Rigidity of Borel Lie algebras

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 $$\mathfrak b$$:

1. $$\mathfrak b$$ 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.]

2. 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?

• By "Borel subalgebra" you mean the same as "minimal parabolic"? (I remember hearing subtle differences in the non-split case but I'm not sure)
– YCor
Mar 3, 2021 at 10:47
• @Ycor for me Borel subalgebra means maximal solvable subalgebra, and parabolic means contains a Borel subalgebra. So yes, for me those would be the same, but I would be glad to understand if there is a better/more correct definition where they differ... Mar 3, 2021 at 10:57
• Oh, I see one issue related to your point. Those are real algebras, not complex ones. The limit algebra đť”ź in 2. is maximal among the subalgebras of $\mathfrak{o}(3,1)$ isomorphic to algebras of upper triangular real matrices, however it is not maximal solvable subalgebra. So it may be that it has $H^2(\mathfrak b, \mathfrak b)$ nonzero. Mar 3, 2021 at 12:24
• OK, YCor's comment got me back on track. I've checked that because I assumed $\mathfrak g$ 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. Mar 3, 2021 at 15:53

So here's the fix thanks to YCor's comment, in the case I managed to confuse anyone.

Let $$\mathfrak g$$ be the split real form of a semisimple complex Lie algebra. There is a difference between:

• Borel subalgebras of $$\mathfrak g$$, that is maximal solvable subalgebras of $$\mathfrak g$$.
• Maximal split-solvable subalgebras of $$\mathfrak g$$; I am not aware of a good terminology for them. They are real forms of Borel subalgebras in $$\mathfrak g\otimes \mathbf C$$.

$$H^2(\mathfrak b_0, \mathfrak b_0)=0$$ holds for every Borel subalgebra $$\mathfrak b_0$$ of $$\mathfrak g$$, but fails for the maximal split-solvable algebra $$\mathfrak b$$ in the question (and I now suspect, many of them).

Use the notation from the question, denote $$(\xi, \eta, \theta)$$ the dual basis to $$(X,Y,T)$$. Then by some computations (I skip the details)

$$\begin{array} & d(\xi\otimes X) = 0 & d(\xi \otimes Y) = 0 & d(\xi \otimes T) = \xi \wedge \eta \otimes Y + \xi \wedge \theta \otimes T \\ d(\eta \otimes X) = 0 & d(\eta \otimes Y) = 0 & d(\eta \otimes T) = - \xi \wedge \eta \otimes X + \eta \wedge \theta \otimes T \\ d(\theta \otimes X)=0 & d(\theta \otimes Y) = 0 & d (\theta \otimes T)=- \eta \wedge \theta \otimes Y - \xi \wedge \theta \otimes X. \end{array}$$

Hence $$B^2(\mathfrak b, \mathfrak b) = \left\{ \xi \wedge \eta \otimes (aY - bX) + \xi \wedge \theta \otimes (aT - cX) + \eta \wedge \theta \otimes (bT - cY) \right\}_{a,b,c \in \mathbf R}$$.

Set $$\gamma = \theta \wedge \eta \otimes X$$, so that along the degeneration $$\mathfrak s \to \mathfrak b$$, $$\mu_{\mathfrak s} = \mu_{\mathfrak b} + e^{-t} \gamma$$ if we put $$\mu_{\mathfrak b}$$, $$\mu_{\mathfrak s}$$ the laws of $$\mathfrak b$$ and $$\mathfrak s$$ respectively on a common vector space. Note that \begin{align} d\gamma ( X \wedge Y \wedge T) & = [X, (\theta \wedge \eta \otimes X)(Y \wedge T)] - [Y, (\theta \wedge \eta) \otimes X (X \wedge T)] \\ & \quad + [T, (\theta \wedge \eta \otimes X) (X \wedge Y) ] - \gamma ([X,Y], T) + \gamma ([X,T], Y) - \gamma([Y,T], X) \\ & = 0 \end{align} hence $$[\gamma]$$ defines a nonzero cohomology class. The same computation should provide the full $$H^2(\mathfrak b, \mathfrak b)$$, and possibly with extra effort $$H^2(\mathfrak b, \mathfrak b)$$ when $$\mathfrak b$$ is maximal split solvable in $$\mathfrak g = \mathfrak o (n,1)$$, $$n>2$$; it is known that only finitely many isomorphism classes of Lie algebra can degenerate to such a $$\mathfrak b$$. [Lauret, Degenerations of Lie algebras and geometry of Lie groups. Differential Geom. Appl. 18 (2003), no. 2, 177-194].