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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!

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  • $\begingroup$ 1. There is an exponential map for each Lie group with $\mathfrak{g}$ as its Lie algebra. You may assume the $\exp$ on the left is precisely the one into $\tilde{G}$. (The one on the right is the one from $\mathfrak{gl}(V)$ into $GL(V)$). $\endgroup$
    – Callum
    Aug 26, 2022 at 18:33
  • $\begingroup$ 2. Try to define a representation $\rho_C:\tilde{G}/C\to GL(V)$ by $\rho_C(gC):=\tilde{\rho}(g)$. What goes wrong if $\tilde{\rho}$ is not trivial on $C$? $\endgroup$
    – Callum
    Aug 26, 2022 at 18:39
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    $\begingroup$ I try to explain what you mean: For $ w \in V$, $ g,c \in \tilde{G}$, we have $\tilde{\rho}(gc)(w)=\tilde{\rho}(g)\tilde{\rho}(c)(w)$ by homomorphism.We mod $\tilde{G}$ by $C$, so for any $c \in C$, we need have $\tilde{\rho}(gc)(w)=\tilde{\rho}(g)(w)\ \Rightarrow \tilde{\rho}(c)(w)=w$. Check $W=\{w\in V: \tilde{\rho}(c)(w)=w\}$ form a representation of $\tilde{G}/C$: $\tilde{\rho}(c)\tilde{\rho}(g)(w)=\tilde{\rho}(g)\tilde{\rho}(c)(w)=\tilde{\rho}(g)(w) \text{, because } C \subset Z(\tilde{G}) \Rightarrow \tilde{\rho}(g)(w) \in W$. $\endgroup$
    – Jino
    Aug 27, 2022 at 3:31
  • $\begingroup$ About the first question, can you tell me more? For $GL_n\mathbb{C}$, $\frac{dg(t)}{dt}$ =${(L_{g(t)})}_{*e}X_{e}$, we solve the ODE to get the exponential map.But $GL_n\mathbb{C}$ is not simply connected, so how to get a exponential map which map lie algebra to its simply connected lie group? $\endgroup$
    – Jino
    Aug 27, 2022 at 5:27
  • $\begingroup$ Well you are already taking it as the solution to an ODE so what is stopping you in this case. Another way to characterise it is: $\exp(X)$ is uniquely defined as $\gamma(1)$ where $\gamma$ is the 1-parameter subgroups whose derivative at the identity is $X$. We are no longer dealing with matrices etc. here so there isn't necessarily a formula for $\exp$ here or anything like the Taylor expansion you see for matrices $\endgroup$
    – Callum
    Aug 27, 2022 at 12:57

1 Answer 1

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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.

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  • $\begingroup$ so, for a same lie algebra, the ODE give a different solution by different pushforward of left action? $\endgroup$
    – Jino
    Aug 27, 2022 at 16:20
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    $\begingroup$ Kinda, yeah. But this is more than a "different solution". The ODE is being constructed over a different codomain. Our group has changed, so left translation is happening in a different space, it's not just a "different pushforward". The notation doesn't change and the resulting properties of $\exp$ that I mention are always true but to be really precise, (I would argue) the formulation of that ODE has changed fundamentally by changing the target of $g(t)$. $\endgroup$
    – Callum
    Aug 27, 2022 at 16:31
  • $\begingroup$ Maybe I will have a deeper understanding after reading this book~ thank you $\endgroup$
    – Jino
    Aug 27, 2022 at 16:43
  • $\begingroup$ "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. " I think if we write the product of two elements in matrix group( in the form of exponential map), then by Campbell-Hausdorff formula it seems map a vector in lie algebra to an element in $GL_n$ . What goes wrong? $\endgroup$
    – Jino
    Sep 2, 2022 at 10:22
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    $\begingroup$ @Jino The BCH formula gives you an infinite series but that series does not have to converge. So the product of two elements which are exponentials does not have to be an exponential itself. If, for example, all the elements are nilpotent then the series must always terminate (so obviously converges) and so in a case like that we can use BCH to show everything is an exponential. $\endgroup$
    – Callum
    Sep 2, 2022 at 12:23

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