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?

  • 1
    $\begingroup$ 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) $\endgroup$
    – YCor
    Mar 3, 2021 at 10:47
  • $\begingroup$ @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... $\endgroup$
    – user102887
    Mar 3, 2021 at 10:57
  • 1
    $\begingroup$ Oh, I see one issue related to your point. Those are real algebras, not complex ones. The limit algebra 𝔟 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. $\endgroup$
    – user102887
    Mar 3, 2021 at 12:24
  • $\begingroup$ 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. $\endgroup$
    – user102887
    Mar 3, 2021 at 15:53

1 Answer 1


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].


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