Weakly transversal gates for the $ [[15,1,3]] $ quantum Reed-Muller code

Question

The $ [[15,1,3]] $ quantum Reed-Muller code is a CSS code famous for implementing logical $ T $ (strongly) transversally. It has stabilizer generators \begin{align*} XXXXXXXXIIIIIII,&ZZZZZZZZIIIIIII \\ XXXXIIIIXXXXIII,&ZZZZIIIIZZZZIII \\ XXIIXXIIXXIIXXI,&ZZIIZZIIZZIIZZI \\ XIXIXIXIXIXIXIX,&ZIZIZIZIZIZIZIZ \\ &ZZZZIIIIIIIIIII \\ &ZZIIZZIIIIIIIII \\ &ZIZIZIZIIIIIIII \\ &ZZIIIIIIZZIIIII \\ &ZIIIZIIIZIIIZII \\ &ZIZIIIIIZIZIIII \end{align*}

In particular, logical $ T $ is implemented using the physical unitary $$ \bigotimes_{i=1}^{15} T^\dagger $$ How do we know that the $ [[15,1,3]] $ code doesn't have weakly transversal implementations of other gates, for example $ \sqrt{T} $? In other words, is it possible that there is some (weakly) transversal physical gate $$ \bigotimes_{i=1}^{15} g_i $$ which implements logical $ \sqrt{T} $ on the codespace? Here all the $ g_i $ are in $ U(2) $ but they are not all assumed to be equal.

The obvious first place to look is the paper The Smallest Code with Transversal T which claims that the smallest code with transversal $ \sqrt{T} $ is the $ [31,1,3] $ quantum Reed-Muller code.

Although this paper does not assume strong transversality (all the $ g_i $ equal) it assumes something that is perhaps even more restrictive. Namely, "Assumption 2" of the paper is that every $ g_i $ for a transversal implementation of logical $ \sqrt{T} $ must be either the single qubit unitary $ \sqrt{T} $ itself or it must be some power of the single qubit unitary $ \sqrt{T} $.

For definitions of transversal (what I call weakly transversal here for emphasis) and strongly transversal we quote the paper above

"Definition 2: A logical single qubit unitary is implemented in a transversal manner if it is implemented by individual operations on each qubit $i$." So $$ \bigotimes_{i=1}^{n} g_i $$ "Definition 3: We say that a gate is strongly transversal if the operation on each set of identically labelled qubits is the same for each and every label." So $$ \bigotimes_{i=1}^{n} g_i $$ with all the $ g_i $ equal.

Answer 1

By Theorem 5/ Example 6 of The disjointness of stabilizer codes and limitations on fault-tolerant logical gates the transversal gates of the $ [[15,1,3]] $ quantum Reed-Muller code are all in the 3rd level of the Clifford hierarchy.

$ \sqrt{T} $ is in the $ 4 $th level of the Clifford hierarchy. So $ \sqrt{T} $ has no weakly transversal implementation for the $ [[15,1,3]] $ quantum Reed-Muller code.

Answer 2

I don't know the answer to your question, but I think I know something relevant. The space of codes with transversal T gates has relationships to the space of circuits that can distill T states. And the distillation circuits don't all look like "and then I apply a big layer of T gates to all qubits leaving out none of them".

The T state distillation circuit I know that uses the fewest non-Clifford gates is one that originally came from block codes but I don't know if it corresponds to a code anymore. It consumes 10 T states with error rate $p$ and produces 2 T states with error rate $O(p^2)$. It's part of a family of circuits that limit to a 3:1 conversion ratio:

So the smallest circuits don't particularly look like codes with strongly transversal T gates. I dunno if that says anything about Reed-Muller codes specifically, but I'd expect the smallest code with a constant depth $\sqrt{T}$ to be something very weird looking.