Gauge color codes: optimal transversal gates and gauge fixing in topological stabilizer codes

Stabilizer codes [20, 21] are a main object of study due to their balance of flexibility and simplicity. Given a system of n qubits, a stabilizer subgroup ${\mathcal{S}}\subseteq {\mathcal{P}}$, with $-{\bf{1}}\unicode{8713}{\mathcal{S}}$, defines a subspace, or code, of states $\psi$ with $s\psi =\psi$ for every $s\in {\mathcal{S}}$. A subsystem stabilizer code [22] is defined by giving in addition a gauge group ${\mathcal{G}}\subseteq {\mathcal{P}}$ such that ${\mathcal{S}}$ is the center of ${\mathcal{G}}$ up to phases. The subspace stabilized by ${\mathcal{S}}$ splits in two subsystems the gauge group generates the full algebra of operators on one of them and acts trivially on the other. Logical qubits (those to be protected) inhabit the later, gauge qubits the former. The elements of ${\mathcal{Z}}{\mathcal{G}}$, i.e. the group of Pauli operators that commute with the elements of ${\mathcal{G}}$, are called bare logical operators they only act on logical qubits. Elements of ${\mathcal{Z}}{\mathcal{S}}$ are their dressed counterpart they may act on gauge qubits. Both quotients ${\mathcal{Z}}{\mathcal{G}}/{\mathcal{S}}$ and ${\mathcal{Z}}{\mathcal{S}}/{\mathcal{G}}$ yield the Pauli group on logical qubits.

Let X_q, Z_q denote the Pauli operators X, Z acting on the qubit q, and similarly

$$\begin{eqnarray}&&{X}_{S}:= \displaystyle \prod _{q\in S}{X}_{q},\qquad {Z}_{S}:= \displaystyle \prod _{q\in S}{Z}_{q},\end{eqnarray} \tag{ 2 }$$

for a set S of qubits. Of interest here are stabilizer codes with (i) a CSS structure [23, 24] ${\mathcal{G}}$ has a generating set ${{\mathcal{G}}}_{0}$ such that each of its elements takes the form X_S or Z_S for some S, and (ii) a single encoded qubit with bare logical operators X_Q, Z_Q, where Q is the set of all physical qubits. For such codes the CNot gate is trivially transversal, and also the Hadamard gate if in addition the code is self-dual, i.e. ${X}_{S}\in {\mathcal{G}}$ if and only if ${Z}_{S}\in {\mathcal{G}}$.

More interesting is the gate ${R}_{n}=\left(\displaystyle \genfrac{}{}{0em}{}{1}{0}\displaystyle \genfrac{}{}{0em}{}{0}{{{\rm{e}}}^{2\pi {\rm{i}}/{2}^{n}}}\right)$. Let $| S|$ denote the cardinality of a set S. As shown in appendix A, ${R}_{n}$ is transversal if the set Q can be divided into two disjoint sets T and T', i.e. $Q=T\bigsqcup T^{\prime }$, such that for every collection of generators ${X}_{{S}_{1}},...,{X}_{{S}_{m}}\in {{\mathcal{G}}}_{0}$, where $1\leqslant m\leqslant n$,

$$\begin{eqnarray}&&\left|T\cap \bigcap _{i=1}^{m}{S}_{i}\right|\equiv \left|T^{\prime } \cap \bigcap _{i=1}^{m}{S}_{i}\right|\mathrm{mod}{2}^{n-m+1}.\end{eqnarray} \tag{ 3 }$$

The idea of the gauge fixing technique [16] is that by fixing some of the gauge degrees of freedom to certain values it might be possible to recover transversal gates that would otherwise be forbidden. A code ${{\mathcal{S}}}_{2},{{\mathcal{G}}}_{2}$ is a gauge fixed version of ${{\mathcal{S}}}_{1},{{\mathcal{G}}}_{1}$ if ${{\mathcal{S}}}_{2}$ extends ${{\mathcal{S}}}_{1}$ by fixing the values of some operators in ${{\mathcal{G}}}_{1}$, i.e.

