Listing all Lie algebras of dimension n

Discussion in 'Physics & Math' started by AlphaNumeric, Aug 29, 2008.

  1. AlphaNumeric Fully ionized Registered Senior Member

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    For reasons too convoluted to go into, I'm looking for a list of all Lie algebras of a particular dimension, namely 12. I know all the 'nice' ones are known, which are semisimple and the main problem is the nilpotent ones. I have a list of all 6 dimensional nilpotent algebras and listing all the 'nice' algebras isn't a problem. Unfortunately I now want to push this to 12 dimensions and it would seem that past n=7, nilpotent algebras are still largely unlisted.

    I have other constraints, such as there being a 6 dimensional subalgebra within that 12 dimensional algebra and the generators have to satisfy particular symmetries. I am half dabbling with the idea of trying to write a Mathematica routine to compute it for me but something tells me it'd grind to a halt at n=12 due to combinatoral considerations. To that end I have found this but how I'd make that into a working algorithm, for even the very restricted system I'm considering, I don't know...

    Anyone know of any work on this? Gilmore talks about it a bit, but nothing really specific.
     
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  3. Guest254 Valued Senior Member

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    I don't think you're going to find anything. And actually compiling such a list would constitute a substantial paper in itself!

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  5. AlphaNumeric Fully ionized Registered Senior Member

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    Yeah, I've realised that I'm not even looking for the 'nice' ones, such as \(\mathfrak{su}(2)^{4}\), I'm looking for the ones which contain particular 6 dimensional subalgebras but cannot be written as the sum of that algebra and another subalgebra! I have particular simplifications to work with, for instance I know there's less than 40 variables in my structure constant, despite the full general structure constant having about 800 variables but I have no idea how to go about listing them.

    In the case of the 6d version, I was looking for the 'nice' ones, like \(\mathfrak{su}(2)^{2}\) or \(\mathfrak{su}(2)\oplus \mathfrak{u}(1)^{3}\). The only nilpotent algebra with the right properties could be read off from the link in my first post, but now I cannot do that for the 12d ones. Ah nuts!

    I have a way of finding all structure constants isomorphic to a given canonical structure constant but I need to know the canonical structure constants first!!
     
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  7. rpenner Fully Wired Valued Senior Member

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    Ok, I've tried to come up to speed on the initial problem space of finding all algebras in vector spaces of \(n\) dimensions, where you have \(n \times \left( \begin{array}{cc} n \\ 3 \end{array} \right)\) equations in \(n \times \left( \begin{array}{cc} n \\ 2 \end{array} \right)\) variables (or \(n^3\) variables if you like, but then you get a lot of equations that set them to zero or equal and opposite to others). Or am I crazy to think that the Jacobi identities of basis vectors written out longhand in terms of the basis expansion vectors to be a natural starting point? So since \(n > 2 + 3\) there are more equations than unknowns, would you tackle this as sort of a computer sudoku problem (up to isomorphism) over a limited subset of \(\mathbf{Q} \left[ \surd , i \right]\)? I'm not sure, but \(\left\{ \left. \sqrt{p} / q \right| p \in \left\{ -n^2 ... n^2 \right\} , q \in \left\{ 1 ... n \right\} \right\}\) looks like a likely starting point to me. But then, I would want to prove that before basing my Ph.D. defense on it.

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    Mind you, 2000+ equations like \(\begin{eqnarray*} 0 & = & K_{aba} K_{aig} & + & K_{abb} K_{big} & + & K_{abc} K_{cig} & + & K_{abd} K_{dig} & + & K_{abe} K_{eig} & + & K_{abf} K_{fig} \\ & & & - & K_{abg} K_{aia} & - & K_{abg} K_{bib} & + & K_{abg} K_{gig} & + & K_{abh} K_{hig} & - & K_{abj} K_{ijg} \\ & - & K_{abk} K_{ikg} & - & K_{abl} K_{ilg} & - & K_{acg} K_{bic} & - & K_{adg} K_{bid} & - & K_{aeg} K_{bie} & - & K_{afg} K_{bif} \\ & & & - & K_{agg} K_{big} & - & K_{ahg} K_{bih} & + & K_{aic} K_{bcg} & + & K_{aid} K_{bdg} & + & K_{aie} K_{beg} \\ & + & K_{aif} K_{bfg} & + & K_{aig} K_{bgg} & - & K_{aig} K_{bii} & + & K_{aih} K_{bhg} & + & K_{aii} K_{big} & + & K_{aij} K_{bjg} \\ & & & + & K_{aik} K_{bkg} & + & K_{ail} K_{blg} & - & K_{ajg} K_{bij} & - & K_{akg} K_{bik} & - & K_{alg} K_{bil} \end{eqnarray*} \) looks like a job for a computer to me. But keeping track of the signs, identifying pivotal unknowns to queue up for guessing and importing the desired 6-dimensional algebra (90 vars?) is exactly where you want the computer assist in finding the remaining 702 vars.

