The algebra for the projection : (U,V,-a) - (0,0,a) = t [ (x,y,z) - (0,0,a)], ie. (U,V,-2a) = (tx,ty,t(z-a) ) , so the 3rd component should be equal, in this case , -2a = t(z-a) , which implies that t= 2a/(a-z). U= tx = 2ax/(a-z) , V=ty = 2ay/(a-z) .
Hello Vidhu, thank you for your comment and the links below should answer your question. www.wikiwand.com/en/Cover_(topology) mathworld.wolfram.com/OpenCover.html www.encyclopediaofmath.org/index.php/Covering_(of_a_set)
May I ask a q? You said in two dimensions Klein bottle is manifold. You know figure eight is not manifold in one dimension. Because is not one to one map. But Klein bottle the same as figure eight has crossing itself that means is not manifold. Can you explain why. Please. Thanks in advance.
Hello Mid, When you run your finger over a figure of 8 drawing in 2D you pass through the centre twice so it is not 1:1 at that point. But the Klein bottle in 3D does not have that problem as you can see by the many images on the web.
IMO, your claim around 6:17 is fatally flawed. The open set U on the sphere is undefined without a pre-existing coordinate system. So, f(U) is undefined since the boundary of U is undefined.
@@TensorCalculusRobertDavie That's not an explanation but an appeal to authority. Does it matter that the location of U is unknown? And if not, why not?
Thank you very much, sir, Good explanation.
Hello Andy and thank you for that.
I love this vid.
Good one.
Thank you.
thank you , good explanation
ikrame daqaq Thanks for that.
The algebra for the projection : (U,V,-a) - (0,0,a) = t [ (x,y,z) - (0,0,a)], ie. (U,V,-2a) = (tx,ty,t(z-a) ) , so the 3rd component should be equal, in this case , -2a = t(z-a) , which implies that t= 2a/(a-z). U= tx = 2ax/(a-z) , V=ty = 2ay/(a-z) .
Thank you for that.
How did you get the algebra at 4:18? Other than that, great introduction to manifolds. It really sets up well for what's to come.
Hello Vidhu, thank you for your comment and the links below should answer your question.
www.wikiwand.com/en/Cover_(topology)
mathworld.wolfram.com/OpenCover.html
www.encyclopediaofmath.org/index.php/Covering_(of_a_set)
Ok. Thank you.
Have you also looked at my original video on Manifolds?
th-cam.com/video/w1FaAu0SL-w/w-d-xo.html
No I haven't. I'll give it a look.
God bless you, 😍
Thank you Elsad and the same to you.
May I ask a q?
You said in two dimensions Klein bottle is manifold. You know figure eight is not manifold in one dimension. Because is not one to one map.
But Klein bottle the same as figure eight has crossing itself that means is not manifold.
Can you explain why. Please.
Thanks in advance.
Hello Mid, When you run your finger over a figure of 8 drawing in 2D you pass through the centre twice so it is not 1:1 at that point. But the Klein bottle in 3D does not have that problem as you can see by the many images on the web.
Hi
Klien bottle is in 4D but it immerse in 3D.
Or you can think we see the shadow of klien bottle in 3d not itself.
IMO, your claim around 6:17 is fatally flawed. The open set U on the sphere is undefined without a pre-existing coordinate system. So, f(U) is undefined since the boundary of U is undefined.
The set U is arbitrary. Functions can be defined on open sets.
@@TensorCalculusRobertDavie That's not an explanation but an appeal to authority. Does it matter that the location of U is unknown? And if not, why not?
Thank you very much, sir, Good explanation.
Thank you Andy.