$$\begin{eqnarray}&&{{\mathcal{S}}}_{1}\subseteq {{\mathcal{S}}}_{2}\subseteq {{\mathcal{G}}}_{1},\qquad {{\mathcal{G}}}_{2}\propto {\mathcal{Z}}{{\mathcal{S}}}_{2}\cap {{\mathcal{G}}}_{1}.\end{eqnarray} \tag{ 4 }$$

The relations (4) become most transparent by choosing canonical generators ${X}_{i},{Z}_{i}$ of the Pauli group, $i=1,...,n$ such that

$$\begin{eqnarray}&&{{\mathcal{S}}}_{1}=\langle {Z}_{1},...,{Z}_{r}\rangle ,\end{eqnarray} \tag{ 5 }$$

$$\begin{eqnarray}&&{{\mathcal{S}}}_{2}=\langle {Z}_{1},...,{Z}_{s}\rangle ,\end{eqnarray} \tag{ 6 }$$

$$\begin{eqnarray}&&{{\mathcal{G}}}_{1}\propto \langle {Z}_{1},...,{Z}_{r},{X}_{r+1},{Z}_{r+1}...,{X}_{t},{Z}_{t}\rangle ,\end{eqnarray} \tag{ 7 }$$

where $r\leqslant s\leqslant t\leqslant n$. Computing centralizers is a trivial task, e.g.

$$\begin{eqnarray}&&{\mathcal{Z}}{{\mathcal{G}}}_{1}\propto \langle {Z}_{1},...,{Z}_{r},{X}_{t+1},{Z}_{t+1}...,{X}_{n},{Z}_{n}\rangle .\end{eqnarray} \tag{ 8 }$$

In particular, as canonical generators of the logical Pauli group for the code ${{\mathcal{S}}}_{1},{{\mathcal{G}}}_{1}$ one can take the following representatives of ${\mathcal{Z}}{{\mathcal{G}}}_{1}/{{\mathcal{S}}}_{1}$

$$\begin{eqnarray}&&{X}_{t+1},{Z}_{t+1},...,{X}_{n},{Z}_{n}.\end{eqnarray} \tag{ 9 }$$

Now, according to (4) we should take

$$\begin{eqnarray}&&{{\mathcal{G}}}_{2}\propto \langle {Z}_{1},...,{Z}_{s},{X}_{s+1},{Z}_{s+1}...,{X}_{t},{Z}_{t}\rangle .\end{eqnarray} \tag{ 10 }$$

This is indeed a valid choice ${{\mathcal{S}}}_{2}$ is the center of ${{\mathcal{G}}}_{2}$ up to phases. Moreover, the operators (9) are clearly also canonical generators of the logical Pauli group for the code ${{\mathcal{S}}}_{2},{{\mathcal{G}}}_{2}$.

Since ${{\mathcal{S}}}_{1}\subseteq {{\mathcal{S}}}_{2}$ and the two codes share logical operators, encoded states of the second code can be regarded as being encoded states of the first. Conversely, given an encoded state of the first code it is possible to make it into an encoded state of the second code without affecting the logical operators. This amounts to fix the eigenvalues of ${Z}_{r+1},...,{Z}_{s}$ to +1, which can be done in two steps. First the ${Z}_{r+1},...,{Z}_{s}$ need to be measured, or equivalently any other generators of ${{\mathcal{S}}}_{2}/{{\mathcal{S}}}_{1}$. Then a product of a subset of the operators ${X}_{r+1},...,{X}_{s}$ is applied, or equivalently any other generators of ${{\mathcal{G}}}_{1}/{{\mathcal{G}}}_{2}$. In particular, X_i is applied if the measurement yields ${Z}_{i}=-1$. All the operations involved commute with the logical operators (9), and therefore are safe to perform.

