Groups that act transitively on $\mathrm{Gr}(k,\Bbb R^n)$ but not transitively on $\mathrm{Gr}(k+1,\Bbb R^n)$ Is it known for which $n, k\in\Bbb N$ there exists a matrix group $\Gamma\subseteq\mathrm{GL}(\Bbb R^n)$ that

*

*acts transitively on $\mathrm{Gr}(k,n)$, i.e., on the $k$-dimensional subspaces of $\Bbb R^n$, but

*acts not transitively on $\mathrm{Gr}(k+1,n)$, i.e., on the $(k+1)$-dimensional subspaces of $\Bbb R^n$?

For example, $\mathrm U(n)$ (acting on $\Bbb C^n\cong\Bbb R^{2n}$) acts transitively on $\mathrm{Gr}(1,\Bbb R^{2n})$ but not on $\mathrm{Gr}(2,\Bbb R^{2n})$.
 A: This is only a partial answer, and it's based on YCor's comment about the groups that act transitively on spheres.  What is missing, as YCor commented, is knowing that if $G\subset\mathrm{GL}(n,\mathbb{R})$ acts transitively on $\mathrm{Gr}(k,\mathbb{R}^n)$ for some $k$ with $1<k<n$, then it acts transitively on $\mathrm{Gr}(1,\mathbb{R}^n)$.
Here is an argument that, at least when $k = 2$ or $3$, this statement is true.
I will assume, as YCor does, that $G$ is closed in $\mathbb{GL}(n,\mathbb{R})$
and hence is a connected Lie group.
First, one might as well use the oriented Grassmanians $\mathrm{Gr}^+(k,\mathbb{R}^n)$ since $G$ acts transitively on $\mathrm{Gr}^+(k,\mathbb{R}^n)$ if and only if it acts transitively on $\mathrm{Gr}(k,\mathbb{R}^n)$.  (After all, if $G$ acts transitively on $\mathrm{Gr}(k,\mathbb{R}^n)$, then its orbits in $\mathrm{Gr}^+(k,\mathbb{R}^n)$ are all open, and $\mathrm{Gr}^+(k,\mathbb{R}^n)$ is connected.)
Now let $E_k\to \mathrm{Gr}^+(k,\mathbb{R}^n)$ be the tautological oriented $k$-plane bundle, i.e., $$E_k = \{\ (e,v)\ |\ v\in e\in \mathrm{Gr}^+(k,\mathbb{R}^n)\ \}.$$
When $k=2$ and $n>2$, the oriented $2$-plane bundle $E_2$ has nonzero Euler class in $H^2(\mathrm{Gr}^+(k,\mathbb{R}^n),\mathbb{Z})\simeq \mathbb{Z}$.  In particular, it is not the sum of two line bundles (which would be trivial).
Now suppose that $G$ acts transitively on $\mathrm{Gr}^+(2,\mathbb{R}^n)$ and let $H\subset G$ be the subgroup that fixes $e_0\in \mathrm{Gr}^+(2,\mathbb{R}^n)$.  Then $H$ acts on the $2$-plane $e_0\subset\mathbb{R}^n$ and hence on $\mathrm{Gr}^+(1,e_0)\simeq S^1$.
If the orbits of $H$ on $\mathrm{Gr}^+(1,e_0)$ are open, then there is only one orbit, so that $H$ acts transitively on $\mathrm{Gr}^+(1,e_0)$.  Since $e_0$ was arbitrary, it follows that $G$ acts transitively on $\mathrm{Gr}^+(1,\mathbb{R}^n)$, as desired.  Meanwhile, if there were a non-open orbit, then fixing one such $H$-orbit $X_0\subset \mathrm{Gr}^+(1,e_0)\subset \mathrm{Gr}^+(1,\mathbb{R}^n)$ and looking at its $G$-orbit, i.e.,
$$
X = \{ ([g v], g(e_0)\ |\ [v]\in X_0, g\in G\ \}
$$
gives a bundle $X\to\mathrm{Gr}^+(2,\mathbb{R}^n)$ with discrete fibers over the simply-connected base $\mathrm{Gr}^+(2,\mathbb{R}^n)$.  Hence it is topologically trivial, so that there exists a section $S\subset X$ that is the projectivization of a line bundle $L\subset E_2$, contradicting the fact that $E_2$ has no rank $1$-subbundles.  Thus, this cannot occur, and the proof for $k=2$ is complete.
In the case $k=3$, one has to deal with the cases $n=4$ and $5$ separately. When $n=4$, the group $G = \mathrm{SU}(2)$ acts transitively on $\mathrm{Gr}^+(3,\mathbb{R}^4)\simeq S^3$, but does not act transitively on $\mathrm{Gr}^+(2,\mathbb{R}^4)$.  When $n=5$, applying duality and the above argument, one sees that if $G$ acts transitively on $\mathrm{Gr}^+(3,\mathbb{R}^5)$, then it acts transitively on $\mathrm{Gr}^+(4,\mathbb{R}^5)$, but then, by the paper YCor cites, this says $G$ must contain a copy of $\mathrm{SO}(5)$ and hence acts transitively on $\mathrm{Gr}^+(k,\mathbb{R}^5)$ for all $k$.
