tetrahedra (i.e. the compact polyhedra obtained by truncating hyperideal tetrahedra as explained in Sect. 1.2). We now recall this formula.

With the notation of Example 2.2 for the dihedral angles of a tetrahedron, let A, B, C and $A',B',C'$ be respectively $-\exp (-\mathbf i\alpha ),-\exp (-\mathbf i\beta ),-\exp (-\mathbf i\gamma )$ and $-\exp (-\mathbf i\alpha ')$, $-\exp (-\mathbf i\beta ')$, $-\exp (-\mathbf i\gamma ')$.¹ Let $\mathrm {Li}_2(z):=\int _0^z \frac{\log (1-t)}{-t}\mathrm {d}t=\sum _{n>0}\frac{z^n}{n^2}$ be the dilogarithm function (well-defined on $\mathbb {C}\setminus [1,\infty )$) and $\Lambda (x):=\int _0^x -\log |2\sin t|\, \mathrm {d}t$. Define

$$\begin{aligned}&\displaystyle U(z):=\frac{1}{2}\big (\mathrm {Li}_2(z)+\mathrm {Li}_2(zABA'B')+\mathrm {Li}_2(zACA'C')+\mathrm {Li}_2(zBCB'C')\nonumber \\&\displaystyle \qquad \qquad \qquad \qquad \quad -\,\mathrm {Li}_2(-zABC)-\mathrm {Li}_2(-zAB'C')-\mathrm {Li}_2(-zA'BC')-\mathrm {Li}_2(-zA'B'C) \big )\nonumber \\ \end{aligned}$$

(14)

$$\begin{aligned}&\displaystyle \Delta (x,y,z) := \frac{-1}{4}\big (\mathrm {Li}_2\left( -\frac{xy}{z}\right) +\mathrm {Li}_2\left( -\frac{yz}{x}\right) +\mathrm {Li}_2\left( -\frac{zx}{y}\right) \nonumber \\&\displaystyle \qquad \qquad \qquad \qquad \qquad \quad +\,\mathrm {Li}_2\left( -\frac{1}{xyz}\right) +(\log (x))^2+(\log (y))^2+(\log (z))^2 \big )\\&\displaystyle V(z):=\Delta (A,B,C)+\Delta (A,B',C')+\Delta (A',B,C')+\Delta (A',B',C)\nonumber \\&\displaystyle \qquad \qquad \qquad +\,\frac{1}{2}(\log (A)\log (A')+\log (B)\log (B')+\log (C)\log (C'))+U(z).\nonumber \end{aligned}$$

(15)

For later purposes, recall also that for all $\theta \in (0,2\pi )$ one has $\mathfrak {I}(\frac{\mathrm {Li}_2( \exp {\mathbf i\theta })}{2})=\Lambda (\frac{\theta }{2})$, so replacing the values of $A,B,C,A',B',C'$ and setting $z=\exp (\mathbf is)$ in (14) we get

$$\begin{aligned}&\mathfrak {I}(U(\exp (\mathbf is)))=\Lambda \left( \frac{s}{2}\right) +\Lambda \left( \frac{s-(\alpha +\alpha '+\beta +\beta ')}{2}\right) \nonumber \\&\quad + \Lambda \left( \frac{s-(\alpha +\alpha '+\gamma +\gamma ')}{2}\right) +\Lambda \left( \frac{s-(\gamma +\gamma '+\beta +\beta ')}{2}\right) -\Lambda \left( \frac{s-(\alpha +\beta +\gamma )}{2}\right) \nonumber \\&\quad -\Lambda \left( \frac{s-(\alpha +\beta '+\gamma ')}{2}\right) -\Lambda \left( \frac{s-(\alpha '+\beta +\gamma ')}{2}\right) -\Lambda \left( \frac{s-(\alpha '+\beta '+\gamma )}{2}\right) . \end{aligned}$$

(16)

