3
$\begingroup$

On Wikipedia:

Since the group SU(2) is simply connected, every representation of its Lie algebra can be integrated to a group representation (Hall 2015 Theorem 5.6).

From Hall's book, we see his Theorems ---

Theorem 5.6. Let G and H be matrix Lie groups with Lie algebras $g$ and $h$; respectively, and let $\phi: g \to h$ be a Lie algebra homomorphism. If G is simply connected, there exists a unique Lie group homomorphism $\Phi: G \to H$ such that $\Phi(e^x) = e^{\phi(x)}$ for all $X \in g$. This result has the following corollary.

Corollary 5.7. Suppose G and H are simplyconnected matrix Lie groups with Lie algebras $g$ and $h$, respectively. If $g$ is isomorphic to $h$; then G is isomorphic to H .

Question: Would this fact fail when the group is not simply connected? Like SO(3)? or others PU(n) =PSU(n)? Why? Or is there a modified theorem?

$\endgroup$

1 Answer 1

4
$\begingroup$

Would this fact fail when the group is not simply connected? Like SO(3)? or others PU(n) =PSU(n)?

Yes, it fails (miserably) : just take a non simply connected group $H$ and its universal cover $G$. Then $\mathfrak{g} =\mathfrak{h}$ but obviously $H$ and $G$ are not isomorphic.

Why?

Morally, this is because, Lie algebras are infinitesimal in nature, they depend only on how the group looks like around the identity element. Therefore, they can't "see" the $\pi_1$.

Or is there a modified theorem?

A modified theorem, could be : a representation of a lie algebra $\mathfrak{g}$ of a group $G$ can be integrated in a group representation of its universal cover $\tilde{G}$.

$\endgroup$
8
  • $\begingroup$ That is a simple answer - but it seems true... +1 $\endgroup$ Aug 27, 2021 at 12:53
  • $\begingroup$ But my question was: representation of its Lie algebra can be integrated to a group representation (Hall Theorem 5.6) of not simply connected Lie group? $\endgroup$ Aug 27, 2021 at 12:55
  • $\begingroup$ @МаринаMarinaS In general, the answer is no : if $G$ is the universal cover of $H$ (non simply connected), then $\mathfrak{h}=\mathfrak{g}$. Yet, clearly, the identity $id : \mathfrak{h} \to\mathfrak{g}$ doesn't integrate to a group representation $H\to G$. $\endgroup$
    – Ayoub
    Aug 27, 2021 at 12:58
  • $\begingroup$ @МаринаMarinaS: I think you may be missing this fact: an $n$-dimensional (complex, quaternionic, real) representation of a group $H$ can be thought of as a homomorphism $H\rightarrow G$ where $G = U(n), Sp(n), SO(n)$, respectively. So, if in Ayoub's last comment, if you take $G = SU(2)$ and $H = SO(3)$, then you find examples of representations of $\mathfrak{g}\cong \mathfrak{h}$ which don't integrate to representations of $H$. $\endgroup$ Aug 27, 2021 at 13:55
  • $\begingroup$ Dear Jason, yes, 𝐺=𝑆𝑈(2) and 𝐻=𝑆𝑂(3) are examples in my mind. $\endgroup$ Aug 27, 2021 at 17:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.