Hello John. Thank you for your comment and request which I will deal with in the future at some stage, just not sure when? Are you looking for more examples?
Hi, just started watching your excellent videos. I got confused with Differentiable Manifold video on unit circle S1 (video 3 of Manifolds playlist). Now f alpha is given as root (1-x^2) and f gamma as root (1-y^2) but I could not see how f alpha gamma as being root (1-x^2) computed.
Hello David and thank you for your comment. I am going to remake this video with a better explanation of the section in question. In the meantime, I hope the following helps. We cover the circle S1 with four open sets U𝛼, U𝛽, U𝛾, and U𝛿. These are the upper and lower red curves with the same for the two left and right blue curves. From the point of view of topology a small part of a circle is the same as a small part of a straight line. Now any point on the top half of the circle (upper red part) can be described uniquely by its ‘x’ coordinate. We can project this part of the circle onto the ‘x’ coordinate by a continuous map or homeomorphism onto the open interval (-1, 1) of the x-axis. The same is true for the lower red part of the circle. These two mappings are our first two charts, f𝛼(x, y) = x = f𝛽(x, y). We now repeat this process for the two blue parts of the circle to produce two more charts, f𝛾(x, y) = y = f𝛿(x, y). Here, y is a homeomorphism onto the open interval (-1, 1) of the y-axis. Now we have covered our circle with four charts. Next we limit ourselves to the intersection of U𝛼 ⋂ U𝛾 which is that part of the circle that lies in the first quadrant. The two charts here, f𝛼 and f𝛾, map this part of the circle to the open interval (0, 1). Let x = n be a number on this interval, then the inverse f𝛼^-1, takes us from this number ‘n’ back to a point on the circle √(1 - n^2). So our transition map or coordinate change is, f𝛼𝛾 = f𝛾◦ f𝛼-1 = f𝛾(f𝛼-1) = f𝛾(n, √(1 - n^2)) = f𝛾(x, y) = y = √(1 - n^2) So, in general, we have, f𝛼𝛾 = f𝛾◦ f𝛼-1 = y = √(1 - x^2).
Clear picture, Clear explanation.
Excellent!
Thank you!
Could you explain more on differential manifold.please.
Excellent video.
Many thanks
Hello John. Thank you for your comment and request which I will deal with in the future at some stage, just not sure when? Are you looking for more examples?
Great introduction!
Thank you.
Absolutely amazing video! Which book(s) do you recommend for further studies?
Thank you. There are many free sources on the web including TH-cam videos.
what is class C-infinity?
Infinitely many times differentiable.
Hi, just started watching your excellent videos. I got confused with Differentiable Manifold video on unit circle S1 (video 3 of Manifolds playlist). Now f alpha is given as root (1-x^2) and f gamma as root (1-y^2) but I could not see how f alpha gamma as being root (1-x^2) computed.
Hello David and thank you for your comment. I am going to remake this video with a better explanation of the section in question. In the meantime, I hope the following helps.
We cover the circle S1 with four open sets U𝛼, U𝛽, U𝛾, and U𝛿. These are the upper and lower red curves with the same for the two left and right blue curves. From the point of view of topology a small part of a circle is the same as a small part of a straight line.
Now any point on the top half of the circle (upper red part) can be described uniquely by its ‘x’ coordinate. We can project this part of the circle onto the ‘x’ coordinate by a continuous map or homeomorphism onto the open interval (-1, 1) of the x-axis. The same is true for the lower red part of the circle. These two mappings are our first two charts, f𝛼(x, y) = x = f𝛽(x, y).
We now repeat this process for the two blue parts of the circle to produce two more charts, f𝛾(x, y) = y = f𝛿(x, y). Here, y is a homeomorphism onto the open interval (-1, 1) of the y-axis.
Now we have covered our circle with four charts.
Next we limit ourselves to the intersection of U𝛼 ⋂ U𝛾 which is that part of the circle that lies in the first quadrant. The two charts here, f𝛼 and f𝛾, map this part of the circle to the open interval (0, 1). Let x = n be a number on this interval, then the inverse f𝛼^-1, takes us from this number ‘n’ back to a point on the circle √(1 - n^2).
So our transition map or coordinate change is,
f𝛼𝛾 = f𝛾◦ f𝛼-1 = f𝛾(f𝛼-1) = f𝛾(n, √(1 - n^2)) = f𝛾(x, y) = y = √(1 - n^2)
So, in general, we have, f𝛼𝛾 = f𝛾◦ f𝛼-1 = y = √(1 - x^2).
@@TensorCalculusRobertDavie Thankyou Robert
Thanks sir 😊
Glad it was helpful to you.
@@TensorCalculusRobertDavie ryt. It's so helpful for me. ..love form india
😍😍😍😍
Thank you!