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Let $K$ be a number field. We say that an $\ell$-adic Galois representation is of pure weight $n$ if for almost all primes $\mathfrak{p}$ the eigenvalues of $\text{Frob}_{\mathfrak{p}}$ are of absolute value $p^n$. A good source of these is the cohomology of smooth proper $K$-schemes, where $H^n(X_{\bar{K}},\mathbb{Q}_\ell)$ is of weight $n$. My question is how does the weight behave with respect to the dual? Since $\mathbb{Q}_\ell(1):=H^2(\mathbb{P}^1)^*$ is of weight $2$, you would expect that the weight of the dual $\rho^*$ is minus the weight of $\rho$. However I don't know how to take an eigenvector of $\rho$ and get an eigenvector of $\rho^*$....

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  • $\begingroup$ If you have a basis of eigenvectors, then you can consider the dual basis. In general, an endomorphism has the same characteristic polynomial (hence the same eigenvalues) as its dual, because that’s just transposing the matrix. $\endgroup$
    – Aphelli
    May 20, 2021 at 8:59
  • $\begingroup$ @Mindlack if you copy your coment into an answer, I'd be happy to accept it... $\endgroup$ May 20, 2021 at 9:03

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In any finite-dimensional $K$-vector space $V$, an endomorphism $u$ has the same characteristic polynomial as its dual. It’s easy to see that, for instance, by considering the matrix of $u$ in a certain basis $\mathcal{B}$, and the matrix of the dual of $u$ In the basis dual to $\mathcal{B}$ – and see that they are transposes one of the other.

However, as David Loeffler points out, the dual representation is considered by taking the inverse of the dual of the group action (indeed, $u \in End(V) \longmapsto u^* \in End(V^*)$ is contravariant, so to restore the correct fonctoriality we need to take the inverse as well).

In particular, this means that the dual of a representation of weight $n$ is of weight $-n$.

However, this doesn’t mean that it’s easy to construct an eigenvector of the dual representation directly from an eigenvector of the original representation. Instead, we need to use a full diagonalization (or at least a trigonalization) of the Frobenius, and then take the dual basis to really “see” the weights acting on the dual representation.

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    $\begingroup$ There's a sign missing here, since the matrix of Frobenius on $\rho^*$ (wrt the dual basis) is not the transpose of the matrix on $\rho$, but the inverse transpose. So if $\rho$ has weight $n$ then $\rho^*$ has weight $-n$, as in the question, not $n$. [PS: edited for typo] $\endgroup$ May 20, 2021 at 17:10
  • $\begingroup$ Oops. Excellent point, thank you very much. $\endgroup$
    – Aphelli
    May 20, 2021 at 17:39

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