    Or is it rubbish? I certainly don't feel brilliant in this, my inaugural post. But I look forward to associating with you two as best I can. Someday I might actually be able to read the physics thesis.

    -RP

    TeX is fun.
     
    Last edited: Sep 1, 2008
  8. Guest254 Valued Senior Member

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    Hi rpenner,

    It's possible I've misunderstood your post, but are you sure you are addressing the same problem?
    The dimension of the representation need not be the same as the dimension of the Lie algebra itself.
     
  9. AlphaNumeric Fully ionized Registered Senior Member

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    Thanks Rpenner, but as Guest says, there's additional complications. There are ways to get a representation for an n dimensional Lie algebra which is in an n x n matrix format for the generators, the adjoint (or fundamental, depending on textbook or lecturer) representation, where \((T_{a})^{b}_{c} = f^{b}_{ac}\) (or something like that), so you only need to know the canonical structure constant to automatically get one explicit representation for the Lie algebra. It's the canonical structure constants which are the problem.

    To explain what I mean about the simplifications, if we're working in 3 dimensions (so there's 3 generators), you can have Lie algebras like the Heisenberg algebra, generated by x, p and 1, so [x,1] = [1,p] = 0 and [x,p] = 1. All consistent and satisfies the requirements for it to be a Lie algebra. But it's not semisimple. Another 3 generator Lie algebra is su(2), the algebra of SU(2). In this case you have that \([T_{1},T_{2}] = T_{3}\) and similarly for permutations of 1,2,3, so \([T_{i},T_{j}] = \epsilon_{ijk}T_{k}\). Since \(\epsilon_{ijk}\) is the only isotropic rank 3 tensor, if we say "Our generators must come in 3's and must be isotropic" then we have reduced all our considerations to just algebras of the form \([T_{i},T_{j}] = k \epsilon_{ijk}T_{k}\). If \(k \not= 0 \) then the algebra is isomorphic to su(2), via a rescaling, if k=0 then it's u(1)^3, ie abelian.

    Thankfully, such a simplification exists in our considerations (well, we impose it or my computer has a heart attack when trying to do change of bases!) and so we know the generators come in triplets and the structure constants are sums of epsilons. When you work out all the allowable triplets and epsilons mixing them in 6 dimensions, you have 6 variables and in 12 dimensions its 40, since if d = 3n then number of variables = \(\frac{n^{2}(n+1)}{2}\), since it's n lots of triplets and you can combine the triplets in Triangular(n) ways, including with themselves.

    6 variables in a 6x6x6 structure constant is enough for Mathematica, Singular and a 2.4Ghz computer to do pretty much instantly. 40 in a 12x12x12 structure constant gives my computer a hernia.
     
  10. rpenner Fully Wired Valued Senior Member

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    Sorry, Guest254. Do you think I should have spoken about n generators? I really haven't internalized most of Lang's Algebra or all of Georgi's Lie Algebras in Particle Physics. Neither of which tackles general Lie algebras in depth....

    Well, obviously a lot of education separates us, so it very well might be that I am confused. But my original thought was that \([ u_a T_a , v_b T_b ] = K_{abc} u_a v_b T_c\) should hold for any Lie algebra, with \([ T_a , T_a ] = 0\) implying \(K_{aac} = 0\), \([ T_a, T_b] = -[ T_b, T_a]\) implying \( K_{abc} + K_{bac} = 0\) and the Jacobi identity giving us \({ {n^2(n-1)(n-2)} \over {6} } = n \times \left( n \\ 3 \right)\) additional constraints on the remaining \({ {n^2(n-1)} \over {2} } = n \times \left( n \\ 2 \right)\) unknown elements of \(K_{abc}\). I'm aware that the general Lie algebra could be over some finite field or space of rational functions, but I assumed that we were talking complex numbers. (The choice of K seems to be a poor one, now that I have seen the draft paper, but I didn't want to use the convention of Georgi, \([ T_a , T_b ] = i f_{abc} T_c\), which conviniently results in real f when the algebra is unitary. So lets drop K in favor of f. ) But since I was under the impression that \(n > 2 + 3 \Rightarrow n \times \left( n \\ 3 \right) > n \times \left( n \\ 2 \right)\) that it should be possible to automate solution over a small number of algebraic numbers of low degree. (I'm not saying give it to Mathematica's Solve[], but write custom software to help explore it, and ideally solve it completely. Solve[] is impressive but when it fails or takes to you get no closer to a solution. Also, I don't have it.

