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Aug 20, 2018 at 8:42 vote accept Ali Taghavi
S Aug 20, 2018 at 8:42 history bounty ended Ali Taghavi
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S Aug 14, 2018 at 8:02 history bounty started Ali Taghavi
S Aug 14, 2018 at 8:02 history notice added Ali Taghavi Authoritative reference needed
Aug 14, 2018 at 7:53 comment added Ali Taghavi @DeaneYang Dear prof. Yang I am sorry if my question is elementary. To be honnest I did not understand a relation between my question and the part of your statement that "......this familly consist of either whole morphisms or 0 or identity.." I would appreciate if you more explain.
Aug 11, 2018 at 18:18 answer added Deane Yang timeline score: 8
Aug 11, 2018 at 9:29 comment added Ali Taghavi @DeaneYang so my question can be read, when the shape operator generate a lie algebra morphism on the space of vector fields on our surface?When its range is a Lie algebra?
Aug 11, 2018 at 3:36 comment added Deane Yang that is correct. The eigenvalues of the shape operator are the principal curvatures. And the shape operator of the cylinder is neither the identity map nor zero.
Aug 11, 2018 at 3:06 comment added Ali Taghavi @DeaneYang I think the eigenvalues of $dN$ are principal curvatures, then for cylinder it is neither identically zero nor "identity". Right?
Aug 10, 2018 at 21:38 comment added Deane Yang $dN$ is known as the shape operator and is a $(1,1)$-tensor, i.e. an endomorphism of $T_*S$. Based on a quick back-of-the-envelope calculation, I believe that $dN$ has to be either the identity map or the zero map. In the first case, it is a sphere. In the second, it is a flat surface.
Aug 10, 2018 at 17:58 history edited Ali Taghavi CC BY-SA 4.0
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Aug 10, 2018 at 13:58 history edited Ali Taghavi CC BY-SA 4.0
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Aug 10, 2018 at 12:18 history edited Ali Taghavi
I add a tag.
Aug 10, 2018 at 11:54 history asked Ali Taghavi CC BY-SA 4.0