# The codimension of a parabolic subalgebra of a semisimple Lie algebra

Given a complex semisimple Lie algebra $$\mathfrak g$$ and a subalgebra $$\mathfrak h$$. If we are given that the complex vector space $$\mathfrak g/\mathfrak h$$ has dimension $$1$$ over $$\mathbb C$$. Is $$\mathfrak h$$ a parabolic subalgebra. i.e., contains a Borel subalgebra?

Is the statement above still true when $$\dim_{\mathbb C}\mathfrak g/\mathfrak h=2$$?

• For your second question, for any $0 \neq x \in \mathfrak{g}:=\mathfrak{sl}_2(\Bbb C)$, $\mathfrak{h}:= \Bbb Cx$ is a counterexample. (I suspect it is essentially the only one though, i.e. if $\mathfrak{g}$ contains no direct summand $\simeq \mathfrak{sl}_2$, the statement might be true; but I'm not at all sure about this.) – Torsten Schoeneberg Nov 18 '18 at 21:03
• The first assertion is true and actually the only possibility (up to direct product of both $\mathfrak{g}$ and $\mathfrak{h}$ by some other semisimple algebra) is when $\mathfrak{g}$ is $\mathfrak{sl}_2$. – YCor Nov 18 '18 at 22:26

The first assertion is true and actually the only possibility (up to direct product of both $$\mathfrak{g}$$ and $$\mathfrak{h}$$ by some other semisimple algebra) is when $$\mathfrak{g}$$ is $$\mathfrak{sl}_2$$.

(Edit: switched to an algebraic proof)

In $$\mathfrak{sl}_2$$ it is easy to check that all codimension 1 subalgebras are conjugate to the Borel (parabolic) one.

It is enough to prove that if $$\mathbf{g}$$ is simple of rank $$\ge 2$$, then it has no codimension 1 subalgebra $$\mathfrak{h}$$.

Choose a Cartan subalgebra $$\mathfrak{h}_0$$ of $$\mathfrak{h}$$. It induces a grading $$(\mathfrak{g}_\alpha)$$ of $$\mathfrak{g}$$, which has to be a quotient of its own Cartan grading.

If $$\mathfrak{h}_0=\mathfrak{g}_0$$, then $$\mathfrak{h}_0$$ is a Cartan subalgebra of $$\mathfrak{g}$$, so $$(\mathfrak{g}_\alpha)$$ is the Cartan grading of $$\mathfrak{g}$$. In this case, it follows that $$\mathfrak{h}$$ is a graded subalgebra containing $$\mathfrak{g}_0$$, so there exists a nonzero root $$\alpha$$ such that $$\mathfrak{h}=\bigoplus_{\beta\neq\alpha}\mathfrak{g}_\beta$$. Using that $$\mathfrak{g}$$ has rank $$\ge 2$$ and is simple, there exist two nonzero roots summing to $$\alpha$$, and this implies that such $$\mathfrak{h}$$ is not a subalgebra, contradiction.

Next, if $$\mathfrak{h}_0\neq\mathfrak{g}_0$$, then being a quotient of the Cartan grading $$(\mathfrak{g}_{(\gamma)})$$ of $$\mathfrak{g}$$, we have $$\mathfrak{g}_0$$ reductive and containing a Cartan subalgebra of $$\mathfrak{g}$$. If $$\mathfrak{g}\neq\mathfrak{g}_0$$, we can argue as follows: $$\mathfrak{g}_0$$ is the sum of $$\mathfrak{g}_{(\gamma)}$$ where $$\gamma$$ ranges over some proper subspace $$M$$ of the space of roots. Since $$\mathfrak{g}$$ is simple, the set of roots $$\gamma$$ not in $$M$$ generates the space of roots (the set of roots is not contained in the union of two proper subspaces), and for each such $$\gamma$$, we have $$\mathfrak{g}_{(\pm\gamma)}\in\mathfrak{h}$$ and hence $$h_\gamma\in\mathfrak{h}$$. Hence $$\mathfrak{h}$$ contains a Cartan subalgebra of $$\mathfrak{g}$$, and this implies $$\mathfrak{h}_0=\mathfrak{g}_0$$ contradiction.

So we have $$\mathfrak{g}=\mathfrak{g}_0$$: in particular $$\mathfrak{h}=\mathfrak{h}_0$$. Hence $$\mathrm{ad}(h)$$ is nilpotent for every $$h\in\mathfrak{h}=\mathfrak{h}_0$$. Letting $$\mathfrak{c}$$ be a Cartan subalgebra of $$\mathfrak{g}$$, this implies that every element of $$\mathfrak{h}\cap\mathfrak{c}$$ is in the kernel of every root. Since the intersection of kernels of roots is zero in a semisimple Lie algebra, this forces $$\mathfrak{c}$$ to have dimension $$\le 1$$. Hence $$\mathfrak{g}$$ has rank $$\le 1$$, a contradiction.