An alternative characterization of (4) is possible in terms of logical operators. Suppose that the two codes share a set ${\mathcal{L}}$ of representatives of bare logical operators. Then (4) holds (up to irrelevant phases in ${{\mathcal{G}}}_{1}$) if and only if

$$\begin{eqnarray}&&{{\mathcal{S}}}_{1}\subseteq {{\mathcal{S}}}_{2}.\end{eqnarray} \tag{ 11 }$$

Alternatively, under the same assumption (11) holds (up to a choice of signs in the stabilizers and irrelevant phases in ${{\mathcal{G}}}_{1}$) if and only if

$$\begin{eqnarray}&&{{\mathcal{G}}}_{2}\subseteq {{\mathcal{G}}}_{1}.\end{eqnarray} \tag{ 12 }$$

To check the statements (11), (12), notice first that for any code ${\mathcal{S}},{\mathcal{G}}$ a set ${\mathcal{L}}$ of representatives of bare logical operators satisfies

$$\begin{eqnarray}&&{\mathcal{S}}\propto {\mathcal{Z}}{\mathcal{L}}{\mathcal{G}},\qquad {\mathcal{G}}\propto {\mathcal{Z}}{\mathcal{L}}{\mathcal{S}}.\end{eqnarray} \tag{ 13 }$$

The equivalence of (11) and (12) under the assumption that the two codes share such a set ${\mathcal{L}}$ follows from the basic properties (13). Indeed, (12) implies that

$$\begin{eqnarray}&&{{\mathcal{S}}}_{1}\propto {\mathcal{Z}}{\mathcal{L}}{{\mathcal{G}}}_{1}\subseteq {\mathcal{Z}}{\mathcal{L}}{{\mathcal{G}}}_{2}\propto {{\mathcal{S}}}_{2},\end{eqnarray} \tag{ 14 }$$

and conversely (11) implies

$$\begin{eqnarray}&&{{\mathcal{G}}}_{2}\propto {\mathcal{Z}}{\mathcal{L}}{{\mathcal{S}}}_{2}\subseteq {\mathcal{Z}}{\mathcal{L}}{{\mathcal{S}}}_{1}\propto {{\mathcal{G}}}_{1}.\end{eqnarray} \tag{ 15 }$$

According to the above discussion, if (4) is satisfied there exist such a shared set ${\mathcal{L}}$ and (11) holds. Conversely, if ${\mathcal{L}}$ is shared and (11) holds then up to phases

$$\begin{eqnarray}&&{{\mathcal{S}}}_{2}\propto {{\mathcal{G}}}_{2}\cap {\mathcal{Z}}{{\mathcal{G}}}_{2}\subseteq {{\mathcal{G}}}_{2}\subseteq {{\mathcal{G}}}_{1},\end{eqnarray} \tag{ 16 }$$

which used (12) and completes the first part of (4), and

$$\begin{eqnarray}&&{\mathcal{Z}}{{\mathcal{S}}}_{2}\cap {{\mathcal{G}}}_{1}\propto {\mathcal{Z}}{{\mathcal{S}}}_{2}\cap {\mathcal{Z}}{\mathcal{L}}{{\mathcal{S}}}_{1}={\mathcal{Z}}{\mathcal{L}}{{\mathcal{S}}}_{1}{{\mathcal{S}}}_{2}={\mathcal{Z}}{\mathcal{L}}{{\mathcal{S}}}_{2}\propto {{\mathcal{G}}}_{2},\end{eqnarray} \tag{ 17 }$$

which is the second part of (4).

Color codes require lattices called colexes [12], but here the focus is shifted to their dual lattices. For those readers already familiar with colexes, the correspondence is clarified in appendix B.

For simplicity we focus on the 2D case. The key to the construction are triangles (2-simplices) with the vertices (0-simplices) labeled with 3 given colors (thus each side or 1-simplex is labeled by 2 colors). We are interested in simplicial lattices (triangulations) with the overall shape of such a colored triangle M. The lattices must have 3-colored 0-simplices such that (i) each 2-simplex is properly colored and (ii) the sides of M have the same coloring as the 1-simplices that lie on them. In figure 1 the first examples of an infinite sequence of such lattices is given: they are obtained by cutting triangles of different sizes out of the same infinite triangular lattice.

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From every such triangulation one can get a 3-colored triangulation of a sphere just by adding 3 extra 0-simplices as shown in figure 2: there is one new 2-simplex for each 1-simplex on the boundary of M, one for each of the vertices of M, and one formed by the 3 new 0-simplices. It is out of this triangulation of the sphere that a color code can be built, in particular via certain sets ${\Delta }_{d}$: ${\Delta }_{d}$ contains all d-simplices in the new triangulation except those that do not contain any of the original 0-simplices of M. Notice in particular that the elements of ${\Delta }_{2}$ correspond one to one to the following elements of M: all 2-simplices, 1-simplices lying on the sides of M, and 0-simplices that are also vertices of M.

