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Let $ \mathcal{C}^{(t)} $ denote the $ t $ level of the $ n $ qubit Clifford hierarchy.

Let $ \mathcal{F}^{(t)} $ denote the collection of all diagonal gates in $ \mathcal{C}^{(t)} $. $ \mathcal{C}^{(t)} $ is not a group for $ t \geq 3 $ see

Is there a closure property for the entire Clifford hierarchy?

However, $ \mathcal{F}^{(t)} $ is a group by proposition 4 of Semi-Clifford operations, structure of $ \mathcal{C}^{(t)} $ hierarchy, and gate complexity for fault-tolerant quantum computation.

The introduction to Climbing the Diagonal Clifford Hierarchy claims that the group $ \mathcal{F}^{(t)} $ is generated by $ C^i Z^{1/2^j} $ with $ i+j=t-1 $ ($ C^i $ means $ i $ control qubits so also $ i \leq n $). The reference given is Semi-Clifford operations, structure of $ \mathcal{C}^{(t)} $ hierarchy, and gate complexity for fault-tolerant quantum computation. However as far as I can tell Semi-Clifford operations, structure of $ \mathcal{C}^{(t)} $ hierarchy, and gate complexity for fault-tolerant quantum computation only proves the claim for $ n=3, t=3 $ (Proposition 9) and not for general $ n $ and $ t $.

Am I missing something? Is there any easy way to show directly that this forms a generating set?

Or perhaps there is an easy way to read this claim out of the results in Diagonal gates in the Clifford hierarchy ?

Ok I think I must be missing something because page 4 of https://arxiv.org/pdf/2212.05398.pdf also attributes this same result to the same paper.

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First note that any diagonal gate on qubits can be written as some product of $U=\exp (i\theta_j \vec{Z}_j)$ where $\vec{Z}_j$ is any qubit Pauli $Z$ string.

In Diagonal gates in the Clifford Hierarchy (https://arxiv.org/pdf/2110.11923.pdf) they show that a diagonal gate on qubits is in the Clifford Hierarchy iff it can be written as a product of $U=\exp (i\theta_j \vec{Z}_j)$ with $\theta_j = \frac{a\pi}{2^k}$ for integers $a,k$. If you include the Identity string (which is just a global phase) there are $2^n$ $Z$ strings which is the same number of entries in an $n$-qubit diagonal gate ($2^n - 1$ if you prefer you gates in $SU(2^n)$ vs $U(2^n)$).

Now, on $n$ qubits there are $n$ single-qubit Pauli $Z$ rotations, $\binom{n}{2}$ diagonal gates with 2-qubit support ($CZ$ like), $\binom{n}{3}$ diagonal gates with 3-qubit support ($CCZ$ like), etc. If you add up all the $C^iZ$-like gates on $n$ qubits (which includes diagonal gates at various levels in $\mathcal{CH}$) and include the identity you get $2^n$ gates. You just have to show that these gates are independent and then they also form a basis for all diagonal gates in $\mathcal{CH}$.

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