Assume that M is a compact manifold with fixed point property. Let N be a compact submanifold of M x M, with dim (N)=dim (M). Assume that $\pi_{1}:N \to M$ is a surjective map, where $\pi_{1}$ is the projection on the first component.
Is it true that N has non empty intersection with the diagonal {(x,x)\in M x M}?
Hello Ali! Maybe people do not answer because the question is not clear enough. What do you exactly mean by compact manifold with fixed point property? Maybe with the exact definition and without needing to guess, we will think more about this. The problem seems to be nice and interesting..
Hello Mihai Thank you very much for your comments on this question;
"A compact manifold with fixed point property" is a compact manifold M such that for every continuous map f:M \to M, there exist a point p in M with f(p)=p.
I see, very nice. So the interval [0,1] is such a variety, isn't it? (take a function f : [0,1] ---> [0,1] and consider the function f(x) - x. If this function changes the sign, it must have a fixpoint by the intermeiate value property. If it does not changes the sign, then f(0) = 0 or f(1) = 1 is a fixpoint.)
So on one hand, the fix point property for [0,1] has a proof which is quite similar to your problem, but for a special family of submanifolds N. On the other hand [0,1] is a variety with bord, and this helps the following construction: Take N to be the union of two closed straight line intervals of the plane, which do not intersect the diagonal of the square. One of them connects the point (1,0) with the point (1/3, 1/3 - epsilon), the other one connects the point (0,1) with the point (2/3, 2/3 + epsilon). Of course N projects surjectively on M, has dimension 1 as a manifold with bord (as M also) but does not intersect the diagonal.
So the property might be true, but possibly needs more conditions: that M and N have no singularities (so both must be smooth) and possibly that both are connected.
S^2 has not the fixpoint property because there is an antipodal map which is continuous and has no fixpoint. Also the torus S^1 x S^1 and the circle S^1 do not have the fixpoint property, because they have various rotations. Which example of variety with fixpoint property do you know excepting [0,1]? (it must be a smooth variety with local dimension = n for all points, unlike [0,1]...) For smooth surfaces of genus g \geq 2 I can also imagine kind of symmetries (similar to antipodal mappings) which have no fixpoint.
Dear Mihai thank you for your interesting example. As you said, we should consider more condition on N, that is connceted manifold without boundary.
real and complex projective spaces of even dimension, satisfy the fixed point property.(Rp^2n and CP^2n)
BTW the following post contains some related question:
http://mathoverflow.net/questions/226577/a-property-stronger-than-the-fixed-point-property
Thanks again for your attention to my question.
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