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What type of algebraic structure is given by the following? \begin{align} [X, Y] &= iH,\\ [H, X] &= -i\{H, Y\},\\ [H, Y] &= i\{H, X\}, \end{align} where $[ \cdot,\cdot ]$ is a commutator and $\{ \cdot,\cdot \}$ is an anticommutator and $i$ is the imaginary unit.

aporete
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    What do you want to know about this thing? – Peter Kravchuk May 28 '17 at 22:56
  • I'm curious, how did this crop up? (Also, what's "$H$"?) – Noah Schweber May 29 '17 at 00:04
  • Thanks for comments. Dr NoahSchweber : This algebra arose from work on the "classical limit" of deformed q-bosons. If H is a function of X^2 + Y^2, then the algebra obeys the Jacobi Identity and thought it may be a disguised Lie algebra, or perhaps a Lie superalgebra. Prof PeterKravchuk: I wanted to know if it was a standard structure because it looks quite symmetric and interesting and I have an interested in deformed uncertainty relations. Prof @CosmasZachos : I'm afraid that I can't immediately see that H^3 = 0. Do I multiply out the commutator for H? – aporete May 29 '17 at 21:23
  • There are at least two ways you can interpret "algebraic structure" here: you might be asking whether this is like an algebra or a Lie algebra or what (the answer is that it is an algebra: in the presence of a commutator and an anticommutator you can recover a multiplication, at least if you're allowed to divide by $2$), or second, you might already know the answer to this question and want to know the name of this particular algebra or Lie algebra or whatever. – Qiaochu Yuan May 30 '17 at 03:35
  • Thank you Dr @QiaochuYuan. I appreciate that the set of elements with these relations form an algebra and I have some knowledge of Lie algebras and QUEAs (and not so much experience of super-versions of these). I was just interested as the formal structure seemed to be very simple and exhibit nice symmetry properties. I thought perhaps it was a simple example of a codified mathematical structure of which I am ignorant. – aporete May 30 '17 at 20:41

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I'm merely replying to your comment question: Apologies, $H^3$ does not vanish.

It is evident from your three relations that you may take H,X,Y hermitian.

It is also evident that the structure of your algebra simplifies upon defining $$ Z\equiv X+iY, \qquad Z^\dagger =X-i Y, $$ so that $$ H Z=0, \qquad Z^\dagger H=0 , \qquad [Z,Z^\dagger ]=2H ~. $$

So, then , multiplying the last relation (commutator) by Hs on the left and the right yields $HZ^\dagger ZH =-2H^3$, instead. It might not be too hard to find a simple matrix realization of the algebra.

Generically, as per your comment, it appears like a deformed QUE (quantized universal enveloping) algebra of an SU(2) Lie algebra. Ideally, you'd seek the singular map from the Lie algebra to it, as in this 1991 mini-review of mine.

Cosmas Zachos
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  • As you probably surmised, the algebra {X, Y, H} came from looking at the q-deformation of the complex plane. It looks like a "classical" counterpart of the q-boson algebra introduced by Kempf, back in the 90s. Many thanks for your input. – aporete May 30 '17 at 20:34
  • Yes, I did try to find a quick deformer for it in this talk but without ready success. – Cosmas Zachos May 30 '17 at 20:44