Does every code have transversal Pauli group?

Question

A transversal logical gate for an $ n $ qubit code is a gate from the group of local unitaries $$ \bigotimes_{i=1}^n U(2) $$ which also preserves the codespace. For an $ ((n,K,d)) $ code we say a particular logical gate $ \tilde{g} \in U(K) $ can be implemented transversally if there exists some gate $$ g=\bigotimes_{i=1}^n g_i \in \bigotimes_{i=1}^n U(2) $$ which implements $ \tilde{g} $ when restricted to the codespace.

My question is the following: For every $ [[n,1,d]] $ code does there exist some basis for the codespace with respect to which logical $ X $ and logical $ Z $ can be implemented transversally? Equivalently, does there always exists a pair of two transversal operators on the codespace which anticommute and both square to the identity?

I know this is possible for all stabilizer codes. Indeed the transversal implementations of logical $ X $ and $ Z $ can even be chosen from the normalizer $ N(S) $, so all the $ g_i $ are themselves single qubit Paulis. The question is whether this can always be done for non stabilizer codes as well.

Also note that we could have asked this for $ k $ greater than $ 1 $ but it seemed good to start small and the answer shouldn't depend on $ k $.

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Clarification:

In my definition of transversality I do not require that all $ g_i $ are equal. There are many definitions of transversality. For example what I would call "strict" or "strong" transversality (e.g. arxiv.org/pdf/2210.14066.pdf Definition 3) is what some people use as the definition of transversal.

The [[4,1,2]] code is an example of a code with logical Paulis transversal but not strongly transversal.

Consider a [[4,1,2]] surface code, for example with stabilizer $ S=<XXXX,ZZII,IIZZ> $. Then logical $ X $ can be implemented transversally as $ XXII $ and logical $ Z $ can be implemented transversally as $ ZIZI $. Indeed, as I noted above, the Pauli group can be implemented transversally for every stabilizer code using Pauli gates from the normalizer $ N(S) $.

Note: This question has already been answered for the strongly transversal case here Does every code have a strongly transversal Pauli group?

Answer 1

I don't think so - consider e.g. the 'diagonal' representation $$\phi:SU(2) \rightarrow GL((\mathbb{C}^2)^{\otimes 4})$$ $$U \mapsto U^{\otimes 4}. $$ The Clebsch-Gordan series tells us that, for spin representations $S_i$, this decomposes as $(\mathbb{C}^2)^{\otimes 4} \cong 2S_0 \oplus 3S_1 \oplus S_2$. Take your code to be $2S_0$, i.e. spanned by the two 1-dimensional spin-0 irreps. Since the irreps are 1-dimensional, this code is stabilized by every diagonal operator $U^{\otimes 4}$. In particular, it's also stabilized by $X^{\otimes 4}$ and $Z^{\otimes 4}$, so it's distance-2. Intuitively, the logical operators should commute with all $U^{\otimes 4}$, so they're all going to look like linear combinations of permutations of qubits. I'm not Schur (heh) of the most direct proof, but this should give you an example of a $((4,2,2))$ code whose transversal logical operators form only the logical identity.

Answer 2

Every code that can be implemented by a stabilizer circuit (this includes stabilizer codes, gauge codes, floquet codes, etc) has this type of subset-transversal Pauli gate.

In such a code, the X, Y, and Z logical observables will correspond to a triplet of anticommuting Pauli products. Applying those Pauli products as gates applies the correspond observable, and this process is fault tolerant. It's fault tolerant because Pauli gates in this context can be tracked entirely in the classical control system. You can apply any Pauli gate to the quantum system, and cancel it out by also applying the same Pauli gate to the Pauli frame in the control system. The rest of the computation and correction then plays out isomorphically to not applying the gates and is therefore fault tolerant.