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The general D-dimensional case is entirely analogous, with D-simplices instead of triangles and $D+1$ colors instead of 3. In particular, every d-simplex is labeled by $d+1$ colors. For a formal exposition see appendix B.

The 3D case is the most important, so here is a construction where a tetrahedron is carved out of a suitably colored BCC lattice. Vertices are placed at points ${\bf{x}}$ or ${\bf{x}}+{{\bf{l}}}_{0}$, where ${\bf{x}}$ has integer coordinates and ${{\bf{l}}}_{0}:= (\frac{1}{2},\frac{1}{2},\frac{1}{2})$. Red, green, blue and yellow vertices have positions ${\bf{x}}$ such that ${\bf{x}}\cdot {{\bf{l}}}_{0}$ is 0, $\frac{1}{2}$, $\frac{3}{4}$ and $\frac{1}{4}$ modulo 1, respectively. Any 3-simplex has vertices of the form

$$\begin{eqnarray}&&{\bf{x}},\;{\bf{x}}+{\bf{a}},\;{\bf{x}}+\displaystyle \frac{1}{2}({\bf{a}}+s{\bf{b}}+{\bf{c}}),\;{\bf{x}}+\displaystyle \frac{1}{2}({\bf{a}}+s{\bf{b}}-{\bf{c}}),\end{eqnarray} \tag{ 18 }$$

where $s=\pm 1$ and the triad $({\bf{a}},{\bf{b}},{\bf{c}})$ is a permutation of the triad $({\bf{i}},{\bf{j}},{\bf{k}})$ of canonical vectors. Given a positive integer n, retain those simplices with vertices ${\bf{x}}$ such that

$$\begin{eqnarray}&&{\bf{x}}\cdot {{\bf{l}}}_{k}\leqslant \displaystyle \frac{k}{4}+(n-1){\delta }_{k0},\qquad k=0,1,2,3,\end{eqnarray} \tag{ 19 }$$

where ${{\bf{l}}}_{1}:= (\frac{1}{2},-\frac{1}{2},-\frac{1}{2})$, ${{\bf{l}}}_{2}:= (-\frac{1}{2},\frac{1}{2},-\frac{1}{2})$, ${{\bf{l}}}_{3}:= (-\frac{1}{2},-\frac{1}{2},\frac{1}{2})$. This gives a tetrahedron: each constraint produces a face with coloring dictated by k, see figure 3.

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Consider sets of d-simplices ${\Delta }_{d}$ from a D-dimensional construction as above, $D\geqslant 2$. Attach a physical qubit to each D-simplex in ${\Delta }_{D}$, also denoted Q, and define the groups of operators

$$\begin{eqnarray}&&C(d,e):= \langle \{{X}_{{S}_{\delta }}\;| \;\}\delta \in {\Delta }_{d-1}\cup \{{Z}_{{S}_{\delta }}\;| \;\}\delta \in {\Delta }_{e-1}\rangle ,\end{eqnarray} \tag{ 20 }$$

where $e,d$ are positive integers, $d,e\lt D$, ${S}_{\delta }$ is the set of D-simplices that contain δ, and $\langle A\rangle$ denotes a group with generators A. The groups (20) satisfy

$$\begin{eqnarray}&&C({d}_{1},{e}_{1})\subseteq C({d}_{2},{e}_{2})\ \Longleftrightarrow \ {d}_{1}\leqslant {d}_{2},{e}_{1}\leqslant {e}_{2},\end{eqnarray} \tag{ 21 }$$

and their centralizer takes the form

$$\begin{eqnarray}&&{\mathcal{Z}}C(d,e)\propto \langle {X}_{Q},{Z}_{Q}\rangle \cdot C(\bar{e},\bar{d}),\end{eqnarray} \tag{ 22 }$$

where ∝ denotes equality up to global phase [14]. Moreover, $| Q|$ is odd and the group $\langle {X}_{Q},{Z}_{Q}\rangle$ has trivial intersection with any of the groups $C(d,e)$.

