I wanted to classify all irreducible complex representations of the following basic abelian Lie groups: $\mathbb{S}^1$ the circle in the complex plane, $\mathbb{R}_{>0}$ the positive real numbers, $\mathbb{C}^\times$ the complex numbers without zero, $\mathbb{R}^\times$ the real numbers without zero and $\mathbb{R}^\times\times\mathbb{S}^1$.

I know, since this are abelian groups and $\mathbb{C}$ is algebraically closed, from Schur's Lemma, that the irreducible complex representations I'm searching for are one dimensional.

I also found searching on the internet that for the first three groups I should have: $$\rho_m:\mathbb{S}^1\rightarrow \mathbb{C}^\times \quad | \quad \rho_m(z)=z^m \quad , \quad \quad \forall m\in\mathbb{Z}$$ $$\rho_z:\mathbb{R}_{>0}\rightarrow \mathbb{C}^\times \quad | \quad \rho_z(t)=e^{zt} \quad , \quad \quad \forall z\in\mathbb{C}$$ $$\rho_n:\mathbb{C}^\times\rightarrow \mathbb{C}^\times \quad | \quad \rho_n(z)=z^n \quad , \quad \quad \forall n\in\mathbb{Z}$$ but I don't really understand why. I think I could maybe use the Lie functor and factor this maps on the Lie algebra of the respective Lie group... I'm lost. Any help would be much appreciated.


1 Answer 1


These groups are all products of the groups $\mathbb{R}, S^1, C_2$. Using the fact that if the group of characters of $A$ is $\widehat{A}$ and the group of characters of $B$ is $\widehat{B}$ then the group of characters of $A \times B$ is $\widehat{A} \times \widehat{B}$, and then using an argument to reduce the case of $S^1$ to the case of $\mathbb{R}$, it suffices to understand the characters of $\mathbb{R}$ (the characters of $C_2$ are very straightforward).

From here the argument depends slightly on whether you mean continuous representations or smooth representations. The answer is the same either way (continuous representations turn out to be automatically smooth) but the argument in the continuous case is a little longer. I will assume smoothness for simplicity. So, let $\rho : \mathbb{R} \to \mathbb{C}^{\times}$ be a smooth character of $\mathbb{R}$. By differentiating $\rho(s + t) = \rho(s) \rho(t)$ we see that $\rho$ satisfies a differential equation

$$\rho'(t) = \rho'(0) \rho(t), \rho(0) = 1.$$

Here $\rho'(0) \in \mathbb{C}$. Various standard arguments can be used to show that this differential equation has unique solution $\rho(t) = \exp(\rho'(0) t)$. (We don't need any general facts about Lie algebras and derivatives of maps between Lie groups; the abelian case is easy enough to handle directly.) Conversely the usual properties of the exponential imply that all exponentials of this form are smooth characters. (This is the only part of the argument where any "real work" is happening; everything from here is bookkeeping.) So we conclude that the $1$-dimensional complex irreducible representations of $\mathbb{R}$ are given by

$$\rho_z(t) = \exp(zt), z \in \mathbb{C}.$$

Now let $\rho : S^1 \to \mathbb{C}^{\times}$ be a smooth character of $S^1$. By precomposing with the exponential map $\exp(2 \pi i t) : \mathbb{R} \to S^1$ we get a character of $\mathbb{R}$, which must have the form $\exp(wt)$. But the further condition that this character factors through the quotient map $\mathbb{R} \to \mathbb{R}/\mathbb{Z} \cong S^1$ means that we must have $\exp(w(t+1)) = \exp(wt)$, which is equivalent to $\exp(w) = 1$. This gives that $w = 2 \pi i k$ must be an integer multiple of $2 \pi i$, and we conclude that the $1$-dimensional complex irreducible representations of $S^1$ are given by (writing $z = \exp(2 \pi i t)$)

$$\boxed{ \rho_k(z) = z^k, k \in \mathbb{Z} }.$$

This is the first group on your list. The next is $\mathbb{R}_{+}$, which is isomorphic to $\mathbb{R}$ via the exponential map $t \mapsto \exp(t)$. So the classification is the same as $\mathbb{R}$ but must be written differently (it is not what you wrote which is the classification for $\mathbb{R}$); writing $r = \exp(t)$ this gives

$$\boxed{ \rho_z(r) = r^z, z \in \mathbb{C} }.$$

Next, polar decomposition gives $\mathbb{C}^{\times} \cong \mathbb{R}_{+} \times S^1$. So a character of $\mathbb{C}^{\times}$ is a pair consisting of a character of $\mathbb{R}_{+}$ and a character of $S^1$. This gives the representations

$$\boxed{ \rho_{z, k}(re^{i \theta}) = r^z e^{ik \theta}, z \in \mathbb{C}, k \in \mathbb{Z} }.$$

(So the classification you gave in this case is incorrect. It gives the holomorphic representations, which correspond to the case $z = k$ of the above classification, whereas here we consider all smooth representations.)

Next, we have $\mathbb{R}^{\times} \cong \mathbb{R}_{+} \times C_2$ (where the $C_2$ factor is given by the sign). The irreducible representations of $C_2$ are easy to classify by thinking about eigenvalues: there are two of them, the trivial representation and the sign representation, which is just the inclusion of $C_2 = \{ \pm 1 \}$ into $\mathbb{C}^{\times}$. This gives the irreducible representations

$$\boxed{ \rho_{z, k}(\pm r) = (\pm)^k r^z, z \in \mathbb{C}, k = 0, 1 }.$$

Here $r$ is positive and $\pm$ is the sign.

Finally $\mathbb{R}^{\times} \times S^1$ directly combines two cases we've done already.

  • $\begingroup$ thank you very much! very clear now $\endgroup$ Commented Jun 15 at 18:31
  • 1
    $\begingroup$ Yes, exactly, the mean value theorem gives uniqueness, once we have differentiability of merely-assumed-continuous characters. And, indeed, the latter is a special case of something equivalent to differentiation-under-an-integral (in a parameter). To my mind the latter simplification/reduction is the truly subtle part of this. $\endgroup$ Commented Jun 15 at 22:05

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .