# Does every code have a strongly transversal Pauli group?

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. If all the $$g_i$$ are equal then we say $$\tilde{g}$$ is strongly transversal.

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 $$Z$$ have a strongly transversal implementation?

• I think there is a misunderstanding here. In my definition of transversality I do not require that all $g_i$ are equal. Although I gave my definition in the question I perhaps should have stated more emphatically that the $g_i$ do not need to all be equal. Your definition is what I would call "strict" or "strong" transversality. For example see this paper arxiv.org/pdf/2210.14066.pdf Definition 3. And indeed the [[4,1,2]] code is the first example that comes to mind for me as well of a code with Paulis not strongly transversal. However this is not the question I am asking. Mar 21 at 14:39
• Take the [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 note in my question, the Pauli group can be implemented transversally for every stabilizer code using Pauli gates from the normalizer $N(S)$. I'll add these comments to my question to clarify. Mar 21 at 14:44