Approximating the Turaev-Viro Invariant of Mapping Tori is Complete for One Clean Qubit

Abstract

The Turaev-Viro invariants are scalar topological invariants of three-dimensional manifolds. Here we show that the problem of estimating the Fibonacci version of the Turaev-Viro invariant of a mapping torus is a complete problem for the one clean qubit complexity class (DQC1). This complements a previous result showing that estimating the Turaev-Viro invariant for arbitrary manifolds presented as Heegaard splittings is a complete problem for the standard quantum computation model (BQP). We also discuss a beautiful analogy between these results and previously known results on the computational complexity of approximating the Jones Polynomial.

Appendices

Equivalence Between One Clean Qudit Models

Given a quantum circuit on $a$-dimensional qudits we wish to construct a quantum circuit on $b$-dimensional qudits that has the same trace. If $b=ca$ for some integer $c$ then this is easy. We just consider each $b$-dimensional qudit to be an $a$-dimensional qudit plus a $c$-dimensional “gauge” qudit that we ignore. Similarly, if $b^d = ca$ for some integers $d,c$ then we can treat $d$-tuples of $b$-dimensional qubits as an $a$-dimensional qudit plus a $c$-dimensional gauge qudit. For these encodings, the encoded circuit is easy to construct gate by gate. Given a gate acting on $n$ $a$-dimensional qudits, we can write down a unitary acting on $dn$ $b$-dimensional qudits equal to the original gate tensored with the $c$-dimensional identity on the gauge system. This $dn$-dimensional gate can be exactly decomposed into a product of $O(b^{2dn})$ 2-qudit gates using the standard construction from Section 4.5 of [21]. Because $d$ and $n$ are constants, this is sufficiently efficient. The normalized trace of the encoded circuit is exactly equal to the normalized trace of the original circuit.

The harder case is when there do not exist integers $c$ and $d$ such that $b^d = ca$. In this case we find $c,d \in {\mathbb {Z}}$ such that $b^d \simeq ca$. Specifically, suppose we achieve

$$\begin{aligned} \frac{ca}{b^d} = 1 - \delta \end{aligned}$$

(4)

for some $\delta \ll 1$. Then we can encode one $a$-dimensional qudit plus a $c$-dimensional gauge qudit into $d$ $b$-dimensional qudits with a few (namely $\delta b^d$) noncoding states left over. We can define our encoded gates to act as the identity on these noncoding states. If we make sure the noncoding states are a small fraction of all $b^{dn}$ states, the normalized trace of the encoded circuit will approximately match the normalized trace of the original circuit.

Let $U_a$ be the original unitary acting on $n$ $a$-dimensional qudits and let $U_b$ be the unitary acting on $dn$ $b$-dimensional qudits, in which we encode $U_a$ as described above. Then, $U_b$ acts on $b^{dn}$ states, of which $(ca)^n$ encode states of the original circuit,

$$ \frac{\mathrm {Tr}[U_b]}{b^{dn}} = \frac{c^n \mathrm {Tr}[U_a] + (b^{dn}-(ca)^n)}{b^{dn}}. $$

The magnitude of the discrepancy $\varDelta$ between the normalized traces of $U_b$ and $U_a$ is thus

$$\begin{aligned} \varDelta&= \left| \frac{c^n \mathrm {Tr}[U_a] + (b^{dn}-(ca)^n)}{b^{dn}} - \frac{\mathrm {Tr}[U_a]}{a^n} \right| \\&= \left| \left( \left( \frac{ca}{b^d} \right) ^n - 1 \right) \frac{\mathrm {Tr}[U_a]}{a^n} + 1 - \left( \frac{ca}{b^d} \right) ^n \right| \\&\le \left| \left( \frac{ca}{b^d} \right) ^n - 1 \right| \cdot \left| \frac{\mathrm {Tr}[U_a]}{a^n} \right| + \left| 1-\left( \frac{ca}{b^d} \right) ^n \right| \\&\le \left| \left( \frac{ca}{b^d} \right) ^n - 1 \right| + \left| 1 - \left( \frac{ca}{b^d} \right) ^n \right| \\&= 2 \left| (1 - \delta )^n - 1 \right| . \end{aligned}$$

