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Jul 10, 2016 at 15:02 comment added Tom Goodwillie My guess above is simply wrong.
Jun 18, 2016 at 5:20 vote accept Abdoul
Jun 18, 2016 at 5:19 vote accept Abdoul
Jun 18, 2016 at 5:20
May 20, 2016 at 18:06 comment added Tom Goodwillie The Euler class that I mean is an element of $H^2(S^2)$, the Euler class of a rank two oriented vector bundle on $S^2$. Such a bundle remains non-trivial when added to the tangent bundle of $S^2$ if and only its Euler class maps to the nontrivial element of $H^2(S^2;\mathbb Z/2)$.
May 20, 2016 at 11:23 comment added Abdoul Sorry, I need some explanation about $M$( I refer to $M$ of the example), please. I understand that $M$ is the total space of vector bundle of rank two. My problem is that: it's possible to have a four manifold with odd Euler class? Thanks
May 18, 2016 at 11:12 comment added Tom Goodwillie rank means dimension of the fibers
May 17, 2016 at 9:58 comment added Abdoul Thank you, I will try to prove that. But, the mean of "M to be a rank two vector bundle over $S^2$" is not very clear for me. Could you explain it to me please? Thank you again. Abdoul
May 16, 2016 at 12:27 comment added Tom Goodwillie I'm pretty sure that if you take $M$ to be a rank two vector bundle over $S^2$ with odd Euler class then $Imm(M,\mathbb R^{4+k})$ is inequivalent to $Imm(S^2\times \mathbb R^2,\mathbb R^{4+k})$ for large $k$.
May 15, 2016 at 17:20 comment added Abdoul Thank you very much for this example. It is possible to have an example for simply connected manifolds? Because in this case the action of the base on the fiber is always trivial. Thanks
May 13, 2016 at 11:19 vote accept Abdoul
May 31, 2016 at 20:38
May 13, 2016 at 1:01 history answered Tom Goodwillie CC BY-SA 3.0