For positive integers $d,e$ with $d+e\leqslant D$, the (d, e) gauge color code is defined by

$$\begin{eqnarray}&&{\mathcal{S}}:=C(d,e),\quad {\mathcal{G}}:=C(\bar{e},\bar{d})\end{eqnarray} \tag{ 23 }$$

where $\bar{x}:=D-x$. According to the above properties the definition is valid: ${\mathcal{S}}$ is abelian and the center of ${\mathcal{G}}$. Whatever the values of $d,e$, there is a single encoded qubit with logical Pauli operators ${X}_{Q},{Z}_{Q}$. When $e=\bar{d}$ the color code is conventional [14]. Conventional color codes yield all the possible quotient groups appearing in the construction of logical operators. In particular, it follows from the results on conventional color codes of [12] that logical operators are always non-local: those of the form X_S are $\bar{d}$-brane-net like if bare and e-brane-net like if dressed, and those of the form Z_S are $\bar{e}$-brane-net like if bare and d-brane-net like if dressed.

Some properties of the above 2D and 3D examples can be extracted from the first instances of each family. For example, in the 2D case the number of qubits is $1+3n+3{n}^{2}$, and in the 3D case it is $1+4n+6{n}^{2}+4{n}^{3}$. Also, in 2D the generators of $C(1,1)$ have support on at most 6 qubits, and the same holds for $C(2,2)$ in 3D.

For gauge color codes the CNot gate is transversal. For self-dual codes, d = e, the Hadamard gate is transversal. What about the gate ${R}_{n}$? As shown in appendix D, a set T as in (3) exists if

$$\begin{eqnarray}&&D\geqslant n\bar{e}.\end{eqnarray} \tag{ 24 }$$

In particular, ${R}_{D}$ can be implemented by taking $\bar{e}=1$. The result (24) was obtained for $T=\varnothing$ and $d=\bar{e}$ in [14], but an important limitation there was the lack of a general recipe to construct lattices satisfying (3).

As it turns out the set T only depends on the lattice, not the specific (d, e) parameters of the code. It must be such that the operator X_T, regarded as an error in the corresponding $(1,D-1)$ color code, commutes with a Z stabilizer generator if and only if the generator has support on a number of qubits that is a multiple of 4.

For the above 2D and 3D families of color codes we can give T explicitly. In the 2D case it is simplest to depict it, see figure 1. As for the 3D case, first notice that qubits correspond to the following elements of the triangulation of a colored tetrahedron M: all 3-simplices, the 2-simplices on the faces of M, the 1-simplices on the edges of M, and the 0-simplices that are the vertices of M. A valid choice for the set T is the union of (i) the set T₃ of all 3-simplices with $({\bf{a}},{\bf{b}},{\bf{c}})$ as in (18) an even permutation of the triad $({\bf{i}},{\bf{j}},{\bf{k}})$, (ii) the set T₂ of all 2-simplices that are not a subsimplex of any element of T₃, see figure 3, and (iii) the set T₁ containing all 1-simplices that are not a subsimplex of any element of T₂.

The different gauge color codes that can be defined on a given, fixed lattice are related by gauge fixing, either directly or via some other code in the family. Indeed, all the codes have a shared group ${\mathcal{L}}$ of bare logical operators (namely, those with support in all qubits) and the condition (12) is satisfied when ${d}_{1}\leqslant {d}_{2}$, ${e}_{1}\leqslant {e}_{2}$. Take e.g. D = 3. For a given geometry we might consider both the code $(1,1)$ or the gauge fixed version $(1,2)$. Within $(1,1)$ CNot and Hadamard gates are transversal. Fixing the gauge we can move into $(1,2)$ to apply transversal ${R}_{3}$ gates, completing a universal set of transversal gates. The ideal strategy is to transition to $(1,2)$ only momentarily. This way there is no need to ever measure directly the large stabilizer generators of the (1, 2) code. Instead, it is enough to measure the gauge generators in $C(2,2)$, which only involve up to 6 qubits. Notice that the only non-transversal element of the procedure is the classical computation to find the gauge operator that will fix the gauge as desired. This is entirely analogous to error correction.