Thus if

$$\begin{aligned} \delta = \frac{\epsilon }{n} \end{aligned}$$

(5)

we have, for small $\epsilon$,

$$\begin{aligned} \lim _{n \rightarrow \infty } \varDelta = 2 \left| e^{-\epsilon } -1 \right| \simeq 2 \epsilon . \end{aligned}$$

(6)

Comparing (4), (5), (6), we see that in the limit of large $n$ and small $\epsilon$, in order to achieve error upper bounded by $\varDelta$ it suffices to obtain

$$ \frac{b^d - ca}{b^d} \le \frac{\varDelta }{2n}. $$

For given $b,d,a$ there always exists an integer $c$ such that $b^d - c \le a$. So we just need to choose $d$ sufficiently large that

$$ \frac{a}{b^d} \le \frac{\varDelta }{2n}. $$

Equivalently,

$$ d \ge \log _b \left( \frac{2na}{\varDelta } \right) . $$

A $k$-qudit gate from $U_a$ thus gets encoded as a $dk$-qudit gate in $U_b$. This encoded gate acts on a $b^{dk}$-dimensional space. We have just shown that it suffices to choose $d = \log _b \left( \frac{2na}{\varDelta } \right)$. Thus the encoded $k$-qudit gate acts on a $\left( \frac{2na}{\varDelta } \right) ^k$-dimensional space. Using the construction from section 4.5 of [21], we can implement an arbitrary $D$-dimensional unitary exactly with $O(D^2)$ 2-qudit gates. Thus each $k$-qudit gate in $U_a$ gets encoded by $O \left( \left( \frac{2na}{\varDelta } \right) ^{2k} \right)$ elementary gates in $U_b$. By gate universality, we can assume $k \le 2$, so our encoding has an overhead quartic in $n$ and $1/\varDelta$. This is perhaps not very efficient, but is nevertheless polynomial, and thus suffices to prove the equivalence of DQC1 defined with qudits of any constant dimension.

Estimating the Absolute Trace is DQC1-Hard

In this section we slightly adapt the proof from [24] to show that estimating the absolute value of the trace of a quantum circuit to within $\pm 1/24$ is a DQC1-complete problem. Consider an arbitrary DQC1 computation. We start with the state $\left| {0}\right\rangle \left\langle {0}\right| \otimes \frac{1\!\!1}{2^n}$, apply an arbitrary quantum circuit $U$, and then measure the first qubit in the $\left| {0}\right\rangle ,\left| {1}\right\rangle$ basis. Changing the initial state of the pure qubit, or changing the measurement basis does not add generality, as these changes can be subsumed into $U$. The probability of measurement outcome $\left| {0}\right\rangle$ is

$$\begin{aligned} p_0 = \mathrm {Tr}\left[ (\left| {0}\right\rangle \left\langle {0}\right| \otimes 1\!\!1) U (\left| {0}\right\rangle \left\langle {0}\right| \otimes 1\!\!1/2^n)U^\dag \right] . \end{aligned}$$

(7)

Let $U'$ be the unitary implemented by the following quantum circuit on $n+2$ qubits.

Thus, $p_0 = 2 \frac{\mathrm {Tr}U'}{2^{n+2}}$, as one can see by writing out the trace as a sum over diagonal matrix elements in the computational basis. Because $p_0$ is real it is also true that $p_0 = 2 \frac{\left| \mathrm {Tr}U' \right| }{2^{n+2}}$. Hence estimating the absolute value of the normalized trace of quantum circuits to suffices to predict the outcome of any DQC1 experiment.

As is standard in the complexity theory of probabilistic computation, “yes” instances of DQC1 are defined to have acceptance probability 2/3 and “no” instances are defined to have acceptance probability 1/3. Thus, deciding DQC1 is equivalent to estimating the normalized trace of a quantum circuit to within $\pm 1/6$. The reduction here has a factor of four overhead in normalization, thus estimating the absolute trace to within $\pm 1/24$ is DQC1-complete.

Efficiently Computing Thresholds

Consider the standard spine of the genus-$g$ surface, numbered as in Fig. 5. Suppose edges 1 through $i$ have been labeled in a fusion-consistent manner with anyon types $s_1,\ldots ,s_i$. We wish to compute how many completions there are to this partial labelling. That is, we wish to compute the number of fusion-consistent strings of $3g-3$ labels, whose first $i$ labels are given by $s_1,\ldots ,s_i$.

