At 2:29 the map h you defined should a chart of R^n right ? I'm a bit confused because I thought that R^n with the standard smooth structure only had the identity as a chart, but clearly the identity doesn't map the circle of R^2 to a line in R^1 for instance
Ok I think I've figured out, the atlas that comes with R^n is not just equipped with the identity but it's the whole maximal extension. So it contains all the compatible charts which are smoothly compatible with the identity so that's pretty much "all" the smooth homeomorphisms of R^n, which h is definitely part of (although I guess it's quite hard to describe such a chart for an arbitrary submanifold).
So just to be clear, \psi^{-1}(p=f(x,y)) in the last example is really just (x,y) - where in the jacobian calculation in the example it was implicitly plugged in?
@@brightsideofmaths oh sorry i meant to ask if \psi^{-1}(p)=(x,y) (not \phi) you found the span at the end but just wondering how you got to the span from the jacobian? And if you used the equality above - given if it is correct ofc haha
These concepts are super intimidating when reading about them in books, but you make them so much more concrete and accessible !
Thanks a lot :)
I always was confused about the term 'submanifold of R^n' ... until I learned about abstract manifolds.
Nice! I am glad you learnt something here :)
you're so good at this
Thanks a lot!
At 2:29 the map h you defined should a chart of R^n right ? I'm a bit confused because I thought that R^n with the standard smooth structure only had the identity as a chart, but clearly the identity doesn't map the circle of R^2 to a line in R^1 for instance
Ok I think I've figured out, the atlas that comes with R^n is not just equipped with the identity but it's the whole maximal extension. So it contains all the compatible charts which are smoothly compatible with the identity so that's pretty much "all" the smooth homeomorphisms of R^n, which h is definitely part of (although I guess it's quite hard to describe such a chart for an arbitrary submanifold).
So just to be clear, \psi^{-1}(p=f(x,y)) in the last example is really just (x,y) - where in the jacobian calculation in the example it was implicitly plugged in?
I don't understand your question. What do you mean by psi?
@@brightsideofmaths oh sorry i meant to ask if \psi^{-1}(p)=(x,y) (not \phi) you found the span at the end but just wondering how you got to the span from the jacobian? And if you used the equality above - given if it is correct ofc haha
Yes, the span is just given by the columns of the Jacobian because they describe tangent vectors :)