# Example: O(n)

## The orthogonal group

### Definition

Let $O(n)$ denote the group of $n$ -by-$n$ orthogonal matrices. Recall that a matrix $M$ is called orthogonal if $M^{T}M=I$ . Geometrically, this means that $M$ preserves dot products. This is because $v\cdot w=v^{T}w$ , so $(Mv)\cdot(Mw)=v^{T}M^{T}Mw$ , which equals $v^{T}w$ for all $v$ and $w$ if and only if $M^{T}M=I$ . Preserving dot products means you preserve lengths of and angles between vectors, so orthogonal matrices represent things like rotations and reflections.

### Lie algebra

By definition, the Lie algebra $\mathfrak{o}(n)$ of $O(n)$ is $\mathfrak{o}(n)=\{X\in\mathfrak{gl}(n,\mathbf{R})\ :\ \exp(tX)\in O(n)\forall t% \in\mathbf{R}\}.$ This means $(\exp(tX))^{T}\exp(tX)=I$ . We can take the transpose term-by-term in the exponential power series, and, since $(MN)^{T}=N^{T}M^{T}$ , we get $(M^{n})^{T}=(M^{T})^{n}$ , so $(\exp(tX))^{T}=\exp(tX^{T})$ .

Therefore $\exp(tX^{T})=(\exp(tX))^{-1}$ and we've seen that $(\exp(tX))^{-1}=\exp(-tX)$ , so we deduce that $X\in\mathfrak{o}(n)\mbox{ if and only if }\exp(tX^{T})=\exp(-tX)\ \forall t\in% \mathbf{R}.$

Certainly if $X^{T}=-X$ then $\exp(tX^{T})=\exp(-tX)$ , so $X\in\mathfrak{o}(n)$ . Therefore any antisymmetric matrix is in the Lie algebra of the orthogonal group. It will turn out that this is everything. To see this, note that by taking $t$ to be sufficiently small we can assume that $tX^{T}$ and $-tX$ lie in our favourite neighbourhood of the zero matrix and $\exp(tX^{T})$ and $\exp(-tX)$ are in our favourite neighbourhood of the identity. Our favourite neighbourhood of the identity is the one on which the local logarithm is defined, so we can take the logarithm of each side and deduce that $tX^{T}=-tX$ for all sufficiently small $t$ , hence $X^{T}=-X$ .

Remark:

This is why we require that $\exp(tX)\in O(n)$ for all $t$ : we want to be able to put ourselves near to the identity to take logs.

We have now shown that:

Lemma:

$\mathfrak{o}(n)$ is the set of antisymmetric matrices, that is the matrices $X$ such that $X^{T}=-X$ .

### Example

Example:

We saw that $\exp\begin{pmatrix}0&-\theta\\ \theta&0\end{pmatrix}=\begin{pmatrix}\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\end{pmatrix}$ . Here, $X=\begin{pmatrix}0&-\theta\\ \theta&0\end{pmatrix}$ is antisymmetric and its exponential is a rotation matrix (hence orthogonal).

Remark:

Another, nicer, way to deduce that $\exp(tX^{T})=\exp(-tX)$ for all $t$ implies $X^{T}=-X$ , is to differentiate the equation $\exp(tX^{T})\exp(tX)=I$ with respect to $t$ . The right-hand side is constant, so its derivative is zero. The left-hand side can be evaluated using the product rule, and we get: $0=\frac{d}{dt}(\exp(tX^{T})\exp(tX))=X^{T}\exp(tX^{T})\exp(tX)+\exp(tX^{T})X% \exp(tX)$ for all $t$ . If we now set $t=0$ we get $0=X^{T}+X,$ so $X^{T}=-X$ .

Remark:

Note that $O(n)$ is topologically closed. To see this, consider the map $F\colon\mathfrak{gl}(n,\mathbf{R})\to\mathfrak{gl}(n,\mathbf{R})$ defined by $F(M)=M^{T}M$ . We have $O(n)=F^{-1}(I)$ by definition. Note that $F$ is continuous because the matrix entries of $M^{T}M$ are polynomials in the matrix entries of $M$ (and polynomials are continuous). Therefore if $M_{k}\in O(n)$ is a sequence of orthogonal matrices converging to a matrix $M$ then $F(M_{k})=I$ for all $k$ , so $\lim_{k\to\infty}F(M_{k})=I$ , and since $F$ is continuous we have $F(M)=\lim_{k\to\infty}F(M_{k})=I$ , so $M\in O(n)$ .

Remark:

Whenever your group is cut out by a continuous equation like this it will be a topologically closed group of matrices.

## Pre-class exercise

Exercise:

If $X,Y\in\mathfrak{o}(n)$ then $[X,Y]\in\mathfrak{o}(n)$ .