(xournal_134)
Given a manifold $M$ we can think of a "procedure" that takes a point $p\in M$, a real number $t\in \mathbb{R}$ and a vector field $X$ on $M$ and returns a new point of $M$, working at this way:
1. we start at $p$ and move a little along $X_p$. That is, we compute $q_1=p+tX_p/10$. This is obviously not defined, but we take a blind eye. Also, we denote it by $q_1=p(Id+tX/10)$. This notation is justified because $X$ is, loosely speaking, an element of the Lie algebra of $\text{Dif}(M)$ so we can think of $Id+tX/10$ as an element of $\text{Dif}(M)$ near to $Id$.
2. we move a little from $q_1$ along the direction of $X_{q_1}$, that is, we compute $q_2=q_1+tX_{q_1}/10=q_1(Id+tX/10)$.
3. We finish after 10 steps.
Observe that, abusing of notation,
$$ q\approx q_{10}=q_9(Id+tX/10)=q_8(Id+tX/10)^2= $$ $$ =p(Id+tX/10)^{10} $$Of course, we have had better precision if we do this in more steps. In the limit (steps $\to \infty$) we would obtain the true output that we will denote by
$$ q=p \lim_{n\to \infty}(Id+tX/n)^{n} \equiv pe^{tX} $$and is nothing but the flow of $X$. (See also Lie algebra action. And Lie series.)
This is a very general construction, which allows you to find curves with a prescribed velocity.
What has to do with the usual exponential function and, moreover, with the number $e$? Suppose the manifold $M=\mathbb{R}$ and the vector field $X$ given in canonical coordinates by $X_x=x$.
Then the number $e$ appears when we move from 1 a time 1 by means of that particular vector field.
That is, if you consider a curve $\alpha: \mathbb{R} \to \mathbb{R}$ such that $\alpha(0)=1$ and its velocity at $x$ is precisely $x$, then at time $t=1$ you will have arrived to position $e$.
This general construction can also be particularized to the case of matrix exponential. Suppose that the manifold is $M=GL(2,\mathbb{R})$ (it s valid for an arbitrary $GL(n)$, of course). So given any vector field $X\in \mathfrak{X}(GL(2,\mathbb{R}))$ we can define $e^{tX}$ as above.
But since in this case $M$ is a Lie group we can suppose that the chosen vector field $X$ is a left invariant vector field. Then it takes the form
$$ X_A=B\cdot A, $$being $A\in GL(2)$ (an invertible matrix) and $B\in T_{Id} GL(2)$ (an arbitrary matrix). When we compute $e^B$ what we are doing is to consider the corresponding left invariant vector field $X$ and move along its flow a time $t=1$ from the point $Id$.
Consider a Lie group $G$, and an element $v$ in the Lie algebra $\mathfrak{g}$. Observe that, when we consider $G$ acting on itself, every vector $v\in T_{Id} G$ gives rise to a left invariant vector field $V\in \mathfrak{g}$ associated to $v$:
$$ V_g=d(L_g)_{Id}(v). $$Then, if we consider the flow $\phi_V$ of this vector field on $G$, we have a one-parameter subgroup of $\mbox{Diff}(G)$ (we are assuming that it is a complete vector field, which is something proven here.
With this set up, the exponential map is defined by
$$ exp(v)=\phi_V(1,Id) $$and is usually also denoted by $e^v$.
Since $G\subseteq \mbox{Diff}(G)$ we can wonder: is it, in fact, a one-parameter subgroup of $G$? Or in other words, does every diffeomorphism $\phi_V(t,-)$ correspond to left translation by an certain element $g_t \in G$?
The answer is yes. We can take $g_t=\phi_V(t,e)\in G$, and define $\tilde{\phi}(t,m)=m\cdot g_t$. But we can show that
$$ \phi_V=\tilde{\phi} $$because of the uniqueness of the flow, since:
If we have a Riemannian manifold $(N,g)$ we have the exponential map for Riemannian manifolds:
$$ exp_p: T_pN \to N $$ $$ V\mapsto exp_p(V) $$How does it fit in this framework? If we consider the manifold $M=TN$ and the vector field $X\in \mathfrak{X}(TN)$ called the geodesic spray then we have, abusing of notation, $P=(p,V) \in TN$ and it turns out that
$$ P\cdot e^X=exp_p(V). $$It is important to note that $\phi_V(t,e)=e^{tv}$. But this is obvious from flow theorem for vector fields#Important property. Indeed
$$ e^{tv}=\phi_{tV}(1,e)=\phi_V(t,e). $$Therefore we have
$$ v=\frac{d}{dt}|_{t=0} e^{tv} $$expression that I like to call inverse relation between exponential and differentiation at the origin. In a sense, is like saying that exponentiation is like an "integration process" (see the picture of the section Motivation).
Consider the exponential map
$$ exp: \mathfrak{g} \mapsto G $$along with its differential at $0$:
$$ d (exp)_0: T_0 \mathfrak{g} \mapsto \mathfrak{g} $$The differential is the identity map, which implies, by the inverse function theorem, $exp$ acts as a diffeomorphism in a neighbourhood of $0\in \mathfrak{g}$ and $e\in G$. This neighbourhood is denoted as $U_e$. This ensures that the exponential map is always a local diffeomorphism at the origin of the Lie algebra. As a result, it sets up a local chart for a neighbourhood of the identity in the Lie group.
It's essential to note that while the map acts as a diffeomorphism locally, its behavior can be different when viewed globally. Specifically:
According to Proposition 1.24 in @olver86, any element $g\in G$ can be expressed as a product of elements $g_i$ that belong to $U_e$. Formally, for any $g\in G$:
$$ g=g_1\cdot g_2\cdot \cdots = exp(v_1)\cdot exp(v_2)\cdot \cdots $$where each $v_i$ is an element of $\mathfrak{g}$.
Proposition
For a compact connected Lie group $G$, the exponential map is surjective.
Proof
The proof is intricate and is provided in an external source. For a detailed understanding, refer to:
https://terrytao.wordpress.com/2011/06/25/two-small-facts-about-lie-groups/
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Author of the notes: Antonio J. Pan-Collantes
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