Thus, we can assume that $n\ge 6$.  In this case, we have that $H^p(\mathrm{Gr}^+(3,\mathbb{R}^n),\mathbb{Z})=0$ for $p = 1, 2, 3$, but that, not only is $H^4(\mathrm{Gr}^+(3,\mathbb{R}^n),\mathbb{R})\not=0$, we have that $p_1(E_3)\not=0$.  By a characteristic class argument, it follows that the bundle $E_3\to\mathrm{Gr}^+(3,\mathbb{R}^n)$ does not split as the sum of a line bundle and a $2$-plane bundle.
Now, I claim that if $G\subset\mathrm{GL}(n,\mathbb{R})$ acts transitively on $\mathrm{Gr}^+(3,\mathbb{R}^n)$ then it acts transitively on $\mathrm{Gr}^+(2,\mathbb{R}^n)$.  To see this, suppose that $G$ acts transitively on $\mathrm{Gr}^+(3,\mathbb{R}^n)$, and let $H\subset G$ be the stabilizer of $e_0\in \mathrm{Gr}^+(3,\mathbb{R}^n)$, and consider the action of $H$ on $e_0\simeq\mathbb{R}^3$.
First, suppose that $H$ has only open orbits in $\mathrm{Gr}^+(1,e_0)\simeq S^2$.  Then $H$ acts transitively on $\mathrm{Gr}^+(1,e_0)$ and hence $G$ acts transitively on $\mathrm{Gr}^+(1,\mathbb{R}^n)$.
Second, suppose that $H$ has a 0-dimensional orbit $X_0\subset\mathrm{Gr}^+(1,e_0)$. Then, constructing the bundle $X\to\mathrm{Gr}^+(3,\mathbb{R}^n)$ with discrete fibers in the projectivization of $E_3$ as was done above for $E_2$, we see that $X$ has a section and that this can be used to construct a splitting of $E_3$ as the sum of a line bundle and a $2$-plane bundle, which is known to be impossible.
Finally, suppose that all of the orbits of $H$ on $\mathrm{Gr}^+(1,e_0)$ are either $1$- or $2$-dimensional.  Not all of the orbits can be $1$-dimensional because $S^2$ has no foliation, and there cannot be more than a finite number of components of the union of the $1$-dimensional orbits (and there must be at least 1 component, otherwise we would be in the first case already dealt with).  Thus, there are only a finite number of components of the union of the $2$-dimensional orbits.  Each such component, which is homogeneous under a connected, finite dimensional Lie group, must have Euler characteristic either $0$ or $1$.  Thus, there must exist exactly two components that have Euler characteristic 1, i.e., are contractible disks, and these must map to a single component $D_0$ in $\mathrm{Gr}(1,e_0)$.  Using the contractibility of this component, we can now construct a rank 1 subbundle $L$ of $E_3\to \mathrm{Gr}^+(3,\mathbb{R}^n)$ with the property that the projectivization of $L_e$ lands in the $e$-fiber of the $G$-orbit of the contractible orbit $D_0$.  Again, this is impossible because $E_3$ does not split nontrivially as a sum.
Remark:  As far as I know, the only time that $G$ acts transitively on $\mathrm{Gr}(k,\mathbb{R}^n)$ with $4\le k\le n/2$ is when $G$ contains a copy of $\mathrm{SO}(n)$.  When $k=3$, there is $\mathrm{Spin}(7)\subset\mathrm{SO}(8)$
and, when $k=2$, we have $\mathrm{G}_2\subset\mathrm{SO}(7)$ and $\mathrm{Spin}(7)\subset\mathrm{SO}(8)$.  Neither $\mathrm{Spin}(9)$ nor $\mathrm{Spin}(9,1)$
in their $16$-dimensional representations act transitively on any $\mathrm{Gr}^+(k,\mathbb{R}^{16})$ except when $k=0,1,15,16$.
I believe that one could continue the above line of argument, showing that, if $G$ acts transitively on $\mathrm{Gr}^+(4,\mathbb{R}^n)$ for $n\ge 8$, then it must act transitively on $\mathrm{Gr}^+(3,\mathbb{R}^n)$, and so on.  However, I think that this argument will get more complicated as $k\le n/2$ increases, and it's probably not the right approach for general $k$.
