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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?

Update: The answer to

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)

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  • $\begingroup$ @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! $\endgroup$ Oct 27, 2022 at 21:46

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If you take the Reed-Muller code of 15 qubits, this is a distance 3 CSS code (so has transversal c-NOT, Z and X) but it also has transversal T (and transversal controlled-S and controlled-controlled-Z). What it doesn't have is transversal Hadamard. You'll find this code properly defined in a bunch of places, but, for example, here is the first one that Google threw at me!

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  • $\begingroup$ 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? $\endgroup$ Mar 2, 2022 at 14:07
  • $\begingroup$ @IanGershonTeixeira I guess that is true, but I don't know if anyone has ever proven it. $\endgroup$
    – DaftWullie
    Mar 2, 2022 at 14:29
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    $\begingroup$ @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. $\endgroup$
    – DaftWullie
    Mar 3, 2022 at 9:49
  • $\begingroup$ 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. $\endgroup$ Oct 27, 2022 at 21:00

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