The "Motivation" section is a cute story, and may be skipped; the "Definitions" section establishes notation and background results; my question is in "My Question", and in brief in the title. Some of my statements go wrong in non-zero characteristic, but I don't know that story well enough, so you are welcome to point them out, but this is my characteristic-zero disclaimer.


In his 1972 talk "Missed Opportunities" (MR0522147), F. Dyson tells the following story of how mathematicians could have invented special and much of general relativity long before the physicists did. Following the physicists, I will talk about Lie groups, but I really mean Lie algebras, or maybe connected Lie groups, ...

The physics of Galileo and Newton is invariant under the action of the Galileo Group (indeed, this is the largest group leaving classical physics invariant and fixing a point), which is the group $G_\infty = \text{SO}(3) \ltimes \mathbb R^3 \subseteq \text{GL}(4)$, where $\text{SO}(3)$ acts as rotations of space, fixing the time axis, and $\mathbb R^3$ are the nonrelativistic boosts $\vec x \mapsto t\vec u$, $t\mapsto t$. This group is not semisimple. But it is the limit as $c\to \infty$ of the Lorentz Group $G_c = \text{SO}(3,1)$, generated by the same $\text{SO}(3)$ part but the boosts are now $$ t \mapsto \frac{t + c^{-2}\vec u \cdot \vec x}{\sqrt{1 - c^{-2}u^2}} \quad \quad \vec x \mapsto \frac{\vec x + t\vec u}{\sqrt{1 - c^{-2}u^2}} $$ Since semisimple groups are easier to deal with than nonsemisimple ones, by Occam's Razor we should prefer the Lorentz group. Fortunately, Maxwell's equations are not invariant under $G_\infty$, but rather under $G_c$, where $c^{-2}$ is the product of the electric permititivity and magnetic permeability of free space (each of which is directly measurable, giving the first accurate measurement of the speed of light). Actually, from this perspective, $c^{-2}$ is the fundamental number, and we should really think of the Galileo Group as a limit to $0$, not $\infty$, of something.

But of course from this perspective we should go further. Actual physics is invariant under more than just the Lorentz group, which is the group that physics physics and a point. So Special Relativity is invariant under the Poincare Group $P = G_c \ltimes \mathbb R^{3+1}$. Again this is not semisimple. It is the limit as $r \to \infty$ of the DeSitter Group $D_r$, which in modern language "is the invariance group of an empty expanding universe whose radius of curvature $R$ is a linear function of time" (Dyson), so that $R = rt$ in absolute time units. Hubble measured the expansion of the universe in the first half of the twentieth century.

Anyway, I'm curious to know if it's true that every Lie group is a limit of a semisimple one: how typical are these physics examples? To make this more precise, I'll switch to Lie algebras.


Let $V$ be an $n$-dimensional vector space. If you like, pick a basis $e_1,\dots,e_n$ of $V$, and adopt Einstein's repeated index notation, so that given $n$-tuples $a^1,\dots,a^n$ and $b_1,\dots,b_n$, then $a^ib_i = \sum_{i=1}^n a^ib_i$, and if $v \in V$, we define the numbers $v^i$ by $v = v^ie_i$; better, work in Penrose's index notation. Anyway, a Lie algebra structure on $V$ is a map $\Gamma: V \otimes V \to V$ satisfying two conditions, one homogeneous linear in (the matrix coefficients) of $\Gamma$ and the other homogeneous quadratic: $$ \Gamma^k_{ij} + \Gamma^k_{ji} = 0 \quad\text{and}\quad \Gamma^l_{im}\Gamma^m_{jk} + \Gamma^l_{jm}\Gamma^m_{ki} + \Gamma^l_{km}\Gamma^m_{ij} = 0 $$ Thus the space of Lie algebra structures on $V$ is an algebraic variety in $V \otimes V^{*} \otimes V^{*}$, where $V^{*}$ is the dual space to $V$.

