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I got stucked in this problem:

Show that:
i) Every embedded closed hypersurface $S$ is orientable.
ii) Every differentiable hypersurface defined by a regular cartesian equation $\ g(x_1,..., x_n)=0$ is orientable.
iii) A hypersurface is orientable if and only if it admits an orientable atlas $A={(\phi_i, U_i)}$, i.e. such that $\ det[d(\phi_i(\phi_j^{-1}))] >0 $ in every point where the composition is defined.

Now.. I only have a clue on few points (for instance the "if part" of ii) is trivial: the gradient gives the normal vector field..). My main issue is that here we don't have the cross product.. I know there is a way to exted the definition of orientability to higher dimensions by using differential volume forms but being this an exercise for a first course in differential geometry I am still a novice, so we're not meant to use very fancy tools! I was thinking of something more basic, like Gram–Schmidt orthogonalization for instance. Any help or hint would be great, my friends!!

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    $\begingroup$ I guess what you want is "closed hypersurfaces in $\mathbb R^n$ are orientable". Otherwise, it's false: $\mathbb{RP}^2$ is a non-orientable submanifold of the orientable $\mathbb{RP}^3$. $\endgroup$ – user98602 Jun 22 '15 at 16:03
  • $\begingroup$ yes indeed, closed! $\endgroup$ – yeahyeah Jun 22 '15 at 17:27
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This is confusing. For i), I suppose that you speak about embeddings into an orientable manifold, such as $\Bbb R^n$? However, the Mobious strip embedds into $\Bbb R^3$ and is clearly not orientable.

For closed hypersurfaces in $\Bbb R^n$, you may find these links useful: here, here and here.. Not sure if it can be proved by a completely elementary and intuitive argument, however.

For ii), it is not completely clear what you mean by regular, but I suppose that 0 is a regular value of $g$. Then there is a unique vector field $V$ on the surface, in which $g$ is increasing; you can define an orientation in $x\in g^{-1}(0)$ so that tangent vectors $(v_1,\ldots, v_{n-1})$ are positively oriented iff $(v_1, \ldots, v_{n-1}, V_x)$ is positively oriented.

For iii), I suppose that you can show that some atlas exists. If you have a notion of "positively oriented" for a basis of tangent vectors on the hypersurface, then the subatlas consisting of all maps that take positively oriented frames on Euclidean space to positively orientated frames on the hypersurface, has the above property. Conversely, if an atlas has the above property, you can define a basis of the tangent space to be positively oriented, if it is the image of a canonical basis in $\Bbb R^k$ under some (and hence any) coordinate map.

But maybe you have a different definition of orientation than what I'm using here (existence of a continuous section of "positively oriented" on tangent frames).

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  • $\begingroup$ Oh sorry for being sloppy! For i) I did mean closed hypersurfaces! Also we simply say that a surface in R^n is orientable if you can find a continuous, normal, non-zero vector field on it. $\endgroup$ – yeahyeah Jun 22 '15 at 17:09
  • $\begingroup$ For (i), you may find the answers here useful, but the answer is not completely elementary.. I would like to see myself a good "beginners" and intuitive answer. math.stackexchange.com/questions/879334/… $\endgroup$ – Peter Franek Jun 22 '15 at 17:12
  • $\begingroup$ Also here... google.cz/… $\endgroup$ – Peter Franek Jun 22 '15 at 17:29

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