Let now $z_\pm$ be the two nontrivial solutions of the equation (which reduces to a degree 2 polynomial equation):

$$\begin{aligned} \frac{\mathrm {d}}{\mathrm {d}z} U(z)\in \frac{\pi \mathbf i}{z} \mathbb {Z}. \end{aligned}$$

As shown by Ushijima [22] these can be expressed as:

$$\begin{aligned} z_\pm = -2\frac{\sin (\alpha )\sin (\alpha ')+\sin (\beta )\sin (\beta ')+\sin (\gamma )\sin (\gamma ')\pm \sqrt{\det (G)}}{AA'+BB'+CC'+ABC'+A'BC+AB'C+A'B'C'+ABCA'B'C'} \end{aligned}$$

(17)

where

$$\begin{aligned} G=\begin{pmatrix} 1 &{} \cos (\alpha ) &{} \cos (\beta ) &{}\cos (\gamma ')\\ \cos (\alpha ) &{} 1 &{} \cos (\gamma ) &{}\cos (\beta ')\\ \cos (\beta ) &{} \cos (\gamma ) &{} 1 &{} \cos (\alpha ')\\ \cos (\gamma ') &{} \cos (\beta ') &{} \cos (\alpha ') &{}1 \end{pmatrix} \end{aligned}$$

(18)

Then the following was first proved by Murakami and Yano [17] for compact hyperbolic tetrahedra and was later shown by Ushijima [22] to hold for the case of a truncated hyperideal tetrahedron:

Theorem 5.5

[17, 22] The volume of the truncated hyperbolic tetrahedron whose exterior dihedral angles are as in Example 2.2, is

$$\begin{aligned} \mathrm {Vol}(Tet)=\mathfrak {I}\left( V(z_-)\right) =-\mathfrak {I}\left( V(z_+)\right) =\mathfrak {I}\left( \frac{U(z_-)-U(z_+)}{2}\right) . \end{aligned}$$

1.2 Proof of Theorem 1.4

We will need the following:

Lemma 5.6

Let $\alpha \in (0,1)$ and let $(a_n)_{n\in \mathbb {N}}$ be a sequence of integers such that $\lim _{n\rightarrow \infty } \frac{a_n}{n}=\alpha$. Then $\{a_n\}$ is meromorphic, for n sufficiently large it has no pole at $\exp \left( \frac{\mathbf i\pi }{2n}\right)$ and the following holds:

$$\begin{aligned} \lim _{n\rightarrow \infty } \frac{\pi }{n}\log \left( \frac{\mathrm{ev}_n(\{a_n\}!)}{\mathbf i^{a_n}}\right) =-\Lambda (\pi \alpha ) \end{aligned}$$

where we imply that the argument of the $\log$ is real positive.

Similarly, if $\alpha \in (1,2)$ and $(a_n)_{n\in \mathbb {N}}$ is a sequence of half-integers such that $\lim _{n\rightarrow \infty } \frac{a_n}{n}=\alpha$, then $\{a_n\}!$ has a simple zero at $\exp \left( \frac{\mathbf i\pi }{2n}\right)$ for n big enough and the following holds:

$$\begin{aligned} \lim _{n\rightarrow \infty } \frac{\pi }{n}\log \left( \frac{\mathrm{ev}_n(\{a_n\}!)}{(-1)^{n+a_n}2n^2 \mathbf i^{a_n+1}\exp \left( -\frac{\mathbf i\pi }{2n}\right) }\right) =-\Lambda (\pi \alpha ) \end{aligned}$$

where, again, we imply that the argument of the $\log$ is real positive.