The result (24) was already proven in [14] for conventional color codes in the special case $T=\varnothing$, i.e. when the same rotation is applied to all physical qubits. Generalizing slightly to the subsystem case, this means that the result hinged on gauge generators ${X}_{{S}_{i}}$ satisfying

$$\begin{eqnarray}&&\left|\bigcap _{i=1}^{m}{S}_{i}\right|\equiv 0\mathrm{mod}{2}^{n-m+1}.\end{eqnarray} \tag{ C1 }$$

These are local constraints on the D-colex, because the generators ${X}_{{S}_{i}}$ are local.

Fortunately conditions (C1) can be simplified a lot. This is due to the following property of colexes [14]:

Given a collection of cells ${c}_{i}$ with color sets ${\kappa }_{i}$, their intersection is a collection of cells with color label κ:

$$\begin{eqnarray}&&\kappa =\bigcap _{i}{\kappa }_{i}.\end{eqnarray} \tag{ C2 }$$

The cells of the collection must be disjoint, because they have the same color. Of interest here is the case where the m cells c_i have all dimension $e+1$ (as they correspond to gauge generators). Suppose that $| \kappa | =d^{\prime }$, so that the cells in the intersection have dimension d'. Since there are a total of $D+1$ colors, m cannot have an arbitrary value, but rather

$$\begin{eqnarray}&&(e+1-d^{\prime } )m\leqslant (m-1)(D+1-d^{\prime } ).\end{eqnarray} \tag{ C3 }$$

Indeed, consider the collection of pairs (i, r) where $i=1,...,m$ and r is one of the colors in ${\kappa }_{i}-\kappa$. The left hand side is the total number of such pairs. The right hand side is the maximum number of pairs that we could form with such r, which cannot be shared by the m cells c_i and therefore can appear at most $m-1$ times. Thus the inequality, which can be restated as

$$\begin{eqnarray}&&d^{\prime } \geqslant D-m\bar{e}+1.\end{eqnarray} \tag{ C4 }$$

Going back to (C1), the case m = n will be satisfied if and only if the intersection is composed of cells of dimension at least 1. For this to hold, the right hand side of (C4) hast to be greater or equal than one, i.e. the inequality (24) must hold. Naturally, the conditions (C1) must be satisfied also for $m\lt n$, but when (24) holds

$$\begin{eqnarray}&&d^{\prime } \geqslant D-m\bar{e}+1\geqslant (n-m)\bar{e}+1\geqslant n-m+1.\end{eqnarray} \tag{ C5 }$$

Thus the conditions (C1) are satisfied if all d-cells have a number of vertices multiple of ${2}^{d}$. This motivates the following definition:

A D-colex is perfect if every cell has $0\mathrm{mod}{2}^{d}$ vertices, where d is the dimension of the cell.

In [14] the problem of the existence of perfect colexes was left open. The rest of this appendix is devoted to show that one can always obtain a perfect colex from a generic one by transforming the lattice in the neighborhood of a collection of vertices T. The next appendix in turn shows that the actual substitution of the generic colex by its perfect counterpart is unnecessary. Rather, a transversal ${R}_{n}$ is recovered by inverting the rotation at those qubits in the set T.

Crucial to the construction of perfect colexes are the 'simplest' colexes described in the previous appendix, which are perfect. Recall that the boundary of a $(D+1)$-hypercube is a D-colex, from which the 'simplest' punctured D-colex is obtained by removing a vertex and all the cells containing it. Here instead we consider removing a small neighborhood of a vertex, so that we are left with a D-dimensional ball that contains ${2}^{D+1}-1$ vertices and in which all 'complete' d cells have ${2}^{d}$ vertices, while the 'incomplete' ones are lacking the removed vertex and thus keep only ${2}^{d}-1$. The procedure to make a generic colex into a perfect one involves substituting the neighborhood of each of the vertices in a collection T by such D-balls, as in figure C1 (this is actually a connected sum with a neat combinatorial description, see [12]). Therefore, a given d cell c with vertex set V_c in the original colex gives rise to a d-cell c' with vertex set ${V}_{{c}^{\prime }}$ such that

$$\begin{eqnarray}&&| {V}_{{c}^{\prime }}| =| {V}_{c}| +({2}^{d}-2)| T\cap {V}_{c}| .\end{eqnarray} \tag{ C6 }$$

All the new d-cells added to the colex have ${2}^{d}$ vertices.

