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

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 $mathfrak{h}=bigoplus_{betaneqalpha}mathfrak{g}_beta$.
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 $mathfrak{h}_0neqmathfrak{g}_0$, 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 $mathfrak{g}neqmathfrak{g}_0$, 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 $mathfrak{g}_{(pmgamma)}inmathfrak{h}$ and hence $h_gammainmathfrak{h}$. 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 $hinmathfrak{h}=mathfrak{h}_0$. Letting $mathfrak{c}$ be a Cartan subalgebra of $mathfrak{g}$, this implies that every element of $mathfrak{h}capmathfrak{c}$ 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 $rge 1$. 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 $Phisubset V_1cup V_2$ where $V_i$ are subspaces. Then there exist subsets $Phi_1,Phi_2$ such that $Phi_isubset V_i$, $Phi_1cupPhi_2=Phi$, and $langlePhi_1,Phi_2rangle=0$.

Proof of sublemma. This is vacuously true in dimension $0$, and more generally if $V_2=V$. In dimension $rge 1$, write $Psi_1=Phismallsetminus V_2$, $Psi_2=Phismallsetminus V_1$, and $Psi_{12}=Psicap V_1cap V_2$. Clearly $Psi$ is the disjoint union $Psi_1sqcupPsi_2sqcupPsi_{12}$. Also, $Psi_1,Psi_2$ are orthogonal: indeed otherwise, we can find $alphainPsi_1$, $betainPsi_2$ with $langlealpha,betarangle<0$, so $alpha+betainPhi$ 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 $W_1=V_1cap V_2$, and $W_2$ the orthogonal of $V_1$ in $V_2$, and $Phi'=Phicap V_2=Psi_2sqcup Psi_{12}$, with $Psi_2subset W_2$ 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 $Phicap V_2=Phi'_1cupPhi'_2$ with $langlePhi'_1,Phi'_2rangle =0$ and $Phi'_isubset W_i$. Then $Phi=Psi_1sqcupPhi'_1sqcupPhi'_2$, with $Phi'_2subset V_2$ orthogonal to $Psi_1sqcupPhi'_1subset V_1$. 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 $mathfrak{h}=mathfrak{g}_0oplusbigoplus_{alphainPhismallsetminus F}mathfrak{g}_alpha$, 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 $alphain F$, and $V_2$ its orthogonal. Then by the lemma, there exists a root $betanotin V_1cup V_2$. Let $P$ be the plane generated by $alpha$ and $beta$. So $Phicap P$ is an irreducible root system in $P$, and we can find in $P$ two roots, not collinear to $alpha$, and avoiding the $V_2cap P$ (which has dimension $le 1$), with negative scalar product and summing to $alpha$. This shows that $mathfrak{g}_alphasubsetmathfrak{h}$, a contradiction.

In the other case $mathfrak{g}_0neqmathfrak{h}_0$, we almost only need to copy the previous proof.