Deformations of semisimple Lie algebras In the questions Is "semisimple" a dense condition among Lie algebras? and What is the Zariski closure of the space of semisimple Lie algebras?, something equivalent to the following is mentioned: if you have a smoothly varying family of semisimple Lie algebras, all the Lie algebras in the family are isomorphic. e.g. the following quote:
"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."
I can't see how this follows just from the discreteness of the classification. Can anyone explain why it's true or give a counterexample?
e.g. could you not have a $\mathbb{P}^1$ of semisimple Lie algebras which are generically isomorphic to $\mathfrak{d}_7 \oplus \mathfrak{a}_1$ say, but at one point you get $\mathfrak{e}_8$, or something similar?
 A: I think that the explanation 
"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"
is a bit simplistic. The real reason, as many people have already mentioned, is Weyl's theorem on complete reducibility, which of course fails in charatcteristic $p$. And it shouldn't come as a  surpise that over an algebraically closed field $K$ of characteristic $p>3$ one encounters situations where there are finitely many isoclasses of simple Lie algebras of dimension $N$ and, at the same time, there exist algebraic families of simple $N$-dimensional Lie algebras 
{$\mathfrak{g}_t|\ t\in K$} over $K$ such that $\mathfrak{g}_t\cong L$ for all $t\ne 0$ and $\mathfrak{g}_0\not\cong L$ for some simple Lie algebra $L$.
Indeed, let $N=p^2-2$. Then it follows from the the classification theory that there are finitely many isoclasses of simple $N$-dimensional Lie algebras over $K$. Now consider the associative $K$-algebra $A$ generated by two elements $x,y$ such that $x^p=y^p=0$ and $[x,y]=1$. This is a fake modular version of the first Weyl algebra, and it is easy to see that it is simple and has dimension $p^2$. It has a finite increasing algebra filtration (with $x,y$ living in degree $1$) such that the corresponding graded algebra $P:={\rm gr}(A)$ is the truncated polinomial ring in $x,y$ with induced Poisson bracket satisfying {$x,y$} $=1.$ Then the Lie algebra $[A,A]/K1$ is isomorphic to $\mathfrak{psl}_p(K)$ whilst the Lie algebra {$P,P$}$/K1$ is nothing but the simple Cartan type Lie algebra $H(2;\underline{1})^{(2)}$. Both Lie algebras are simple of dimension $N$ and the latter is a contraction of the former.  Moreover, they are not isomorphic when $p>3$.
A: Locally in any continuous family of semisimple Lie algebras, you can choose a (constant) subfamily of Cartan subalgebras etc., and so get a continuous map sending a point in the base to a Cartan matrix, which has integer coefficients.
A: While I think that what Johannes wrote is right, it misses out on an underlying principle.  There's a one sentence explanation for this fact, which is "the tangent space to a Lie algebra in the moduli space of Lie algebras is $H^2(\mathfrak{g},\mathfrak{g})$, the second cohomology valued in the adjoint representation."  This cohomology group can also be identified with the space of extensions of Lie algebras of $\mathfrak{g}$ by a copy of the adjoint representation. 
So, the slick proof is to either prove that this is 0 using the action of the Casimir as in Weibel's book, or to note by the Levi complement theorem that all abelian extensions of a semi-simple Lie algebra split.
After all, if you had a family $\tilde{\mathfrak{g}}$ of such Lie algebras over $\mathbb{A}^1$, then you could look at $\tilde{\mathfrak{g}}/h^2\tilde{\mathfrak{g}}$ (where $h$ is the parameter on $\mathbb{A}^1$), and see that that is an extension of $\mathfrak{g}$ by the adjoint representation.
A: The experts should correct me if there is a fatal mistake in the argument, I am neither an algebraic geometer nor a Lie theorist. I am working over $\mathbb{C}$.


*

*Let $\mathfrak{g}$ be a semisimple Lie algebra, $G$ be the adjoint group, $Aut(\mathfrak{g})$ be the automorphism group. Let 
$Aut_0 (\mathfrak{g})$ be the subgroup of automorphisms that preserve the decomposition of $\mathfrak{g}$ into simple ideals.
It has finite index in the whole automorphism group, since the decomposition into ideals is unique and an automorphism has to 
map a simple ideal into a simple ideal. The group $Aut_0(\mathfrak{g}$ is the product of the automorphism groups of the simple factors.
The automorphism group of any simple Lie algebra has the adjoint group as a finite index subgroup; the quotient is the automorphism group of 
the Dynkin diagram. The upshot of this discussion is: $Aut(\mathfrak{g})$ has $G$ as a finite index subgroup. In particular, the dimension of 
$Aut(\mathfrak{g})$ only depends on the dimension of $\mathfrak{g}$!


EDIT: there is a better proof of this step in the literature, e.g. in Procesi's book, page 301.


*

*Let $V$ be the variety of all Lie algebra structures on $C^n$; it has an action of $GL_n$ on it, the stabilizers are the 
automorphism groups, the orbits are the isomorphism classes. Now let $\mathfrak{g} \in V$ be semisimple and let 
$O \subset V$ be its $GL_n$-orbit. Now use the orbit closure theorem (Borel, Linear 
Algebraic Groups, page 53). It says that any $GL_n$-orbit $O$ is open in its closure and that $\bar{O} \setminus O$ consists 
of orbits of smaller dimension. Hence all Lie algebras in the closure of $O$ which are not isomorphic to $\mathfrak{g}$
must have a larger-dimensional automorphism group. By part 1, they are not semisimple. 

A: If you find non-standard arguments more intuitively appealing you might be interested in a non-standard proof of this result which appears in Springer LNM 881 `Nonstandard analysis' by Lutz and Goze, p.140.
