Non-orientable surfaces---the Mobius band | Algebraic Topology 6 | NJ Wildberger

[@chaoli3401]

Fantastic illustration! Thank you so much for your generous sharing of these knowledge and insights with the world.

[@thinking1st]

You are really a good teacher. Thanks kindly.

[@trevorjenglish]

Fantastic video. I’m an working on an article about Möbius strips and topology and this lecture was incredibly helpful!

[@jjhimself6830]

This topic fascinates me, I'm filled with joy when I hear people talk about it. is that weird?

[@Mystic-Lemon]

No it’s not

[@VishalKumar-yd4tq]

Thank u sir

[@temp8420]

Excellent

[@YoungMeasures]

I came here to praise his accurate presentation of non-orientability. Despite the convenient but incorrect argument using a pin on the Mobius band, it is mathematically not a proof of nonorientability. It uses the ambient space R^3, whereas orientability (or lack there of) is an inherent property of the space itself, independent of any embedding. On a 2-D manifold, THERE IS NO NOTION OF SIDE AND THE OTHER SIDE. As his translucent model depicts perfectly, a 2D manifold does not have thickness, like a sheet of paper does, to be able to talk about "side". It is about the intrinsic, within manifold itself, notion of an ordered choice of coordinate bases -- in case of 2D, a left versus right rotations, i.e. handedness. Now, where does the thing with pin come from? The answer is that in the UNIQUE situation of 2D manifold in R^3, there is a one-to-one correspondence between a choice of rotation at a point on the manifold and a choice between the two of the unit normal vectors to the the manifold at that point. Once you choose your rotation on (and within) the manifold at a point, only one of the unit normal vectors matches the right-hand gesture of positive orientation in R^3. Conversely, choosing your normal (your "side") goes well with either right or left rotations on the manifold, not both. Under this correspondence, a continuous choice of orientations is possible if and only if a continuous choice of a unit normal is. That is why it works after all for embedded 2-D manifolds in R^3. The problem: It all falls apart if you embed the same exact manifold in R^4. But many of the cheap animations on KZbin fail to make that point.

[@RichardSouthwell]

Thank you for your wonderful videos Professor. I'm especially interested in finding a homeomophism between the cross cap (cc) and Mobius strip (ms). I found your sketched ms->cc conversion very hard to do physically, so I've tried to make my own ms->cc homoemorphism, using white tack. A New Homoeomorphism From The Mobius Strip To The Cross-Cap A New Homoeomorphism From The Mobius Strip To The Cross-Cap I would be honored if you would give me your opinion about whether this method really corresponds to a homeomphism. I'm a little unsure about my final step.I'm wondering if it could be fixed to give an easier way to demonstrate the equivalence of cc and ms,

[@Pygmygerbil88]

11:28 also happy 🤣 as well as a fantastic professor, norman has a very great sense of humor