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).



Isotopy of positively curved surfaces of revolution in $\mathbb{R}^3$
1. The two boundary components of positively curved surface might be linked.
In this case it is not isotopic to the surface of revolution.
2. 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.)
3. 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.


