We consider CP^{2n+1} with its natural Riemannian metric(The quotient of S^{4n+3} under the natural action of S^{1}. We denote by "d" the metric which is induced by this Riemannan metric.
We define F on CP^{2n+1} with F(x)=the (unique) farthest point to x(with respect to the above metric d).
Is F(x) well defined?(Is the point F(x) unique?). Is F \ circ F=Id (that is F^{2}=Id?).
Now for every x, we consider M(x)=All points y \in CP^{2n+1} such that d(y,x)=d(y,F(x)).
Is M(x) a ( codimenion one )totally geodesic submanifold of CP^{2n+1}? Does M(x) separate CP^{2n+1} into two disjoint part with the same area?
How many geodesics join x tp F(x)? Is it true to say that any geodesic which joints x to F(x) must be perpendicular to M(x)?
All of the above questions are motivated by n=0(the sphere).In this case the answer to all of the above questions are affirmative.
Note that we exclude CP^{even} since it has the fixed point property, so obviously the above F does not exist.
The answer is no. For example, for CP^2, CP^2 =C^2 UCP^1, where CP^1 corresponds to the infinite points of C^2. All the points in CP^1 have the same distance to the origin (0, 0)! They are not unique. For the CP^1 case, you only have one infinite point. That was why your process works for CP^1. But it does not work anymore for CP^2 and hence CP^n with n>1.
The distance function on CP^n is the angle between the lines that the points in CP^n represent. i.e.
d([u], [v]) = arccos( ||/|u|,|v|)
It folows that @Daniel Guang's example is exactly to the point: the points furthest away from a point [u] are simply the orthogonal complement, i.e. the complex codimension 1 hyperplane { = 0}.
Hyperplanes are totally geodesic submanifolds.
I am sorry to tell you that the question is not correctly posed, since the dimensions do not fit. Indeed, the famous Hopf fibration is the natural projection $\pi: S^{2n+1} \rightarrow CP^n$, with $\dim_R (CP^n)= 2n$.
On the other hand, given a point $x\in CP^n$, the set $F_x=\{ p\in CP^n : d(x,p) is maximal\}$ is a totally geodesic hyperplane $CP^{n-1}$. Then, any geodesic starting at x will meet the totally geodesic hyperplane $CP^{n-1}$. You may have a look to the following paper: Cecil, Thomas E.; Ryan, Patrick J. Focal sets and real hypersurfaces in complex projective space. Trans. Amer. Math. Soc. 269 (1982), no. 2, 481–499. From the point of view of this paper, the focal set of a totally geodesic hyperplane $CP^{n-1}$ is just a point, and viceversa.
Dear Miguel Ortega
I do not underestand why dimensions do not fit. In my question I put n:=2n+1. right?
I thank you all, for your answers. I think the subject of this question isa motivation to pose the following question:
What can be said about (the curvature of) a compact Riemannian manifold with following property:
For every x there is a UNIQUE point y=F(x) which maximize d(y,x). And F^{2}=Id.
Is it true to say that the spheres are the only possibility(Up to isometry)? Moreover, is the last condition F^{2}=Id, redundant?
Any way, existence of such map F is an obstruction for fixed point property.
Again, I thank you all for your interesting answers .
As i learn from 3 answers to this question the farthest point F(x) is not a single valued map,but it is a set valued map. so a natural question: "Is there a continuous selection for F"?(A continuos section)?
I think yes. Basically, you have the CP^{n-1} bundle over CP^n induced by the (holomorphic) tangent bundle. At each point, you have the standard SU(n) action on the holomorphic tangent bundle. That induces the SU(n) structure on the CP^{n-1} bundle. Since SU(n) is semisimple for n>1, all the SU(n) bundle is trivial.
Dear Daniel Guan. Thank you very much for your answer.I think that this bundle involvs the canonical line bundle on projective space rather than tangent bundle. i think it is the projectivization of the orthogonal bundle to the canonical bundle. I do not see how the tangent bundle would appear in your consideration.
On the other hand I doubt that this fibre bundle is trivial for n even. since CP^{even} has the fixed point property. In fact if the bundle were trivial we have aglobal section. That is a continuos map F on CP^{even} such that F(x) is the farthest point. Of course this contradict to Fixed point property. Am I mistaken?
