  A popular universal gate set for quantum computing with qubits is Clifford+T,
as this can be readily implemented on many fault-tolerant architectures. For
qutrits, there is an equivalent T gate, that, like its qubit analogue, makes
Clifford+T approximately universal, is injectable by a magic state, and
supports magic state distillation. However, it was claimed that a better gate
set for qutrits might be Clifford+R, where R=diag(1,1,-1) is the metaplectic
gate, as certain protocols and gates could more easily be implemented using the
R gate than the T gate. In this paper we show that when we have at least two
qutrits, the qutrit Clifford+R unitaries form a strict subset of the Clifford+T
unitaries, by finding a direct decomposition of $R \otimes \mathbb{I}$ as a
Clifford+T circuit and proving that the T gate cannot be exactly synthesized in
Clifford+R. This shows that in fact the T gate is at least as powerful as the R
gate, up to a constant factor. Moreover, we additionally show that it is
impossible to find a single-qutrit Clifford+T decomposition of the R gate,
making our result tight.