If $\Gamma$ is a Lie algebra structure on $V$, the corresponding Killing form $\beta$ is the symmetric bilinear pairing $\beta_{ij} = \Gamma^k_{im}\Gamma^m_{jk}$. Then $\Gamma$ is semisimple if and only if $\beta$ is nondegenerate. Nondegeneracy is a Zariski-open condition on bilinear forms, since $\beta$ is degenerate if and only if a certain homogeneous-of-degree-$n$ expression in $\beta$ vanishes (namely $\sum_{\sigma \in S_n} (-1)^\sigma \prod_{k=1}^n \beta_{i_k,j_{\sigma(k)}} = 0$ as a map out of $V^{\otimes n} \otimes V^{\otimes n}$, where $S_n$ is the symmetric group on $n$ objects and $(-1)^\sigma$ is the "sign" character of $S_n$). Since $\beta$ is expressed algebraically in terms of $\Gamma$, semisimplicity is a Zariski-open condition on the variety of Lie algebra structures on a given vector space.

Incidentally, Cartan classified all semisimple Lie algebras (at least over $\mathbb C$ and $\mathbb R$) up to isomorphism, and the classification is discrete. So any two semisimple Lie algebras in the same connected component of the space of semisimple structures are isomorphic.

My Question

Is the space of semisimple Lie algebra structures on $V$ dense in the space of all Lie algebra structures on $V$? (I.e. if $\Gamma$ is a Lie algebra structure on $V$ and $U \ni \Gamma$ is an open set of Lie algebra structures, does it necessarily contain a semisimple one?) This is really two questions. One is whether it is Zariski-dense. But we can also work over other fields, e.g. $\mathbb R$ or $\mathbb C$, which have topologies of their own. Is the space of semisimple Lie algebra structures on a real vector space $V$ dense with respect to the usual real topology?

(The answer is no when $\dim V = 1$, as then the only Lie algebra structure is the abelian one $\Gamma = 0$, which is not semisimple, and it is no when $\dim V = 2$, as there are nontrivial two-dimensional Lie algebras but no semisimple ones. So I should ask my question for higher-dimensional things.)

Edit: I have posted the rest of these as this follow-up question.

If the answer is no in general, is it possible to (nicely) characterize the Lie algebra structures that are in the closure of the semisimple part?

A related question is whether given a nonsemisimple Lie algebra structure, are all its nearby semisimple neighbors isomorphic? Of course the answer is no: the abelian Lie algebra structure $\Gamma = 0$ is near every Lie algebra but in general there are nonisomoprhic semisimple Lie algebras of the same dimension, and more generally we could always split $V = V_1 \oplus V_2$, and put a semisimple Lie algebra structure on $G_1$ and a trivial one on $V_2$. So the converse question: are there any nonsemisimple Lie algebras so that all their semisimple deformations are isomorphic? Yes, e.g. anyone on the three-dimensional vector space over $\mathbb C$. So: is there a (computationally useful) characterization of those that are?

If the answer to all my questions are "yes", then it's probably been done somewhere, so a complete response could consist of a good link. The further-further question is to what extent one can deform representations, but that's probably pushing it.

  • 2
    $\begingroup$ That talk by Dyson is really interesting! Do you know if the d = 26 case of the identity he refers to in the beginning has been explained? $\endgroup$ Dec 24, 2009 at 9:32
  • 1
    $\begingroup$ Well, might as well ask it: mathoverflow.net/questions/9716 $\endgroup$ Dec 24, 2009 at 23:50
  • 1
    $\begingroup$ @QiaochuYuan: I'm a simple guy: I see $26$, I know the answer is bosonic string theory. $\endgroup$
    – Alex M.
    Apr 7, 2019 at 20:41
  • $\begingroup$ This reminds as well the birth of quantum groups (Drinfeld's 83 ICM paper, maybe?) when the mantra was "Lie algebras are rigid, but here we are, with our weird quantum groups, to deform them". But that's old stories.. $\endgroup$
    – Richard
    Apr 7, 2019 at 20:57
  • $\begingroup$ A possibly relevant article is [MR0981620 (90e:17008) Goze, Michel: Perturbations of Lie algebra structures. Deformation theory of algebras and structures and applications (Il Ciocco, 1986), 265–355, NATO Adv. Sci. Inst. Ser. C Math. Phys. Sci., 247, Kluwer Acad. Publ., Dordrecht, 1988.] and related papers. $\endgroup$ May 30, 2021 at 6:24

