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The 3D index of an ideal triangulation and angle structures

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Abstract

The 3D index of Dimofte–Gaiotto–Gukov is a partially defined function on the set of ideal triangulations of 3-manifolds with r tori boundary components. For a fixed 2r tuple of integers, the index takes values in the set of q-series with integer coefficients. Our goal is to give an axiomatic definition of the tetrahedron index and a proof that the domain of the 3D index consists precisely of the set of ideal triangulations that support an index structure. The latter is a generalization of a strict angle structure. We also prove that the 3D index is invariant under 3–2 moves, but not in general under 2–3 moves.

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Notes

  1. The variables (me) are named after the magnetic and electric charges of [6].

  2. The sum of the 3 angles around each vertex is traditionally \(\pi \).

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Acknowledgments

The author wishes to thank Greg Blekherman, Nathan Dunfield, Christoph Koutschan, Henry Segerman, and Josephine Yu for enlightening conversations. The author wishes to especially thank Tudor Dimofte for explaining the 3D index, and for his generous sharing of his ideas. The work was initiated during a Clay Conference in Oxford, UK. The author wishes to thank the Clay Institute and Oxford University for their hospitality.

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Correspondence to Stavros Garoufalidis.

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With an appendix by Sander Zwegers

This study was supported in part by NSF Grant DMS-11-05678.

Appendices

Appendix A: \(I_{\Delta }\) satisfies the pentagon identity

There are several proofs of the key pentagon identity of the tetrahedron index \(I_{\Delta }\). The proofs may use an integral representation of the quantum dilogarithm, or q-holonomic recursion relations, or algebraic identities of generating series of q-series of Nahm type [10].

1.1 A generating series proof of the pentagon identity

In this section, we will prove that \(I_{\Delta }\) satisfies the pentagon identity using generating series. We will abbreviate the Pochhammer symbol

$$\begin{aligned} (x;q)_\infty =\prod _{n=0}^\infty (1-xq^n) \end{aligned}$$

by \((x)_\infty =(x;q)_\infty \). The proof

  • starts from an associativity identity

    $$\begin{aligned}&\frac{(z_1z_2)_\infty }{(z_1)_\infty (z_2)_\infty } \cdot \frac{(x_1z_1^{-1}q)_\infty (x_2z_2^{-1} q)_\infty }{ (x_1 x_2 z_1^{-1}z_2^{-1}q)_\infty }\\ \nonumber&\quad =\frac{(x_1z_1^{-1}q)_\infty }{(z_1)_\infty } \cdot \frac{(x_2z_2^{-1}q)_\infty }{(z_2)_\infty } \cdot \frac{(z_1z_2)_\infty }{(x_1 x_2 z_1^{-1}z_2^{-1}q)_\infty } \end{aligned}$$

    that uses four additional variables \(\{x_1,x_2,z_1,z_2\}\) in addition to the other four variables \(\{m_1,m_2,e_1,e_2\}\),

  • extracts coefficients with respect to \((z_1,z_2)\), and

  • specializes \((x_1,x_2)=(q^{-m_1},q^{-m_2})\). This last part is not algebraic and required to show convergence. The latter follows from Corollary 2.7.

Let us now give the details. Consider

$$\begin{aligned} F_e(x)=\sum _n (-1)^n \frac{q^{\frac{1}{2}n(n+1)} x^n}{(q)_n(q)_{n+e}} \in \mathbb Z[[x,q]]\, . \end{aligned}$$
(6.10)

Here and below, summation is over the set of integers, with the understanding that \(1/(q)_n=0\) for \(n < 0\).

We will show that

$$\begin{aligned} q^{e_1 e_2} F_{e_1}(q^{e_2} x_1)F_{e_2}(q^{e_1} x_2) =\sum _{e_3} (x_1 x_2 q)^{e_3} F_{e_1+e_3}(x_1) F_{e_2+e_3}(x_2) F_{e_3}(x_1 x_2)\nonumber \\ \end{aligned}$$
(6.11)

in the ring \(\mathbb Z((x_1,x_2,q))\). Since

$$\begin{aligned} F_e(q^{-m})=q^{\frac{e m}{2}}I_{\Delta }(m,e) \,, \end{aligned}$$

the substitution \((x_1,x_2)=(q^{-m_1},q^{-m_2})\) (which converges by Corollary 2.7) implies the pentagon identity of Eq. 3.6.

