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It is a known fact that for simply connected Lie groups, each representation of the Lie algebra comes from a representation of the Lie group [See this]. Consequently, we study representations of the Lie algebra. Then, if the group is compact, I can build a representation of the Lie group by exponentiating a representation of the Lie algebra. I have two questions:

  • (1) Does every representation of the Lie group come from a representation of the Lie algebra? My guess is that, if $\rho\colon G\to GL(V)$ is a representation of the Lie group, then $\rho_*\colon \mathfrak{g}\to \mathfrak{gl}(V)$ is a Lie algebra representation. So all group representations are Lie algebra representations, and the converse is also true because the group is simply connected. Hence I do have a 1-1 correspondence between representations of the group and representations of the algebra.

  • (2) If the group is non compact, then $\exp(\cdot)$ is not surjective, so I cannot come up with a representation $\tilde{\rho}$ for the group given a representation $\rho$ of the algebra. Namely, if I have an element $g\in G$ such that there is no $x\in\mathfrak{g}$ such that $\exp(x)=g$, how do I define $\tilde{\rho}(g)$? If $g=\exp(x)$ then I may define $\tilde{\rho}(g):= \exp(\rho(x))$, but what if this is not the case?

For instance take $SL(2,\mathbb{C})$, which is simply connected and non-compact. The matrix (see this question)

$$ g=\begin{pmatrix} -1 & 1 \\ 0 & -1 \end{pmatrix} \not\in\exp(\mathfrak{sl}(2,\mathbb{C})) $$

Say that I have a representation of $\mathfrak{sl}(2,\mathbb{C})$. How do I build a representation of $SL(2,\mathbb{C})$ that is well defined on $g$? My thought on this is that if the representation is two-dimensional it can either be the defining representation or the conjugate representation, and in this case the definition is natural (multiplication by $g$). What if the dimension is not 2?

I am aware of this answer which says that any element may be written as a product of exponentials. I would appreciate any deeper insight into this, namely a proof of this fact and how to do it explicitly in case the proof is non-constructive.

Elementarium
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  • Take a closer look at the linked Wikipedia article. – Moishe Kohan Nov 28 '23 at 18:09
  • @MoisheKohan would you be so kind to be more specific? Clearly, if I have asked the question, it is because I have skipped something... – Elementarium Nov 28 '23 at 18:14
  • The answer to your question is literally in the link, where they discuss the Lie-group-Lie-algebra correspondence for homomorphisms. – Moishe Kohan Nov 28 '23 at 18:46
  • @MoisheKohan "The extension is done by defining f along a path and then using the simple connectedness of G to show that the definition is independent of the choice of path." If this is what you mean by a "literal" answer to my question(s), I believe we have a very different idea of what a good answer to a meditated and elaborate question is. Also, neither is this a challenge nor did I explicitly ask for hints. You could have pointed out the sentence you referred to from the beginning, since the article is quite long. – Elementarium Nov 28 '23 at 23:42

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