# Does this condition make a manifold smooth/analytic?

Question 1

Specifically, if we have an orientable $n$-dimensional manifold $\mathcal{M}$ embedded in $\mathbb{R}^{n+1}$ and we require that the class of all outward-facing normal vectors $\mathcal{N}_\mathcal{M}$ at all points of $\mathcal{M}$ form a vector space, does this force the manifold to be smooth or analytic? Being closed under vector addition seems like it would make the manifold smooth, and the existence of inverses makes me think that the surface would have to be closed in the $2$-manifold case.

EDIT: I believe this can be phrased also as: if the image of the Gauss map for an $n$-dimensional surface in $\mathbb{R}^{n+1}$ has no 'holes' in it, does this impose any particularly nice conditions on the surface? It seems like the Gauss-Bonnet theorem might come into play here?

Question 2

If we drop the requirement that inverses exist but maintain all the other structure requirements for a vector field, does this allow for open manifolds like a (finite) plane?

EDIT: I believe that this is equivalent to allowing for a non-symmetric image for the the surface under the Gauss map, but this might not be what I want to say. (is there a 'canonical' Gauss mapping?)

Does this change for non-orientable manifolds? Any references to appropriate literature would be greatly appreciated.

• What are outward-facing normal vectors? Is your $M$ inside some euclidean space (may be $\mathbb R^{n+1}$)? – user99914 Jul 30 '17 at 0:49
• I don't understand what inverses you are referring to. – gary Jul 30 '17 at 1:08
• Fixed the question, thanks, @gary I'm referring to the requirement that all elements of a vector space have an additive inverse under the addition operation in the vector space, in this case addition of normal vectors. If we require that all normal vectors must have an inverse normal vector in the sense that they sum to the zero vector, then I think the surface has to be closed. – Alec Rhea Jul 30 '17 at 1:12
• Alec, what do you mean by an outward-facing normal vector to a manifold embedded in $\mathbb{R}^{n+1}$ not necessarily smoothly? – Qiaochu Yuan Jul 30 '17 at 1:19
• @AlecRhea I'm not thinking of anything particularly interesting about the normal vectors and their multiples forming a vector space. The fact that the "normal vectors point in all possible directions" (i.e. the Gauss map is surjective), or even better (and more stable) the fact that the Gauss map has nonzero degree is what I'd suggest is closer to the intuition you might be having, and does have quite interesting facts related to it. Look up the Gauss map and its relationship to Euler characteristic. – aes Jul 30 '17 at 4:48