Lie Theory

Geometry of complex numbers and the Quaternions

The main aim of this section is to study groups of linear transformations (the orthogonal transformations) by representing them as matrix groups. The group of all rotations on $\mathbb{R}^2$ is denoted $SO(2)$, for the "special orthogonal" linear transformations from $\mathbb{R}^2$ to $\mathbb{R}^2$. When represented as matrices, they have the property that their determinant is 1, and that $AA^t=A^tA=I$, which also means that they admit a basis of eigenvectors.

We consider the rotation matrix on $\mathbb{R}^2$.

Defining the first matrix to be $\mathbf{1}$, and the second to be $\mathbf{i}$, we get the regular identities fro complex numbers. We also see that for any complex number $z=a\mathbf{1}+b\mathbf{i}$, we can write

and

We can try a similar process over $\mathbb{R}^4$, One thought is to associate them with a $2× 2$ matrix over $\mathbb{C}$. We want all of the matrices to rotate the vector $(1,0)^t$ in orthogonal ways (under the standard inner product on $\mathbb{R}^n$)

Now, consider the purely imaginary quaternions $\rho = b\mathbf{i}+c\mathbf{j}+d\mathbf{k}$, which are the orthogonal compliments to the real quaternions $a\mathbf{1}$. For a purely imaginary product, we see that

We notice that this product is algebraically closed when $\mathbf{u}\cdot \mathbf{v}=0$, meaning $\mathbf{u}$ and $\mathbf{v}$ are orthogonal. There is some interesting matrix algebra related to the representation of complex numbers, and the quaternions. Notice that if we take the conjugate of $a+bi$ we get $a-bi$, which has the matrix representation

And for the quaternion conjugate $\overline{z}=a\mathbf{1}-b\mathbf{i}-c\mathbf{j}-d\mathbf{k}$, we have

This means that for quaternions (as they are not commutative) $\overline{z_1z_2}=\bar{z_2}\bar{z_1}$. Consider unit quaternions in the form $z=\cos(θ)+u\sin(θ)$, where $u$ is a pure complex quaternion. A rotation through the angle $θ$ can be defined as $t^{-1}zt$, where $t=\cos(θ/2)+u\sin(θ)$. This is a rotation by theta through the axis $u$ in $\mathbb{R}^3$. The same rotation is induced by $-t$, as $(-t)^{-1}z(-t)=t^{-1}zt$. Thus rotations in $\mathbb{R}^3$ are induced by pairs of antipodal unit quaternions.

It is also useful to be able to talk about reflections in terms of quaternions as well. A Reflection in the hyperplane $O$ which is orthogonal to the unit quaternion $u$ is a map which sends each $q∈ \mathbb{R}^4$ to $-u\bar{q}u$. A linear map is called orientation preserving if its matrix representation has a positive determinant, and orientation reversing otherwise. Based on this map, reflections are orientation reversing. Rotations about $O$ are orientation preserving isometries which fix $O$. By the Cartan-Dieudonné theorem, every rotation in an $n$ dimensional symmetric bilinear space can be represented by the product of at most n reflections. Since the determinant of a reflection is negative, this shows that in $\mathbb{R}^4$ every rotation is the product of 0, 2, or 4 reflections. If the form of the reflection is considered, it is easy to see that a rotation is a map $q\mapsto vqw$ for unit quaternions $v,w$.

Groups

The unit quaternions are usually denoted by the set $\mathbb{S}^{3}$, as they define a sphere in $\mathbb{R}^4$. Each unit quaternion was shown to be equivalent to the set of all complex matrices in the form

This group is usually denoted $\textrm{SU}(2)$, for the $2× 2$ special unitary matrices. The related group $\textrm{SO}(3)$, or the group of all rotations in $\mathbb{R}^3$, is somewhat of a refinement on $\textrm{SU}(2)$. Unit quaternions correspond to rotations of $\mathbb{R}^3$, with antipodal pares of quaternions giving rise to the same rotation in $\mathbb{R}^3$. This suggests that it is possible to construct a 2-1 homomorphism from $\textrm{SU}(2)$ to $\textrm{SO}(3)$.

A useful property of groups is that many of them may be represented as a direct product of groups $A$ and $B$ under the product of pairs operation $(a_1,b_1)(a_2,b_2)=(a_1a_2,b_1,b_2)$. An example of this kind of group is $\mathbb{R}^n$ with vector addition. Another interesting example is the group of $n× n$ matrices. If $A,B∈ \mathbb{M}_{n× n}(\mathbb{F})$, then We can write all elements in $A× B$ in the form

Since the elements of $\mathbb{R}$ are isomorphic to the group of matrices

It follows that $\mathbb{R}^n$ is isomorphic to a group of $2n× 2n$ matrices. Replacing $\mathbb{R}$ by $\mathbb{S}^2$ gives a similar isomorphism with $θ$ in place for $x$. The group $\mathbb{S}^1× \mathbb{S}^1$ is called the unit torus, as for $θ, ϕ∈ \mathbb{S}^1$, we can specify a point $(θ, ϕ)$ which corresponds to a point on the surface of the torus moving $θ$ radians across along the outside edge, and $ϕ$ along the inside edge. This group is denoted $\mathbb{T}^2$ (figure 1).

$\mathbb{S}^2×\mathbb{S}^1=\mathbb{T}^2$

Since $\mathbb{R}$ and $\mathbb{S}^1$ are abelian groups, their products are abelian as well. It can be shown with some work that $\mathbb{R}^m× \mathbb{T}^n$ for $m,n∈ \mathbb{N}$ include all connected abelian matrix Lie groups. (Note that $\mathbb{S}^n$ is not associated with a Cartesian product the way that $\mathbb{R}^n$ is, and $\mathbb{T}^n\ncong\mathbb{S}^n$).

Now, we can try classifying rotations in $\mathbb{R}^4$ (the group SO(4)) by associating them with the group SU(2). Recall that all rotations can be represented by the form $q\mapsto vqw$, but since if $v$ is a quaternion, then $v^{-1}$ exists, so there is no harm in defining the map to instead be $q\mapsto v^{-1}qw$. With this in mind, consider the group $\textrm{SU}(2)× \textrm{SU}(2)$. There is a homomorphism from $\textrm{SU}(2)× \textrm{SU}(2)$ to $\textrm{SO}(4)$ which takes $(v,w)∈ \textrm{SU}(2)× \textrm{SU}(2)$ to $f(q)=v^{-1}qw$. We again encounter the antipodal pares which produce the same rotation, so the homomorphism has a kernel of two elements. It is possible to find a nontrivial subgroup of $\textrm{SO}(4)$ which is normal by considering the image of the subgroup of points $(v,1)$ in $\textrm{SU}(2)× \textrm{SU}(2)$. This means that $\textrm{SO}(4)$ is not simple.

Why are we talking about this?

In 1870, Sophus Lie was studying the symmetries of differential equations, which generally form "continuous" groups. The analogous problem for polynomials was solved by Galois previously, so there was incentive to solve the related problem for these new kinds of groups. This lead to a study of simplicity in these continuous groups. Lie understood these groups as groups generated by infinitesimal elements, which lead him to believe that $\textrm{SO}(n)$ should be generalised to consider infinitesimal rotations. Today we separate the infinitesimal rotations out into a group $\mathfrak{so}(n)$ called the Lie Algebra of $\textrm{SO}(n)$. Similarly, the infinitesimal elements of a group G form a Lie algebra $\mathfrak{g}$, which captures most of the important structures of G, but is easier to handle. Using this notion of a generalise group, Lie was able to prove simplicity for all but a finite number of Lie algebras — namely those of the exceptional groups.