So yes you are right we only really need one R^n, each chart map just maps to its image which is an open set, and we can take those to be of the same R^n. I said it was two copies probably sloppily but more to emphasise the two coordinate systems, either way every copy of R^n is isomorphic, which is what allows us to construct the transition function between the two coordinate systems! Good observation thank you!
But if we map the charts to the same Rn won't it signify that we're working with the same set of coordinates? If that is so, how can we really talk about a transition function?
@@shubhamgothwal2427 So we are mapping the chart (manifold subset) to it's image under the chart map, the image is a subset (or could be all) of R^n. The separate coordinate systems are then realised as different subsets of the same R^n, and the transitions functions are between the subsets. Hope that helps!
My appreciation for this video's existence is immeasurable. This stuff is completely incomprehensible from the perspective of someone self-studying from a textbook. Having these illustrations makes these functions as simple as breathing, while the textbook is just an alien language of random functions and subscripts strung together.
Thankyou! As someone who struggled self studying these topics myself it makes me very happy to hear they are helping others in the same situation, as this was my main motivation for making these videos!
Ive been trying to teach myself this stuff from a book. So having someone writing down what im seeing in the book and explaining the notation was awesome
Most introductory script I found on manifolds found their content so basic that they were lacking in visual material. I wished your approach was standard, it really pushed the point for me (still in manifold of course ;-)
@WHYBmaths how would we represent p (the point on the manifold)? I get how we can give it values and describe it on the two different charts using either (x1,x2) or (y1,y2) but what about while its "on" the manifold. Like how do you talk about the elements on the abstract manifold before you map them into a chart?
Makes it clear how those arrow and text diagrams can actually be trusted! But isn't it more conventional to write the 'composed' form of φᵥ(φᵤ⁻¹(p)) as φᵤ⁻¹⚬φᵥ, i.e. the two φₖ (in this case k being x & y) swap places?
Quick question: wouldn’t the transition function have its domain as the image of Ux, not the entire R^2 space? I’m referring to what you have written at the bottom of the board around 14:53, just above the dog.
Yes that is correct, except it would be the intersection of U_x and U_y. The point I was making was that since both U_x and U_y are (open) subsets of R^d the map behaves accordingly and the notions of smoothness and continuity are the standard notions for R^d. Hope that helps!
I might have missed in the video: Is it implicitly required that, for a (smooth) transition map to exist, each chart map must have an (smooth also?) inverse?
Yes I might have forgotten to mention it or maybe it was mentioned in a previous videos, but yes we require the chart functions to have inverses in order to construct such transition functions. For a general topological manifold the chart functions must only be continuously invertible, but if we then restrict the atlas of all possible chart functions down from continuous to k-times differentiable to eventually C^infty, i.e smooth chart functions with smooth inverses (and hence smooth transition functions) we arrive at the structure of a differentiable (or smooth) manifold. This is usually what people actually mean when they say 'a manifold' since the `differentiable' structure (i.e tangent spaces etc) that such smooth chart functions admit is so rich and widely studied. Hope that helps!
@WHYBmaths Why can't we just map both the open sets to the same Rn? Why did we need two separate Rn's?
So yes you are right we only really need one R^n, each chart map just maps to its image which is an open set, and we can take those to be of the same R^n. I said it was two copies probably sloppily but more to emphasise the two coordinate systems, either way every copy of R^n is isomorphic, which is what allows us to construct the transition function between the two coordinate systems! Good observation thank you!
But if we map the charts to the same Rn won't it signify that we're working with the same set of coordinates? If that is so, how can we really talk about a transition function?
@@shubhamgothwal2427 So we are mapping the chart (manifold subset) to it's image under the chart map, the image is a subset (or could be all) of R^n. The separate coordinate systems are then realised as different subsets of the same R^n, and the transitions functions are between the subsets. Hope that helps!
@@WHYBmaths Thank You so much. This silly doubt had been bothering me for days. Now I can visualise it.
@@shubhamgothwal2427 No problem! Glad I could help!
My appreciation for this video's existence is immeasurable. This stuff is completely incomprehensible from the perspective of someone self-studying from a textbook. Having these illustrations makes these functions as simple as breathing, while the textbook is just an alien language of random functions and subscripts strung together.
Thankyou! As someone who struggled self studying these topics myself it makes me very happy to hear they are helping others in the same situation, as this was my main motivation for making these videos!
Ive been trying to teach myself this stuff from a book. So having someone writing down what im seeing in the book and explaining the notation was awesome
Thank you! I'm glad you found it useful!
Same
He is either exceptionaly good at teaching or exceptionally good at maths.
Most introductory script I found on manifolds found their content so basic that they were lacking in visual material. I wished your approach was standard, it really pushed the point for me (still in manifold of course ;-)
Wonderful set of lectures … just superb
Great visualizations! Really help explain the content in a more intuitive way
Glad it was helpful!
Brilliant series mate. Helping me a lot.
OUTSTANDING!!! I'm loving this!
I just noticed that the bottom of your board has a drawing of a dog. How cute!
Yes she keeps me company when the real pup isn't around
@@WHYBmaths that's sweet!
Thanks very much! Very helpful.
Thanks for making these videos man. Their good and i found this info very useful.
Hey this is really good. I really appreciate this explanation. Keep up the videos.
I really love your vids man! Thanks so much for making these
@WHYBmaths how would we represent p (the point on the manifold)? I get how we can give it values and describe it on the two different charts using either (x1,x2) or (y1,y2) but what about while its "on" the manifold. Like how do you talk about the elements on the abstract manifold before you map them into a chart?
Great work!.however I do have a question: why do we need to go to Euclidean space R^n in order to do calculus on manifolds?
thanks a lot!!!
This series seems a bit similiar to the book Geometry and Topology by Mikio Nakahara, are you basing this series on it?
Makes it clear how those arrow and text diagrams can actually be trusted!
But isn't it more conventional to write the 'composed' form of φᵥ(φᵤ⁻¹(p)) as φᵤ⁻¹⚬φᵥ, i.e. the two φₖ (in this case k being x & y) swap places?
Quick question: wouldn’t the transition function have its domain as the image of Ux, not the entire R^2 space? I’m referring to what you have written at the bottom of the board around 14:53, just above the dog.
Yes that is correct, except it would be the intersection of U_x and U_y. The point I was making was that since both U_x and U_y are (open) subsets of R^d the map behaves accordingly and the notions of smoothness and continuity are the standard notions for R^d. Hope that helps!
I might have missed in the video: Is it implicitly required that, for a (smooth) transition map to exist, each chart map must have an (smooth also?) inverse?
Yes I might have forgotten to mention it or maybe it was mentioned in a previous videos, but yes we require the chart functions to have inverses in order to construct such transition functions. For a general topological manifold the chart functions must only be continuously invertible, but if we then restrict the atlas of all possible chart functions down from continuous to k-times differentiable to eventually C^infty, i.e smooth chart functions with smooth inverses (and hence smooth transition functions) we arrive at the structure of a differentiable (or smooth) manifold. This is usually what people actually mean when they say 'a manifold' since the `differentiable' structure (i.e tangent spaces etc) that such smooth chart functions admit is so rich and widely studied. Hope that helps!
Does the 8nverse transition function have to be continuous?
You said at the end. Transition functions are continuous. Thank you, fascinating lecture.
I feel like I joined a cult. I love your vids :)