Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Here is a construction of a very broad class of "Lie-like" algebras, and I want to know more about them.

Here is the main definition: Suppose $\mathfrak{g}$ is a complex semsimple Lie algebra and $\Gamma$ is a finite abelian group. Define a "hybrid algebra" over $(\mathfrak{g},\Gamma)$ as a pair $(V,\Phi)$ where $V$ is a collection of $\mathfrak{g}$-modules $V_a$ indexed by $\Gamma$ and $\Phi$ is a collection of intertwining maps $\phi_{a,b}:V_a\otimes V_b\rightarrow V_{a+b}$ indexed by $\Gamma\times\Gamma$.

What is known about these things? Here are a few things that I know. This is a generalization which encompasses semisimple Lie algebras and classical Lie superalgebras. Thus, for example, a Lie superalgebra $\mathfrak{g}$ is a particular type of hybrid algebra over the pair $(\mathfrak{g}_0,\mathbb{Z}/2\mathbb{Z})$, where $\mathfrak{g}_0$ is the even component of $\mathfrak{g}$. Similarly, one may realize the exceptional $52$-dimensional Lie algebra $\mathfrak{f}_4$ as a hybrid algebra over the pair $(\mathfrak{d}_4,V)$, where $V$ is the Klein four group and using the triality representations of $\mathfrak{d}_4\cong\mathfrak{so(8)}$. There are similar (and similarly elegant) constructions for other semisimple Lie algebras.

Beyond Lie algebras and Lie superalgebras, what are the interesting classes of hybrid algebras? What else can we say?

share|improve this question
    
For the case of $\mathbb{Z}/2\mathbb{Z}$ a related question is this one: mathoverflow.net/questions/2828/… –  José Figueroa-O'Farrill Dec 3 '10 at 22:24
    
Could you please provide a reference for the $\mathfrak{f}_4$ construction? –  Vít Tuček Dec 3 '10 at 23:00
    
I would also love to see some references! If this is new work, I'd love to see the paper when it's written. My e-mail adress is on my Berkeley website (linked from my profile here). –  Theo Johnson-Freyd Dec 4 '10 at 3:30
    
r0b0t: Have you ever heard of Freudenthal's magic square? The construction of $\mathfrak{f}_4$ in this framework, using triality, is fairly standard. There is a Wikipedia article about it, for example. As far as I know, the concept of a hybrid algebra is new. (Someone correct me if I am wrong.) Thus, I don't know of a printed reference which explains $\mathfrak{f}_4$ in this context. –  David Richter Dec 4 '10 at 14:35
add comment

2 Answers 2

There was an investigation of semi-simple Lie algebras graded with finite-generated abelaian groups. These algebras are the examples of your hybrid algebras. There is a good invariant theory of representation $\mathfrak{g}\colon V_a$ (and $G\colon V_a$) for any $a\in\Gamma$. Here is the useful referrence: http://www.math.msu.su/department/algebra/staff/timashev/metabel.ps

share|improve this answer
add comment

The general idea smells like coloured Lie superalgebras. In a nutshell, take the category of $\Gamma$-graded vector spaces and skew braiding by a bicharacter of $\Gamma$. Now consider Lie algebras in this category.

If I understand correctly, things reduce to either graded Lie algebras or graded Lie superalgebras, and you don't get any new interesting theory or examples. There is a book by Bakhturin-Mikhalev-Petrogradsky-Zaicev on them but it costs like a second-hand car...

share|improve this answer
1  
I have seen this construction for color Lie superalgebras. Hybrid algebras are much more general because there is a priori no specific conditions on the intertwining maps. That is, the reason you don't get any particularly new theory with color Lie superalgebras is that one also requires a twisted version of the Jacobi identity. Thus, a significant problem for hybrid algebras is to seek weaker forms of the Jacobi identity (additional conditions on the intertwining maps) which may lead to interesting structures. –  David Richter Dec 4 '10 at 13:22
    
@David, if you do not require any conditions on the intertwining maps, then the theory of these hybrid algebras is going to be quite arid... –  Mariano Suárez-Alvarez Dec 4 '10 at 16:36
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.