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Finite dimensional complex simple Lie algebras are classified using Cartan matrices. One of the main ingredients is Serre Relations. Lets call this Cartan-Killing theory.

I have the following questions.

Let $\mathfrak{g}$ be a basic classical simple Lie superalgebra.

Is there a similar theory for $\mathfrak g $? i.e., can $\mathfrak g$ be associated a Cartan matrix?

and

is there Serre relation in Super setting to get back the algebra from the matrix?

Kindly share your thoughts.

Thank you.

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The Serre relations (some authors also call them Serre-Chevalley relations) for the finite dimensional, complex, basic, classical, simple Lie superalgebras -in analogy with the Lie algebra case- read: $$ (ad E_i^\pm)^{1-\tilde{a}_{ij}}E_j^\pm=\sum_{n=0}^{1-\tilde{a}_{ij}}(-1^n)\binom{1-\tilde{a}_{ij}}{n}(E_i^\pm)^{1-\tilde{a}_{ij}-n}E_j^\pm (E_i^\pm)^n=0 $$ where, the $E_i^\pm$ are the raising/lowering operators associated to the simple root system $\Delta^0$ and the matrix $\tilde{A}=(\tilde{a}_{ij})$ is derived from the corresponding Cartan matrix of the LS by replacing all its positive off-diagonal entries by $-1$.

However, the above relations -in contrast to the Lie algebra case- are not enough: The superalgebra generated by the usual Cartan-Kac relations, together with the above Serre-Chevalley relations is generally a bigger superalgebra than the one under consideration.

It can be shown that, the above relations need to be supplemented by higher order relations, generally involving more than two generators, in order to quotient the bigger superalgebra and recover the initial one. These supplementary relations (too complicated to be included in this post) generally depend on the different kinds of vertices which appear in the corresponding Dynkin diagram of the LS under consideration. You can find more details on these, and the explicit form of the supplementary relations at: Dictionary on Lie superalgebras, 1996, by L. Frappat, A. Sciarrino, P. Sorba, sect. 43, p. 62-63 (the supplementary relations are those in p. 63).
(There, you can also find the Cartan matrices and the Dynkin diagrams for all fin. dim., complex, basic, classical, simple Lie superalgebras).

There exists significant research on this topic in the mathematical physics literature (where frequently such constructions have been extended to the $q$-deformed case). See for example, for the case of $osp(1/2n)$ Lie superalgebras, the article: The quantum superalgebra $U_q\big(osp(1/2n)\big)$: deformed para-Bose operators and root of unity representations, 1995, by T.D. Palev, J. Van der Jeugt. The corresponding Serre-Chevalley relations, for the undeformed case, are eq. (2.11), p. 4.

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    $\begingroup$ Nice explanation. Thank you. the corresponding Cartan matrix of the LS by replacing all its positive off-diagonal entries by −1. What is LS ? please explain. $\endgroup$ – GA316 Feb 19 '19 at 3:29
  • $\begingroup$ Lie Superalgebra $\endgroup$ – Konstantinos Kanakoglou Feb 19 '19 at 3:31
  • $\begingroup$ Thank you 🙂 again $\endgroup$ – GA316 Feb 19 '19 at 4:18
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An analogue of Serre's relations is established for finite dimensional simple Lie superalgebras in https://arxiv.org/abs/1101.3114 (Serre presentations of Lie superalgebras, by R. B. Zhang).

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