Lemma 6.4

For \(|q|<1\), we have

$$\begin{aligned} \frac{1}{(x)_\infty }= & {} \sum _n \frac{x^n}{(q)_n}, \qquad \qquad |x|<1 ,\\ (xq)_\infty= & {} \sum _n (-1)^n \frac{q^{\frac{1}{2}n(n+1)} x^n}{(q)_n}, \\ \frac{(xy)_\infty }{(x)_\infty }= & {} \sum _n \frac{(y)_n x^n}{(q)_n}, \qquad \qquad |x|<1, \\ \frac{(xy)_\infty }{(x)_\infty (y)_\infty }= & {} \sum _{r,s} \frac{q^{r s} x^r y^s}{(q)_r(q)_s },\qquad \qquad |x|<1,\,|y|<1, \\ \frac{(xq)_\infty (yq)_\infty }{(xyq)_\infty }= & {} \sum _{r,s} (-1)^{r+s} \frac{q^{\frac{1}{2}(r-s)^2+\frac{1}{2}(r+s)} x^r y^s}{(q)_r(q)_s }, \qquad \qquad |xyq|<1. \end{aligned}$$

Proof

The first three identities are well known and appear in [29, Prop. 2]. The last two follow from the first three:

$$\begin{aligned} \sum _{r,s} \frac{q^{r s} x^r y^s}{(q)_r(q)_s }= & {} \sum _r \frac{x^{r}}{(q)_r} \sum _s \frac{(q^r y)^s}{(q)_s }= \sum _r \frac{x^{r}}{(q)_r} \frac{1}{(q^ry)_\infty } \\= & {} \frac{1}{(y)_\infty } \sum _r \frac{(y)_r x^r}{(q)_r}= \frac{(xy)_\infty }{(x)_\infty (y)_\infty }\, ,\\ \sum _{r,s} (-1)^{r+s} \frac{q^{\frac{1}{2}(r-s)^2+\frac{1}{2}(r+s)} x^r y^s}{(q)_r(q)_s }= & {} \sum _r \frac{(-1)^r q^{\frac{1}{2}r^2+\frac{1}{2}r} x^r}{(q)_r} \sum _s \frac{(-1)^r q^{\frac{1}{2}s^2+\frac{1}{2}s} (q^{-r}y)^s}{(q)_s} \\= & {} \sum _r \frac{(-1)^r q^{\frac{1}{2}r^2+\frac{1}{2}r} x^r}{(q)_r} (q^{1-r}y)_\infty \\= & {} (yq)_\infty \sum _r \frac{(y^{-1})_r (xyq)^r}{(q)_r} = \frac{(xq)_\infty (yq)_\infty }{(xyq)_\infty } \, . \end{aligned}$$

\(\square \)

Remark 6.5

The identities of Lemma 6.4 also hold in the ring \(\mathbb Z((x,y,q))\).

Observe that \(F_e(x)\) is an analytic function of (xq) when \(|q|<1\) and \(x \in \mathbb C\). With \(|q|<1\) and \(|y|<1\), Lemma 6.4 gives

$$\begin{aligned} \sum _e F_e(x)y^e= & {} \sum _n \frac{(-1)^n q^{\frac{1}{2}n^2+\frac{1}{2}n} x^n}{(q)_n} \sum _e \frac{y^e}{(q)_{n+e}} \\= & {} \frac{1}{(y)_\infty } \sum _n \frac{(-1)^n q^{\frac{1}{2}n^2+\frac{1}{2}n} (xy^{-1})^n}{(q)_n} = \frac{(x y^{-1}q)_\infty }{(y)_\infty } \,. \end{aligned}$$

