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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.

alvarezpaiva
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  • 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. – Misha Aug 02 '13 at 05:57
  • 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. – alvarezpaiva Aug 02 '13 at 06:18
  • True, but, still, might be worth checking. – Misha Aug 02 '13 at 06:24
  • @Misha: See section 1.3 in Vladimir"s paper. – alvarezpaiva Aug 02 '13 at 07:39
  • Juan-Carlos, there were two mainstreams in the classical (<1990) theory of projectively equivalent metrics: the french'' (+ Lie) andjapanese'' studied mostly infinitesimal projective transformations and ``soviet'' mostly projectively equivalent metrics. In the case your metric has a big group of symmetries, the results of both groups can be used by the following simple observation that I also explained in my answer: if the metrics are projectively equivalent, then an isometry of the first is projective transformation of the second. – Vladimir S Matveev Aug 02 '13 at 08:14
  • @Vladimir: Thanks !! I get the argument now. However, one can also make this argument for the real projective space and yet there are lots of projective transformations that are not isometries. – alvarezpaiva Aug 02 '13 at 08:30
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    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. – Ben McKay Aug 02 '13 at 09:50
  • Juan-Carlos, projective Lichnerowicz-Obata conjecture says that on a closed Riemannian manifold of nonconstant sectional curvature every infintesimal projective symmetry is a Killing vector field. It therefore can not be applied to the standard sphere or to the standard real projective space because they do have constant sectional curvature – Vladimir S Matveev Aug 02 '13 at 10:19
  • Thanks for you patience. I think I'm really thick today. @Ben: maybe it has to do with me being really thick today, but I can't see a clear-cut, easy argument characterizing the complex lines among the totally geodesic submanifolds of $\mathbb{CP}^n$. See my comments to Anton's answer. – alvarezpaiva Aug 02 '13 at 10:31

2 Answers2

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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.

Vladimir S Matveev
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  • 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? – alvarezpaiva Aug 02 '13 at 07:46
  • 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 – Vladimir S Matveev Aug 02 '13 at 07:55
  • 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? – alvarezpaiva Aug 02 '13 at 08:08
  • 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. – Vladimir S Matveev Aug 02 '13 at 08:15
  • thanks again. I think I need to see this theorem of Sinjukov (is there a proof somewhere? Dokl. is just for announcements, right?). – alvarezpaiva Aug 02 '13 at 08:39
  • You would not love the initial proof which is bases on tensor calculations. It is quite short though and could be put on the 2-page Doklady paper which I do not have by hand and can not check therefore. Actually, the review in MathReviews already contain some information. If you want a conceptual proof, a symmetric space is einstein so the Thomas cone connection is actually the Levi-Civita connection of a certain metric on the cone over the manifold. Projectively related metrics correspond to parallel (0,2) tensors for this cone metric and because of many symmetries they do not exist – Vladimir S Matveev Aug 02 '13 at 09:54
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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.)

Anton Petrunin
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  • Anton: You need more work for the 1st part of the proof, since "and the other way around" is clearly false. – Misha Aug 02 '13 at 04:51
  • @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 !! – alvarezpaiva Aug 02 '13 at 05:07
  • Hmm, I think I was a bit quick in my previous comment. I'm taking for granted that the only totally geodesic two-dimensional submanifolds in the complex projective plane are the complex lines and the Lagrangian real projective planes, but I do not really know if this is true. – alvarezpaiva Aug 02 '13 at 05:56
  • @alvarezpaiva: Yes, this is what my comment is about. – Misha Aug 02 '13 at 05:58
  • @Misha: we're just at the point where we wait for Robert to give us the answer ... ;-) – alvarezpaiva Aug 02 '13 at 06:25
  • This is a comment on the comment of Mischa about "and the other way around is clearly faulse". The small correction of Anton that makes the argument correct is ``every totally geodesic two-dimensional submanifold of CP(n) is a projective line''. – Vladimir S Matveev Aug 02 '13 at 07:48
  • @Vladmir: that's not true: there are (at least ?) the Lagrangian real projective planes as well. – alvarezpaiva Aug 02 '13 at 08:26
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    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. – Ben McKay Aug 02 '13 at 10:42
  • @Ben, Thank you --- that is exactly what I mean (and by the way my statement is correct; one only has to read it carefully). – Anton Petrunin Aug 02 '13 at 11:04
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    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. – Ben McKay Aug 02 '13 at 11:15
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    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. – Ben McKay Aug 02 '13 at 11:21