3
$\begingroup$

Let $G$ and $H$ be two compact Lie groups with isomorphic Lie algebras $\frak{h} \simeq \frak{g}$, but which are non-isomorphic as topological spaces. From the isomorphism assumption it (should) follows that we have a bijection between the irreducible representations of $G$ and $H$. From this it should also follow that the Peter–Weyl decomposition of the space matrix coefficient functions of both groups are isomorphic as $\frak{g}$-modules. Now the completion of the space of matrix coefficients of both groups to their spaces of continuous functions cannot be isomorphic since $G$ and $H$ are not isomorphic as topological spaces. So where does this failure of isomorphism come from? Does it come from the algebra structure of the space of matrix coefficients? Or might it come from the completion being different, i.e. the $\|\cdot\|_{\infty}$ norm is different in both cases.

Edit: Following the comments below, the example I am thinking about is $U_2$ and $SU_2 \times U_1$. Does $SU_2 \times U_1$ have more irreducible representations than $U_2$?

$\endgroup$
2
  • $\begingroup$ Surely you mean "non-isomorphic as topological groups"? "Isomorphism as topological spaces" is a strange property to study for Lie groups. (But, come to that, I don't actually know whether this is one of those instances of rigidity of Lie groups where homeomorphic Lie groups are automatically isomorphic Lie groups.) $\endgroup$ – LSpice Dec 22 '20 at 19:43
  • 2
    $\begingroup$ $\operatorname{SU}_2 \times \operatorname U_1 \to \operatorname U_2$ is a 2-fold cover, so you would expect it to have, loosely—*not* formally—speaking, twice as many representations. (This is the same kind of situation as $\operatorname{SU}_2 \to \operatorname{SO}_3$ mentioned in @QiaochuYuan's answer.) For example, the projection $\operatorname{SU}_2 \times \operatorname U_1 \to \operatorname U_1$ on the second factor does not factor through $\operatorname U_2$. $\endgroup$ – LSpice Dec 22 '20 at 19:59
9
$\begingroup$

It's just not true that having isomorphic Lie algebras implies a bijection between the irreducibles (presumably you mean a bijection compatible with the isomorphism between the Lie algebras). For example when $G = SU(2), H = SO(3)$ only half of the irreducibles of $SU(2)$ come from irreducibles of $SO(3)$.

Continuing, the natural double cover $SU(2) \to SO(3)$ identifies the algebra $C(SO(3))$ of continuous functions on $SO(3)$ with a subalgebra of $C(SU(2))$ (exactly the subalgebra of "even functions" with respect to $-1 \in SU(2)$) and in the Peter-Weyl decomposition this subalgebra corresponds to the irreducibles of $SU(2)$ which descend to irreducibles of $SO(3)$ ("integer spin"), while there's a whole other subspace of odd functions corresponding to the irreducibles which don't ("half-integer spin").

Generally, if $G$ and $H$ are compact connected Lie groups related by a covering map $p : G \to H$ then they fit into a short exact sequence

$$1 \to Z \to G \to H \to 1$$

where $Z = \text{ker}(p)$ is a finite central subgroup of $G$, which when $G$ is simply connected can be identified with $\pi_1(H)$. When $G = SU(2), H = SO(3)$ we have $Z = \{ \pm 1 \}$. The pullback $p^{\ast} : \text{Rep}(H) \to \text{Rep}(G)$ then identifies the representations of $H$ with the representations of $G$ on which $Z$ acts trivially; meanwhile there must be other representations of $G$ on which $Z$ doesn't act trivially (by Peter-Weyl). This pullback is compatible with the pullback $p^{\ast} : C(H) \to C(G)$ on continuous functions, which identifies $C(H)$ with the subalgebra of $C(G)$ on which $Z$ acts trivially (the generalized "even subalgebra").

$\endgroup$
6
  • $\begingroup$ So the example I am thinking about is $U_2$ and $SU_2 \times U_1$. Does $SU_2 \times U_1$ have more irreducible representations than $U_2$? $\endgroup$ – Piet Bongers Dec 22 '20 at 19:57
  • 2
    $\begingroup$ @PietBongers, addressed in a comment on the main question (although maybe @‍QiaochuYuan will have more to say). $\endgroup$ – LSpice Dec 22 '20 at 20:02
  • 1
    $\begingroup$ @Piet: it's exactly as LSpice says, you get a double cover so again there are "twice" as many irreducibles. $\endgroup$ – Qiaochu Yuan Dec 22 '20 at 20:03
  • 1
    $\begingroup$ @QiaochuYuan: Thanks a lot for the extra comments. So if I have understood correctly, for a Lie group $G$, will the irreps of $G$ coincide with the irreps of $\frak{g}$ when $\pi(G)$ is trivial? $\endgroup$ – Piet Bongers Dec 22 '20 at 21:00
  • 2
    $\begingroup$ @Piet: yes, that's right (for finite-dimensional representations). $\endgroup$ – Qiaochu Yuan Dec 22 '20 at 22:50

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.