Let $ M $ be the Moebius band. In other words, the total space of the nontrivial line bundle over the circle. Can we equip $ M $ with a metric such the the isometry group acts transitively?


My intuition is that yes we can because $ M $ is the total space of a vector bundle over a compact Riemannian homogenous space (the circle).


However thanks to Ben McKay for the argument that $ M $ does not have a transitive isometry group. Also this fact is stated in
"Isometries of 2-Dimensional Riemannian Manifolds into Themselves" by
Sumner Byron Myers.



Is the Moebius strip Riemannian homogeneous?
Suppose by contradiction it is. Write it as $G/K$ where $G$ is the identity component of the isometry group and $K$ is compact, and $G$ acts faithfully on $G/K$. Since $G$ is connected, maximal compact subgroups are connected. Since $G/K$ is not contractible, $K$ is not maximal compact and it follows that maximal compact subgroups have codimension $1$. Since in noncompact simple Lie groups, maximal compact subgroups have codimension $\ge 2$, it follows that $G$ has no noncompact simple factor.


Suppose by contradiction that $G$ is not solvable. Then it has a simple compact connected subgroup $S$. Let $K'$ be a maximal compact subgroup. Then $K'\cap S$ has codimension $\le 1$ in $S$. Since a simple compact group has no subgroup of codimension $1$, it follows that $S\subset K'$. In turn, since $K$ has codimension $1$ in $K'$, by the same argument we deduce that $S\subset K$. Hence $S$ is contained in the intersection of all conjugates of $K$, which is trivial. Contradiction.


So $G$ is a connected solvable Lie group. Let $M$ be a closed connected normal subgroup of codimension $1$. If $MK=G$ then $M$ has smaller dimension and acts transitively, so we can argue by induction. Hence, assuming that $G$ has minimal dimension, we have $MK\neq G$, so $K\subset M$. Since in an abelian connected Lie group the intersection of all codimension 1 closed connected subgroups is trivial, we deduce that $K\subset \overline{[G,G]}$. The latter is nilpotent, and hence $K$ is a central torus in $\overline{[G,G]}$. Since the action of $G$ on the maximal torus of $\overline{[G,G]}$ is trivial, we deduce that $K$ is central, hence trivial. So $G$ is a 2-dimensional Lie group, and in particular is orientable.


(We have proved that if a noncompact connected surface can be endowed with a homogeneous Riemannian metric, then it is diffeomorphic to a Lie group, and hence to the plane, the cylinder. Of course various approaches to this result exist and some have already been mentioned in the comments.)


