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We had a lecture a few weeks back, looking at Gauss' divergence theorem, and in the definition of the theorem, it specified that the boundary of the volume under consideration, S, had to be a 'piecewise smooth, orientable, closed surface'.

What bothers/intrigues me is that I cannot understand how a closed surface in 3D space CANNOT be orientable. Surely every closed surface is orientable!

My highly non-rigorous, intuitive argument runs as follows:

1) As the surface is closed, we can define two regions, one inside the surface, and one outside

2) We can construct a normal to the surface at any point P that is pointing towards the inside region. Thus the direction of the normal is defined for every point.

3) As the surface is piecewise continuous, this normal will vary continuously.

4) Coupling (2) (defined direction of normal) with (3) (continuously changing normal) gives us an orientation for the closed surface.

5) Therefore every closed surface is orientable.

But of course, the precise wording of the statement for Gauss' Law strongly suggests that people far smarter than me have discovered some exotic non-orientable, closed surface. Is this true?

When I asked my lecturer about this, he just smiled and said he didn't know any examples, but that they do exist, and then said something even more tantalising about 'reflections of higher dimensional objects'

I would love it if anyone could shed some light on my situation.

Thanks

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The Klein bottle en.wikipedia.org/wiki/Klein_bottle is an example. You cannot embed it in $\mathbb{R}^3$ though. –  Thomas Rot Jun 5 '11 at 12:40
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I think you interpret 'closed' incorrectly. I don't quite understand your reasoning in (1) that a surface being topologically closed implies it has an 'inside' and 'outside'. –  wildildildlife Jun 5 '11 at 13:35
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@wildildildlife: In differential geometry, a closed surface is one which is compact and without boundary. It's an unfortunate coincidence of terminology. –  Zhen Lin Jun 5 '11 at 15:08
    
@wildildildlife Mmm, no you're exactly right. I suppose it's because all of the basic applications for closed surfaces (Gauss' Divergence theorem, the relevant Maxwell equations, etc.) all talk about closed surfaces CONTAINING stuff, it's easy to ignore these special cases. But now I see that closed surfaces need not divide 3D space (unlike closed curves in the 2D plane I suppose) –  tom Jun 6 '11 at 9:45
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up vote 11 down vote accepted

You're right that if the surface has to fit into 3D, and if it is non-self-intersecting, then it has to be orientable.

But if you allow the 2D surface to self-intersect or go to the 4th dimension, there are many counterexamples. Klein Bottle is the simplest one.

enter image description here

Take a bottle, heat it up, and push the throat through the body, to connect it with a hole that you prepared at the bottom of the bottle. This manifold is equivalent to a $Z_2$ orbifold of a torus - where the $Z_2$ generator shifts by half a period in one direction of the torus and reflects the other direction of the torus.

Your rule (1) fails because there is no well-defined interior and exterior. Indeed, if you make trips around the Klein bottle that reverse the orientation, it inevitably exchanges the interior with the exterior as well. Note that if you enter the hole at the bottom of the bottle (see the picture), you're still outside, but as you travel through the throat, it becomes clear that you have gotten inside the "object" as well, so there's no distinction between the exterior and interior.

The Klein bottle is self-intersecting if embedded into three dimensions. Alternatively, you may avoid the intersections if one of the pieces of the surface that would intersect each other are shifted in a new, fourth dimension of space.

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Wow, thankyou so much for that, that's a fantastic little surface. It's funny, running the normal around it in your head, to see it change orientation, feels just like with a mobius strip. –  tom Jun 5 '11 at 12:46
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Dear @Tom, it was a pleasure. The Möbius strip is nothing else than a properly chosen one half of the Klein Bottle. Place a topologically nontrivial circle in the Klein bottle - for example, run from the bottom of the bottle along the throat to the left, and return to the same place. And then cut it along this circle by scissors. You will get an unorientable surface with one boundary - the circle - and it's the Mobius strip. It's safer to cut a strip along the circle I just mentioned, this one is manifestly a Mobius strip. Unlike the Klein bottle, the Mobius strip has a boundary. –  Luboš Motl Jun 5 '11 at 13:20
    
Haha, you're exactly right :) Takes a bit to visualise though. I also see what you mean about this being related to a torus. If you were to cut a circle around a torus as you did for this, you'd just get circular stip without the half-turn. BTW, would this be (very) basic Topology, looking at questions like this? –  tom Jun 6 '11 at 9:49
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