First of all, I am interested in the general case of a non-orientable manifold but let's for now consider the projective plane $\mathbb{R}P^2.$ In short, I am curious if there is any relation between the diffeomorphism group $\text{Diff}(\mathbb{R}P^2)$ of the projective plane and the diffeomorphism group $\text{Diff}(S^2)$ of its orientation double cover.


As I understand, any diffeomorphism $\mathbb{R}P^2\to \mathbb{R}P^2$ can be lifted to a diffeomorphism $S^2\to S^2$ of the orientation bundle (same in the general case). That means that we can consider $\text{Diff}(\mathbb{R}P^2)$ as a subgroup in $\text{Diff}(S^2).$ I know that there are a lot of results on geometry and topology of this group and I wonder if some results remain true for $\text{Diff}(\mathbb{R}P^2).$ I tried to search for some literature on this topic but didn't find anything useful for me (partially because I am not sure what I am looking for). My final goal for now is to compute (or find results on) the curvature for a right-invariant metric on the group of volume-preserving diffeomorphisms $\text{SDiff}(\mathbb{R}P^2).$ I would be very much interested to learn anything on this topic.


Thanks.



Diffeomorphism group of the projective plane
Two different answers using almost identical techniques! Allen's response got me to think through my response more carefully. Let me edit in a comment to point out my sloppiness, as it points out a useful detail in the machinery we are using.


$\DeclareMathOperator\Diff{Diff}\DeclareMathOperator\Emb{Emb}\DeclareMathOperator\SO{SO}$One approach to studying $\Diff(\Bbb RP^2)$ would be to look at the bundle


$$\Diff(D^2) \to \Diff(\Bbb RP^2) \to \Emb(S^1, \Bbb RP^2).$$


**Edit:** technically this is not the fiber. Since the $D^2$ is not embedded, the fiber is the diffeomorphisms of $\Bbb RP^2$ that restrict to the identity on the embedded curve. If we blow up the curve, you can think of this as the group of diffeomorphisms of $D^2$ that are either the identity on the boundary, or the antipodal map on the boundary. Sticking with the convention that $Diff(D^2)$ is diffeomorphisms of the $2$-disc that are the identity on the boundary, the fiber would be $Diff(D^2) \times \Bbb Z\_2$. So Smale's theorem tells us $Diff(\Bbb RP^2)$ is (up to a homotopy-equivalence) a $2$-sheeted covering space of the component of $\Bbb RP^1$ in $Emb(S^1, \Bbb RP^2)$. A more systematic way to see this would be to consider the space of embeddings of a tubular neighbourhood of $\Bbb RP^1$ in $\Bbb RP^2$. There is the embedded circle it contributes, but there are also the automorphisms of the tubular neighbourhood, switching the directions of the fibers of the M"obius band. The fiber of the restriction map $Diff(\Bbb RP^2) \to Emb(M, \Bbb RP^2)$ where $M$ is the M"obius band, now is literally a copy of $Diff(D^2)$.


The bundle is not onto the base space, it is onto the subspace of embeddings whose complement is a disc, i.e. embedded curves whose normal bundles are Moebius bands.


I believe this embedding space has the homotopy-type of the subspace of linear embeddings. The linear embeddings has the homotopy-type of the unit tangent bundle of $\Bbb RP^2$, which could be thought of as


$$ UTS^2 / \Bbb Z\_2 $$


where in the action one negates the base-point and the tangent vector simultaneously.


$UTS^2$ can be thought of as a copy of $\SO\_3$, which I believe would make the quotient a copy of the lens space $L\_{4,1}$ — the action of $\Bbb Z\_2$ on $\SO\_3$ performs a rotation by $\pi$ on the first two column vectors. **Edit:** Passing to the covering space, we recover that $Diff(\Bbb RP^2)$ is homotopy-equivalent to $\SO\_3$.


Let me know if that makes sense or not. Sometimes I am a little rusty when writing answers in the morning.