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    )

    Since then, I have realized that \(ab + cd = 0\) does not imply that any of a,b,c or d is algebraic, but it still branchs into two cases: \(a = 0 \mathrm{ and } cd = 0\) or \(a \neq 0 \mathrm{ and } b = { -cd \over a}\) and with so many more equations than unknowns, I thought computer methods had real potential, even if I can't prove that. (That's the type of identification of cases and small theorems that remind me of sudoku.)

    I think that additional information like the isometries and equation 8.1 from the draft paper just makes things easier. But, as I'm still struggling to read the paper and have not been able to set up the problem in a form which is entirely comphrehensible for myself. For example, I don't know how to express my 792 unknowns in terms of AlphaNumeric's 40 unknowns. And I didn't know the result AlphaNumber quotes where [a,b] = c is the same algebra as [a,b] = 2c.

    I'm not disagreeing with Guest254 that this is a daunting problem whose solution would be worthy of a paper of its own. I'm just interested in exploring of how to get man and machine to work together on this problem. Sorry if I've damaged myself in your view. One never feels so stupid as when one is learning.

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  11. rpenner Fully Wired Valued Senior Member

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    And now, I begin to suspect that there is no such list or magic computer solution for the general solution.

    Chong-Yun Chao. "Uncountably Many Nonisomorphic Nilpotent Lie Algebras" Proceedings of the American Mathematical Society 13, 6 (Dec., 1962), pp. 903-906

    Lawrence J. Corwin and Frederick P. Greenleaf. Representations of Nilpotent Lie Groups and Their Applications Cambridge University Press, 1990, Page 210

    There are uncountably many nilpotent Lie algebras of dimension >= 10. (Do I know what that means? Only that there is no explicit list. There might be a parameterized list of families or it might be horrible.)

    But since you know a 6-d subalgebra and the vastly simplifying isomorphism solutions, and already have done tons of heavy lifting, perhaps the specific solution(s) will turn out to be tractable.

    Wecht's paper helps, but I'm not at all up to speed on the notation and concepts.
     
  12. AlphaNumeric Fully ionized Registered Senior Member

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    Demonstration that Lie algebras with proportional structure constants are isomorphic :

    Suppose you have the Lie bracket \([T_{a},T_{b}] = f^{c}_{ab}T_{c}\). Let \(T_{a} = \lambda S_{a}\) then \([\lambda S_{a},\lambda S_{b}] = f^{c}_{ab}\lambda S_{c}\) which simplifies to \([S_{a},S_{b}] = \frac{1}{\lambda} f^{c}_{ab} S_{c}\). Therefore, provided \(\lambda\) is invertible (ie non-zero and finite) the two algebras are the same, just a rescaling of the generators.

    This is why the algebra su(2), when derived from SU(2), is \([T_{a},T_{b}] = 2 \epsilon^{c}_{ab}T_{c}\) but physicists prefer \([\sigma_{a},sigma_{b}] = i \epsilon^{c}_{ab}\sigma_{c}\). Also, having your generators pick up a factor of i often takes them from a compact setup to a non-compact setup or vice versa. It's the 'Weyl trick', as discussed in Gilmore. For instance, the usual generators of so(1,1) give you elements of SO(1,1) of the form \(\left( \begin{array}{cc} \cosh n & \sinh n \\ \sinh n & \cosh n \end{array} \right)\), but if you'd exponentiated the i times the generators of so(1,1) you'd get \(\left( \begin{array}{cc} \cos n & -\sin n \\ \sin n & \cos n \end{array} \right)\), which are elements of SO(2), which is compact, since the entries in the elements are bounded.
     
  13. Guest254 Valued Senior Member

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    No, I think I take blame for this one! For some reason, I read your quote as "on vector spaces" not "in vector spaces", hence my talk of representations! Had I read the rest of your post, this might have been more apparent - let that be a lesson to me.

    Sadly, I still don't think this problem is going to be all that tractable, purely because there seems to be a fair amount on this type of classification problem for n much less than 12! I'm not at all up to speed with the computer-type-attack, but I imagine if it was straight forward to write up a piece of software to sort the problem, it wouldn't be much of a problem to start with!
     
