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Take the Klein four group $V_4 = \mathbb{Z}_2 \times \mathbb{Z}_2$: $$ \{e,a,b,ab\}, \quad \mathrm{with} \quad a^2=b^2=(ab)^2 = e $$ Over an algebraically closed field of characteristic 0, being Abelian it has only one dimensional irreducible representations. In fact there are exactly four up to isomorphism: $$ \begin{split} \rho_1(\{e,a,b,ab\})&=\{1,1,1,1\}\\ \rho_A(\{e,a,b,ab\})&=\{1,1,-1,-1\}\\ \rho_B(\{e,a,b,ab\})&=\{1,-1,1,-1\}\\ \rho_{AB}(\{e,a,b,ab\})&=\{1,-1,-1,1\} \end{split} $$ I have very suggestively named these representations, due to the following observation: under tensor products of these representations, the representations themselves behave under the action of the Klein four group: $$ \rho_1 \otimes \rho_A = \rho_{A}, \quad\rho_A \otimes \rho_B = \rho_{AB}, \quad \rho_{AB} \otimes \rho_A = \rho_{B}, \quad \mathrm{etc.} $$ Is there a name for this phenomenon? Is it true for all representations of Abelian groups (even Lie groups) over algebraically closed fields of characteristic zero?

Craig
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The representation theory of a finite group over an algebraically closed field of characteristic zero doesn't depend on which of those fields we take, so we might as well work over $\Bbb C$.

If $G$ is finite abelian, then all irreps of $G$ are one-dimensional and may hence be identified with group homomorphisms $G \to \mathrm{GL}_1(\Bbb C)=\Bbb C^\times$. The image is always contained in $S^1$, so we might as well consider homomorphisms $G \to S^1$. Under this identification, tensor products correspond to pointwise multiplication of functions. The group of all such functions (or all irreps) is called the Pontryagin dual group of $G$. For any (locally compact Hausdorff, e.g. discrete) abelian group $H$, the Pontryagin dual group $\widehat{H}$ is defined as $\mathrm{Hom}(H,S^1)$.

Now it is a well-known fact that the Pontryagin dual of a finite abelian group is isomorphic to the group itself, but non-canonically.

So to answer your actual questions: Yes, this is true for all finite Abelian groups over an algebraically closed field of characteristic zero. The example $G=\Bbb Z$ shows that it doesn't need to hold for infinite groups. (There are uncountably many non-isomorphic irreps of $\Bbb Z$ over $\Bbb C$) The name for this phenomenon I'd suggest is just that Finite Abelian groups are self-dual, where we can interpret self-dual either via the classical Pontryagin dual or via the "representation theoretic dual", which would be one-dimensional representations under the tensor product (these two duals are isomorphic for finite Abelian gruops, as sketched above).

Lukas Heger
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    Great answer! By this same token may we argue the same statement holds for irreducible complex representations of $U(1) \cong S^1$? – Craig Oct 02 '24 at 17:06
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    @Craig there are only countably many irreducible representations of $S^1$, so $S^1$ can't be self-dual. For any integer $n$, there's the representation $z \mapsto z^n$, corresponding to the fact that the dual group of $S^1$ is $\Bbb Z$ – Lukas Heger Oct 02 '24 at 17:10
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    Using continuous homs to $S^1$ to define the dual of a locally compact abelian group, the dual of a compact abelian group is discrete and the dual of a discrete abelian group is compact. Thus if a compact or discrete abelian group $G$ is self-dual then $G$ is compact and discrete. A topological space that is compact and discrete is finite, so $G$ is a finite abelian group. Thus an infinite self-dual loc. compact abelian group is neither compact nor discrete, such as $\mathbf R^n$ where $n \geq 1$. – KCd Oct 02 '24 at 17:37
  • I see now, thank you! – Craig Oct 02 '24 at 18:12