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Mar 1, 2022 at 21:59 comment added Vincent To be clear: this is not meant as a rhetorical question. I truly don't know. I just started thinking about this because of your question and as I said my intuition is not much help to me here
Mar 1, 2022 at 16:17 comment added Vincent Can I ask a naive counterquestion: is there any subalgebra that you expect to be more 'truly' the Lie algebra of n qubits? What I mean is this. $\mathfrak{su}(2^n)$ acts on $\mathbb{C}^{2^n}$ but from the phrase 'n qubits' I conclude that you think about this space as $\mathbb{C}^2 \otimes \ldots \otimes \mathbb{C}^2$. But Lie alg $\mathfrak{su}(2^n)$ doesn't know about this tensor product structure and also doesn't care. Are you looking for a subalgebra that respects or at least in some vague sense 'acknowledges' the tensor structure? Does such a thing even exist? My intuition fails me here
Mar 1, 2022 at 13:46 comment added André Henriques This contains, as a sub-problem, the question of enumerating all ways of writing $2^n$ as a sum of dimensions of irreps of $\mathfrak g$, for $\mathfrak g$ some semisimple Lie algebra. If $\dim(\mathfrak g)$ is small compared to the number $2^n$, then this looks intractable. If $\mathfrak g$ is "big", then there might be some hope (that's the beginning of the "downward induction" that Robert Bryant suggested using).
Mar 1, 2022 at 12:56 comment added Robert Bryant I would think that you could answer this question using the Dynkin tables, which list the maximal subalgebras of every simple Lie algebra. Then, by 'downward induction', you should be able to list all the subalgebras in any particular case. Of course, as $n$ increases, the length of the 'induction chains' will increase, so it could become combinatorially complicated.
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