tags: math quaternion freya-holmer

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These are my notes from Freya Holmér's excellent
[presentation from Dutch Game Day 2023](https://www.youtube.com/watch?v=htYh-Tq7ZBI).

# Why Can't We Multiply Vectors
Let's start with the requirement that
\[
\mathbf{v}^2=\left\|\mathbf{v}^2\right\|
\]
For example
\[
(1,2,3)^2 = 1^2+2^2+3^2 = 14
\]

## What about our basis vectors

![]{cs small}(FH_Basis_Vectors_1.jpg) 
\[
\begin{align*}
\mathbf{x}\mathbf{x} &= \mathbf{x}^2 = 1^2 = 1\\
\mathbf{y}\mathbf{y} &= \mathbf{y}^2 = 1^2 = 1\\
\mathbf{z}\mathbf{z} &= \mathbf{z}^2 = 1^2 = 1
\end{align*}
\]

## Expanding the product
![]{cs}(PH_Expand_Vector_Product.jpg) 

Writing \(\mathbf{a}=a_x\mathbf{x} + a_y\mathbf{y} + a_z\mathbf{z}\), and naively expanding:
\[
\begin{align*}
\mathbf{ab} &= a_xb_x\mathbf{x}\mathbf{x} + a_xb_y\mathbf{x}\mathbf{y} + a_xb_z\mathbf{x}\mathbf{z} \\
            &+ a_yb_x\mathbf{y}\mathbf{x} + a_yb_y\mathbf{y}\mathbf{y} + a_yb_z\mathbf{y}\mathbf{z} \\
            &+ a_zb_x\mathbf{z}\mathbf{x} + a_zb_y\mathbf{z}\mathbf{y} + a_zb_z\mathbf{z}\mathbf{z}
\end{align*}
\]
and then \(\mathbf{x}\mathbf{x} = 1\), so this simplifies to
\[
\begin{align*}
\mathbf{ab} &= a_xb_x + a_xb_y\mathbf{x}\mathbf{y} + a_xb_z\mathbf{x}\mathbf{z} \\
            &+ a_yb_x\mathbf{y}\mathbf{x} + a_yb_y + a_yb_z\mathbf{y}\mathbf{z} \\
            &+ a_zb_x\mathbf{z}\mathbf{x} + a_zb_y\mathbf{z}\mathbf{y} + a_zb_z
\end{align*}
\]
So collecting the scalars
\[
\begin{align*}
\mathbf{ab} &= a_xb_x + a_yb_y  + a_zb_z
            &+ a_xb_y\mathbf{x}\mathbf{y} + a_xb_z\mathbf{x}\mathbf{z} \\
            &+ a_yb_x\mathbf{y}\mathbf{x} + a_yb_z\mathbf{y}\mathbf{z} \\
            &+ a_zb_x\mathbf{z}\mathbf{x} + a_zb_y\mathbf{z}\mathbf{y}
\end{align*}
\]
So what are \(\mathbf{x}\mathbf{z}\) etc.?
Consider the diagonal between \(\mathbf{x}\) and \(\mathbf{y}\):

![]{cs small}(FH_Diag_XY_1.jpg) 

It's length is \(\left\|\mathbf{x}+\mathbf{y}\right\|=\sqrt{2}\).
But then we can square it, using \(\mathbf{v}^2=\left\|\mathbf{v}\right\|^2\),
\[
\begin{align*}
(\mathbf{x}+\mathbf{y})^2 &= \mathbf{x}\mathbf{x} + \mathbf{x}\mathbf{y} + \mathbf{y}\mathbf{x} + \mathbf{y}\mathbf{y} \\
                          &= 1 + \mathbf{x}\mathbf{y} + \mathbf{y}\mathbf{x} + 1 = 2
\end{align*}
\]
and so
\[
                          \mathbf{x}\mathbf{y} + \mathbf{y}\mathbf{x} = 0
\]
or
\[
                          \mathbf{x}\mathbf{y} = -\mathbf{y}\mathbf{x}
\]
So we get

