# Eastin Knill Theorem and groups of transversal gates

The Eastin-Knill Theorem shows that the transversal gates always form a group and that moreover this group is a finite subgroup of the group of all unitaries.

For many codes, for example all self dual CSS codes, the group of transversal gates is exactly the Clifford group.

https://quantumcomputing.stackexchange.com/a/22226/19675

Does anyone know of codes whose group of transversal gates is significantly different from the Clifford group? For example, does anyone know a code on $$n$$ qubits whose group of transversal gates is not isomorphic to a subgroup of the Clifford group on $$n$$ qubits?

Or failing that, does anyone know any restrictions on which finite groups can occur as the group of transversal gates of a code?

Exotic transversal gate group

seems to show that the transversal gate group of an $$[[n,1,d]]$$ stabilizer code, for $$d \geq 2$$, must be generated by Clifford gates and/or $$T_k$$ gates. So by the classification of finite subgroups of $$PU_2$$ we can conclude that the transversal gate group must be either a dihedral 2 group $$D_{2^k}$$ or $$A_4$$ or $$S_4$$ (cyclic transversal gate group is not possible because we have chosen to specialize to stabilizer codes and thus $$X$$ and $$Z$$ are both transversal and so they generate a noncyclic Klein 4 subgroup and thus the whole transversal gate group must be non cyclic).

Moreover all these groups can indeed be realized of the transversal gate group of some $$[[n,1,d]]$$ , $$d \geq 2$$, stabilizer code. Each dihedral 2-group $$D_{2^k}$$ arise as the transversal gate group of the corresponding $$[[2^{k+1}-1,1,3]]$$ quantum Reed-Muller code. $$A_4$$ is the transversal gate group of the perfect $$[[5,1,3]]$$ code see

What are the transversal gates of the [[5,1,3]] code?

And $$S_4$$ (the single qubit Clifford group) is the transversal gate group of the $$[[7,1,3]]$$ Steane code see

Transversal logical gate for Stabilizer (or at least Steane code)

• @unknown that's absolutely right! Ya I tend to think of all my transversal gates in the projective unitary group (otherwise every code has infinitely many global phase gates which are transversal), that's why I mentioned $PU_2$ above. But it's good that you clarified since I did not! Oct 27, 2022 at 21:46

• Wow this is a great example! Follow-up question: Let $Cl_n$ be the (logical) Clifford group on $n$ (logical) qubits (say encoded with respect to the $[7,1,3]$ code so that all Clifford gates are transversal). Any set of generators for $Cl_n$ together with any one gate outside of $Cl_n$ forms a universal gate set. Let $G_n$ be the group of transversal gates on $n$ qubits encoded with respect to the $[15,1,3]$ code. Is it true that any set of generators of $G_n$ together with any one gate outside of $G_n$ forms a universal gate set? Mar 2, 2022 at 14:07
• @IanGershonTeixeira Let me revoke my previous affirmation as it's clearly not true. The group of unitaries created by $G_n$ is a subgroup of the permutation matrices with phases on each entry. Other phase gates and other multi-controlled phase gates extend the subgroup, but keep it within that larger group (which essentially corresponds to reversible classical computation). I guess the condition you would need is any unitary for which at least one row has more than one non-zero entry. Mar 3, 2022 at 9:49
• your answer here quantumcomputing.stackexchange.com/a/28699/19675 is nice because it highlights that every dihedral 2-group $D_{2^k}$ can be realized as the transversal gates set of the $[[2^{k+1}-1,1,3]]$ quantum Reed-Muller code. This follows from the fact that $T_{k}=diag(1,e^{2 \pi i/2^k})$ is transversal for the $k$th quantum Reed-Muller code. It also seems to provide a restriction on transversal gate groups for $[[n,1,d]]$ stabilizer codes, they must be generated by Cliffords and/or $T_k$ gates. Oct 27, 2022 at 21:00