Proof

We limit ourselves to a sketch, see [3] for a detailed proof. Clearly $\mathrm{ev}_n(fg)=\mathrm{ev}_n(f)\mathrm{ev}_n(g)$, and if k is not a multiple of n, $\mathrm{ev}_n(\{k\})=2\mathbf i\sin (\frac{\pi k}{n})$ while $\mathrm{ev}_n(\{n\})=-2n\exp (\frac{-\mathbf i\pi }{2n})$ and thus if $a_n\in [0,n)$ (which is true for the first statement),

$$\begin{aligned} \mathrm{ev}_n(\{a_n\}!)=\prod _{k=1}^{a_n} 2\mathbf i\sin \frac{\pi k}{n} \sim \mathbf i^{a_n}\exp \frac{-n \left( \Lambda \left( \frac{\pi a_n}{n} \right) -\Lambda \left( \frac{\pi }{n} \right) \right) }{\pi }; \end{aligned}$$

while if $a_n\in [n,2n)$,

$$\begin{aligned} \mathrm{ev}_n(\{a_n\}!)= & {} \left( \prod _{k=1}^{n-1} 2\mathbf i\sin \frac{\pi k}{n} \right) \left( -2n\exp \frac{-\mathbf i\pi }{2n} \right) \left( \prod _{k=n+1}^{a_n} 2\mathbf i\sin \frac{\pi k}{n} \right) \\= & {} \left( -2\mathbf i^{n-1}n^2\exp \frac{-\mathbf i\pi }{2n} \right) \prod _{k=n+1}^{a_n} -\mathbf i\left| 2 \sin \frac{\pi k}{n} \right| \sim \\&-\,(-\mathbf i)^{a_n+1}(-1)^n2n^2\exp \frac{-\mathbf i\pi }{2n} \exp \frac{-n \left( \Lambda \left( \frac{\pi a_n}{n}\right) -\Lambda \left( \pi + \frac{\pi }{n}\right) \right) }{\pi }. \end{aligned}$$

$\square$

We can now prove Theorem 1.4. For the sake of self-containedness we start by sketching the proof of the following proved in [3] for the skein normalization of the tetrahedron:

Theorem 5.7

Let $\Gamma$ be the 1-skeleton of a hyperideal tetrahedron Tet whose exterior dihedral angles are as in Example 2.2 and let $(c_n)_{n\in \mathbb {N}}$ be a sequence of colorings on $\Gamma$ such that $c_n\in [\frac{n}{2},n)$ and $\lim _{n\rightarrow \infty } 2\pi (1-\frac{c_n}{n})$ equals the corresponding exterior dihedral angles of Tet. Then,

$$\begin{aligned} \lim _{n\rightarrow \infty }\frac{\pi }{n}\log \big (| \mathrm{ev}_n \langle \Gamma , c_n\rangle ^{\mathrm{U}}|\big )=\mathrm {Vol}(Tet^{\mathrm {trunc}}). \end{aligned}$$

and Conjecture 1.3 holds in this case.

Proof

The idea of the proof is to first identify the leading term in (5), then to recognize it as the volume of the tetrahedron using Theorem 5.5. Despite the signs present in (5), there will be no cancellation. Let us start by computing the evaluation at $A=\exp \left( \frac{\mathbf i\pi }{2n}\right)$ of $\Delta (a,b,c)$ (as in (5)) using Lemma 5.6:

$$\begin{aligned}&\lim _{n\rightarrow \infty }\frac{\pi }{n}\log \left( |\mathrm{ev}_n(\sqrt{\Delta (a,b,c)})|\right) =\frac{1}{2}\left( \Lambda \left( \frac{\alpha +\beta -\gamma }{2}\right) \right. \\&\quad \left. +\,\Lambda \left( \frac{\alpha -\beta +\gamma }{2}\right) +\Lambda \left( \frac{-\alpha +\beta +\gamma }{2}\right) -\Lambda \left( \frac{\alpha +\beta +\gamma }{2}\right) \right) \end{aligned}$$

and similarly for $\Delta (a,e,f),\Delta (d,b,f),\Delta (d,e,c)$. Indeed for instance we have $\lim _{n\rightarrow \infty } \frac{a_n}{n}=1-\frac{\alpha }{2\pi }$ (and similarly for the other ratios) and so

$$\begin{aligned} \lim _{n\rightarrow \infty } \frac{\pi }{n} \log (\mathrm{ev}_n([a_n+b_c-c_n]!))=-\Lambda \left( \pi -\frac{\alpha +\beta -\gamma }{2}\right) =\Lambda \left( \frac{\alpha +\beta -\gamma }{2}\right) . \end{aligned}$$

Observe that the above formula equals the imaginary part of Eq. (15) (recall that for all $\theta \in (0,2\pi )$ one has $\mathfrak {I}(\frac{\mathrm {Li}_2( \exp {\mathbf i\theta })}{2})=\Lambda (\frac{\theta }{2})$).

