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Mar 22, 2015 at 13:41 vote accept mfl
Mar 22, 2015 at 8:37 comment added valeri @mfl I did not included the (obvious) proof that 3-dim P(t) of positive curvature, simply connected is S^3, is it ok?
Mar 22, 2015 at 0:09 comment added valeri One more way - prove that P(t) is simply connected. In order to do this, note that the parallel transport of vertical V=p(0)p(t) along big geodesic l(s) bounding half-sphere B in zero section S^2 gives again V. Now take the center O of B and transport parallel V first along l till l(s) and then along radius to O. The obtained family V*(s) gives you a circle S^1(t) over O, proving that vertical S^1 id homotopic to l in S^2 - null homotopic...
Mar 22, 2015 at 0:01 comment added valeri curvature of P(t) in 2-direction (X_p, W) (W - direction of the upper side p(t)q(t)) vanishes only if the holonomy of X along the triangle p(0)p(t)q(t) vanishes - but the last is given as t->0 by the same curvature component R(X,Y)V,W as the holonomy of the tangent bundle (see Ambrose-Singer formula), which is constant non-zero. Mean, that P(t) is 3-sphere for small t. QED. Yet another way - you may employ [O'Neil ... 1966?] formulas for the submersion directly.
Mar 21, 2015 at 23:51 comment added valeri @mfl P(t) is known as Berger sphere (kind of), may be [Cheeger, Some examples of manifolds ... J Diff G v8] consider it. To prove that P has positive curvature, take two points p(t) and q(t) over p(0) on zero section S^2 and two vectors X_p(t) and X_q(t) (horizontal lifts of X tangent to S^2 at p(0)) Now "move" with a parameter s the triangle p(0)p(t)q(t) in directionS X, X_p, X_q along three geodesics - you have a prism = family of isometric triangles due to the construction. Since the upper side p(t,s)q(t,s) has constant length on s - dirst and second variation formulas imply that the
Mar 21, 2015 at 21:40 comment added mfl Could you explain me a bit why the boundary $P(t)$ is diffeomorphic to a $3$-dimensional sphere?
Mar 21, 2015 at 21:25 history answered valeri CC BY-SA 3.0