In reading the Harris-Taylor book, I encounter expressions like "Let $\xi$ be an algebraic representation of $G$ over $\mathbb{C}$". What does this mean? Here $G$ is a reductive group over $\mathbb{Q}$, namely a group scheme over $\mathbb{Q}$. It seems to me that what they really mean is that $\xi$ is a representation of the real points $G(\mathbb{R})$. Am I understanding correct? Also what does "algebraic" mean?
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3$\begingroup$ Without reading the text you mention, I would guess it's homomorphisms of algebraic groups over $\mathbb{C}$ from the basechange $G_{\mathbb{C}}:=G \times_{\mathrm{Spec}(\mathbb{Q})} \mathrm{Spec}( \mathbb{C})$ to $\mathrm{GL}_{n,\mathbb{C}}$. $\endgroup$– QfwfqDec 23, 2015 at 13:38
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$\begingroup$ Usually this is used in the context of an automorphic representation $\pi$ being "algebraic at infinity", which means that $\pi_\infty$, the product of the factors at the infinite places, is a tensor product of algebraic representations. Since at the infinite places $v|\infty$, $G(Q_v)=G(R)$ or $G(C)$, this has the usual meaning as a morphism of algebraic groups. $\endgroup$– PL.Dec 24, 2015 at 16:07
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