Now let us concentrate on the summation in Formula (5). Let $S_k$ be the $k^{th}$ summand and remark that k ranges in $[\mathrm{max}\{T_i\}, \mathrm{min}\{Q_j\}]$. We claim that $S_k$ has a simple zero at $A=\exp (\frac{\mathbf i\pi }{2n})$ if $k\in [\mathrm{max}\{T_i\},n-2]$ and a zero of higher order otherwise, so only the summands $S_k$ with $k\in [\mathrm{max}\{T_i\},n-2]$ need be considered for the purpose of computing $\mathrm{ev}_n$.

Indeed by the inequalities of Example 3.1 translated in terms of $\gamma \sim 2\pi (1-\frac{c_n}{n})$, for n large enough, using the notation of Definition 3.1, one has $n<\max (\{T_i\})< 2n$; moreover by hypothesis $c(e_i)\in [\frac{n}{2},n)$ for every $e_i\in E(\Gamma )$ so $c(e_i)+c(e_j)-c(e_k)< n$ for every 3-uple of edges touching a common vertex (in whatever order). Since the summation index k ranges from $\max (\{T_i\})$ to $\min (\{Q_j\})$ the differences $k-T_i$ and $Q_j-k$ are bounded above by a term of the form $c(e_i)+c(e_j)-c(e_k)< n$ (for some triple of edges sharing a vertex) and hence the quantum factorials in the denominator of the quantum binomials forming the summands in Formula (5) have arguments $<n$ and are nonzero at $A=\exp \left( \frac{\mathbf i\pi }{2n}\right)$. By contrast, the numerator has a simple zero $\exp \left( \frac{\mathbf i\pi }{2n}\right)$ when k ranges in $[\max (\{T_i\}),n-2]$ and a double zero when k ranges in $[n-1,\min (\{Q_j\})]$. The claim is thus proved.

Our second claim is now that for all $k\in [\max (\{T_i\}),n-3)$, the ratio $\mathrm{ev}_n\big (\frac{S_{k+1}}{S_k}\big )$ is real positive. Indeed:

$$\begin{aligned} \frac{S_{k+1}}{S_k}=-\frac{\{k+1\}\left\{ Q^{(n)}_1-k\right\} \left\{ Q^{(n)}_2-k\right\} \left\{ Q^{(n)}_3-k\right\} }{\left\{ k+1-T^{(n)}_1\right\} \left\{ k+1-T^{(n)}_2\right\} \left\{ k+1-T^{(n)}_2\right\} \left\{ k+1-T^{(n)}_4\right\} } \end{aligned}$$

where we let $Q^{(n)}_j$ and $T^{(n)}_i$ be the “squares and triangles” associated to the coloring $c_n$ as in Definition 3.1. Since $\{k\}=2\mathbf i\sin (\frac{\pi k}{n})$, by the same estimates as in the previous claim the ratio is then positive real as $\{k+1\}$ is a negative multiple of $2\mathbf i$. So $\mathrm{max}\{|S_k|\}\le |\mathrm{ev}_n\big (\sum _k S_k)|\le (n-2-\mathrm{max}\{T_i\}) \mathrm{max}\{ |S_k|\}$, and we are left to find the terms $S_k$ for which $|S_k|$ is maximal.