  14. AlphaNumeric Fully ionized Registered Senior Member

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    Am I right in saying that the main problem in a comprehensive list of 12d algebras are the nilpotent ones? The semisimple ones must fall into sums of algebras from groups in the classical Cartan classification, right?

    For instance, if I know my algebra has 8 generators and isn't nilpotent then it's a combination of various classical algebras, like su(3) or say su(2)+su(2)+u(1)+u(1) and stuff like that, right?

    Because those I can compute in a pretty quick and straight forward manner and I'm sure would have been done before anyway. Infact, once you exclude all the algebras which are of the format G = g+h where g is our 6d algebra, there cannot be that many 12 dimensional semisimple algebras. After all, su(3) is 8 dimensional and su(4) is 15 dimensional so already you're looking at a pretty restricted list. At least that gives us something to work with and the fact such algebras are non-degenerate Killing forms means you can do a ton of stuff to do with transformations and metrics etc, which is something I think has a lot of potential in terms of our work. The one nilpotent algebra in 6 dimensions which meets our symmetry restrictions is generally phenomenologically useless for reasons too convoluted to go into just yet.
     
  15. Guest254 Valued Senior Member

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    Yes. For the semisimple case, just compute the killing form \(\kappa\), find an ideal \(\mathcal{I} \subset \mathcal{L}\) and compute its orthogonal compliment wrt to \(\kappa\) then you have \(\mathcal{L}= \mathcal{I}\oplus \mathcal{I}^{\perp}\). Now repeat...
     
  16. AlphaNumeric Fully ionized Registered Senior Member

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    Okay, here's some thinking out loud and if someone sees a mistake in my logic, point it out:

    I want a list of the 12d algebras, g, which are isotropic (so the generators come in triplets) and have a 6d subalgebra, call it (following Gilmore) t and call the vector space compliment of t in g p. But I cannot split the 12d ones which are of the form 6+6 or 6+3+3, because I know the algebras mingle : [t,t]=t, [p,p] = t, [p,t] = p, so, in a Lie algebra context \(g \not= t \oplus p\). I also know that (t,p)=0, where the inner product is defined via the Killing form, even when t is not nilpotent.

    There are no simple 12d Lie algebras, only semisimple or nilpotent ones. The semisimples could work only if they split as 9+3, since I can still hide t within the 9d algebra. There's no 9 dimensional simple algebras so I'm left looking at either 12d nilpotent algebras or 9d nilpotent algebras with a 3d simple algebra tagged on, so su(2) or u(1)^3.

    I can express the Killing form for the 12d algebra in terms of the structurs constants in [t,t]=t, [p,p] = t, [p,t] = p, which comes out to be block diagonal and one block is proportional to the Killing form of t. Since in the 12 or 9+3 cases the nilpotent algebra should have identically zero Killing form, can I then surmise that the only algebras t can be, if I'm embedding it within this 12d algebra with the triplet nature of it's generators, are themselves nilpotent, ie you cannot have nilpotent algebra and find there's a semisimple or simple subalgebra in there.

    Does that make sense to anyone? Or haven't I explained it very well (which is quite likely).
     
    Last edited: Sep 4, 2008
  17. pasha Registered Member

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    1. Your problem, as I understand it (though I do not fully understand what "isotropic" and "comes in triplets" mean), seems to be hopelles. Classification of all nilpotent Lie algebras of dimension >7 (perhaps >8) is out of reach. The condition that algebra is not decomposable into the sum of subalgebras does not help much, as such algebras still comprise "almost all" algebras (in terms of either Zariski or ordinary topology on the variety of all structure constants).

    2. GAP (or SAGE which includes GAP) and probably Magma have Lie algebras capabilities which largely surpass those of Mathematica.

    3. Probably you will get a much better and elaborated response at liealgebras yahoo group (the ingenious system here prevents me from posting a link, you may google for it).
     
  18. AlphaNumeric Fully ionized Registered Senior Member

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    I've been scribbling away at some thoughts I've had over the last day or two (typical thing, I do no work during the day, then from 6pm I work furiously) and I've got a few restrictions. I'm going to type it up into LaTeX and I'll put it online in the not too distance future to get some feedback.

    Pasha, thanks for the comments. I've heard of Magma but haven't looked into it. I had to learn Singular last year and now I'm pretty proficient at it and getting it and Mathematica to make sweet maths love with one another but it took a while (I used to be unable to get a decent nights sleep due to Singular and Mathematica code going around in my head from the day's work

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    ) and if I can avoid it I'd prefer not to go down that route just yet. I'll keep it in mind though.