3 Answers 3


The answer to the question in the title is "no". Semisimplicity is an open condition; however, it is not a dense open condition. Indeed, the variety of Lie algebras is reducible. There is one equation which nonsemisimple and only nonsemisimple Lie algebra structures satisfy, namely, that the Killing form Tr(ad(x)ad(y)) is degenerate, just as stated in the question. But there is also a system of equations which all semisimple Lie algebra structures satisfy, as do also all reductive and nilpotent Lie algebra structures, but solvable Lie algebra structures in general don't. These are the unimodularity equations Tr(ad(x))=0 for all x in the Lie algebra. These mean that the top exterior power of the adjoint representation is a trivial representation of the Lie algebra, which is obvious for any Lie algebra that coincides with its commutator subalgebra. But the nonabelian 2-dimensional Lie algebra is not unimodular. Hence in any dimension n, the direct sum of the nonabelian 2-dimensional Lie algebra with the abelian (n-2)-dimensional Lie algebra does not belong to the Zariski closure of semisimple Lie algebras.

  • 2
    $\begingroup$ Awesome. Do you know if the semisimples are dense among the unimodulars? $\endgroup$ Dec 24, 2009 at 18:38
  • 9
    $\begingroup$ I didn't, but now Leonid Rybnikov has just explained me in my blog ( posic.livejournal.com/356508.html , in Russian) that they aren't. For any Lie algebra admitting a nondegenerate invariant bilinear form, in particular for any reductive Lie algebra, the equations Tr(ad(x)^n) = 0 hold for all x in the Lie algebra and any odd n, as the operators ad(x) are skew-symmetric. It is easy to construct a solvable Lie algebra satisfying these equations for any fixed subset of n's and not satisfying the remaining ones. $\endgroup$ Dec 24, 2009 at 19:54
  • 11
    $\begingroup$ It seems that it is always interesting to watch the Leonids. en.wikipedia.org/wiki/Leonids $\endgroup$ Dec 24, 2009 at 20:21
  • 1
    $\begingroup$ Cool. Now I really want a useful description of the Zariski closure of the semisimples! $\endgroup$ Dec 24, 2009 at 20:46
  • 5
    $\begingroup$ А further counterexample from L.R.: not even all nilpotent Lie algebra structures belong to the Zariski closure of the reductive ones. To any Lie algebra structure, one assigns a Zariski semi-continuous numerical invariant: the rank of the skew-symmetric form f([x,y]), where f is the generic linear function on the Lie algebra. For the (2n+1)-dimensional Heisenberg algebra (the generic central extension of the 2n-dimensional abelian Lie algebra) this invariant is equal to 2n, which exceeds its value for any (2n+1)-dimensional reductive Lie algebra, assuming n\ge2. $\endgroup$ Dec 24, 2009 at 21:26

I studied deformations of Lie algebras a lot and was able to construct a moduli space and a universal family for subgroups of an algebraic group:


(I am very sorry to seemingly promote my work, please be sure I am not aware of any similar reference.)

I think you are interested by Lie algebras at large, while I can only give you informations on algebraic subalgebras of the Lie algebra of an algebraic group— i.e. subalgebras that are Lie subalgebras of a subgroup, the terminology is of Chevalley. You might however find some these results interessant or useful.

In front of all, this simple but striking consequence of a deformation theorem of Richardson—and the classification of semi-simple algebras:

3.18 COROLLARY Let $\mathfrak{g}$ be a Lie algebra and $k \le \dim\mathfrak{g}$. The map from the variety of $k$-dimensional subalgebras to the set of isomorphism class of semi-simple Lie algebras sending a Lie algebra to the isomorphism class of its semi-simple factors is lower semi-continuous, in the Zariski topology.