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The procedure is generically valid only for either spherical D-colexes, which yield trivial codes, or punctured D-colexes, which are the ones of interest here. Notice however that in order for the above to make sense we need to be sure that every vertex has the right kind of neighborhood, which is true in the bulk but not on the boundary for the punctured colex. Fortunately this is not an important issue. Indeed, it suffices to make the colex into an spherical one by adding a 0-cell in the usual way, perform the changes at T vertices, and then remove the added vertex together with all the cells that contain it.

It might be worth describing the dual picture, in which T is a set of D-simplices. Each simplex in T is divided in ${2}^{D+1}-1$ pieces: there are $D+1$ new vertices, and for each color set κ with $0\lt | \kappa | \lt D+1$ there is a new D-simplex that has as vertices (i) the new vertices with labels κ and (ii) the vertices of the removed simplex with colors in $\bar{\kappa }$. Figure C1 illustrates the dual picture for D = 2. Notice that this is nothing but the construction that was already used in the main text to 'close' M to form a sphere.

It is convenient to first consider the D = 2 case. Figure C2 illustrates the procedure in a 2-colex for which T is composed of a single element, i.e. it is enough to substitute the neighborhood of a single vertex. In the original 2-colex there are 3 faces that have 6 vertices. In the transformed one, each of these faces has gained 2 vertices, so that they have $8\equiv 0\mathrm{mod}4$ vertices, as required.

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It is apparent what the T set of vertices should satisfy in the D = 2 case. 'Good' faces, those with $0\mathrm{mod}4$ vertices, must have an even number v of vertices in T, so that they end up with

$$\begin{eqnarray}&&0+2v\equiv 0\mathrm{mod}4\end{eqnarray} \tag{ C7 }$$

vertices and stay good. 'Bad' faces, those with $2\mathrm{mod}4$ vertices, must have an odd number v of vertices in T, so that they end up with

$$\begin{eqnarray}&&2+2v\equiv 0\mathrm{mod}4\end{eqnarray} \tag{ C8 }$$

vertices and become good. But, is there always such a set T? How do we find it?

At this point it is useful to consider a different problem: error correction. In the D = 2 case, there is a single color code: $(1,1)$. Its stabilizer generators are attached to faces. Given an error of the form X_T, where T is some set of qubits/vertices, its syndrome amounts to the collection of faces f such that X_T and Z_f anticommute, where Z_f stands for the Z stabilizer at f. These syndrome faces f are simply those that contain an odd number of vertices from the set T. I.e. the syndrome of X_T corresponds to those faces that gain $2\mathrm{mod}4$ vertices when the vertices of T are subject to the above geometrical transformation. This allows to answer both questions above. Regarding existence, in the case of punctured $2D$ color codes all syndrome sets are possible [10]. To find T it suffices to solve a set of binary linear equations, just the same way that one can find the possible errors from the error-syndrome.

The stage is set to show how the construction works for general punctured D-colexes. There are two steps to this. First, it is possible to find a set of vertices T such that the resulting D-colex has only good 2-cells as above. Second, it turns out that the resulting colex is perfect.

The existence of T is not obvious. The reason for this is that for $D\gt 2$ not every syndrome is possible. In particular, consider the $(D-1,1)$ conventional color code on the given D-colex: it has Z stabilizer generators attached to 2-cells. These 2-cell operators Z_f are not independent. Instead, in a punctured code they satisfy only local constraints that have their origin in the structure of 3-cells [12]. Namely, every 3-cell c has 3 colors, and thus is composed of 2-cells with 3 different color sets ${\kappa }_{i}$, $i=1,2,3$. Let ${F}_{\kappa }$ be the set of 2-cells of c with color κ, where $\kappa \in \{{\kappa }_{1},{\kappa }_{2},{\kappa }_{3}\}$. By definition, every vertex of c belongs exactly to one 2-cell in each ${F}_{\kappa }$. A trivial consequence is

$$\begin{eqnarray}&&\displaystyle \prod _{f\in {F}_{\kappa }}{Z}_{f}=\displaystyle \prod _{f\in {F}_{{\kappa }^{\prime }}}{Z}_{f},\qquad \kappa ,{\kappa }^{\prime }\in \{{\kappa }_{1},{\kappa }_{2},{\kappa }_{3}\}.\end{eqnarray} \tag{ C9 }$$