    And the 'isotropic' and 'come in triplet' things are the result of particular symmetries I'm imposing in a string theory model I've been researching. This then filters through to the Lie algebras via a series of steps I hope to soon get published (I'm sitting on a paper and have been for months waiting for some other people to pull their finger out and publish their work first and then I follow up quickly with the generalisation). Because string theory needs 6 extra dimensions, you end up with spaces which have spaces which have symmetries with 6 generators. The isotropic thing is a constraint on the space to be even more symmetric. Thus the generators go from 6 independent things to 2 lots of three 'coupled' generators.

    To give an example, suppose you have 3 directions, x, y and z and your 6 generators split into 2 types, generators of translations in the x, y and z directions by \(\delta_{x}e_{x}\), \(\delta_{y}e_{y}\) and \(\delta_{z}e_{z}\) and 3 generators which do rotations about the x, y and z axis , written as say \(\theta_{x}e_{x}\), \(\theta_{y}e_{y}\) and \(\theta_{z}e_{z}\). The isotropy just says that \(\delta_{x}=\delta_{y}=\delta_{z}\) and similarly for the rotations. You're making their effect the same, in a manner of speaking.

    I can go into more detail if you wish.
     
  19. lucifers angel same shit, differant day!! Registered Senior Member

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    i have already admitted in anouther thread that i dont know much about algebra, so can someone show me basic algebra equasions?
     
  20. AlphaNumeric Fully ionized Registered Senior Member

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    'Algebra', in the way schools use it, just means you're using letters to represent quantities rather than specific numbers and often you try to rearrange things to work out what those quantities are. For instance, 2x-3 = 5. Rearranging gives x=4.

    The 'algebras' I've been talking about are a more technical notion than that, http://en.wikipedia.org/wiki/Algebra_over_a_field . If you're unfamiliar with the school concept of algebra, you're not going to understand that link I'm afraid and any equation we give you relating to this thread topic will be beyond you too.
     
  21. pasha Registered Member

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    Yes, please.
     
  22. AlphaNumeric Fully ionized Registered Senior Member

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    A colleague and I have been working in the area of string theory which is discussed in detail by Wecht here (Rpenner and I have talked about it before, hence why he mentioned Wecht earlier in the thread). The background to the theory is mostly in that paper. The orbifold we're considering is constructed in such a way that the compact space in the model is not \(T^{6}\) but actually the product of 3 different \(T^{2}\). However, if we impose isotropy, we are saying it's no longer \((T^{2})_{1} \times (T^{2})_{2} \times (T^{2})_{3}\) but just \((T^{2})^{3}\). So we end up cutting the number of variables in the system down by about 1/3. It brings it within the ability of computers to make quick attacks on the problems at hand using the methods we've come up with.

    In the link, Wecht talks about X and Z generating a 12 dimensional Lie algebra. Under certain conditions, the X's form a subalgebra, so we examined that first, used our ideas and solved the problem in full generality. I then worked out how to include an additional thing which Wecht doesn't talk about (S duality, discussed here) and that's the paper I'm currently sitting on.

    But that was all for the 6d subalgebra. I'm now trying to work out how it all fits within the 12d (X,Z) algebra, hence some of the questions I've been asking. I didn't take any pure courses in Lie algebras as a student, I only did what amounted to "Lie algebra for physicists" and even simple questions like "Can a nilpotent Lie algebra have a non-nilpotent subalgebra" aren't immediately addressed in such a course, you spend 90% of the time talking about su(2) and su(3), hence I have a tendency to ask what turns out to be pretty stupid questions sometimes. Plus I use either sloppy physicist notation or worse, stuff I've cobbled together from various sources, so when I ask someone whose a dab hand at Lie algebras they have no idea what I'm talking about because my notation is non-standard.
     
  23. AlphaNumeric Fully ionized Registered Senior Member

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    http://www.hep.phys.soton.ac.uk/~g.j.weatherill/nilpotent.pdf

    Anyone got any comments? I manage to narrow down the number of unknowns in the structure constant considerably, particularly in the 9+3 case, all before applying the requirement it's a Lie algebra but there's still the matter of checking each combination and then seeing how many unique ones I get are.

    I'm wondering if this isn't isomorphic to some lower dimensional system, because the generators always come in triplets, because considering 3 or 4 dimensional systems would be much simpler (and orders of magnitude quicker on a computer).
     
    Last edited: Sep 7, 2008

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