Now remark that since the arguments of the factorials in the denominators of $S_k$ all belong to [0, n), their evaluations (by Lemma 5.6) grow respectively like $\mathbf i^{k-T_i}\exp \left( -\frac{n}{\pi }\Lambda \left( \pi \frac{k-T_i}{n}\right) \right)$ and $\mathbf i^{Q_j-k}\exp \left( -\frac{n}{\pi }\Lambda \left( \pi \frac{Q_j-k}{n}\right) \right)$. By contrast the numerator grows like

$$\begin{aligned} \mathbf i^{k+2}(-1)^{k+1+n}2n^2\exp \left( \frac{-\mathbf i\pi }{2n}\right) \exp \left( -\frac{\pi }{n}\Lambda \left( \pi \frac{k}{n}\right) \right) , \end{aligned}$$

so, taking into account the sign $(-1)^k$ in front of the multinomial, the $k^{th}$-summand grows like:

$$\begin{aligned} \frac{(-1)^n2n^2 \exp \left( -\frac{\mathbf i\pi }{2n}\right) }{\{1\}} \exp \left( \frac{n}{\pi }\left( -\Lambda \left( \pi \frac{k}{n}\right) +\sum _i\Lambda \left( \pi \frac{k-T_i}{n}\right) +\sum _j \Lambda \left( \pi \frac{Q_j-k}{n}\right) \right) \right) . \end{aligned}$$

Now let us define numbers $\tau _i, \nu _j$ by

$$\begin{aligned} \lim _{n\rightarrow \infty }\pi \frac{T^{(n)}_i}{n}= & {} 3\pi -\frac{\gamma (e)+\gamma (e')+\gamma (e'')}{2} =:3\pi -\frac{\tau _i}{2}\\ \lim _{n\rightarrow \infty }\pi \frac{Q^{(n)}_j}{n}= & {} 4\pi -\frac{\gamma (e)+\gamma (e')+\gamma (e'')+\gamma (e''')}{2} =:4\pi -\frac{\nu _j}{2} \end{aligned}$$

for the edges $e,e',e'',e'''$ naturally associated to $T_i$ or $Q_j$. Letting $\pi \frac{k}{n}=2\pi -\frac{s}{2}$ and

$$\begin{aligned} f(s)=-\Lambda \left( 2\pi -\frac{s}{2}\right) +\sum _i\Lambda \left( \frac{-s+\tau _i}{2}-\pi \right) -\sum _j \Lambda \left( -2\pi +\frac{\nu _j-s}{2}\right) \end{aligned}$$

and comparing with Eq. (16), we see that $f(s)=\mathfrak {I}\left( U\left( \exp \left( \mathbf is\right) \right) \right)$ and that the growth rate of the logarithm of the norm of the evaluation of the sum is given by:

$$\begin{aligned} \max _{s\in [0,\min (\{\tau _i-2\pi \})]} \frac{n}{\pi }f(s). \end{aligned}$$

We thus search for the maximum of f and imposing $f'(s)=0$ we get an equation of the form:

$$\begin{aligned} \frac{|\sin \left( 2\pi -\frac{s}{2}\right) \sin \left( \frac{\nu _1-s}{2}\right) \sin \left( \frac{\nu _2-s}{2}\right) \sin \left( \frac{\nu _3-s}{2}\right) |}{\left| \sin \left( \frac{\tau _1-s-2\pi }{2}\right) \sin \left( \frac{\tau _2-s-2\pi }{2}\right) \sin \left( \frac{\tau _3-s-2\pi }{2}\right) \sin \left( \frac{\tau _4-s-2\pi }{2}\right) \right| }=1 \end{aligned}$$

Recalling that $\nu _j-\tau _i>0$ for all i, j (see Example 3.1), the above equation is equivalent to:

$$\begin{aligned} \frac{\sin \left( 2\pi -\frac{s}{2}\right) \sin \left( \frac{\nu _1-s}{2}\right) \sin \left( \frac{\nu _2-s}{2}\right) \sin \left( \frac{\nu _3-s}{2}\right) }{\sin \left( \frac{\tau _1-s-2\pi }{2}\right) \sin \left( \frac{\tau _2-s-2\pi }{2}\right) \sin \left( \frac{\tau _3-s-2\pi }{2}\right) \sin \left( \frac{\tau _4-s-2\pi }{2}\right) }=1 \end{aligned}$$

