Let $\mathfrak b$ be a Borel subalgebra of dimension $n$ in a real semisimple Lie algebra $\mathfrak g$. I am trying to reconcile two facts about $\mathfrak b$:

$\mathfrak b$ is rigid, that is, the orbit of its law is open (even, Zariski-open) in the variety $\mathcal R_n$ of solvable Lie algebras of dimension $n$. This can be proven as follows: $H^2(\mathfrak b, \mathfrak b)=0$ (where $\mathfrak b$ acts in $\mathfrak b$ via the adjoint representation), see e.g. [Leger and Luks, Cohomology theorems for Borel-like solvable Lie algebras in arbitrary characteristic,

*Canadian J. Math.***24**(1972), 1019-1026, Corollary 5.6]. And then this vanishing is sufficient (though not necessary) to ensure that $\mathfrak b$ is rigid, see e.g. [Carles, On the structure of rigid Lie algebras,*Ann. Inst. Fourier*(Grenoble)**34**(1984), no. 3, 65-82.]There seemingly exist solvable Lie algebras that can degenerate to $\mathfrak b$. Here is a minimal example where $\mathfrak g$ is rank one and $n=3$: consider $\mathfrak s = \mathfrak t + \mathfrak a$, where $\mathfrak a =\operatorname{span}(X,Y)$ is a $2$-dimensional abelian ideal and $\mathfrak t$ is a $1$-dimensional torus of derivations of $\mathfrak a$ generated by $T$ with $$ \operatorname{ad} T_{\mid \mathfrak a} = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix},$$ that is $[T,X] = X$ and $[T,Y] = X+Y$. Now, for $t \in (0, + \infty)$ let $\varphi(t) \in \operatorname{GL(\mathfrak s)}$ be diagonal and such that $\varphi_t(X) = X, \varphi_t(Y) = e^{-t}Y, \varphi_t(T) = T$. Then $\mathfrak s$ degenerates through $(\varphi_t)$ to the Borel subalgebra of $\mathfrak o (3,1)$ when $t \to + \infty$. (And $\mathfrak t + \mathfrak a$ becomes the Cartan decomposition of $\mathfrak b$ in the limit.)

So where is the snag?

real, $\mathfrak b$ in 2. is not a Borel subalgebra, neither is it Borel-like as in Leger and Luks' terminology from the paper cited above, so their vanishing theorem does not hold for $\mathfrak b$. I also computed by hand part of $H^2(\mathfrak b, \mathfrak b)$ on the $3$-dimensional example, the degeneration I pointed comes from a $2$-cocycle which is not a coboundary. I do not post this as an answer yet because someone may have a more general/insightful view on this. $\endgroup$ – GPallier Mar 3 at 15:53