1.1. Lecture 1

We fix some notation:

•

$k$ denotes either the field $\mathbb{R}$ or $\mathbb{C}$ ;
•

$\mathfrak{gl}_{n,k}=M_{n}(k)$ is the vector space of all $n\times n$ matrices over $k$ .

1.1.1. The exponential map

Definition 1.1.1.

Let $X\in\mathfrak{gl}_{n,k}$ . We define

\exp(X)=\sum_{k=0}^{\infty}\frac{X^{k}}{k!}.

This series is convergent for all $X\in\mathfrak{gl}_{n,k}$ . Let $||\cdot||$ be the matrix norm

||X||=\left(\sum_{i,j}|x_{ij}|^{2}\right)^{1/2}.

This satisfies the triangle inequality and also $||XY||\leq||X||||Y||$ — this can be proved using Cauchy–Schwarz. Then for any $X\in\mathfrak{gl}_{n,\mathbb{C}}$ with $||X||\leq M$ , we have

||\exp(X)||\leq\sum_{k=0}^{\infty}\frac{||X||^{k}}{k!}\leq\exp(M).

In particular, we see that $\exp$ is uniformly absolutely convergent on all compact subsets of $\mathfrak{gl}_{n,k}$ . It follows that $\exp$ is a continuous function.

Lemma 1.1.2.

For all $X,Y\in\mathfrak{gl}_{n,k}$ , $s,t\in k$ and $g\in\operatorname{GL}_{n}(k)$ , we have:

(i)

$\exp(0)=\operatorname{Id}$ .
(ii)

$\exp(X+Y)=\exp(X)\exp(Y)$ if $XY=YX$ . (This is NOT true in general).
(iii)

$\exp(X)$ is invertible, with inverse $\exp(-X)$ .
(iv)

$\exp(sX)\exp(tX)=\exp((s+t)X)$ .
(v)

$g\exp(X)g^{-1}=\exp(gXg^{-1})$ .

Proof.

The first point is obvious. Let’s prove (ii) from which (iii) and (iv) follow. By definition,

$\displaystyle\exp(X+Y)$	$\displaystyle=\sum_{k=0}^{\infty}\frac{(X+Y)^{k}}{k!}$
	$\displaystyle=\sum_{k=0}^{\infty}\sum_{l=0}^{k}\frac{\binom{k}{l}X^{l}Y^{k-l}}% {k!}$	(using that $X$ and $Y$ commute!)
	$\displaystyle=\sum_{k=0}^{\infty}\sum_{l=0}^{k}\frac{X^{l}Y^{k-l}}{l!(k-l)!}$
	$\displaystyle=\left(\sum_{l=0}^{\infty}\frac{X^{l}}{l!}\right)\left(\sum_{j=0}% ^{\infty}\frac{Y^{j}}{j!}\right),$	(putting $j=k-l$ )

which is equal to the right hand side. Rearranging the sums is valid by absolute convergence. Finally, (v) follows from $gX^{k}g^{-1}=(gXg^{-1})^{k}$ . ∎

In fact the exponential map is differentiable as a function of $X$ . For this, recall that a function $f:\mathbb{R}^{N}\rightarrow\mathbb{R}^{M}$ is differentiable at a point $p\in\mathbb{R}^{N}$ if there is a (necessarily unique) linear map $D_{p}f:\mathbb{R}^{N}\to\mathbb{R}^{M}$ such that

\lim_{h\to 0}\frac{||f(p+h)-f(p)-D_{p}f(h)||}{||h||}=0,

and in this case $D_{p}f$ is called the derivative of $f$ at $p$ . (This definition is independent of the choice of norms on $\mathbb{R}^{N}$ and $\mathbb{R}^{M}$ ).

Proposition 1.1.3.

The exponential map is differentiable at the origin (zero matrix), and its derivative at the origin is the identity map from $\mathfrak{gl}_{n,\mathbb{C}}$ to itself.

Proof.

In the above definition we have, $N=M=2n^{2}$ , $f=\exp$ , $p=0$ , and we claim $D_{0}\exp$ is the identity. Thus we need to show

\lim\limits_{||X||\to 0}\frac{||\exp(X)-\exp(0)-X||}{||X||}=\lim\limits_{||X||% \to 0}\frac{||\exp(X)-\operatorname{Id}-X||}{||X||}=0,

which follows from the definition of the exponential map. Indeed,

\frac{||\exp(X)-\operatorname{Id}-X||}{||X||}=\frac{||\sum_{k=2}^{\infty}\frac% {X^{k}}{k!}||}{||X||}\leq||X||\cdot\sum_{k=0}^{\infty}\frac{||X||^{k}}{(k+2)!}% <||X||e^{||X||},

which tends to zero as $||X||\to 0$ . ∎

Remark 1.1.4.

In fact, the exponential function has derivatives to all orders at all points; this follows from the fact that it is given by power series that converge absolutely at all points and all of whose (formal) derivatives also converge absolutely at all points.

By the inverse function theorem, it follows from the remark that

Corollary 1.1.5.

The exponential map is a local diffeomorphism at $0$ : there exist neighbourhoods $U_{0}\subseteq\mathfrak{gl}_{n,k}$ containing $0$ and $V_{0}\subseteq\operatorname{GL}_{n}(k)$ containing $\operatorname{Id}$ such that $\exp|_{U_{0}}$ is a smooth homeomorphism onto $V_{0}$ with smooth inverse.

Remark 1.1.6.

In fact we can take $V_{0}=\{X\in\operatorname{GL}_{n}(\mathbb{C})\,|\,||X-\operatorname{Id}||<1\}$ . The inverse of $\exp$ in this neighbourhood is

\log(X)=\sum_{k=0}^{\infty}(-1)^{k}\frac{(X-\operatorname{Id})^{k+1}}{k+1},

which is convergent when $||X-\operatorname{Id}||<1$ .

Of course, $\exp$ is not injective in general. For example, $\exp(2\pi ik)=1$ for $k\in\mathbb{Z}$ .

1.1.2. Exercises

Problem 1.

(a)

Compute $\exp(X)$ for $X$ equal to $\begin{pmatrix}t&0\\ 0&s\end{pmatrix}$ , $\begin{pmatrix}0&t\\ -t&0\end{pmatrix}$ , and $\begin{pmatrix}0&t\\ t&0\end{pmatrix}$ (where $s,t\in\mathbb{R}$ ).
(b)

Let $E_{a,b}$ be the elementary $n\times n$ matrix with $1$ in the $(a,b)$ -entry and $0$ elsewhere. Compute $\exp(tE_{a,b})$ for $a\neq b$ and $a=b$ .

Problem 2. Show that

\exp(tX)\exp(tY)=\exp\left(t(X+Y)+\frac{t^{2}}{2}[X,Y]+O(t^{3})\right)

as $t\to 0$ , where

[X,Y]=XY-YX.