Motivation for tangent space definitions
Note: different to Motivation for tangent space - that page talks about tangents, and going between manifolds. THIS page will talk about the reason for definitions. Like a study guide.
Contents
Why have geometric tangent space?
Take a sphere, [ilmath]\mathbb{S}^2[/ilmath] - anyone doing A-levels can define the tangent plane at a point! It's the plane through [ilmath]p[/ilmath] with normal the same as the normal to the surface of the sphere at that point (which is the direction from the origin to [ilmath]p[/ilmath])
This is crap for manifolds, there isn't really an origin - we have no (given) ambient Euclidean space to put the plane into - tangent line or plane or whatever just makes no sense.
First steps
Seeing that a tangent depends on the map
Tangents section of motivation for tangent space shows that a map between two smoothly compatible charts gives rise to a map between directions. That is given a tangent vector in one space, where it ends up is linear - regardless of the function to which it is a tangent of.
This becomes clear with the dimensional analysis of the transition (stated again here)
[math]\begin{pmatrix} \frac{\delta r}{\delta x} & \frac{\delta r}{\delta y} \\ \frac{\delta \theta}{\delta x} & \frac{\delta \theta}{\delta y} \end{pmatrix}\times\begin{pmatrix} \delta x \\ \delta y \end{pmatrix}= \begin{pmatrix} \frac{\delta r}{\delta x}\delta x + \frac{\delta r}{\delta y}\delta y \\ \frac{\delta \theta}{\delta x}\delta x + \frac{\delta \theta}{\delta y}\delta y \end{pmatrix}=\begin{pmatrix} \delta r \\ \delta \theta \end{pmatrix}[/math]
Given a direction - our [ilmath]\delta x[/ilmath] and [ilmath]\delta y[/ilmath] - the mapping that maps these to corresponding changes in a totally different chart is linear and based ONLY on the point the tangents are at.
Defining tangent space
So far on smooth manifolds all we have are the following definitions:
Okay, well we can do directions! Let us make our first definition
Geometric tangent space
Given a chart [ilmath](U,\varphi)[/ilmath] we know by definition that [math]\varphi(U)\subseteq\mathbb{R}^n[/math] - and is indeed an open subset. Given a point in [ilmath]\varphi(U)[/ilmath] there's an open ball around it, so we can go in all directions.
Furthermore, we can go in any direction we like. Because it's a vector space if we can define the arrow inside [ilmath]\varphi(U)[/ilmath] we can scale it up! So lets have a go at defining the "geometric tangent space" at [ilmath]p\in\varphi(U)[/ilmath] as:
[math]G_p(\mathbb{R}^n)=\{(p,v)|v\in\mathbb{R}^n\}[/math]
To get tangents we need to be able to differentiate in directions, given a map [ilmath]f:\mathbb{R}^2\rightarrow\text{Whatever} [/ilmath] in the form [ilmath]f(x,y)[/ilmath] we can differentiate it in two directions, but [ilmath](x,y)[/ilmath] is a vector! So differentiating with respect to [ilmath]x[/ilmath] say is just a direction - with this in mind:
Derivations
Having decided we want to be able to differentiate in directions, we need a notion of directional derivative. The easiest being:
[math]D_v\cdot|_p:C^\infty(R^n)\rightarrow\mathbb{R}[/math] where [ilmath]\cdot[/ilmath] is where we'd put a [ilmath]C^\infty[/ilmath] function (eg [math]D_vf|_p[/math])
and defining this as follows:
[math]D_v\cdot|_p=\frac{d}{dt}[\cdot(p+tv)]\Bigg|_{t=0}[/math] - it should already be obvious where this is going.
Automatically (by the product rule of calculus) this satisfies the Leibniz rule (which is good to know - the more we can say the better)
It's also a linear map over [ilmath]\mathbb{R} [/ilmath] as:
[math]D_v(af+bg)\Big|_p[/math][math]=\frac{d}{dt}[(af+bg)(p+tv)]\Bigg|_{t=0}=[/math][math]a\frac{d}{dt}[f(p+tv)]\Bigg|_{t=0}+b\frac{d}{dt}[g(p+tv)]\Bigg|_{t=0}[/math][math]=aD_vf|_p+bD_vg|_p[/math]
Going between tangent vectors and derivations
So [ilmath]T_p(\mathbb{R}^n)[/ilmath] is linear, [ilmath]G_p(\mathbb{R}^n)[/ilmath] is linear - there's something at play here.
Taking [math]\alpha:G_p(\mathbb{R}^n)\rightarrow T_p(\mathbb{R}^n)[/math] with [math]\alpha:v_p\mapsto D_v\cdot|_p[/math] we actually have an Isomorphism which is both surprising and not surprising.
TODO: Link to isomorphism proof
Note that [ilmath]T_p(\mathbb{R}^n)[/ilmath] has VERY LITTLE actually to do with [ilmath]\mathbb{R}^n[/ilmath]!
TODO: Finish
