Relation between the representation of lie algebra and lie group

Question

In the book Representation Theory: A First Course by Fulton and Harris (page 369): Representations of a complex lie algebra $\mathfrak{g}$ will correspond exactly to representations of the associated simply connected Lie group $\tilde{G}$: specifically, for any representation:

\begin{align*} \rho:\mathfrak{g}\to\mathfrak{gl}(V) \end{align*} of $\mathfrak{g}$, setting \begin{align*} \tilde{\rho}(\text{exp}(X))=\text{exp}(\rho(X)) \end{align*} determines a well-defined homomorphism \begin{align*} \tilde{\rho}:\tilde{G}\to \text{GL}(V) \end{align*}

For any other group with algebra $\mathfrak{g}$, given as the quotient $\tilde{G}/C$ of $\tilde{G}$ by a subgroup $C \subset Z(\tilde{G})$, the representations of $G$ ($\tilde{G}/C$) are simply the representations of $\tilde{G}$ trivial on $C$.

Question:

1.Exponential map doesn't map lie algebra to its simply connected lie group every time, so why $\tilde{\rho}$ is a map of $\tilde{G}$ which is simply connected.

2.Why the representations of $G$ ($\tilde{G}/C$) are simply the representations of $\tilde{G}$ trivial on $C$?

related https://math.stackexchange.com/q/2316410

Thank you!

Answer 1

There wasn't enough space for this as a comment so I've expanded into an answer. As I have said there is no formula for $\exp$ in the general situation. Indeed look at the formula for a linear Lie group: $$ \exp(X) = I + X + \frac{1}{2}X^2 + \frac{1}{3!}X^3 + \cdots$$

What does $X^2$ mean? That is not something available in an abstract Lie algebra, so what's going on? The answer is that we have implicitly picked a representation. $X$ is really being used here to mean $\rho(X)$ and since $\rho(X)$ lives in $\mathfrak{gl}(V)$ which is a associative algebra as well as a Lie algebra $\rho(X)^2$ makes sense. Of course a representation is necessary to write something as matrices so this step is often overlooked when you only work with matrix Lie groups.

But a given Lie algebra does not just have one Lie group (indeed it doesn't even have to have just one matrix Lie group since we can pick different representations and get different groups that way). What it does have is a single simply connected Lie group. Such a group is often awkward to handle since, although it has all the same representations as the Lie algebra, they don't have to be faithful and only faithful representations can be visualised with matrices. As an example the Spin group has no faithful irreducible representations. So how can we find a way to represent elements?

The answer: don't bother. This is already hard enough for simple examples like the universal cover of $SL(2,\mathbb{R})$. However, we know how representations on this simply connected group work. You have written the formula down in your question: $\tilde{\rho}(\exp(X)) := \exp(\rho(X))$. Of course, not every element can be written as an exponential in general (unless your group is compact or nilpotent, for example) but it can be written as the product of exponentials. So you can piece together the whole action of the group that way.

So sacrificing our desire to have a "name" (i.e. a matrix) for each element. We can still understand how they interact, how they act on representations and how the exponential map behaves.

In short, $\gamma(t) := \exp(tX)$ defines a curve through $e$ called a 1-parameter subgroup whose derivative at $e$ is $X$. This curve is exactly the integral curve for the left-invariant (or right-invariant, both work) vector field generated by $X$. The ODE you referred to is simply conveying this idea. So instead of a formula we have a characterisation: it is the unique map that does this.