Every semisimple Lie algebra is rigid I am trying to understand why every (complex, finite-dimensional) semisimple Lie algebra is rigid. I understand there is a cohomological proof, but I would like to understand a more direct argument, namely via the Killing form.
Definition: $GL(n,\mathbb{C})$ acts on the set of all $n$-dimensional Lie algebra laws $\lambda$ by $(g\cdot\lambda)(x,y)=g(\lambda(g^{-1}x,g^{-1}y))$ for all $x,y\in\mathbb{C}^n$. We can denote the orbit under this action by $O(\lambda)$. If $O(\lambda)$ is open in the Zariski topology, then the Lie algebra with law $\lambda$ is called rigid.
Now, I want to prove that every semisimple Lie algebra is rigid. The argument seems to be that "Killing form being non-degenerate is an open condition" (see [1]). 
One can identify an $n$-dimensional Lie algebra (given a basis $\{e_1,...,e_n\}$) with law $\lambda$ by its structure constants $\lambda(e_i,e_j)=\sum\limits_{k=1}^{n}C_{ij}^ke_k$, so we can view $\lambda$ as the vector $(C_{ij}^k)\in\mathbb{C}^{n^3}$. So, I (believe I) want to show that the set
$\{a=(a_1,...,a_{n^3})\in\mathbb{C}^{n^3}:\text{ the Lie algebra law corresponding to $a$ is semisimple} \}$ is open in the Zariski topology. How do I do this? 
We have
$\{(a_1,...,a_{n^3})\in\mathbb{C}^{n^3}:\text{ the corresponding Lie algebra law is semisimple} \}=\{a=(a_1,...,a_{n^3})\in\mathbb{C}^{n^3}:\text{ the corresponding Killing form } \kappa_a \text{ is non-degenerate} \}$.
So I need to find a set of polynomials in $n^3$ variables such that its zeros are precisely the vectors $a$ where $\kappa_a$ is non-degenerate.
This is where I fail. I suspect that these polynomials will be very similar to the Killing form itself (like, the Killing form viewed as a polynomial for fixed $x$ and $y$), but I cannot create these polynomials.
Moreover, I don't know what to do next. So suppose I can show that the set $A$ of all semisimple Lie algebra laws is open, how do I show that each seperate orbit $O(\mu)$ (which consists of the Lie algebras isomorphic to $\mu$) is open?
[1] Yu. A.  Neretin, An estimate of the number of parameters defining an n-dimensional algebra, Math. USSR Izvestiya, Vol. 30 (1988), No. 2
 A: The "more direct way" with the Killing form is not enough as Mariano already explained. 
I think, the natural way to see that every semisimple Lie algebra is rigid, over a field of characteristic zero, is indeed via $2$-cocycles for the adjoint representation. One could also argue that this is the most direct way, because a formal deformation of a Lie algebra $L$ directly leads to $2$-cocycles. 
This is due to Gerstenhaber, see On the Deformation of Rings and Algebras: A formal one-parameter deformation of $L$ is a power series
$$
[g,h]_t := [g,h] + \sum_{k\ge 1}\phi_k(g,h)t^k
$$
such that the Jacobi identity for $[ \;, \;]_t$ holds, with $2$-cochains
$\phi_k\in C^2(L,L)$ and $g,h\in L$. The Jacobi identity implies in particular, that the maps $\phi_k$ are $2$-cocycles for the adjoint representation, i.e., 
$$
\phi_k\in Z^2(L,L)
$$
Two formal deformations of $L$ are called equivalent, if
the resulting Lie algebras are isomorphic. The equivalence classes are represented by a cohomology class from $H^2(L,L)$. Now
$L$ is called formally rigid, if every formal deformation
is trivial.
If $H^2(L,L)=0$, then $L$ is obviously formally rigid. The Whitehead Lemma says that every semisimple Lie algebra over characteristic zero satisfies $H^2(L,L)=0$, hence is formally rigid.
Gerstenhaber and Schack proved that formally rigid is equivalent to geometrically rigid, which means that $L$ has open orbit in the "variety of all Lie algebra structures of dimension $\dim (L)$".
