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.

Why have geometric tangent space?

Take a sphere, $\mathbb{S}^2$ - anyone doing A-levels can define the tangent plane at a point! It's the plane through $p$ with normal the same as the normal to the surface of the sphere at that point (which is the direction from the origin to $p$)

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)

$$\begin{pmatrix} \frac{\delta r}{\delta x} & \frac{\delta r}{\delta y} \\ \frac{\delta \theta}{\delta x} & \frac{\delta \theta}{\delta y} \end{pmatrix}\times\begin{pmatrix} \delta x \\ \delta y \end{pmatrix}= \begin{pmatrix} \frac{\delta r}{\delta x}\delta x + \frac{\delta r}{\delta y}\delta y \\ \frac{\delta \theta}{\delta x}\delta x + \frac{\delta \theta}{\delta y}\delta y \end{pmatrix}=\begin{pmatrix} \delta r \\ \delta \theta \end{pmatrix}$$

Given a direction - our $\delta x$ and $\delta y$ - 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:

• Smooth manifold
• Smooth function
• Smoothly compatible charts

Okay, well we can do directions! Let us make our first definition

Geometric tangent space

Given a chart $(U,\varphi)$ we know by definition that

$$\varphi(U)\subseteq\mathbb{R}^n$$ - and is indeed an open subset. Given a point in $\varphi(U)$ 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 $\varphi(U)$ we can scale it up! So lets have a go at defining the "geometric tangent space" at $p\in\varphi(U)$ as:

$$G_p(\mathbb{R}^n)=\{(p,v)|v\in\mathbb{R}^n\}$$

To get tangents we need to be able to differentiate in directions, given a map $f:\mathbb{R}^2\rightarrow\text{Whatever}$ in the form $f(x,y)$ we can differentiate it in two directions, but $(x,y)$ is a vector! So differentiating with respect to $x$ 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:

$$D_v\cdot|_p:C^\infty(R^n)\rightarrow\mathbb{R}$$ where $\cdot$ is where we'd put a $C^\infty$ function (eg $$D_vf|_p$$ )

and defining this as follows:

$$D_v\cdot|_p=\frac{d}{dt}[\cdot(p+tv)]\Bigg|_{t=0}$$ - 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 $\mathbb{R}$ as:

$$D_v(af+bg)\Big|_p$$ $$=\frac{d}{dt}[(af+bg)(p+tv)]\Bigg|_{t=0}=$$ $$a\frac{d}{dt}[f(p+tv)]\Bigg|_{t=0}+b\frac{d}{dt}[g(p+tv)]\Bigg|_{t=0}$$ $$=aD_vf|_p+bD_vg|_p$$

Going between tangent vectors and derivations

So $T_p(\mathbb{R}^n)$ is linear, $G_p(\mathbb{R}^n)$ is linear - there's something at play here.

Taking

$$\alpha:G_p(\mathbb{R}^n)\rightarrow T_p(\mathbb{R}^n)$$ with $$\alpha:v_p\mapsto D_v\cdot|_p$$ we actually have an Isomorphism which is both surprising and not surprising.

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TODO: Link to isomorphism proof

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Note that $T_p(\mathbb{R}^n)$ has VERY LITTLE actually to do with $\mathbb{R}^n$!

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TODO: Finish

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