In words: the number of 2-cell operators Z_f with $f\in {F}_{\kappa }$ and negative syndrome is even if and only if the number of 2-cell operators Z_f with $f\in {F}_{{\kappa }^{\prime }}$ and negative syndrome is even.

Fortunately, the constraints (C9) are no obstacle for the present purpose because bad 2-cells satisfy them. Namely, if among the 2-cells in ${F}_{\kappa }$ there are ${m}_{\kappa }$ bad ones, then $| {V}_{c}|$, the total number of 0-cells in c, satisfies

$$\begin{eqnarray}&&| {V}_{c}| \equiv 2{m}_{\kappa }\equiv 2{m}_{{\kappa }^{\prime }}\mathrm{mod}4.\end{eqnarray} \tag{ C10 }$$

This follows again from the fact that every vertex of c belongs exactly to one 2-cell in each of the sets ${F}_{\kappa }$: the elements of V_c can be counted by adding the number of vertices in each 2-cell of ${F}_{\kappa }$ for any given κ: the result is always the same, but (C10) yields

$$\begin{eqnarray}&&{m}_{\kappa }\equiv {m}_{{\kappa }^{\prime }}\mathrm{mod}2,\end{eqnarray} \tag{ C11 }$$

which means that a syndrome such that 2-cells have eigenvalue $+1/-1$ if they are respectively good/bad does exist because the constraints (C9) are satisfied. There exists T such that the set of 2-cell operators Z_f that anticommute with X_T is the same as the set of bad 2-cells.

It only remains to show that every D-colex with no bad 2-cells is perfect. This can be done recursively. Assume that for a given D-colex with no bad 2-cells every d-cell has $0\mathrm{mod}{2}^{d}$ vertices. The aim is to show that every $(d+1)$-cells has $0\mathrm{mod}{2}^{d+1}$ vertices.

First, every such $(d+1)$-cell has a boundary that is a spherical d-colex. From this spherical d-colex one can build a new 'shrunk' d-dimensional lattice by shrinking all the d-cells with a given color set κ to a point [12]:

• The 0-cells of the shrunk lattice correspond to d-cells with color set κ, or κ-cells, of the d-colex.
• The 1-cells of the shrunk lattice correspond to 1-cells of color r of the d-colex, or r-cells, where r is the only color with $r\;\unicode{8713}\;\kappa$.
• The 2-cells of the shrunk lattice correspond to 2-cells with color sets $\kappa ^{\prime }$ such that $r\in \kappa ^{\prime }$. In particular, the shrunk 2-cell has an even number of 1-cells in its boundary because the original 2-cell has $0\mathrm{mod}4$ 1-cells and half of them are r-cells.

Since the homology of the sphere is trivial and all the 2-cells have an even number of edges, the graph formed by the 0-cells and 1-cells of the shrunk lattice is bipartite. Indeed, any closed path must have an even number of vertices and edges because (i) as a ${{\bf{Z}}}_{2}$ 1-chain it has no boundary and thus must be a sum of boundaries of 2-cells and (ii) the sum of 1-chains with an even number of edges must have an even number of edges.

Therefore, the set of κ-cells in the d-colex is the disjoint union of two sets A and B, and every r-cell in the d-colex shares a vertex with exactly one element of the set A. Since every κ-cell in A has $0\mathrm{mod}{2}^{d}$ vertices, it follows that there are $0\mathrm{mod}{2}^{d}$ r-cells and therefore $0\mathrm{mod}{2}^{d+1}$ vertices in the d-colex (every vertex belongs exactly to one r-cell, and each r-cell has 2 vertices).