Thus, the generating function of the left-hand side of Eq. (6.11) is

$$\begin{aligned}&\sum _{e_1,e_2} q^{e_1 e_2} F_{e_1}(q^{e_2} x_1)F_{e_2}(q^{e_1} x_2) z_1^{e_1} z_2^{e_2} \\&\quad =\sum _{n_1,n_2} \frac{(-1)^{n_1+n_2}q^{\frac{1}{2}n_1^2+\frac{1}{2}n_2^2 + \frac{1}{2}n_1+ \frac{1}{2}n_2}x_1^{n_1} x_2^{n_2}}{(q)_{n_1}(q)_{n_2}} \sum _{e_1,e_2} \frac{q^{e_1 e_2 + n_2 e_1 + n_1 e_2}z_1^{e_1}z_2^{e_2} }{(q)_{n_1+e_1}(q)_{n_2+e_2}} \\&\quad =\frac{(z_1 z_2)_\infty }{(z_1)_\infty (z_2)_\infty } \sum _{n_1,n_2} \frac{(-1)^{n_1+n_2}q^{\frac{1}{2}(n_1-n_2)^2 + \frac{1}{2}n_1+ \frac{1}{2}n_2}(x_1 z_1^{-1})^{n_1} (x_2 z_2^{-1})^{n_2}}{(q)_{n_1}(q)_{n_2}} \\&\quad =\frac{(z_1z_2)_\infty }{(z_1)_\infty (z_2)_\infty } \cdot \frac{(x_1z_1^{-1}q)_\infty (x_2z_2^{-1} q)_\infty }{ (x_1 x_2 z_1^{-1}z_2^{-1}q)_\infty } \,. \end{aligned}$$

Likewise, the generating function of the right-hand side of Eq. (6.11) is the same

$$\begin{aligned}&\sum _{e_1,e_2} \left( \sum _{e_3} (x_1 x_2 q)^{e_3} F_{e_1+e_3}(x_1) F_{e_2+e_3}(x_2) F_{e_3}(x_1 x_2)\right) z_1^{e_1} z_2^{e_2} \\&\quad = \left( \sum _{e_1} F_{e_1}(x_1) z_1^{e_1}\right) \left( \sum _{e_2} F_{e_2}(x_2) z_2^{e_2}\right) \left( \sum _{e_3} F_{e_3}(x_1x_2) (x_1 x_2 z_1^{-1} z_2^{-1} q)^{e_3}\right) \\&\quad =\frac{(x_1z_1^{-1}q)_\infty }{(z_1)_\infty } \cdot \frac{(x_2z_2^{-1}q)_\infty }{(z_2)_\infty } \cdot \frac{(z_1z_2)_\infty }{(x_1 x_2 z_1^{-1}z_2^{-1}q)_\infty }\, . \end{aligned}$$

The above identities for each side of Eq. (6.11) hold when \(|q|<1\), \(|z_1|<1\), \(|z_2|<1\), and \(|x_1x_2z_1^{-1}z_2^{-1}q|<1\). Remark 6.5 implies that they also hold in the ring \(\mathbb Z((x_1,x_2,z_1,z_2,q))\). Extracting the coefficient of \(z_1^{e_1} z_2^{e_2}\) from the above concludes the proof of Eq. (6.11). \(\square \)

1.2 A second proof of the pentagon identity

In this section, we give a second proof of the pentagon identity using

$$\begin{aligned}&\frac{1}{(q)_m (q)_n} =\underset{s+t=n}{\underset{r+s=m}{\sum _{r,s,t}}} \frac{q^{rt}}{(q)_r(q)_s(q)_t}\, , \nonumber \\&\frac{q^{mn}}{(q)_m (q)_n} =\underset{s+t=n}{\underset{r+s=m}{\sum _{r,s,t}}} \frac{(-1)^s q^{\frac{1}{2}s^2-\frac{1}{2}s}}{(q)_r(q)_s(q)_t}= \sum _s \frac{(-1)^s q^{\frac{1}{2}s^2-\frac{1}{2}s}}{(q)_{m-s}(q)_{n-s}(q)_s}\, . \end{aligned}$$
(6.12)

The first identity is well known [29, Eqn. (13)], and the second follows from the first by replacing q with \(q^{-1}\) and multiplying both sides by \((-1)^{m+n} q^{-\frac{1}{2}(m-n)^2 -\frac{1}{2}(m+n)}\).

