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Are there non-trivial diffeomorphisms (i.e., different from isometries) of the complex projective plane that map geodesics (for the canonical Riemannian metric) to geodesics?

Same question for all other rank one symmetric spaces different from spheres and real projective spaces.

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  • $\begingroup$ There is a vast literature on "projective maps", going back to Sophus Lie, see e.g. www.minet.uni-jena.de/~matveev/Datei/lichnerowicz.ps and references there. $\endgroup$
    – Misha
    Commented Aug 2, 2013 at 5:57
  • $\begingroup$ Misha: thanks for the reference. However, most and probably all (?) of this classic work relies on the existence of "infinitesimal" projective transformations. Here I would like to know if there is just one non-isometric transformation that maps geodesics to geodesics. $\endgroup$
    – alvarezpaiva
    Commented Aug 2, 2013 at 6:18
  • $\begingroup$ True, but, still, might be worth checking. $\endgroup$
    – Misha
    Commented Aug 2, 2013 at 6:24
  • $\begingroup$ @Misha: See section 1.3 in Vladimir"s paper. $\endgroup$
    – alvarezpaiva
    Commented Aug 2, 2013 at 7:39
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    $\begingroup$ I once wrote a paper called Smooth projective planes, which proved that the continuous maps which preserve orientation and take lines to lines are diffeomorphisms. It turned out that this was already known in the literature of topological projective planes. Using the fact that the geodesics of the complex projective plane lie on the projective lines, you can easily show that the homeomorphisms preserving orientation and geodesics are complex projective transformations. But then preserving geodesics is actually stronger, so they must be isometries as indicated below. $\endgroup$
    – Ben McKay
    Commented Aug 2, 2013 at 9:50

2 Answers 2

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The answer is no. The explanation of Anton is of course correct but there exist stronder statements in the literature: for example by Sinjukov (Dokl. Akad. Nauk SSSR (N.S.) 98, (1954) 21--23) any symmetric space is locally \emph{geodesically rigid} is the sense that any other metric having the same (unparameterized) geodesics with it is affinely equivalent to if (i.e., the Levi-Civita connections coincide) which in the irreducible case means that the metrics are proportional.

Actually, stronger statements hold. For example from the Lichnerowicz-Obata conjecture arXiv:math/0407337 it follows that compact Riemannian homogeneous metrics such that sectional curvature is not constant and positive are also geodesically rigid. Indeed, a Killing vector field for the initial metric is a infinitesimal projective transformation for the second, which must be also Killing by the projective Lichnerowicz-Obata conjecture. Then, the isometry algebras of the metrics are the same and therefore their volume forms are the same and these already implies (short tensor calculations, see for examples eqns. (1), (4), (5) of arXiv:0806.3169) that the metrics are affinely equivalent. I do not know whether homogeneous metrics of nonconstant curvature are geodesically rigid locally but all examples indicate that probably they are.

Now, in the case your metric is Kähler and not flat,
then if it is not geodesically rigid then it is locally a cone over a (sasakian) manifold which in particular implies that the manifold is not compact. This statement is pretty nontrivial and follows from Theorem 4.6. of Mikes (Journal of Mathematical Sciences 78(1996) 311-333) combined with the Splitting Lemma from arXiv:0904.0535 and combined with the following statement which was explained to me by Kiosak and which is probably not published: Warped product Kähler nonflat metric is a locally a cone over a sasakian manifold.

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  • $\begingroup$ In the question there is no connected Lie group, no infitesimal projective transformations. Could you please specify how much of what you are saying carries through to this situation? $\endgroup$
    – alvarezpaiva
    Commented Aug 2, 2013 at 7:46
  • $\begingroup$ I do not understand your comment. If you are asking whether in my answer I assumed that the projective transformation is actually an infinitesimal projective transformation, I did not do it and spoke about projectively equivalent metrics only. If you would like me to tell you more assuming the existence of an infinitesimal projective transformation, then globally (in the riemannian case) it is in arXiv:math/0407337 and locally in Solodovnikov 1956 in dimensions >2 and in arXiv:0802.2344 + arXiv:0705.3592 in dimension =2. In the case I misunderstood your comment, please explain $\endgroup$
    – Vladimir S Matveev
    Commented Aug 2, 2013 at 7:55
  • $\begingroup$ In your paper on the Lichnerowiz-Obata conjecture you explicitly warn that you need the hypothesis of a connected Lie group of projective transformations. Here I'm asking for just the existence of one non-trivial projective transformation that can be very far from the identity. In your comments do you assume that we have a connected Lie group of projective transformations? $\endgroup$
    – alvarezpaiva
    Commented Aug 2, 2013 at 8:08
  • $\begingroup$ No, I dont do it in my answer. The result of Sinjukov is local, works in all signatures, and is about projectively equivalent metrics and not about projective transformations or connected groups of projective transformations. The arguments in my asnwer using Lichnerowciz-Obata relies on the existence of a big group of isometries of the metrics in your question. If there exists a (noninfinitesimal) projective transformation, then the pullback of the killing vector fields are projective vector fields and one can use Licherowicz-Obata. The Kähler result is also about projective equivalence. $\endgroup$
    – Vladimir S Matveev
    Commented Aug 2, 2013 at 8:15
  • $\begingroup$ thanks again. I think I need to see this theorem of Sinjukov (is there a proof somewhere? Dokl. is just for announcements, right?). $\endgroup$
    – alvarezpaiva
    Commented Aug 2, 2013 at 8:39
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For complex projective plane with the canonical metric you get only isometries.

Indeed, note that such map has to send complex lines to the complex lines. It follows since, any complex line is a union of an infinite family of geodesics passing through two points and the other way arround.

It remains to check which complex projective maps send geodesics to geodesics.

Note that a Möbius transformation of a sphere (=complex projective line) sends geodesic to geodesic if and only if it is an isometry. Hence the result follows.

(The same might follow from the projective curvature tensor, but I do not know how one calculates it.)

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  • $\begingroup$ Anton: You need more work for the 1st part of the proof, since "and the other way around" is clearly false. $\endgroup$
    – Misha
    Commented Aug 2, 2013 at 4:51
  • $\begingroup$ @Anton: I thought about this line of attack, but I got stuck in proving that complex lines get mapped to complex lines. Oops, never mind: they are the only totally geodesic submanifolds in their homology class. They have to be mapped to each other. Thanks !! $\endgroup$
    – alvarezpaiva
    Commented Aug 2, 2013 at 5:07
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    $\begingroup$ If you take any two points of the complex projective plane, there is either a unique geodesic connecting them, or else the union of the geodesics is a projective line. Therefore geodesic preserving homeomorphisms preserve projective lines, and therefore are smooth projective transformations (from my work, or earlier work on smooth projective planes). Then Anton's argument finishes the proof: all geodesic preserving homeos of the complex projective plane are isometries. $\endgroup$
    – Ben McKay
    Commented Aug 2, 2013 at 10:42
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    $\begingroup$ I keep referring to earlier work that proves smoothness of homeomorphisms preserving projective lines. The proof: Bödi, Richard; Kramer, Linus On homomorphisms between generalized polygons. Geom. Dedicata 58 (1995), no. 1, 1–14. $\endgroup$
    – Ben McKay
    Commented Aug 2, 2013 at 11:15
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    $\begingroup$ The same proof then proves that any homeomorphism of the octave projective plane or quaternionic projective plane which preserves geodesics is an isometry. It is nice how the proof breaks for the real projective plane. $\endgroup$
    – Ben McKay
    Commented Aug 2, 2013 at 11:21

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