Let$f\in End(X)$ for an abelian variety X and let $f'=\phi_H^{-1}\hat{f}\phi_H$ be the usual Rosati involution. An endomorphism of abelian varieties is called symmetric if $f=f'$.

Now I want to show that symmetric endomorphisms have real eigenvalues. This is supposed to be similar to the proof that symmetric matrices only posses real eigenvalues.

Any advice appreciated.

  • 1
    $\begingroup$ I assume you are working over $\mathbb{C}$, your $X$ is the quotient of a vector space $V$ by a lattice, you have a polarization on $X$ given by a positive hermitian form $H$ on $V$, and the eigenvalues you are considering are those of the endomorphism $\tilde{f} $ of $V$ induced by $f$. Then $f$ is symmetric if and only if $\tilde{f} $ is self-adjoint w.r.t. $H$, in which case it has real eigenvalues. $\endgroup$ – abx Aug 3 at 19:47
  • $\begingroup$ Yes your assumptions are correct. If we denote the analytic representation of $\tilde f$ by A and by using the fact that symmetric endomorphisms are self-adjoint w.r.t to H we get hat $A=A'=(H^t)^{-1}\bar A^t H^t$. Then we get for an eigenvalue $\lambda$ of A to the eigenvector x:$\lambda x^t x = x^t(\lambda x) = x^t(Ax) = (A^t x)^t x = (H\bar A H^{-1}x)^t x$ If I can now show that $\bar \lambda$ is an eigenvalue of $H \bar A H^{-1}$ we get $\lambda x^tx= (\bar \lambda x)^t x = \bar \lambda x^t x$ and therefore $\lambda = \bar \lambda$. Is there any theorem about eigenvalues of conjugations? $\endgroup$ – Christopher Bunse Aug 4 at 8:33

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