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> Is there any way to distinguish between orientable or nonorientable surfaces when you're simply walking on it?

Draw a circle on the surface and orient it. If the surface is non-orientable then there's a way to take a long walk, return to the circle, and find that the direction of the orientation has changed.

This depends upon the circle you chose. Nonorientable manifolds still have many orientable circles (any sufficiently small loop is orientable).

You confuse me.

If the surface is non-orientable then I can draw a very small circle, put an orientation on it, then there's a path I can take that brings me back to find the orientation has changed.

Let's be more explicit.

Take a non-self-intersecting embedding of RP^2 in R^4, a point P in our RP^2, and a sufficiently small epsilon e. Then take three points on the circle of size e centred on P, and think of them as going in order, thus defining an orientation. Now take an appropriate walk around RP^2 and return to the circle. For an appropriate walk I will now find that the order of the points on my circle has reversed.

That's a more precise way of saying what I originally intended, and in that context your comment doesn't make sense to me. Can you expand on it?

I think when they say "circle" they're talking about your path, not your orientation reference. That's probably wrong terminology, but it makes some sense if you're thinking about how the path has to be a closed loop.

Yes, I took "circle" to mean any closed path since the difference is immaterial in generic manifold. You wouldn't normally be talking about "orienting a circle" in the manner the original poster did, but I see now what they mean. I took it to mean "choose an orientation along a closed path" which would be impossible to do for some (but not all) closed paths in a non-orientable surface.

A closed path is S^1, which is certainly orientable, no matter what it's embedded into. You just draw an arrow on the line.

ColinWright's point is that drawing such an arrow on a tiny circle is equivalent to drawing the letter R on the 2D manifold. It gives a local orientation to the 2D surface. (If you wish you may think of this as an arrow into some 3D embedding space, but you don't have to.)

If the 2D space is orientable, then when you take a copy of this little circle (or letter R) and go for a long walk, when you get home your copy will always match the original. That's all that orientable means. In the standard usage, it's a property of the 2D manifold, not of the particular walks you take. I think this is the point of confusion here.

Yes, you can always choose an orientation of the path but not necessarily of the manifold along the path. I understand what he was saying now and misunderstood since it is not the usual way one talks about specifying an orientation at a point and in fact had imagined he was specifically specifying a "big" circle, like a nontrivial loop on the torus. You'd more typically talk about choosing a basis of the tangent space at a point - presumably ColinWright wanted to avoid involving extra definitions like tangent space and so chose a visual definition that wasn't quite as precise. I don't think there's any further confusion.

Cool, I think we're on the same page. I know what you mean by "orientation of the manifold along the path" but this isn't precisely the standard usage. (My vague memory is that you can define an orientation without needing a tangent space, but I can't think of an example, so could be wrong here.)

My background is in riemannian geometry so I never had to worry about lack of a tangent space. Certainly you can define orientability of some topological spaces that aren't even manifolds. I'd forgotten but you can even define a local orientation at a point for a general topological manifold in terms of it's top-dimensional homology. I think that's the most general situation it makes sense in, you need a well defined dimension to consider this.

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