Using these equations, we will show here that

$$\begin{aligned}&q^{e_1e_2} \frac{(-1)^{n_1+n_2} q^{\frac{1}{2}n_1^2+\frac{1}{2}n_2^2+\frac{1}{2}n_1+\frac{1}{2}n_2+e_2n_1+e_1n_2}}{ (q)_{n_1} (q)_{n_2} (q)_{n_1+e_1} (q)_{n_2+e_2}} \nonumber \\&\quad = \underset{r_2+r_3+e_3=n_2}{\underset{r_1+r_3+e_3=n_1}{ \sum _{r_1,r_2,r_3,e_3}}} \frac{ (-1)^{r_1+r_2+r_3} q^{\frac{1}{2}r_1^2+\frac{1}{2}r_2^2 +\frac{1}{2}r_3^2 +\frac{1}{2}r_1+\frac{1}{2}r_2 +\frac{1}{2}r_3 +e_3}}{ (q)_{r_1} (q)_{r_1+e_1+e_3}(q)_{r_2}(q)_{r_2+e_2+e_3}(q)_{r_3}(q)_{r_3+e_3}}. \end{aligned}$$
(6.13)

The sum on the right actually only has a finite number of nonzero terms, so there is no issue with convergence. If we multiply both sides by \(x_1^{n_1} x_2^{n_2}\) and sum over all \(n_1\) and \(n_2\), then we again find

$$\begin{aligned} q^{e_1e_2} F_{e_1} (q^{e_2} x_1) F_{e_2} (q^{e_1}x_2) = \sum _{e_3} (x_1x_2q)^{e_3} F_{e_1+e_3}(x_1) F_{e_2+e_3}(x_2) F_{e_3}(x_1x_2)\, . \end{aligned}$$

To prove (6.13), we use Eq. (6.12) which gives

$$\begin{aligned}&\frac{q^{(n_1+e_1)(n_2+e_2)}}{(q)_{n_1} (q)_{n_2} (q)_{n_1+e_1} (q)_{n_2+e_2}}\\&\qquad = \underset{r_2+e_3=n_2}{\underset{r_1+e_3=n_1}{\sum _{r_1,r_2,r_3,e_3}}} \frac{(-1)^{r_3} q^{\frac{1}{2}r_3^2-\frac{1}{2}r_3 +r_1r_2}}{(q)_{r_1} (q)_{r_2} (q)_{n_1+e_1-r_3} (q)_{n_2+e_2-r_3} (q)_{r_3} (q)_{e_3}}\, . \end{aligned}$$

Replacing \(e_3\) by \(e_3+r_3\) in this sum, we get

$$\begin{aligned}&\frac{q^{(n_1+e_1)(n_2+e_2)}}{(q)_{n_1} (q)_{n_2} (q)_{n_1+e_1} (q)_{n_2+e_2}}\\&\quad = \underset{r_2+r_3+e_3=n_2}{\underset{r_1+r_3+e_3=n_1}{\sum _{r_1,r_2,r_3,e_3}}} \frac{(-1)^{r_3} q^{\frac{1}{2}r_3^2-\frac{1}{2}r_3 +r_1r_2}}{(q)_{r_1} (q)_{r_2} (q)_{n_1+e_1-r_3} (q)_{n_2+e_2-r_3} (q)_{r_3} (q)_{r_3+e_3}}\\&\quad = \underset{r_2+r_3+e_3=n_2}{\underset{r_1+r_3+e_3=n_1}{\sum _{r_1,r_2,r_3,e_3}}} \frac{(-1)^{r_3} q^{\frac{1}{2}r_3^2-\frac{1}{2}r_3 +r_1r_2}}{(q)_{r_1} (q)_{r_2} (q)_{r_1+e_1+e_3} (q)_{r_2+e_2+e_3} (q)_{r_3} (q)_{r_3+e_3}}\, . \end{aligned}$$

Now multiplying both sides by \((-1)^{n_1+n_2} q^{\frac{1}{2}(n_1-n_2)^2 +\frac{1}{2}n_1 +\frac{1}{2}n_2}\) gives Eq. (6.13).