![]{cs}(FH_Vector_Product_1.jpg) 

And we recognise the dot and cross products

![]{cs}(FH_Vector_Product_2_DotCross.jpg) 

And then

\[(\mathbf{x}\mathbf{y})^2=\mathbf{x}\mathbf{y}\mathbf{x}\mathbf{y}=-\mathbf{x}\mathbf{x}\mathbf{y}\mathbf{y}=-1\cdot 1=-1\]

And thus

\[(\mathbf{x}\mathbf{y})^2=(\mathbf{y}\mathbf{z})^2=(\mathbf{z}\mathbf{x})^2=(\mathbf{x}\mathbf{y})(\mathbf{y}\mathbf{z})(\mathbf{z}\mathbf{x})=-1\]

And this is familiar, the **quaternions**:

\[i^2=j^2=k^2=ijk=-1\]

So our vector products live in a space with basis

\[\left\{1,\mathbf{x}\mathbf{y},\mathbf{y}\mathbf{z},\mathbf{z}\mathbf{x}\right\}\]

## What about 2D?

We get

\[\mathbf{ab} = a_xb_x + a_yb_y + \mathbf{xy}(a_xb_y - a_yb_x)\]

which has a basis of

\[\left\{1,\mathbf{xy}\right\}\]

with

\[\mathbf{xy}^2=-1\]

which looks like the complex numbers.

# Planes and Clifford Algebras

![]{cs small}(FH_Cross_Product_1.jpg) 

Now the cross product has magnitude equal to the area of the parallelogram constructed
from the two vectors, and is in a direction perpendicular to them. So maybe this has something
to do with planes.

Now we call \(\mathbf{xy}\) a bivector (whereas \(\mathbf{x}\) is a vector). And we consider
basis bivectors:

![]{cs small}(FH_XY_Plane1.jpg) 

![]{cs small}(FH_XYZ_Planes1.jpg) 

But maybe rather than as a vector, we can treat the product as an **oriented area**:

![]{cs}(FH_Oriented_Area_1.jpg) 

It is specified by three components, e.g. \((0.1,0.2,0.3)\), like a vector, but
algebraically behaves differently since it is has a basis of \(\{\mathbf{xy},\mathbf{yz},\mathbf{zx}\}\).

**Bivectors** represent the minimum information required in any given dimension, to store
both a **plane** and a **magnitude**. These show up in **rotations** because rotations happen
in a **plane**. Rotations happen in a plane, rather than around an axis. (In 2D, for example, there
is no axis, merely a centre point.)

## Looking Back At The Cross Product
The **cross product** is a pseudovector, only works in 3D and 7D, and behaves differently to vectors under reflection
of the axes. Whereas there is the **wedge product** has the same components but different basis vectors:

![]{cs}(FH_CrossProduct_WedgeProduct_1.jpg)

So we get

![]{cs}(FH_Vector_Product_3.jpg) 

that is:

\[\mathbf{ab}=\mathbf{a}\cdot\mathbf{b}+\mathbf{a}\wedge\mathbf{b}\]

**The product of two vectors is the sum of the dot product and the wedge product.**

# Example: Curvature

![]{cs}(FH_Curvature_1.jpg) 

Actually we have

\[\kappa = \frac{f^{\prime}\wedge f^{\prime\prime}}{\left\|f^\prime\right\|^3}\]

which generalises to any dimension. In the case of the 2D curvature, the sign comes
out of the wedge product.

# Summing up

![]{cs}(FH_Overall_Picture_1.jpg)

So rather than the messy situation we had before, now everything follows from the axiom that 
\[\mathbf{v}^2=\left\|\mathbf{v}\right\|^2\]

# Geometric Algebras

![]{cs}(FH_2D_and_3D_VGA.jpg) 

![]{cs}(FH_Clifford_Algebra.jpg) 

# What does Multiplication Represent

The product is:
* a rotation by twice the angle between the two vectors (where angle is \(0\leq\theta<\pi\)),
* in the plane containing those two vectors
* with the direction of rotation going from the first (left) vector to the second (right)