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Jan 27, 2022 at 7:41 comment added Alphonse @GeorgLehner : thanks, I forgot that irreps of an abelian group are 1-dimensional... (it should also be true for infinite abelian groups, see math.stackexchange.com/questions/354926, en.wikipedia.org/wiki/…)
Jan 27, 2022 at 7:35 vote accept Alphonse
Jan 26, 2022 at 23:31 comment added Georg Lehner So knowing the $a_i$ which appear in the characteristic polynomial $\rho_V(\phi)$ completely determines the decomposition into irreducibles. The formula for the dimension of $Hom_G$ now follows from Schurs Lemma. (And I hope the same reasoning goes through for the continuous case as well)
Jan 26, 2022 at 23:30 comment added Georg Lehner If $V$ decomposes as a sum of irreducibles $V_i$, then the characteristic polynomial of $\rho_V(\phi)$ is the product of the characteristic polynomials $\rho_{V_i}(\phi)$. Since $G$ is abelian, all irreducible representations are 1-dimensional, therefore each irreducible $V_i$ gives you one eigenvalue $a_i$. Conversely, since $G$ is cyclic, the character of $V_i$ is completely defined by $a_i$ (the values are powers of it) hence $V_i$ is determined as well.
Jan 26, 2022 at 15:30 comment added Alphonse Thank you very much. May I ask how exactly the eigenvalues 𝑎𝑖 give a decomposition of $V$ into irreps? The $a_i$'s might not be all distinct a priori. (Sorry if this is a silly question, I am not very familiar with representation theory...). This has probably to do with the group being (pro)-cyclic?
Jan 26, 2022 at 8:41 comment added Johannes Hahn You are right. Then, we're in the first case again.
Jan 25, 2022 at 19:26 comment added Benjamin Steinberg Doesn't the question demand V,W be semisimple?
Jan 25, 2022 at 17:09 history edited Johannes Hahn CC BY-SA 4.0
Fixed typo
Jan 25, 2022 at 11:51 history answered Johannes Hahn CC BY-SA 4.0