In the questions [Is "semisimple" a dense condition among Lie algebras?](https://mathoverflow.net/questions/9661/is-semisimple-a-dense-condition-among-lie-algebras) and [What is the Zariski closure of the space of semisimple Lie algebras?](https://mathoverflow.net/questions/9719/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?



Deformations of semisimple Lie algebras
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 $\mathbb{C}$.


1. 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.


1. 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.


