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This is just a question about what constitutes standard notation.

In a context where $A$ and $B$ are understood to be Hermitian matrices, what is usually meant by the bracket $[A,B]$ ?

On the one hand, I'd have thought it was standard for $[A,B]$ to denote the commutator $AB-BA$. On the other hand, I'd also have thought it was standard for $[-,-]$ to denote the bracket operation in a Lie algebra, in which case you'd want $[A,B]$ to be something like $i(AB-BA)$. These can't both be right. So is there some standard?

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  • $\begingroup$ I think the point is that for a Lie algebra inside $gl(n)$ there would not be an $i$ as a leading coefficient, ... so there'd be no incompatibility at all. Hermitian matrices are inside $gl(n)$, although not a Lie subalgebra, indeed. $\endgroup$ – paul garrett Aug 12 '15 at 19:40
  • $\begingroup$ @paulgarrett: The point is precisely that the Lie algebra of Hermitian matrices is a subset of $gl(n)$, but not a subalgebra of $gl(n)$. So when $A$ and $B$ are Hermitian matrices, there is ambiguity about whether $[A,B]$ means the Lie bracket in the Lie algebra of Hermitian matrices or the (different) Lie bracket in the larger Lie algebra $gl(n)$. I am asking whether there is an "industry standard" about which of these two things $[A,B]$ ordinarily denotes in this context. $\endgroup$ – WillO Aug 12 '15 at 22:29
  • $\begingroup$ More pointedly: I have never in my life seen $[A,B]=i(AB-BA)$, so I'd never suspect the bracket might refer to that. As far as I know, the issue you describe does not actually exist (in any of the several contexts I've seen). It occurs to me that someone might insist that all their operators be self-adjoint, and $AB-BA$ is skew-adjoint for self-adjoint for $A,B$ self-adjoint... and the insertion of the factor of $i$ sorta-artificially makes the outcome again hermitian. So, if anything, the question is perhaps posed oppositely to reported uses? Still, I'd never tack on that bogus "$i$". $\endgroup$ – paul garrett Aug 12 '15 at 22:44
  • $\begingroup$ ... Oh, and, from your comment, it's not that the vector space of Hermitian matrices is a Lie algebra with that inserted "$i$", although one can insist so, but that skew-Hermitian matrices are, ... with $[A,B]=AB-BA$, as always, ... for the unitary group. I really do think insertion of an $i$ factor is asking for trouble... $\endgroup$ – paul garrett Aug 12 '15 at 22:46
  • $\begingroup$ @paulgarrett: In Dirac's papers on quantum mechanics, $[A,B]$ denotes $i(AB-BA)$ (actually $i\hbar(AB-BA)$ but I'm happy to set $\hbar=1$). $\endgroup$ – WillO Aug 12 '15 at 23:07
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The answer to your question is, that for matrices $A$ and $B$ in $M_n(K)$ the standard Lie bracket is the commutator, i.e., $[A,B]=AB-BA$. The skew-hermitian matrices form a Lie subalgebra under this bracket, the unitary Lie algebra. For the commutator of Hermitian matrices see here. Then the usual commutator does not give a Lie subalgebra (the commutator of two Hermitian matrices is not Hermitian in general).

Every finite-dimensional Lie algebra can be faithfully represented by matrices with the standard commutator $[A,B]=AB-BA$, in some $\mathfrak{gl}(m)$ by Ado's theorem. However, $m$ might be large. In this sense, the usual commutator is standard. The given Lie algebra becomes a Lie subalgebra of $\mathfrak{gl}(m)$ with Lie bracket $[A,B]=AB-BA$.

Denote by $[A,B]$ always the Lie bracket of the Lie algebra, and the commutator of $A$ and $B$ always by $AB-BA$. Consider for example the Lie algebra $\mathfrak{su}(2)$, consisting of Hermitian matrices, with the Pauli matrices as basis. Here the standard commutator does not define a Lie bracket; a Lie bracket $[A,B]$ is rather given by $[A,B]:=i(AB-BA)$. However, $\mathfrak{su}(2)$ can also be represented by matrices inside $\mathfrak{gl}(3)$ by Ado's theorem, such that the Lie bracket is given by $[A,B]=AB-BA$. These matrices are given by the adjoint representation for $\mathfrak{su}(2)$ with the new basis $(e_1,e_2,e_3)$ and brackets $[e_1,e_2]=2e_3$, $[e_1,e_3]=-2e_2$, $[e_2,e_3]=2e_1$ - see also here. For example, $$ ad(e_1)=\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & -2 \\ 0 & 2 & 0\end{pmatrix}. $$

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  • $\begingroup$ The elements of the Lie algebra $su(n)$ are not self-adjoint, so I'm afraid this doesn't help. $\endgroup$ – WillO Aug 9 '15 at 20:06
  • $\begingroup$ I still don't see how this addresses the question. The whole point of the question is that the usual commutator does not give a Lie subalgebra. This is exactly why, in contexts where $A$ and $B$ are understood to be Hermitian, there are two different "natural" interpretations of the bracket $[A,B]$ --- one that is compatible with the Lie algebra structure on $M_n(K)$ and the other of which puts a Lie algebra structure on the Hermitian matrices. I don't see how this answers the question of whether one of these interpretations is in some sense "standard". $\endgroup$ – WillO Aug 10 '15 at 1:15
  • $\begingroup$ Thanks for this, though I'd feel much better if I knew of even one textbook or paper that uses this convention in the Hermitian context --- or more generally, of a textbook or paper that uses $[A,B]$ to mean something other than the Lie bracket of $A$ and $B$ in the context of any Lie algebra. Do you know of one? $\endgroup$ – WillO Aug 11 '15 at 4:20
  • $\begingroup$ I appreciate any help you care to give, but the answer as it stands is not helpful, particularly since your followup comments contradict each other. I'm asking about a situation in which the commutator is not a Lie bracket. In one comment, you say that in such cases $[A,B]$ always represents the commutator. In your most recent comment, you say that you know of cases in which $[A,B]$ represents not the commutator but the Lie bracket. If you care to respond further, I'll be glad to know which of these you now believe, and gladder still if you have a reference. $\endgroup$ – WillO Aug 11 '15 at 13:08
  • $\begingroup$ No, they do not contradict. Perhaps we misunderstand each other. $[A,B]=AB-BA$ does not always represent a Lie bracket, but we can always pass to an isomorphic Lie algebra, by Ado, such that it becomes a Lie subalgebra of some $gl(n)$ with the standard commutator. I think this was exactly your question with Hermitian matrices. $\endgroup$ – Dietrich Burde Aug 11 '15 at 13:31

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