1
$\begingroup$

$\DeclareMathOperator\Hom{Hom}$Let $G$ be a (pro-)cyclic group (topologically) generated by $\phi$, and let $V,W$ be two finite-dimensional (continuous) $k$-linear semisimple representations of $G$, where $k$ is some algebraically closed (complete) field.

Let $$P_V(t) = \prod_{i=1}^n (t - a_i), P_W(t) = \prod_{j=1}^m (t - b_j)$$ be the characteristic polynomials of $\rho_V(\phi), \rho_W(\phi)$ respectively.

I would like to know if $$\dim_k \Hom_G(V, W) = \# \{ (i, j) : a_i = b_j \}$$

I know that $\dim_k \Hom_G(V, W) = \langle \chi_V, \chi_W \rangle$, and that $f \in \Hom_G(V, W)$ iff $f \circ \rho_V(\phi) = \rho_W(\phi) \circ f$. But I don't know how to proceed from there.

(NB: already asked here two weeks ago, with no success).

Improve this question
$\endgroup$
1
  • $\begingroup$ This is (16.22) Lemma (ii) in Geer--Moonen--Edixhoven's book project on Abelian Varieties. $\endgroup$
    – Watson
    Commented Apr 17, 2022 at 8:06

1 Answer 1

Reset to default
7
$\begingroup$

If $k$ is of characteristic zero or characteristic $p$ and $G$ is a pro-$p'$-group, then yes. In that case, everything in sight is semisimple by Maschke's theorem, the eigenvalues $a_i$ and $b_j$ give you the decomposition into irreducibles, and every pair of equal eigenvalues contributes one degree of freedom by Schur's lemma.

Otherwise no, even in the simplest case, because of lack of semiplicity. If $k=\overline{\mathbb{F}_p}$, $G=C_p$, $\rho_V(\phi) = \begin{pmatrix}1&1\\&1\end{pmatrix}$ and $\rho_W(\phi)=\begin{pmatrix}1&\\&1\end{pmatrix}$, then $a_1=a_2=b_1=b_2$ so that your conjecture would predict $\dim_k \operatorname{Hom}_G(V,W) = 4$, but it is only 2, because $V$ is not semi simple, but $W$ is, and therefore every morphism must factor through $V/\operatorname{rad}(V)$.

Improve this answer
$\endgroup$
6
  • $\begingroup$ Doesn't the question demand V,W be semisimple? $\endgroup$
    – Benjamin Steinberg
    Commented Jan 25, 2022 at 19:26
  • $\begingroup$ You are right. Then, we're in the first case again. $\endgroup$
    – Johannes Hahn
    Commented Jan 26, 2022 at 8:41
  • $\begingroup$ 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? $\endgroup$
    – Alphonse
    Commented Jan 26, 2022 at 15:30
  • 1
    $\begingroup$ 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. $\endgroup$
    – Georg Lehner
    Commented Jan 26, 2022 at 23:30
  • 1
    $\begingroup$ 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) $\endgroup$
    – Georg Lehner
    Commented Jan 26, 2022 at 23:31

You must log in to answer this question.

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