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 $\dim_{\mathbb C}\mathfrak g/\mathfrak h=2$?

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    $\begingroup$ 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.) $\endgroup$ – Torsten Schoeneberg Nov 18 '18 at 21:03
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    $\begingroup$ 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 $\mathfrak{sl}_2$. $\endgroup$ – 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 $<r$.

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 $<r$ 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 $<r$.

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.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – Aloizio Macedo Nov 23 '18 at 17:28
  • $\begingroup$ @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? $\endgroup$ – user328669 Dec 9 '19 at 8:10
  • $\begingroup$ 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. $\endgroup$ – YCor Dec 9 '19 at 9:03

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