Appendix B: The tetrahedron index and the quantum dilogarithm

Gukov–Gaiotto–Dimofte came up with the beautiful formula (1.2) for the tetrahedron index from a Fourier transform of the quantum dilogarithm. For completeness, we include this relation here, taken from [6]. The quantum dilogarithm of Faddeev and Kashaev is a fundamental building block of quantum topology [8, 13, 14]. The q-series version of this analytic function is given by

$$\begin{aligned} L(m,x,q)=\frac{(q^{-\frac{m}{2}+1}x^{-1})_\infty }{ (q^{-\frac{m}{2}}x)_\infty } \in \mathbb Z((x))[[q^{1/2}]]. \end{aligned}$$
(6.14)

We claim that

$$\begin{aligned} \sum _e I(m,e)(q) x^e = L(m,x,q) . \end{aligned}$$
(6.15)

To prove this, use the definition of I(me), shift e to \(e-n\), and use the first two identities of Lemma 6.4. We get

$$\begin{aligned} \sum _e I(m,e)(q) x^e&= \sum _{n,e} (-1)^n \frac{q^{\frac{1}{2}n(n+1) -\left( n+\frac{1}{2}e\right) m}x^e}{(q)_n(q)_{n+e}} \\&= \sum _{n,e} (-1)^n \frac{q^{\frac{1}{2}n(n+1)} \left( q^{-\frac{m}{2}} x^{-1}\right) ^n \left( q^{-\frac{m}{2}}x\right) ^e}{(q)_n(q)_{e}} \\&= \frac{(q^{-\frac{m}{2}+1}x^{-1})_\infty }{(q^{-\frac{m}{2}}x)_\infty } . \end{aligned}$$

Each of the recursion relations (3.1a), (3.1b), (3.3a), and (3.3b) is equivalent to the corresponding relations (6.16a6.16d) for the generating series L(mxq):

$$\begin{aligned} (-1+ q^{-\frac{m}{2}} x^{-1}) L(m,x,q) + L(m+1,q^{\frac{m}{2}} x,q) =0, \end{aligned}$$
(6.16a)
$$\begin{aligned} (1- q^{-\frac{m}{2}} x^{-1}) L(m,x,q) + L(m-1,q^{\frac{m}{2}} x,q) =0, \end{aligned}$$
(6.16b)
$$\begin{aligned} (1+x^2 -(q^{\frac{m}{2}}+q^{-\frac{m}{2}})) L(m,x,q) +x q^{\frac{m}{2}} L(m,q x,q) =0, \end{aligned}$$
(6.16c)
$$\begin{aligned}&L(m-2,x,q) -( L(m-1,q^{\frac{1}{2}} x,q)+L(m-1,q^{-\frac{1}{2}} x,q)\\&\quad -q^{1-m} L(m-1,q^{-\frac{1}{2}} x,q) ) + L(m,x,q) =0\nonumber \end{aligned}$$
(6.16d)

Equations (6.16a6.16d) are easy to verify using the fact that L(mqx) is a proper hypergeometric function of (mq). This gives an alternative proof of part (a) of Theorem 3.7.

Observe finally that the recursions (3.1a) and (3.1b) have a solution space of rank 2. On the other hand, the recursions (6.16a) and (6.16b) have a solution space of rank 1.

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Garoufalidis, S. The 3D index of an ideal triangulation and angle structures. Ramanujan J 40, 573–604 (2016). https://doi.org/10.1007/s11139-016-9771-7

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