### Introduction to Lie groups

When you first learned about cyclic groups, the picture in your head was that of the unit circle (complex numbers with norm one). Sure, the unit circle isn't actually a cyclic group, but it really feels like one. When I first motivate group theory, I even base the motivation on the close similarities between the circle group and the modular addition group $\mathbb{Z}/p\mathbb{Z}$. Indeed, the circle group is just the group of real numbers mod $2\pi$.

The solution to this problem can be seen from the quickest proof that the unit circle isn't cyclic -- the fact that it isn't countable (while the integers are). Well, what if we discard the centrality of the integers to our definition of a cyclic group and admit real powers on groups?

Ok, but how? It's easy to construct integer powers on an arbitrary group -- in terms of repeated addition (which defines natural powers) and inverses. But the real numbers are a wholly different beast -- they require a nice and connected "smooth" structure, a geometry on the group. We can certainly visualise this geometry on the unit circle or the positive real numbers (which is also "real-power cyclic"), but it's interesting to think about how one might introduce such a geometry on other groups (groups that admit such a geometry are called Lie groups).

Well, if you think for a while, you might get the idea of defining a group via a real-number paramterisation $\mathbb{R}\to G$. The unit circle can be parameterised as $g(\theta)=\exp i\theta$, the positive real numbers can be parametarised as $g(\xi)=\exp\xi$, etc. This parameterisation would then give $\exp rt =(\exp t)^r$ for real powers $r$ of elements in the group.

But here's the thing -- we could have introduced any sort of ugly and terrible parameterisation for our group. We knew how the parameterisation should look for the unit circle, but we could have as well have created something definitely not smooth -- like mapping $\pi i$ to $-1$ and mapping $(\pi + \varepsilon)i$ to $i$ (sorry, you can't use too much dramatic hyperbole on the unit circle... fine, let's map it to 30 gazillion, which isn't on the unit circle, but whatever), and the real-power would look ridiculous, not at all what we want, and we may not even have a "real-power cyclic" structure.

What exactly do we want from our parameterisation?

Let's think about what a generator looks like with real powers on the unit circle. Well, really any non-identity element $e^{i\theta_0}$ can generate the group (take it to the power of $\theta/\theta_0$), but if we want to emulate the case of the integers under addition $\{...a^{-2},a^{-1},1,a,a^2,a^3,...\}$, we'd like to call the element really close to 1 the generator. Well, there's no element that's really close to 1, so we're talking about some kind of an infinitesimal thing. This is called an infinitesimal generator of the Lie group.

In the first-order approximation, such an element would be of the form $1+i\varepsilon$. By making $\varepsilon$ sufficiently small, the element will be sufficiently close to being "on the unit circle", with an arc length of $\varepsilon$ away from the identity, and its real power $r$ of the element will have an arc length of $r\varepsilon$ away from the identity. So to generate the element with parameter $\theta$, we need to take $1+i\varepsilon$ to a real power of $r=\theta/\varepsilon$. I.e.

$$g(\theta) = \lim\limits_{\varepsilon\to 0} (1+i\varepsilon)^{\theta/\varepsilon}=\lim\limits_{r\to\infty} \left(1+\frac{i\theta}{r}\right)^{r}=\exp i\theta$$
If you were studying calculus for the first time, this is really solid intuition for Euler's formula. Conversely, you can go in the other direction and say it's solid intuition for the compound-interest limit.

$$\lim\limits_{\varepsilon\to 0} (1+\varepsilon\theta t)^{1/\varepsilon} = \exp(t\theta)$$
But here, we can view it in a more general light, and say this is the definition of the exponential map to a Lie group. What exactly is it a map from? I.e. what is the parameterising space? Well, as you can see, it maps an element $i\theta$ to the group parameterised by $\theta$ -- what is $i\theta$? It is

$$\lim\limits_{\varepsilon \to 0} \frac{{(1 + i\varepsilon \theta ) - 1}}{\varepsilon }$$
I.e. these are the elements span the tangent line to the group at 1. In general, one may have more dimensions to this group, i.e. more parameters to put in the smooth parameterisation -- in this case we have:

$$g(\theta ) = \lim\limits_{\varepsilon \to 0} {\left( {1 + \varepsilon ({t_1}{\theta _1} + \ldots {t_n}{\theta _n})} \right)^{1/\varepsilon }} = \exp \vec \theta$$
Where $\vec\theta \in V$, which is a vector space with basis $\langle t_1 \dots t_n \rangle$ -- the tangent space to the group at the identity. This vector space is called the Lie algebra of the Lie group.

Take a moment to appreciate the significance of this -- smoothness tells us (sorta) that a function or structure can be determined by the values of all its derivatives at a point. But when you add the group structure -- when you require an exponential structure for the parameterisation, i.e. (1) $g(\theta_1+\theta_2) = g(\theta_1)g(\theta_2)$; (2) $g(r\theta)=g(\theta)^r$; (3) $g(0)=1$ -- just the first derivative, the tangent plane, determines the entire parameterisation. This is precisely analogous to how given that a given smooth function has an exponential structure $e^{tx}$, it can be determined from its first derivative alone. The structure of a Lie group is "fundamentally exponential".

Here's another way to see how the additivity-multiplicativity condition allows the first derivative to determine the entire parameterisation. The Taylor series of the parameterisation is given by:

$$g(\theta)=\sum\limits_{k=0}^\infty \frac{g^{(k)}(0)}{k!}\theta^k$$
Meanwhile the exponential map is:

$$\exp \left(\theta g'(0)\right) =\sum\limits_{k=0}^\infty \frac{\left(\theta g'(0)\right)^k}{k!}$$
So a sufficient condition for the two to be equal is:

$$g^{(k)}(0)=g'(0)^k$$
This is something that is true for exponential functions, of course, but what's the condition for it to be true in general? Writing both sides in limit form and using the Binomial theorem on the right,

$$\frac{1}{{{h^k}}}\sum\limits_{k = 0}^k {{\binom{n}{k}}{{( - 1)}^k}g\left( {(n - k)h} \right)} = \frac{1}{{{h^k}}}\sum\limits_{k = 0}^k {{\binom{n}{k}}{{( - 1)}^k}g{{(h)}^{n - k}}g{{(0)}^k}}$$
Which is true since $g((n-k)h) = g(h)^{n-k}$ and $g(0)=1$.

(something to note: the "official" word for real-power cyclic is "one-parameter group" or "one-dimensional Lie group". Higher dimensional groups have more generators, i.e. more dimensions)

Show, from the $(1+X/r)^r$ definition of the exponential map, that it can be given by the standard Taylor expansion:

$$\exp X = 1 + X + \frac{X^2}{2!} + \ldots$$
You can't really assume the Binomial theorem (as it is only true on commutative rings, and the ring of $n$-dimensional matrices -- which is the ring that we embed Lie groups and their Lie algebras in -- isn't commutative), but perhaps a weaker result holds? What kind of elements still commute on general rings?