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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.

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    $\begingroup$ No, you cannot treat it as a subgroup, since the lift is not unique. What is a subgroup of $Diff(S^2)$ is the central extension of $Diff(RP^2)$ by $Z/2$. $\endgroup$
    – Moishe Kohan
    Commented Feb 16, 2022 at 19:35
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    $\begingroup$ @MoisheKohan: although the lift is not unique, there is a unique orientation-preserving lift. So I think with some hustle you can ensure it is a subgroup of the orientation-preserving diffeomorphisms of $S^2$, up to a canonical isomorphism. $\endgroup$
    – Ryan Budney
    Commented Feb 16, 2022 at 21:36
  • $\begingroup$ Oh, right, since we are lifting maps $S^2\to \mathbb{R}P^2$. So the ratio of these two lifts is the antipodal map. Hence only one of them preserves the orientation. For example, for the fixed Riemannian metric there is a canonical choice. $\endgroup$
    – Ilia
    Commented Feb 16, 2022 at 21:53

2 Answers 2

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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.

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  • $\begingroup$ Yes, that makes perfect sense! Thanks! So for the sphere the same construction gives us the bundle $\text{Diff}(D^2)\times \text{Diff}(D^2) \to \text{Diff}(S^2) \to \text{Emb}(S^1, S^2)$? $\endgroup$
    – Ilia
    Commented Feb 16, 2022 at 19:18
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    $\begingroup$ Essentially, except for $S^2$ the compliment of a circle is two discs, so in the fiber there is the additional diffeomorphism that switches the two discs, i.e. the group is a $\Bbb Z_2$-extension of what you have written. The target embedding space is homotopy-equivalent to $SO_3$, and the $\Bbb Z_2$ in the fiber is how $Diff(S^2)$ can be homotopy-equivalent to $O_3$. Apologies, it looks like you are editing the comment I replied to. $\endgroup$
    – Ryan Budney
    Commented Feb 16, 2022 at 19:20
  • $\begingroup$ I just realized that my comment is incorrect as stated but you were faster:-) $\endgroup$
    – Ilia
    Commented Feb 16, 2022 at 19:22
  • $\begingroup$ Aren't diffeomorhisms of the disks are supposed to agree on the circle? Do we account for it here? $\endgroup$
    – Ilia
    Commented Feb 16, 2022 at 19:46
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    $\begingroup$ @Ilia: we are using the lazy notation where $Diff(D^2)$ means diffeomorphisms of the disc that are the identity on the boundary. $\endgroup$
    – Ryan Budney
    Commented Feb 16, 2022 at 21:31
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It is a theorem of A. Gramain from 1973 (Annales Sci. E.N.S.) that the diffeomorphism group of the projective plane has the homotopy type of $SO(3)$, the subgroup of isometries of the standard constant curvature metric. This is proved by projecting the diffeomorphism group onto the space of embeddings of a disk into the projective place by restricting diffeomorphisms to embeddings of a disk. This projection is a fibration whose fiber is the diffeomorphism group of the Moebius band fixing the boundary. Gramain showed this fiber is contractible. The base space of the fibration is easily seen to have the homotopy type of $SO(3)$.

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    $\begingroup$ Newer proofs of this follow using Ricci flow. The space of Riemannian metrics is contractible. Under Ricci flow, the metrics flow to the round metric by a result of Bennett Chow. Hence the space of round metrics is contractible too. These act like a homogeneous space for $Diff(S^2)$, since $Diff(S^2)$ acts on the round metrics with stabilizer $SO(3)$. Hence $Diff(S^2)$ is homotopy equivalent to $SO(3)$. $\endgroup$
    – Ian Agol
    Commented Feb 17, 2022 at 23:36
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    $\begingroup$ @Ilia: no, think about the case of the Klein bottle. The fundamental group of $Diff(S^1 \times S^1)$ is free abelian of rank 2, but the fundamental group of the diffeomorphism group of the Klein bottle is infinite cyclic. I think you can also see a difference at the $\pi_0$ level, since the Klein bottle maps Moebius bands to Moebius bands, and these are preferred mod-2 homology classes in the surface. But the diffeomorphism group of the torus acts transitively on the primitive homology classes. $\endgroup$
    – Ryan Budney
    Commented Feb 18, 2022 at 0:56
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    $\begingroup$ $Diff(S^1 \times S^1)$ has the homotopy-type of $GL_2 \Bbb Z$. The Klein bottle, I don't think I've ever worked out the homotopy-type exactly, but I think it's relatively easy to deduce $Diff(K^2)$ has the homotopy-type of $S^1 \times \pi_0 Diff(K^2)$, and $\pi_0 Diff(K^2)$ appears to be $\Bbb Z_2 \times \Bbb Z_2$. The $S^1$ comes from thinking of the Klein bottle as fibered over the circle, and motion corresponds to pushing twice around the base circle. The abelian group of order 4 corresponds to (a) the automorphim reversing the base circle. The other (b) seems to come from... $\endgroup$
    – Ryan Budney
    Commented Feb 18, 2022 at 4:40
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    $\begingroup$ thinking of the Klein bottle as the union of two Moebius bands. There is the diffeomorphism that fixes the boundary circle of the two Moebius bands, but swaps the interiors. I suppose you could say this fixes the base, and is complex conjugation on the fibers. $\endgroup$
    – Ryan Budney
    Commented Feb 18, 2022 at 4:42
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    $\begingroup$ Typo above: $Diff(S^1 \times S^1)$ has the homotopy-type of $S^1 \times S^1 \times GL_2 \mathbb Z$. I forgot the manifold factor. $\endgroup$
    – Ryan Budney
    Commented Feb 18, 2022 at 6:57

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