Denote the horizontal edges of the standard spine from right to left by $e_1,e_2,\ldots$, as shown below.

Let $Z_b^{(k)}$ be the number of completions in which the rightmost labeled edge is $e_k$ and has label $b \in \{0,1\}$. One sees that $Z_0^{(1)} = 2$, and $Z_1^{(1)} = 1$, by the following enumeration of fusion-consistent diagrams.

Furthermore, we have the recurrence relations $Z_0^{(n+1)} = 2 Z_0^{(n)} + Z_1^{(n)}$ and $Z_1^{(n+1)} = 3 Z_1^{(n-1)} + Z_0^{(n-1)}$, by the following enumeration of fusion-consistent diagrams.

Solving these recurrence relations yields

$$ \left[ \begin{array}{l} Z_0^{(n)} \\ Z_1^{(n)} \end{array} \right] = \left[ \begin{array}{ll} 2 &{} 1 \\ 1 &{} 3 \end{array} \right] ^{n-1} \left[ \begin{array}{l} 2 \\ 1 \end{array} \right] . $$

The other two cases—completions starting with an upper curved edge, or a lower curved edge—can be solved similarly. The $n^\mathrm {th}$ power of a matrix may be computed using $O(\log n)$ operations, thus calculating the number of completions for any $i$ in $O(\log g)$ steps. The corresponding thresholds are then immediately obtained by taking ratios of these.

Inchworm

Suppose the spine-labeling contains a segment of the following form.

(8)

Here $a$ can be 1 or 0. We call this configuration the inchworm. We may regard the right instance of as its head, and the left instance as its tail. We next show a sequence of two reversible operations by which we can move the inchworm one handle rightward. In the first step the head moves one handle to the right, leaving the tail in place, and in the second step, the tail catches up, hence the name “inchworm.”

Examination of the above diagram shows that if the fusion rules are obeyed in the initial configuration, they are also obeyed in the intermediate and final configurations. Furthermore, both steps are reversible (i.e. information preserving). Thus, they may be written as permutation matrices acting on the space of allowed configurations, and are therefore unitary. The first unitary transformation can be implemented by local Dehn twists, because the zero in the tail of the inchworm implies density of the Fibonacci representation on the segment to the right of it. The second unitary transformation can be implemented by local Dehn twists because the zero in the head of the inchworm implies density on the segment to the left of it. (In both steps, we are applying density to the twice-punctured genus-4 surface with one puncture labeled zero. There are 75 labelings in which the other puncture is labeled one and 50 labelings in which the other puncture is labeled zero. Thus, the decoupling lemma of [1] implies density jointly on these two subspaces.) Repeating this process and its reverse, we may bring the inchworm to any location within the spine.

To use the inchworm construction, we need to ensure that a segment of the form (8) exists in the first place. We may do this by implementing a reversible operation on the leftmost six handles, so that if the configuration (8) is absent, the matrix is strictly off-diagonal, and does not contribute to the trace. Specifically, we consider the leftmost two handles to be an ancilla system, and the next four handles to be the starting location of the inchworm. If these four handles do not take the form (8) we cyclically permute the (five) basis states of the ancilla system. Because this is done on the leftmost six handles, the segment is only singly-punctured, and thus Theorem 6.2 of [14] implies density without braiding.

The noncontributing labelings decrease the normalized WRT by a constant factor, which correspondingly necessitates decreasing the precision parameter $\epsilon$ by the same factor. More precisely, in the Fibonacci model, there are 325 fusion-consistent labelings for the spine of the genus-four doubly-punctured surface. Among these, there are two inchworm configurations ($a=0$ and $a=1$). Compounding this $2/325$ normalization cost with the precision $\epsilon = 1/24$ obtained in Appendix B for DQC1-hardness of absolute trace, we find that estimating the normalized WRT invariant to within $\pm 1/3900$ is DQC1-hard.

As an aside, we note that the inchworm construction here is simpler than that in [25], in the following sense. The inchworm construction of [25] involved reversible operations on logarithmically large regions. Although the density theorems imply that arbitrary reversible operations can be implemented on these regions, they do not imply that the decomposition into local moves is efficient. Rather this had to be explicitly proven in Appendix D of [25]. In contrast the inchworm construction here involves reversible operations only on $O(1)$ handles, thus no question of efficiency arises.