Replacing $\sin (x)$ by $\frac{e^{\mathbf ix}-e^{-\mathbf ix}}{2\mathbf i}$ and setting $A:=-\exp \left( -\mathbf i\alpha \right)$ (and similarly for all the angles) and $z=-\exp (\mathbf is)$ one gets a polynomial equation of degree 2 in z whose solutions were shown by Murakami and Yano to be $z_+,z_-$ given in Formula (17). (Even if this is not apparent from the formula, as soon as the Gram matrix in Eq. (18) has negative determinant, then $|z_-|=|z_+|=1$.) In particular, in the case of a regular ideal octahedron (i.e. a maximally truncated tetrahedron), whose exterior dihedral angles are all $\pi$, $f(s)=4\Lambda \left( \frac{s}{2}\right) +4\Lambda \left( \frac{\pi }{2}-\frac{s}{2}\right) ,\ s\in [0,\pi ]$, whose extremum is attained at $s=\frac{\pi }{2}$ and its value is $8\Lambda \left( \frac{\pi }{4}\right) >0$. By Theorem 5.5 the corresponding point is then $z_-$ because $\mathrm {Vol}(Tet)=\mathfrak {I}\left( V\left( z_-\right) \right) =-\mathfrak {I}\left( V\left( z_+\right) \right) =\mathfrak {I}\left( \frac{U\left( z_-\right) -U\left( z_+\right) }{2}\right) >0$.

One concludes the proof by putting together all the terms in the asymptotical behavior of Formula (5) and comparing with Theorem 5.5.$\square$

We are now able to prove Theorem 1.4 in full generality. We do this by induction on the number n of moves of the form

producing $\Gamma$ from a tetrahedron. If $n=0$ then $\Gamma$ is the 1-skeleton of a hyperideal hyperbolic tetrahedron whose possible angles are provided by Example 3.1. For such a graph, Theorem 5.7 proves Theorem 1.4. If now $\Gamma$ can be obtained from $\Gamma '$ (for which we assume Theorem 1.4 holds) by a single move, suppose that $\alpha ,\beta ,\gamma$ are the exterior angles on the three edges of $\Gamma '$ involved in the move. Observe that any geometric structure of the hyperbolic polyhedron $P^{\mathrm {trunc}}_{\Gamma }$ can be obtained by gluing two polyhedra $P^{\mathrm {trunc}}_{\Gamma '}$ and $Tet^{\mathrm {trunc}}$, along the face corresponding to the triple of edges of $\Gamma '$ where the move is applied. Conversely if $P^{\mathrm {trunc}}_{\Gamma '}$ and $Tet^{\mathrm {trunc}}$ are equipped with geometric structures such that the dihedral angles along the truncation faces to be matched are the same, then they can be glued to form a hyperideal hyperbolic structure on $P^{\mathrm {trunc}}_{\Gamma }$. Clearly the volumes add up: $\mathrm {Vol}(P^{\mathrm {trunc}}_{\Gamma })=\mathrm {Vol}(P^{\mathrm {trunc}}_{\Gamma '})+\mathrm {Vol}(Tet^{\mathrm {trunc}})$.

On the quantum side, this follows from Eq. (8). For any coloring $c_n$ on $\Gamma$ one has

$$\begin{aligned} \langle \Gamma , c_n\rangle ^{\mathrm{U}}=\langle \Gamma ', c'_n\rangle ^{\mathrm{U}}\langle Tet, c''_n\rangle ^{\mathrm{U}} \end{aligned}$$

where $c'_n$ and $c''_n$ are the restrictions of $c_n$ to the edges of $\Gamma '$ and Tet respectively (here we use the natural injections $E(\Gamma ')\rightarrow E(\Gamma )$ and $E(Tet)\rightarrow E(\Gamma )$ to restrict the $c_n$). Then one concludes by induction using Theorem 5.7.$\square$