View source for Notes:Tangent space

==Motivating example==
Consider the "manifold":
* {{M|1=M:=\{p\in\mathbb{R}^3\ \vert\ p=(x,y,x^2+y^2)\}\subset\mathbb{R}^3}} (with the subspace topology) and the two charts:
*# {{M|1=i:\mathbb{R}^2\rightarrow M}} by {{M|1=i:(x,y)\mapsto(x,y,x^2+y^2)}} (think of it as like the "identity chart" it is what the manifold is)
*# {{M|1=c:(\underbrace{\mathbb{R}_{>0}\times([0,2\pi)\subset\mathbb{R})}_{\text{polar coordinates} })\rightarrow M}} by {{M|1=c:(r,\theta)\mapsto(r\cos(\theta),r\sin(\theta),r^2)}} ({{M|c}} for chart)

Pick a {{M|p\in M}}, suppose we pick {{M|(x,y,x^2+y^2)}} then we see:
# {{M|1=i^{-1}(x,y,x^2+y^2)=(x,y)}} and we'll define {{M|1=p_i:=i^{-1}(p)=(x,y)}}
# {{M|1=c^{-1}(x,y,x^2+y^2)=(\sqrt{x^2+y^2},\arctan(\frac{y}{x}))}} and we'll define {{M|1=p_c:=c^{-1}(p)=(\sqrt{x^2+y^2},\arctan(\frac{y}{x}))}}

Notice that we're sort of using the "identity" chart when we're dealing with {{M|c}} anyway; this is the point.
==Questions==
# The image of the tangent space under the derivative is really important, as that is what makes this manifold more than just a topological one, however I don't see how we can shrug off the manifold being in some {{M|\mathbb{R}^N}} and have MEANINGFUL tangent spaces.
#* {{M|1=\mathrm{d}i\vert_{(x,y)}:T_{(x,y)}(\mathbb{R}^2)\rightarrow T_{i(x,y)}(\mathbb{R}^3)}} and {{M|1=\mathrm{d}c\vert_{(r,\theta)}:T_{(r,\theta)}(\mathbb{R}^2)\rightarrow T_{c(r,\theta)}(\mathbb{R}^3)}}
#* I claim that the image of these too is "the same space with a different basis (if you consider the image of the standard basis)" - not just isomorphic, obviously any two planes are isomorphic, that's too weak.
# Suppose we want to integrate to find the surface area of a region, how do we do this without it just being a standard calculus "change of coordinates" question? 
#* Suppose we pick a region in {{M|\mathbb{R}^2}}, in this example there's a natural choice of chart, but other charts could go much faster, eg {{M|:(x,y)\mapsto(x^3,y^3,x^6+y^6)}}, at {{M|(2,2)}} this is moving much faster than {{M|i}} would! 
#* In a manifold there is no natural chart like our {{M|i}}, it is clear that "1 unit squared" in {{M|i}} is different to 1 unit squared somewhere else!
# There's a lot of talk about immersing manifolds and such, if we move and stretch bits of it around so it doesn't self intersect for example, surely we change the results of certain integrals?
# Take the unit sphere, and take the sphere of double the radius, using charts how can we distinguish these? 
# What is the point? I do see that when I've done calculus exercises and integrated something in chunks that that's sort of like a chart, and that you can change coordinates, but there's a natural choice of coordinates then! "the unit area/volume/length" is defined relative to something
# Where does length come from? Take a circle with 4 charts (that each cover "almost" half of the circle), suppose chart {{M|A}} covers the half above the x-axis.
#* "some length" in the chart that maps to {{M|1=x=0}} and {{M|1=y=1}} is worth less than "some length" that maps to the edge of the chart. (See picture)
[[File:Pitfalls of circle chart.JPG|thumb|150px]]
=="Tangent planes"==
{{M|T_p(V)}} is the tangent space to a point, it's a vector space and consists of [[tuple|tuples]] of the form {{M|(p,v)}} for {{M|v\in V}}. These are quite literally arrows that start at {{M|p}} and point from {{m|p}} in direction {{M|v}}. 

# {{M|1=\mathrm{d}i\vert_{(x,y)}(u,v)=\left(\begin{array}{c} u\\ v \\ 2xu + 2yv\end{array}\right)}}
#* Now we see the "tangent space" is the plane: {{M|1=\left(\begin{array}{c}x\\y\\x^2+y^2\end{array}\right)+\alpha\left(\begin{array}{c}1\\ 0 \\ 2x\end{array}\right)+\beta\left(\begin{array}{c}0\\ 1 \\ 2y\end{array}\right)}} for {{M|\alpha,\beta\in\mathbb{R} }}
# {{M|1=\mathrm{d}c\vert_{(r,\theta)}(u,v)=\left(\begin{array}{c}u\cos(\theta)-r v \sin(\theta)\\ u\sin(\theta)+rv\cos(\theta)\\2ru\end{array}\right)}} ({{M|u}} and {{M|v}} are different now, but can take any value in {{M|\mathbb{R}^2}})
#* The tangent space is the plane: {{M|1=\left(\begin{array}{c}r\cos(\theta)\\r\sin(\theta)\\r^2\end{array}\right)+\alpha\left(\begin{array}{c}\cos(\theta)\\ \sin(\theta)\\ 2r\end{array}\right)+\beta\left(\begin{array}{c}-r\sin(\theta)\\ r\cos(\theta) \\ 0\end{array}\right)}} - {{Caution|I've checked and these are the same plane, provided I haven't copied from my notes wrongly, what I've written here is the same plane!}}
==Moving between charts==
# {{MM|1=\mathrm{d}(c^{-1}\circ i)\vert_{(x,y)}(u,v)=\left(\begin{array}{c}\frac{xu+yv}{\sqrt{x^2+y^2} }\\\frac{-yu + xv}{x^2+y^2}\end{array}\right)}}
# {{MM|1=\mathrm{d}(i^{-1}\circ c)\vert_{(r,\theta)}(u,v)=\left(\begin{array}{c}u\cos(\theta)-rv\sin(\theta)\\u\sin(\theta)+rv\cos(\theta)\end{array}\right)}}
Again I've checked these with a few values and they seem to be correct. For example:
# Take {{M|i^{-1}(p)=(5,0)}} (so {{M|1=x=5}} and {{M|1=y=0}}) going in the direction {{M|(1,0)}} (1 {{M|x}} unit per unit ....time?), then we get 
#* {{M|(1,0)}} - {{M|r}} increasing by 1 per unit time and theta not changing
# Take {{M|i^{-1}(p)=(5,0)}} again but now the direction {{M|(0,1)}} (vertical), we get {{M|(0,\frac{1}{5})}} (note that only at this instant is change in {{M|r}} per unit time {{M|0}}, after that {{M|r}} has to get bigger to keep going vertical)
# I also checked {{M|1=(r=1,\theta=\frac{\pi}{4})}} in direction {{M|(1,0)}} and got {{M|(\frac{\sqrt{2} }{2},\frac{\sqrt{2} }{2})}} as expected. This is elementary calculus though so not a big deal