The Exponential Map

Abstraction of Exponentials

We know exponentials and logarithms as functions \newcommand{\R}{\mathbb{R}}\R\to \R. Here, we shall generalize them in an abstract way to the setting of formal power series over a field.

Exponential and Logarithm in \newcommand{\[}{\left[\!\left[}\newcommand{\]}{\right]\!\right]}F\[t\]

Suppose $F$ is a field, consider the ring of formal power series over $F$: F\[t\]:=\left\{\sum_{n=0}^{\infty} a_{n}t^{n} \mid a_{n}\in F \right\}.Then the exponential and logarithm are elements defined as follows: \exp(t):=\sum_{n=0}^{\infty} \frac{t^{n}}{n!}\in F\[t\], \quad \log(1+t):=\sum^{\infty}_{m=1} \frac{(-1)^{m-1}t^{m}}{m}\in tF\[t\],where tF\[t\] is the ideal of F\[t\] generated by $t$.

Proposition

Suppose $f(t)$ and $g(t)$ are two elements in F\[t\], and g(t)\in tF\[t\], then $f(g(t))$ is well-defined and belongs to F\[t\].

Proof Suppose $f(t)=a_0+a_{1}t+\cdots$ and $g(t)=b_{1}t+b_2{t}^2+\cdots$. Substitute $g(t)$ into $f(t)$, we get$f(g(t))=a_0+a_1(b_{1}t+\cdots )+a_2(b_{1}t+\cdots {)}^2+\cdots .$Note that $(b_{1}t+\cdots {)}^j$ has lowest degree term of degree $j$, so the $j$th coefficient of $f(g(t))$ is always a finite sum, therefore $f(g(t))$ is well-defined and belongs to F\[t\]. $\square$

Corollary

$\exp (\log (1+t))$ is well-defined and equals $1+t$ for all t\in tF\[t\]. $\log (\exp (t))$ is also well-defined and equals $t$ for all t\in F\[t\].

Proof This is a direct consequence of the above proposition. $\square$

Exponential and Logarithm of Complex Matrices

We now define the exponential and logarithm of complex matrices by substituting matrices into the formal power series defined above.

Proposition

The following properties hold for the exponential and logarithm maps for complex matrices:

1. For all $A\in {\text{GL}}_n(\text{C})$ with $\Vert A-I\Vert <1$, we have $\exp (\log (A))=A$.
2. For all $X\in M_n(\text{C})$ with $\Vert X\Vert <\log 2$, we have $\log (\exp (X))=X$.
3. $\exp (0)=I$.
4. $\exp (X{)}^{\ast }=\exp (X^{\ast })$
5. $\exp (X+Y)=\exp (X)\exp (Y)$ if $X$ and $Y$ commute.
6. $\lambda \exp (X){\lambda }^{-1}=\exp (\lambda X{\lambda }^{-1})$.

Lie's Product Formula

For any two complex matrices $X,Y\in M_n(\text{C})$, we have $\exp (X+Y)={\lim }_{m\to \infty }{\left(\exp (X/m)\exp (Y/m)\right)}^m.$

The Lie Algebra of a Lie Group

Every Lie group $G$ has a canonical Lie algebra $\mathfrak{g}$, which has two natural representations, either as the tangent space at the identity element $e$ of $G$, or as the space of left-invariant vector fields on $G$. We will see that the two representations are isomorphic.

Left-Invariant Vector Field

If $G$ is a Lie group and $v$ is a vector in $T_{e}G$, where $e$ is the identity element in $G$. Suppose ${\ell }_g:G\to G$ is the left multiplication map that sends $h$ to $gh$. Then we can use the maps ${\text{d}}_e{\ell }_g$ to define a natural vector field on $G$, for each $g\in G$. We set
$X(g):={\text{d}}_e{\ell }_g(v).$
The resulting vector field $X$ is called a left-invariant vector field.

e.g.

• Let Lie group $G=\text{C}\setminus \{0\}$, with the group operation of complex number multiplication. Fix some $v\in T_{1}G={\mathbb{R}}^2$, the left-invariant vector field $V$ is given by $V(z)={\text{d}}_1{\ell }_z(v)=zv.$

Proposition

The left-invariant vector fields on a Lie group form a Lie algebra. That is, suppose $X$ and $Y$ are smooth left-invariant vector fields on $G$, then $[X,Y]$ is also left-invariant.

