Consider a surface of revolution of positive curvature. My question is, what are the surfaces (with boundary) in $\mathbb{R}^3$ which are isotopic to the surface of revolution, provided each member of the family is positively curved? I am trying to see if we can achieve any positively curved surface (with boundary) diffeomorphic to the surface of revolution (I guess we have to include the restriction that the boundaries of such surfaces will be "unknots" to remove the obvious obstruction).
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The two boundary components of positively curved surface might be linked. In this case it is not isotopic to the surface of revolution.
The topology of surface of revolution is either disc or cylinder. In the first case you can construct an isotopy to a tiny cap in the disc which is isotopic to the rotationally symmetric one. In the second case you can construct an isotropy to a tiny belt. If the borders are not linked you can bend it in a surface of revolution. (H-principle shold be the key words in the proof.)
Note that the positively curved surfaces are locally convex. If in addition, your surface is globally convex* and isometric to a surface of revolution then you can construct a isometric isotopy. Indeed in this case we can close up the surface by two (or one) caps to get a positively curved metric on the sphere. Find a homotopy of such caps which deform in into rotationally symmetric surface. Then applying Alexandrov's embedding theorem we get an isotopy of surfaces.
*i.e., it lies completely on the surface of its own convex hull.
answered Jan 5, 2015 at 23:59
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$\begingroup$ Thanks for your answer! To clarify: regarding #2, you say "In the second case you can construct an isotropy to a tiny belt". Can you do so such that every member of the deformation has positive curvature? Also, I have no idea about the h-principle. Googling in general reveals something that does not seem (to me!) too related to the problem at hand. Is there some specific source that I can learn from? $\endgroup$ Commented Jan 6, 2015 at 7:01
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$\begingroup$ After a little reading, if I understand this at all, I think may be the h-principle allows us to say something like: the space of positive/negative curvature metrics on the surface without boundaries is contractible. Is that right? $\endgroup$ Commented Jan 6, 2015 at 14:08
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$\begingroup$ @user64672: In this case, H-principle says that we can start with any isotopy and deform it (rel. ends) into positively curved one. $\endgroup$ Commented Jan 6, 2015 at 17:07
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$\begingroup$ Just to clarify: could we also say that we could start with the same surface with boundary, but of negative curvature, and isotope it to a surface of revolution of negative curvature, passing through a family of negatively curved surfaces? Also, could you please mention a reference for learning the h-principle from this context? Thanks a lot! $\endgroup$ Commented Jan 6, 2015 at 18:16
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$\begingroup$ @user64672, yes, any open invariant condition will do; for example we could say that the principle curvatures strictly pinched by given constants. $\endgroup$ Commented Jan 6, 2015 at 19:21