$V$ a finite-dimensional real vector space ($\dim V=n$). $$S=\wedge^2V\otimes V=\{b:V\times V\to V,\ \text{skew-symmetric, bilinear maps}\}$$ A Lie bracket on $V$ is an element $\mu$ of $S$ satisfying the Jacobi identity: $$\mu(x,\mu(y,z))+\mu(y,\mu(z,x))+\mu(z,\mu(x,y))=0,\ \forall x,y,z\in V.$$ $\mathcal{G}=\{\text{Lie brackets on}\ V\}\subset S$ and $\mathrm{GL}(n,\mathbb{R})$ acts on $\mathcal{G}$: $$g\cdot\mu(x,y):=g\left(\mu(g^{-1}(x),g^{-1}(y))\right).$$ $(V,\mu)$ is rigid if every Lie bracket $\mu'$ sufficiently close to $\mu$, $(V,\mu')$ is isomorphic to $(V,\mu)$. i.e. it exists $g\in\mathrm{GL}(n,\mathbb{R})$ close to the identity such that $\mu'=g\cdot\mu$ (or the orbit of $\mu$ is open in $\mathcal{G}$).
What is the relevant topology here: $\mathcal{G}$ is a vector subspace of $S$ endowed with the euclidean topology. Is this topology equivalent to the Zariski one (when $\mathcal{G}$ is identified to the family of constant structures, i.e. as an algebraic variety).
I would like to study the following example on a $3$-dimensional real vector space $V=<e_1,e_2,e_3>$ for the Lie bracket $$\mu(e_1,e_2)=e_3,\ \mu(e_1,e_3)=-e_2,\ \mu(e_2,e_3)=0.$$ Since the orbit $O(\mu)$ is isomorphic to the quotient $\mathrm{GL}(n,\mathbb{R})/G_\mu$ where $G_\mu$ is the isotropy group $G_\mu=\{g\in\mathrm{GL}(n,\mathbb{R}),\ g\cdot\mu=\mu\}$. In order to the orbit to be open, I have to show that $\dim O(\mu)=\dim\mathcal{G}$.
What is the dimension of $\mathcal{G}$? $\dim S=3^2(3-1)/2=9$.
$g=\begin{pmatrix} a_{11}&a_{12}&a_{13}\\ a_{21}&a_{22}&a_{23}\\ a_{31}&a_{32}&a_{33} \end{pmatrix}\in G_\mu$ if and only if \begin{align*} 0=&[g(e_2),g(e_3)]=&-(a_{12}a_{33}-a_{13}a_{32})\,e_2+(a_{12}a_{23}-a_{13}a_{22})\,e_3\\ -g(e_2)=&[g(e_1),g(e_3)]=&-(a_{11}a_{33}-a_{13}a_{31})\,e_2+(a_{11}a_{23}-a_{13}a_{21})\,e_3\\ g(e_3)=&[g(e_1),g(e_2)]=&-(a_{11}a_{32}-a_{12}a_{31})\,e_2+(a_{11}a_{22}-a_{12}a_{21})\,e_3 \end{align*} i.e. $a_{12}=0=a_{13}$ and six other independent quadratic equations. So $\dim G_\mu=8$, $\dim O(\mu)=9-8=1$ and $\mu$ is not rigid.
Is this correct?
Here is a slightly different approach. Your Lie algebra $L$ is solvable, sometimes denoted by $\mathfrak{r}_{3,-1}(\Bbb R)$. Every solvable Lie algebra of dimension $2\le n\le 7$ is rigid if and only if it complete (i.e., has zero center and all derivations are inner). Since $\mathfrak{r}_{3,-1}(\Bbb R)$ is not complete, it is not rigid. So you are right. Actually, the derivation algebra has dimension $4$, which is $3+1$. Here $\dim {\rm Inn}(L)=3$. So not all derivations are inner.