Edit: I was a bit frustrated to make such a proof for such a weak result, but indeed it adapts to the following more natural and stronger (and classical) statement:

Let $$\mathfrak{g}$$ be a absolutely simple Lie algebra over a field of characteristic zero, of (absolute) rank $$r$$. Then $$\mathfrak{g}$$ has no proper subalgebra of codimension $$.

Lemma: let $$\Phi$$ be an irreducible root system in dimension $$r\ge 1$$. Then $$\Phi$$ is not contained in the union of two proper subspaces.

This follows from:

Sublemma: let $$\Phi$$ be a root system in dimension $$r$$ (not necessarily generating). Suppose that $$\Phi\subset V_1\cup V_2$$ where $$V_i$$ are subspaces. Then there exist subsets $$\Phi_1,\Phi_2$$ such that $$\Phi_i\subset V_i$$, $$\Phi_1\cup\Phi_2=\Phi$$, and $$\langle\Phi_1,\Phi_2\rangle=0$$.

Proof of sublemma. This is vacuously true in dimension $$0$$, and more generally if $$V_2=V$$. In dimension $$r\ge 1$$, write $$\Psi_1=\Phi\smallsetminus V_2$$, $$\Psi_2=\Phi\smallsetminus V_1$$, and $$\Psi_{12}=\Psi\cap V_1\cap V_2$$. Clearly $$\Psi$$ is the disjoint union $$\Psi_1\sqcup\Psi_2\sqcup\Psi_{12}$$. Also, $$\Psi_1,\Psi_2$$ are orthogonal: indeed otherwise, we can find $$\alpha\in\Psi_1$$, $$\beta\in\Psi_2$$ with $$\langle\alpha,\beta\rangle<0$$, so $$\alpha+\beta\in\Phi$$ and this is a contradiction because $$\alpha+\beta$$ belongs to neither $$V_1$$ nor $$V_2$$.

Next, we consider the subspace $$V_2$$, and its two subspaces $$W_1=V_1\cap V_2$$, and $$W_2$$ the orthogonal of $$V_1$$ in $$V_2$$, and $$\Phi'=\Phi\cap V_2=\Psi_2\sqcup \Psi_{12}$$, with $$\Psi_2\subset W_2$$ and $$\Psi_{12}\subset W_1$$. We argue by induction inside $$V_2$$ (the trivial case $$V_2=V$$ being excluded), to infer that we can write $$\Phi\cap V_2=\Phi'_1\cup\Phi'_2$$ with $$\langle\Phi'_1,\Phi'_2\rangle =0$$ and $$\Phi'_i\subset W_i$$. Then $$\Phi=\Psi_1\sqcup\Phi'_1\sqcup\Phi'_2$$, with $$\Phi'_2\subset V_2$$ orthogonal to $$\Psi_1\sqcup\Phi'_1\subset V_1$$. This finishes the induction.$$\Box$$

Now let us proceed to the proof of the result. It's an adaptation of the previous proof. Only the first case requires a modification, which is the reason for the above lemma. Namely, let $$\mathfrak{h}$$ have codimension $$ and suppose, with the previous notation that $$\mathfrak{h}_0=\mathfrak{g}_0$$. In this case the grading is the Cartan grading of $$\mathfrak{g}$$, so $$\mathfrak{h}=\mathfrak{g}_0\oplus\bigoplus_{\alpha\in\Phi\smallsetminus F}\mathfrak{g}_\alpha$$, where $$F$$ is a subset of the root system $$\Phi$$ of $$\mathfrak{g}$$, of cardinal $$.

Let $$V_1$$ be the subspace spanned by $$F$$ (a proper subspace of $$\mathfrak{g}_0^*$$). Fix a root $$\alpha\in F$$, and $$V_2$$ its orthogonal. Then by the lemma, there exists a root $$\beta\notin V_1\cup V_2$$. Let $$P$$ be the plane generated by $$\alpha$$ and $$\beta$$. So $$\Phi\cap P$$ is an irreducible root system in $$P$$, and we can find in $$P$$ two roots, not collinear to $$\alpha$$, and avoiding the $$V_2\cap P$$ (which has dimension $$\le 1$$), with negative scalar product and summing to $$\alpha$$. This shows that $$\mathfrak{g}_\alpha\subset\mathfrak{h}$$, a contradiction.

In the other case $$\mathfrak{g}_0\neq\mathfrak{h}_0$$, we almost only need to copy the previous proof.

• Comments are not for extended discussion; this conversation has been moved to chat. – Aloizio Macedo Nov 23 '18 at 17:28
• @YCor do you know where I can find the result in the yellow box? Also, is it true that subalgebras (of simple algebras of rank $r$) of codimension $r$ are parabolic? – user328669 Dec 9 '19 at 8:10
• No, I don't know a reference. I guess your second question has a positive answer, but it would require a whole argument, which I don't have in mind now, and would not fit in a comment. – YCor Dec 9 '19 at 9:03