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$\DeclareMathOperator\su{\mathfrak{su}}$I want to find out all the subalgebras of $\su(N)$, in particular, $N=2^n$, which is the Lie algebra of $n$-qubits.

I don't know whether this is a hard question or not. I assume it should not be, but I have no clues to find relevant references. Anyone can recommend some references that either give the subalgebras of $\su(N)$ or $\su(2^b)$ or contain the techniques to find them?

Thanks a lot in advance.

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    $\begingroup$ 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. $\endgroup$
    – Robert Bryant
    Commented Mar 1, 2022 at 12:56
  • $\begingroup$ 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). $\endgroup$
    – André Henriques
    Commented Mar 1, 2022 at 13:46
  • $\begingroup$ 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 $\endgroup$
    – Vincent
    Commented Mar 1, 2022 at 16:17
  • $\begingroup$ 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 $\endgroup$
    – Vincent
    Commented Mar 1, 2022 at 21:59

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