Given Leonid Positselski's excellent answer and comments to this question, I expect that the present one is a hard question. Recall that the Lie algebra structures on a (finite-dimensional over $\mathbb C$, say) vector space $V$ are the maps $\Gamma: V^{\otimes 2} \to V$ satisfying $\Gamma^k_{ij} = -\Gamma^k_{ji}$ and $\Gamma^m_{il}\Gamma^l_{jk} + \Gamma^m_{jl}\Gamma^l_{ki} + \Gamma^m_{kl}\Gamma^l_{ij} = 0$; thus the space of Lie algbera structures is an algebraic variety in $(V^\*)^{\otimes 2} \otimes V$. A Lie algebra structure $\Gamma$ is semisimple if the bilinear pairing $\beta_{ij} = \Gamma^m_{il}\Gamma^l_{jm}$ is nondegenerate; thus the semisimples are a Zariski-open subset of the space of all Lie algebra structures. Because the Cartan classification of isomorphism classes of semisimples is discrete (no continuous families), connected components of the space of semisimples are always contained within isomorphism classes. The semisimples are not dense among all Lie algebra structures: if $\Gamma$ is semisimple, then $\Gamma_{il}^l = 0$, whereas this is not true for the product of the two-dimensional nonabelian with an abelian.

  1. Is there a (computationally useful) characterization of the Zariski closure of the space of semisimple Lie algebra structures? (LP gives more equations any semisimple satisfies.)
  2. Suppose that $\Gamma$ is not semisimple but is in the closure of the semisimples. How can I tell for which isomorphism classes of semisimples is $\Gamma$ in the closure of the isomorphism class? (I.e. $\Gamma$ is a limit of what algebras?)
  3. To what extent can I understand the representation theory of algebras in the closure of the semisimples based on understanding their neighboring semisimple algebras?

For 3., I could imagine the following situation. There is some natural "blow up" of the closure of the semisimples, with at the very least each element of the boundary is a limit of only one isomorphism class in the Cartan classification. Then any representation of the blown-down boundary is some combination of representations of the blown-up parts.


3 Answers 3


As a warmup to this question you might want to think about the closure of semisimple associative algebras inside all finite dimensional asociative algebras of a given dimension. As a warmup to that question you might want to think about the commutative case. As a warmup to that question you might look at a beautiful paper by Bjorn Poonen (in particular, Section 6 and Remark 6.11).

  • $\begingroup$ I'll definitely look at the Poonen paper. I had intended to ask my question dimension-by-dimension; in the "follow-down" question (or whatever it's called that this is the follow-up to) I was more explicit a picked a finite-dimensional vector space over C and asked for Lie algebra structures on it. $\endgroup$ Dec 25, 2009 at 6:59
  • $\begingroup$ Oh, I reread what you wrote: when you say (semisimple) algebra, you mean (semisimple) associative algebra, as opposed to (semisimple) Lie algebra. $\endgroup$ Dec 25, 2009 at 7:02
  • $\begingroup$ Good point, fixed. $\endgroup$ Dec 25, 2009 at 17:59

A (hopefully helpful) comment. I've thought about the problem of finding the closure of one isomorphism class - I haven't got an answer but I had an idea that I hope is helpful towards a solution.

Consider the closure of the set S of Lie algebras isomorphic to a fixed semisimple lie algebra $L$ of dimension $n$, and fix a basis of $L$ which gives you the structure constants. Then there is a surjection from invertible matrices $GL_n(\mathbb{C})$ to $S$ , namely by acting on a fixed standard basis of $V$ with $x \in GL_{n}(\mathbb{C})$ to give you another basis, and now you force this other basis to have the properties of the basis of $L$ fixed above, i.e. the structure constants - then trace this back to get the values of $\Gamma^k_{ij}$ defining this particular Lie algebra.

To be precise with the above, I think it is best described as a transitive action of the algebraic group $GL_{n}(\mathbb{C})$ on the variety $S$. I think there are some matrices which act trivially however, and that these matrices correspond to automorphisms of the Lie algebra (which leave the structure constants invariant) – i.e. the point stabilizers correspond to automorphisms of the Lie algebras, so the homogenous space has the structure of the quotient of $GL_{n}(\mathbb{C})$ by this point stabilizer, which is the Lie group of automorphisms of $L$.

I think this could help getting the closure of a single isomorphism class of Lie algebras (and since there are only finitely many isomorphism classes of semisimple Lie algebras of fixed dimension, should help with that problem too). But I’m not sure how – I tried naively by saying that perhaps this closure consists of the union of isomorphism classes which you get, in an intuitive sense, by replacing the invertible matrix $x$, by allowing singular matrices as well; but what I get from that seems to be some rubbish so I’m sure that path is mistaken.


This is a really a comment to rajamanikkam's answer, but it does not fit in the comment box.

What rajamanikkam is describing in the last paragraph is essentially a Lie algebra contraction of $L$. It is well known that this is how one obtains Lie algebras which are close in some sense to the original Lie algebra. (I'm afraid that I do not speak the right language, so I am not sure whether this is in the Zariski closure.)

One way to define a contraction of a given Lie algebra $L$, say complex of dimension $n$, is to consider a continuous curve $A(t)$ in $\mathfrak{gl}_n(\mathbb{C})$ which for $t$ in some interval, say $[1,\infty)$, lies in $\mathrm{GL}_n(\mathbb{C})$. For $t$ in that interval the Lie algebras $L(t)$ related to $L$ via $A(t)$ will be be isomorphic to $L$, but if $\lim_{t\to\infty} L(t)$ exists -- which is by no means the case for all curves $A(t)$ -- then it will give rise to a Lie algebra which may or may not be isomorphic to $L$.

I'm not sure if contractions are sufficient to generate the full closure, or indeed whether this is the sort of closure that the originally question intended.

  • $\begingroup$ Ah right, thanks! I am fairly sure that they do not generate the full closure - from the constructions I tried above, these seem to be "smaller" homomorphic images of the Lie algebra L in part, combined with some "other stuff" to account for the rest of the dimension; for simple Lie algebras homomorphic images would be trivial so it's entirely "other stuff", and I'm not sure what restrictions there would be on this "other stuff". I'm sure someone with more experience can help. $\endgroup$
    – Vinoth
    Dec 25, 2009 at 6:46

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