Linear combinations of structure constants

Main Question or Discussion Point

I have two Lie algebras with structure constants [tex]f^{a}_{bc}[/tex] and [tex]g^{a}_{bc}[/tex], the number of generators being the same (as will become clear).

Due to a particular symmetry/construct, I have that the system needs to be valid under [tex]g \to af + bg[/tex] (and a similar transform for g), which leads through to the Jacobi constraint [tex]f^{d}_{[ab}g_{c]d}^{e} + g^{d}_{[ab}f_{c]d}^{e} = 0[/tex], once you take into account that f.f = g.g = 0.

Now this new equation is trivially satisfied if f=Ag for some number A. However, I want to know if there's some way of working out what systems are or aren't possible to exist under this symmetry. For instance, if f is from a bunch of U(1)'s then it's trivially true for any g. But what if I have an SU(2) group? To be more specific, I'm working over 6 generators so the list of nice Lie algebras is pretty short but the list of other algebras such as combinations of 2 or 3 generator algebras and the Nilpotent algebras aren't quite so short.

Rather than computing an explicit form of f and g for every pair of possible algebras, can I say "The algebra [something] with structure constant f cannot pair with the algebra [something] because ..." (well I cannot think of any such examples). Does it matter if one of them is rotated away from the canonical form? The model which this relates to doesn't allow for independent rotations of the structure constants.

I find it hard to believe that there isn't some kind of restriction on what two algebras can be combined. But I've no idea how to go about finding such methods.

Even a few pointers, not a complete walk through, would be nice, so I can go through the method on my own to make sure I end up grasping it. Thanks :)
 

Answers and Replies

fresh_42
Mentor
Insights Author
2018 Award
11,983
8,406
If you only have the structure constants, then how should a non algorithmic answer to your question even look like? You can compute many invariants and compare them, e.g. the inner and outer derivations, the nilradical etc.

Every Lie algebra can be written as a semidirect sum of its semisimple part and its radical. For dimension six algebras, the semisimple parts are quite restricted. Hence you should determine radical and semisimple part and then you know which Lie algebra you have and whether f and g are isomorphic or not. If the semisimple part is ##\{\,0\,\}##, however, things quickly become complicated.
 
Wow, you found a post from... (checks date) 11.5 years ago.

Thanks for the reply regardless. Just so you know, there is an algorithmic way to do it IF you know the specific set of Lie algebras f and g can be the structure constants of but it's highly non-trivial - in the case I was interested in you could only apply transformations involving integers (ie an SL(N,Z) transform in the basis of generators) and finding such integer solutions required a TON of computing power and custom written algebraic geometry algorithms in Singular.

It ended up comprising a major section in my 2009 PhD thesis after being the main result in a paper I authored in 2008. If you're REALLY bothered I'll message you a link to the paper.

Got my PhD at least.
 
fresh_42
Mentor
Insights Author
2018 Award
11,983
8,406
Congratulations! What was its title?
Link it. One never knows what it is good for nowadays in a world ruled by links.

If you're REALLY bothered I'll message you a link to the paper.
I know that algorithmic solutions for whatever Lie algebra problem is a mountain of work. I once calculated some lower bounds for the bilinear complexity of Borel subalgebras of the simple ones. I did the exceptionals by a computer program.
I have also seen algorithms for the degree of nilpotency or similar. The further away from simplicity you get the crazier the zoo is.

The only algorithms I would be interested in is to calculate
$$
\mathfrak{A(g)} = \{\,\alpha\, : \,\mathfrak{g}\longrightarrow \mathfrak{g}\,|\,[\alpha(X),Y]+[X,\alpha(Y)]=0 \,\forall \,X,Y\in \mathfrak{g}\,\}
$$
given the multiplication table of ##\mathfrak{g}##. (I haven't MathLab.)
 
Last edited:
https://arxiv.org/abs/0811.2190
I'm George, not Adolfo. By the sounds of it you're more general and abstract than my approach.

We came to the problem by selecting an orientifold in string theory with a particular symmetry, an isotropic 6d torus - this meant all structure constants must have cyclic symmetry and some parity conditions. The dimensionality of the relevant cohomology was 6 so we needed to consider/explore all 6d Lie algebras for the right symmetries - these have been full classified. That took us down to around half a dozen algebras whose mutual f.g + g.f = 0 compatibility conditions under the SL(Z) transforms T and S dualities induced needed solving.

This could be viewed as cohomology deformations on the underlying mirror dual set of orientifolds and f.g + g.f = 0 effectively restricted the GL(6,6) group of T duality transforms down into a set we had to identify. Each pairing then had to be explored individually. The S duality's SL(2,Z) transform was the origin of the f.g+g.f mixing. To get valid and stable flux compactifications we needed integer (quantised flux) solutions of the vacua equations of motion, hence algebraic geometry and Sturm counting and Singular and so many reducible algebraic varieties...

In the end it came down to finding an SL(2,Z) mixing which could glue together the relevant symmetries of each algebra represented by f and g. If you used invariance to get rid of arbitrary transformations on certain solutions for f and g separately you could completely classify the mixed algebras.

If that doesn't make sense it's because I'm writing it off the top of my head a decade later without rereading or even opening the paper.

If you have specific questions about parts of that paper I can try to answer. I think my coauthor is still in academia, though I haven't checked in a few years whilst I went into mathematics research in the private sector in 2010.

Took me, literally, 7 years before I was willing to even open my thesis again.
 

Related Threads for: Linear combinations of structure constants

Replies
3
Views
1K
  • Last Post
Replies
15
Views
682
  • Last Post
Replies
7
Views
5K
  • Last Post
Replies
12
Views
7K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
2
Views
2K
Top