# Which codes have transversal $T$ gate?

A previous post transversal P (phase) gate shows that codes where all stabilizer elements have weights that are multiple of 4 will have a transversal $$P$$ gate. "Transversal" seems to have multiple definitions; here a gate $$G$$ is transversal for the code with stabilizer $$S$$ if it is in the normalizer of $$S$$ in the full unitary group : $$G \in U_n : G' S G = S$$. (note $$G$$ need not be in the clifford group). This is shown using $$P^\dagger X P = \imath XZ, P^\dagger Z P = Z;$$. At first guess I thought that codes where all stabilizer elements have weight that's a multiple of 8 would have a transversal $$T$$ gate. ($$T^2=P$$); but $$T^\dagger X T = w_8 P Z X; T^\dagger Z T=Z$$; where $$w_8^8=1$$. So it's not obvious if a similar argument can be used. Does anyone know how to extend the result for $$P$$ gate to $$T$$?

• I think this is impossible due to the Eastin–Knill theorem: en.wikipedia.org/wiki/Eastin%E2%80%93Knill_theorem Jun 7 at 5:47
• Yes, but it's a part of an active research project of mine that I wouldn't want to give too much away on! In short, look for "triorthogonal codes". Jun 7 at 6:37
• @VictoryOmole this doesn't violate Eastin-Knill. It just tells you that you cannot have transversal Hadamard (which in turn says that the code cannot be a self-dual CSS code). Jun 7 at 6:51
• Ah, I see. Thanks. Jun 7 at 8:41
• Probably very relevant: arxiv.org/abs/2001.04887 Jun 9 at 22:34