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.


### Motivation


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.


### Definitions


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](https://mathoverflow.net/questions/9719/what-is-the-zariski-closure-of-the-space-of-semisimple-lie-algebras).


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.



Is "semisimple" a dense condition among Lie algebras?
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.


