2
$\begingroup$

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?

$\endgroup$
4
  • 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 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... $\endgroup$ – GPallier Mar 3 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. $\endgroup$ – GPallier Mar 3 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. $\endgroup$ – GPallier Mar 3 at 15:53
1
$\begingroup$

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].

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.