Proof It suffices to check $\text{D}{\ell }_g{|}_e$

Theorem

The set of all left-invariant vector fields on a Lie group $G$ is isomorphic to the tangent space at the identity element $e$ of $G$, via evaluation at $e$. In other words, every tangent vector $v\in T_{e}G$ extends uniquely to a left-invariant vector field.

Proof Denote the space of left-invariant vector fields as ${\mathfrak{X}}^{\ell }(G)$. We define $\Phi :{\mathfrak{X}}^{\ell }(G)\to T_{e}G$ by $X\mapsto X{|}_e$, and $\Psi :T_{e}G\to {\mathfrak{X}}^{\ell }(G)$ by $v\mapsto X$ where $X(g)={\text{d}}_e{\ell }_g(v)$. It is clear that $\Phi$ and $\Psi$ are linear maps, and $\Phi \circ \Psi ={\mathrm{id}}_{T_{e}G}$, $\Psi \circ \Phi ={\mathrm{id}}_{{\mathfrak{X}}^{\ell }(G)}$. Therefore, $\Phi$ and $\Psi$ are isomorphisms. $\square$

Lie Algebra of a Lie Group

The Lie algebra of a Lie group $G$, often denoted $\mathfrak{g}$, is up to isomorphic, the Lie algebra of left-invariant vector fields on $G$, or the tangent space at the identity element $e$ of $G$.

Remark

Elements of the Lie algebra represent “infinitesimal transformations” or “directions of motion” within the group starting from the identity.

Exponential Map

For any Lie group $G$ with Lie algebra $\text{g}$, the exponential map $\exp :\text{g}\to G$ is defined by $\exp (X)=\gamma (1),$where $\gamma$ is the unique integral curve of the left-invariant vector field $X$ such that $\gamma (0)=e$, or equivalently, $\gamma$ is the one-parameter subgroup generated by $X$.

e.g. The Lie algebra of $S^1$, denoted $\mathfrak{s}$, is isomorphic to $\mathbb{R}$. The exponential map of $S^1$ satisfies $\exp (tx)=e^{itx},\text{for all }x\in \text{s}\cong \mathbb{R},t\in \mathbb{R}.$

Proposition

Let $G$ be a Lie group with associated Lie algebra $\text{g}$. For any $X\in \text{g}$, $s\mapsto \exp (sX)$ is the one-parameter subgroup of $G$ generated by $X$. That is, $\exp (tX)=\gamma (t)$ for all $t\in \mathbb{R}$.

Proof

Fundamental Vector Field

Suppose $G$ is a Lie group with Lie algebra $\text{g}$, acting on a smooth manifold $M$ with the action $\ast :G\times M\to M$ that $(g,p)\mapsto g\ast p$. The fundamental vector field on $M$ generated by the infinitesimal action of $\xi \in \text{g}$, denoted by ${\xi }_M$ (or $X^{\xi }$), is defined at each point $p\in M$ as ${\xi }_M{|}_p:=\frac{\mathrm{d}}{\mathrm{d}t}{|}_{t=0}\left(\exp (t\xi )\ast p\right).$

e.g. Since the $n$-torus $T^n={\left(S^1\right)}^n$, its Lie algebra $\mathfrak{t}$ is isomorphic to ${\mathbb{R}}^n$. From the previous example of $S^1$, we can deduce that the exponential map of an $n$-torus takes the form $\exp (tX)=\left(e^{it{x}_1},e^{it{x}_2},\dots ,e^{it{x}_n}\right),\text{for all }X=(x_1,\dots ,x_n)\in \text{t}\cong {\mathbb{R}}^n.$Consider a torus $T^n$ acting on ${\text{C}}^n$ by $(e^{i{\theta }_1},\dots ,e^{i{\theta }_n})\ast (z_1,\dots ,z_n):=(e^{i{k}_1{\theta }_1}{z}_1,\dots ,e^{i{k}_n{\theta }_n}{z}_n),$for nonzero constants $k_1,\dots ,k_n\in \mathbb{Z}$. . Then for any $\xi =({\xi }_1,\dots ,{\xi }_n)\in \text{t}$ and $z=(z_1,\dots ,z_n)\in {\text{C}}^n$, there holds ${\xi }_M{|}_z:=\text{ddtz}\left(\exp (t\xi )\ast z\right)=(i{k}_1{\xi }_1{z}_1,\dots ,i{k}_n{\xi }_n{z}_n).$