# Matrix Cayley Transform

Sometime ago, in my research, I came across a matrix map I am interested at. It is a generalisation of the complex map

$z \mapsto \dfrac{z-i}{z+i}$

defined on a complex projective plane $\mathbb{C}\mathrm{P}^2$.

Let $\mathbf{F}$ be a field. Given a skew-symmetric matrix $X \in M_n\left(\mathbf{F}\right)$, i.e. $X^T = -X$, one can define the Cayley transform of $X$, denoted by $C(X)$, as

$C(X) = (I-X)(I+X)^{-1},$

for all $X$ such that $I+X$ is invertible.

If we were somewhat more restrictive in our assumption about $\mathbf{F}$, which in this case is to consider an ordered field, then $I+X$ is always invertible for all skew-symmetric matrix $X \in M_n(\mathbf{F})$. To see this, suppose it is not, then there exists a non-zero $\mathbf{u} \in \mathbf{F}^n$ such that $(I+X)\mathbf{u} = \mathbf{0}$, i.e. $X\mathbf{u} = -\mathbf{u}$. However,

$\mathbf{u}^T\mathbf{u} = \left(-X\mathbf{u}\right)^T\mathbf{u} = -\mathbf{u}^TX^T\mathbf{u} = \mathbf{u}^TX\mathbf{u} = -\mathbf{u}^T\mathbf{u},$

which implies $\mathbf{u}^T\mathbf{u} = 0$. Since we are working in an ordered field, this forces $\mathbf{u} = \mathbf{0}$, a contradiction.

It can be shown that $C(X)$ is an orthogonal matrix, i.e. $C(X)^TC(X) = I$ where $I$ is the identity matrix of the same size. It’s not hard to see this by direct computation:

\begin{aligned} C(X)^TC(X) &= \left(\left(I+X\right)^T\right)^{-1}\left(I-X\right)^T(I-X)(I+X)^{-1} \\ &= \left(I-X\right)^{-1}(I+X)(I-X)\left(I+X\right)^{-1} \\ &= \left(I-X\right)^{-1}(I-X)(I+X)\left(I+X\right)^{-1} \\ &= I. \end{aligned}

In the proof above, we used the fact that $I-X$ commutes with $I+X$. Moreover, $\det C(X) = 1$ because

$\det(I-X) = \det\left(I+X\right)^T = \det\left(I+X\right).$

Proposition. The map $C$ is an involution, i.e. $C^2 = \mathrm{id}.$

Example. If $n = 2$, then

$X = \begin{pmatrix} 0 & a \\ -a & 0 \end{pmatrix}$

and

$C(X) = \dfrac{1}{1+a^2}\begin{pmatrix} 1-a^2 & 2a \\ -2a & 1-a^2 \end{pmatrix},$

for some $a \in \mathbf{F}$ satisfying $a^2 + 1 \neq 0$. If $n = 3$, then

$X = \begin{pmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{pmatrix}$

so

$C(X) = \dfrac{1}{1+a^2+b^2+c^2}\begin{pmatrix} 1-a^2+b^2+c^2 & 2(-a+bc) & 2(c+ab) \\ 2(a+bc) & 1+a^2-b^2+c^2 & 2(-b+ac) \\ 2(-c+ab) & 2(b+ac) & 1+a^2+b^2-c^2\end{pmatrix},$

for all $a,b,c\in\mathbf{F}$ such that $a^2+b^2+c^2+1\neq 0$.

In Lie group and Lie algebra notation,

$C \enspace \colon \enspace \mathfrak{o}(n) \longrightarrow \mathrm{SO}(n) \le \mathrm{O}(n),$

where $\mathfrak{o}(n)$ is the Lie algebra of skew-symmetric matrices equipped with the usual Lie bracket, $\mathrm{O}(n)$ the Lie group of orthogonal matrices, and $\mathrm{SO}(n)$ a subgroup of $\mathrm{O}(n)$ consisting of orthogonal matrices with determinant 1. We can then think of Cayley transformation as a map from $\mathfrak{o}(n)$ into its Lie group $\mathrm{O}(n)$.

This map is nice because it allows us parametrise rotation matrices! Generally, if one wishes to study a Lie group, one might do that by studying its Lie algebra which behaves much nicer because we can do linear algebra stuff there.

One thing I want to mention about $C$ is that it does not require infinite processes. The formula is precise and does not require approximation. Compare this with the more commonly studied exponential map $\exp$, which is also a map from a Lie algebra $\mathfrak{g}$ to its Lie group $G$:

$\exp(X) = \displaystyle\sum_{k\ge 0} \dfrac{X^k}{k!}$

for $X \in \mathfrak{g}$. In this setting, usually the underlying field is taken to be $\mathbf{R}$ so if one is given $X \in \mathfrak{g}$, one needs to take a limit (which requires infinite processes) to exactly compute $\exp(X)$. It shares a similar property with the Cayley transformation: if $X$ is skew-symmetric, then $\exp(X)$ is orthogonal. In some ways, exponential map is stronger because we have things like

$\exp(aX)\exp(bX) = \exp\left((a+b)X\right),$

for example, while we don’t have an analogue for this with the Cayley transformation. What I mean by that is it’s not possible to find $t \in \mathrm{F}$ as a function of $r,s\in\mathrm{F}$ that satisfies

$C(rX)C(sX) = C(tX)$

for all $X \in \mathfrak{o}(n)$. Nevertheless, it does not mean that the Cayley transformation is less interesting. I feel that this map needs to be more well-studied, especially because it might provide an alternate theory of connecting Lie algebras with Lie groups more algebraically.

I will discuss a generalisation of this Cayley transform to a more general geometric setting in my next post. Roughly speaking (I will make this more rigorous in the next post), given a symmetric, non-degenerate matrix $B$, we can define a symmetric bilinear form $\left(\mathbf{x},\mathbf{y}\right) = \mathbf{x}^TB\mathbf{y}$ for all $\mathbf{x}, \mathbf{y} \in \mathbf{F}^n$. This gives a new geometry, with the case $B = I$ giving the Euclidean geometry back. The matrix $B = \mathrm{diag}(1,1,\ldots,1,-1)$ gives relativistic geometry.

This generalised Cayley transform depends on $B$. We will also see that a modification of being skew-symmetric and orthogonal with respect to $B$ in the next post.

Brisbane, 28 June 2019.