# Motivation for tangent space

The isomorphism between tangents and derivations is surprising! As is the fact it is a linear map. However with calculus one is not far from that definition already.

## Motivating example

Let us take (informally, because cases where [ilmath]\theta=\tfrac{\pi}{2}[/ilmath] and [ilmath]r=0[/ilmath] must be treated carefully) the manifold of the plane. The reader should be familiar with polar coordinates (giving things as an angle and a distance from the origin, rather than [ilmath]x[/ilmath] and [ilmath]y[/ilmath])

We will have two ways of looking at points, as [ilmath](x,y)[/ilmath] - traditionally, and [ilmath](r,\theta)[/ilmath] where:

• [ilmath](r,\theta)\mapsto(r\cos(\theta),r\sin(\theta))[/ilmath]
• [ilmath](x,y)\mapsto\left(\sqrt{x^2+y^2},\arctan(\frac{y}{x})\right)[/ilmath]

## The line

Take the line [ilmath]y=mx+c[/ilmath], where [ilmath]m[/ilmath] is the gradient and [ilmath]c[/ilmath] is the intercept with the y axis, writing this we see the line can be given as:

Form First coord Second coord
[ilmath]x,y[/ilmath] [ilmath]x=t[/ilmath] [ilmath]y=mt+c[/ilmath]
[ilmath]r,\theta[/ilmath] [ilmath]r=\sqrt{t^2(m^2+1)+2mct+c^2}[/ilmath] [ilmath]\theta=\arctan\left(m+\frac{c}{t}\right)[/ilmath]
Pure forms
Form map
Cartesian [ilmath]y=mx+c[/ilmath]
Polar[1] $r=|c|\sqrt{\frac{(m^2+1)}{(\tan(\theta)-m)^2}+\frac{2m}{\tan(\theta)-m}+1}$

These formulas are easily found from substitution.

## Lines as seen in the polar coordinate plane

Some lines in polar coordinates, the y axis is the angle and the x radius
all these lines have gradient 1 (the gradient would just affect where r tends off towards infinity, which happens at pi/4 here) and from left to right, c=0.5, 1, 2 and 4

(See diagram on right) - notice that there are some "iffy" parts with lines (as seen above) - there's a discontinuity at the asymptote! - let us see how it is drawn to understand why.

### Drawing the line y=x+1

Using the equation above we see:

• $r=|c|\sqrt{\frac{(m^2+1)}{(\tan(\theta)-m)^2}+\frac{2m}{\tan(\theta)-m}+1}$ becomes
• $r=\sqrt{\frac{2}{(\tan(\theta)-1)^2}+\frac{2}{\tan(\theta)-1} + 1}$ - this isn't that useful though.

Consider drawing the graph, the logic goes as follows:

• [ilmath]\theta=0[/ilmath] - well where can [ilmath]r[/ilmath] actually be now? Undefined? If you graph the above, [ilmath]r=1[/ilmath]
• As [ilmath]\theta[/ilmath] increases, we keep drawing [ilmath]y=x-1[/ilmath] - it grows faster and faster as we get near [ilmath]\theta=\tfrac{\pi}{4}[/ilmath]
• - [ilmath]r[/ilmath] is undefined!
• [ilmath]\theta[/ilmath] increases further - we now start to come back down, and we draw [ilmath]y=x+1[/ilmath], this keeps being drawn until:
• [ilmath]\theta=\tfrac{3\pi}{4}[/ilmath] - we become undefined for [ilmath]r[/ilmath] again
• [ilmath]\theta[/ilmath] increases further - we draw more of [ilmath]y=x-1[/ilmath], until we meet

## Tangents

Let us go from tangents in [ilmath](x,y)[/ilmath] to tangents in [ilmath](r,\theta)[/ilmath]

Using the notation of the charts for the smooth manifold of (upper quadrant) of the plane we can use the transition map:

• $(\beta\circ\alpha^{-1}):\mathbb{R}^2_++\rightarrow\mathbb{R}_+\times(0,\tfrac{\pi}{2})$
• $(\beta\circ\alpha^{-1}):(x,y)\mapsto\left(\sqrt{x^2+y^2},\arctan\left(\frac{y}{x}\right)\right)$

to go from the plane in terms of [ilmath](x,y)[/ilmath] to the [ilmath](r,\theta)[/ilmath] "chart" of the plane.

Let us compute the Jacobian of this map (at the point [ilmath](x,y)[/ilmath]:

$\text{Jac}(\beta\circ\alpha^{-1})=\begin{pmatrix} \frac{\partial r}{\partial x} & \frac{\partial r}{\partial y} \\ \frac{\partial \theta}{\partial x} & \frac{\partial \theta}{\partial y} \end{pmatrix}=\begin{pmatrix} \frac{x}{\sqrt{x^2+y^2}} & \frac{y}{\sqrt{x^2+y^2}} \\ \frac{-y}{x^2+y^2} & \frac{x}{x^2+y^2} \end{pmatrix}$

Unit wise:

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

### Calculating tangents

Where [ilmath](u,v)[/ilmath] is a column vector representing a tangent (in the sense of direction) in the [ilmath](x,y)[/ilmath] sense of the manifold we can compute the following:

$\begin{pmatrix} \frac{x}{\sqrt{x^2+y^2}} & \frac{y}{\sqrt{x^2+y^2}} \\ \frac{-y}{x^2+y^2} & \frac{x}{x^2+y^2} \end{pmatrix}\times\begin{pmatrix} u \\ v \end{pmatrix}=\begin{pmatrix} \frac{x}{\sqrt{x^2+y^2}}u+\frac{y}{\sqrt{x^2+y^2}}v \\ \frac{-y}{x^2+y^2}u+\frac{x}{x^2+y^2}v \end{pmatrix}$

Notice that so far we have not mentioned any sort of "function", and this is linear! (at a point anyway)

#### Specific tangent

Take [ilmath]y=x+1[/ilmath], at [ilmath]x=1[/ilmath] we see [ilmath]\frac{dy}{dx}=1[/ilmath], that is to say the tangent at [ilmath]p=(1,2)[/ilmath] is [ilmath](1,1)_p[/ilmath]

$\text{Jac}(\beta\circ\alpha^{-1})\begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} \frac{3\sqrt{5}}{5} \\ \frac{-1}{5} \end{pmatrix}$ (using graphical calculator)

From this we see:

$\frac{dr}{d\theta}=\frac{\frac{3\sqrt{5}}{5}}{\frac{-1}{5}}=-3\sqrt{5}\approx-6.708\text{ (4sf)}$

Using my graphical calculator I found:

$\left.\frac{d}{dt}\left[ |c|\sqrt{\frac{(m^2+1)}{(\tan(t)-m)^2}+\frac{2m}{\tan(t)-m}+1} \right]\right|_{t=\arctan(\frac{y}{x})}\approx -6.708\text{ (4sf)}$
where
• [ilmath]m=1[/ilmath]
• [ilmath]c=1[/ilmath]
• [ilmath](x,y)=p[/ilmath]

TODO: get back to tangent space