The CP^{n-1} on the infinity is determined by the complex ray of the tangent at the given point and therefore, it is determined by the projected holomorphic tangent bundle!
Best,
DG
Dear Daniel Guan; The fibre bundle is $$ \{(x,y) \in CPn\times CPn \mid d(y,x) is maximal\}$$
right?
So it is isomorphic to projectivization of the orthogonal bundle to the canonical line bundle over CPn
So I can not underestand why you say this bundle is always trivial. Am I missing some thing?
Dear Ali,
It seems to me that you are talking about a manifold. Not a fiber bundle.
For the fiber bundle, for each point x in CP^n, you have a CP^{n-1} which represent the complex lines passing through that point. Each point of that bundle corresponding a point at the infinity. The total space of the bundle might be the same as the manifold you mentioned. The complex dimension is 2n-1. The manifold is compact. The canonical line bundle has a complex dimension n+1 and is open, but does have at least two section after the compactification---the zero section and the infinity section!
Best,
DG
I think that it is a CP^{n-1} bundle but the tangent not the SU(n) bundle. Instead, it is a U(n) bundle.CP^n is not Calabi-Yau. However, the S^1 action is trivial on the CP*{n-1} bundle and therefore you do have a SU(n) CP^{n-1} bundle. My original argument goes through. Therefore, the answer is yes (the fixed point Theorem argument possibly only work for vector fields, e.g., for the CP^1 case you do have the unique map!).
Best,
DG
Dear Daniel Yes that is the total space of the fibre bundle.
By canonical line bundle I mean the one dimensional vector bundle over CPn(As the standard terminology).
The role of fixed point property of CP{2n} is the following:
if the above CP(n-1) fibre bundle were trivial then we have a global section. This would imply that there is a continuos function F:CP{2n} to CP{2n} such that F(x) is the farthest point to x. This contradicts to existence of a fixed point.
Best
AT
Hi, I think it through and changed my earlier comment. Hope that it helps.
Best,
DG
The set of farthest points is a projective hyperplane (the "polar"). The set of points with farthest distance from that hypersurface is only the initial point. You can see this already in the real projective space (the sphere with antipodal point identification). The set of farthest points from the north pole is the equator.
Best regards
Jost Eschenburg
I thank every body for helps on this question. I understand the projective space(unless CP1) is not a good space for existence of such single valued map F(the farthet point).
So I ask an independent question in the following link,to finda nontrivial example of this situation:Any comment or answer to this question is very appreciated)
https://www.researchgate.net/post/To_what_extent_all_manifolds_with_this_property_are_classified
Hi,
I repost my earlier post here in case that you did not get it.
"
I think that it is a CP^{n-1} bundle but the tangent not the SU(n) bundle. Instead, it is a U(n) bundle.CP^n is not Calabi-Yau. However, the S^1 action is trivial on the CP*{n-1} bundle and therefore you do have a SU(n) CP^{n-1} bundle. My original argument goes through. Therefore, the answer is yes (the fixed point Theorem argument possibly only work for vector fields, e.g., for the CP^1 case you do have the unique map!).
Best,
DG"
Hope that this helps.
Best,
DG
As Professor Eschenburg pointed out, F(x) is not a unique point (except for CP^1 = S^2). The cut locus C(x) of a point x in CP^n is a CP^{n-1}. For any y \in C(x), the shortest geodesics fro x to y span a CP^1 = S^2. This is true for any n \geq 1. n doesn't have to be odd. Since F(x) is not unique, you have to modify the definition of "M(x)", but the set of points "half-way" between x and C(x) is diffeomorphic, but not isometric (except in the case of n=1) to a sphere S^{2n-1}, separating CP^n into two disjoint pieces. The piece that contains x is diffeomorphic to a disk D^{2n} and the other piece containing the cut locus C(x) is a disk (D^2) bundle over CP^{n-1) (the corresponding circle bundle is the dual of the Hopf bundle). I do not think the two pieces have the same volume (except in the very special case of n =1). You can find these things out, including computing volumes of the two pieces by looking at the exponential map from the tangent space at x to the manifold and computing Jacobi fields, etc. Remember that the cut locus coincides with the conjugate locus for CP^n.
I hope this clarifies the question.
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Ali Taghavi
Qom University Of Technology
Maung Min-Oo Is the distance we discussed equal to the diameter of $CP^n$?
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Daniel Guan
University of California, Riverside
yes.
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