(Isomorphism classes of semi-simple algebras are ordred by injections.)

I cannot totally answer your question (for deformations of subalgebras):

“If the answer is no in general, is it possible to (nicely) characterize the Lie algebra structures that are in the closure of the semisimple part?”

However it is worth noting, that all what you need to get the description you are looking for, is to understand degenerations on nilpotent algebras. Indeed, consider a group $G$ and two subgroups $H$ and $M$ such that $M$ is in the closure of the conjugacy class of $H$. (The closure is taken in the moduli space, but replacing connected groups by their Lie algebras, you can embed the closure of the orbit of $H$ live in an appropriate Grassmannain variety.) So you have a curve $C$ in $G$ such that $M$ is in the closure of the translated of $H$ under elements of $C$. A consequence of the corollary is that if $M$ contains a semi-simple group $S$ then you can replace $H$ by a conjugate containing $S$ and assume that $C$ is in the centralizer of $S$.

So there is a natural way to enumerate possible degenerations, by looking at possible degenerations of the semi simple part (in the Levi-Malcev decomposition) and then study what can happen when you fix the smallest semi-simple part. In practice, small rank semi-simple examples should be very tractable, because we know very well the centralizers in the large group of maximal tori of the small semi-simple part.

A few words about the moduli space. The moduli space is not an algebraic variety but rather a countable union of (projective) algebraic varieties. This unpleasant phenomenon is indebted to tori: let me explain what happens for tori and for one dimensional subgroups of $SL_3$.

Tori. Let $T$ be a torus of dimension $r$, its Lie algebra ${\mathfrak t}$ contains a lattice $M$, the kernel of the exponential. As we know, the only $k$-dimensional subspaces of ${\mathfrak t}$ that are algebraic algebras are spanned by points of $M$. Hence the set $\mathfrak{P}_k(T)$ of $k$ dimensional subgroups of $T$ is the set of rational points if the Grasmann variety of $k$. Its topology is not exactly the topology induced by the Grassmann variety: it is on each irreducible component—which in this special case just means that $\mathfrak{P}_k(T)$ is discrete.

$SL_3$. Let me describe the set of $\mathfrak{P}_1(SL_3)$ of all one-dimensional subgroups of $SL_3$. We are then looking for $1$-dimensional algebraic subalgebras of the $\mathfrak{sl}_3$. There is the two nilpotent orbits, and the countably many orbits of $SL_3$ through $\mathfrak{P}_1(T)$, once we choosed a torus $T$ (these orbits almost never meet, by the normalizer theorem). Each of these orbits contains a nilpotent orbit in its closure, so from the smallest nilpotent orbit, you can generalise to any semi-simple orbit: subgroups in two distinct semi-simple orbits are abstractly isomorphic but not isomorphic as subgroups of $SL_3$.

Closing remark. I do not know enough scheme theory to write my work in a suitable language, so you may be puzzled by some terminology I use. I am using algebraic varieties as in Shafarevic I or in the book of Hanspeter Kraft (Geometrische Methoden der Invarianten Theorie). It is probably very clumsy! I am quite confident that these results would survive in the world of schemes, there is actually a scheme-theoretic version of Richardson's theorem, that Brian Conrad once showed me.


I would need more time than I have at present to try to write down a reasonable answer, but one quick comment on the question of deformation of representations, is that there is this paper by Nijenhuis and Richardson where they study the question of deformations of homomorphisms, hence in particular of representations, of Lie algebras and Lie groups. Nijenhuis and Richardson have several seminal papers on the deformations of Lie algebras, which are probably worth a look.

The inverse process to deformations is that of Lie algebra contractions, as in the original paper of Inonu and Wigner. These are limits which allow you to go from semisimple Lie algebras to their nonsemisimple "boundary".


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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