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(This might be an easy quesiton but I'm new to representations and could use a helpful pointer or two)

Let $G_{\mathbb{Q}}= \text{Gal}(\overline{\mathbb{Q}} / \mathbb{Q})$ be the absolute Galois group of the rationals and let $K$ be a totally real field. Then the embeddings of $K$ induce a map from $\text{Gal}(\mathbb{C}/\mathbb{R})$ to $\text{Gal}(\overline{K} / K)$. Call $C$ a complex conjugation if it is an image of one of these maps of the generator of $\text{Gal}(\mathbb{C}/\mathbb{R})$ (ie an image point of one of these induced maps of the complex conjugation map).

Next, let $\rho: G_{\mathbb{Q}} \to \text{GL}_{2}(\mathbb{F})$ be a representation over a finite field $\mathbb{F}$ with characteristic $p$. We say that such a representation is odd if there is a complex conjugation $c \in G_{\mathbb{Q}}$ such that $\det(\rho(c))=-1$.

My question is the following. Suppose that a representation is odd. Why must $\det(\rho(c))=-1$ be true for $\textit{all}$ complex conjugations?

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1 Answer 1

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All complex conjugations are conjugate in $G_{\mathbb{Q}}$ (because $\mathbb{Q}$ itself has only one infinite place). So if $\sigma_1$ and $\sigma_2$ are any two complex conjugations, $\rho(\sigma_1)$ and $\rho(\sigma_2)$ are conjugate matrices, so they have the same eigenvalues.

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