Let $K$ be a number field, $\rho\colon \mathrm{Gal}\_K\to \mathrm{GL}\_n(\overline{\mathbf{Q}\_p})$ a geometric (i.e.: unramified a.e., de Rham above $p$) irreducible Galois representation. One piece of the global Langlands conjectures is that there exists a cuspidal automorphic representation $\pi$ of $\mathrm{GL}\_n(\mathbf{A}\_K)$ such that (up to a twist) the Satake parameters of $\pi$ correspond with the Frobenius eigenvalues of $\rho$ at the unramified places. 


Even better, for each $v\nmid p$, we can associate to the local Galois representation $\rho\_v\colon \mathrm{Gal}\_{K\_v}\to \mathrm{GL}\_n(\overline{\mathbf{Q}\_p})$ a representation $\pi\_v$ of $\mathrm{GL}\_n(K\_v)$ via the local Langlands correspondence. The restricted tensor product $\pi^p=\bigotimes\_{v\nmid p}' \pi\_v$ is a representation of $\mathrm{GL}\_n(\mathbf{A}\_{K,f}^p)$ (finite adeles away from $p$), and by multiplicity one theorems, is the away-from-$p$ part of at most one automorphic representation. Call it $\pi$, if it exists. 



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> Question: how does one "read off" the expected $\pi\_\infty$ from $\rho$? In other words, how does one see the infinity-type of the expected automorphic representation on the Galois side? 
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Surely $\rho\_\infty\colon \mathrm{Gal}\_{K\_\infty}\to \mathrm{GL}\_n(\overline{\mathbf{Q}\_p})$ does not "know" $\pi\_\infty$. For example, when $K=\mathbf{Q}$ and $n=2$, $\rho\_\infty$ only knows the "parity" of $\pi\_\infty$. 



The infinity-type of automorphic representations in the Langlands correspondence
Given $\pi$, Langlands' philosophy in its original form would predict the existence of a representation $\sigma$ from the (conjectural) global Langlands group of $\mathbf{Q}$ to $GL(n,\mathbf{C})$. For this representation there is "perfect local-global" -- the local behaviour of $\pi$ at a place $v$ (finite or infinite) should be related to the local behaviour of $\sigma$ at $v$ via the local Langlands correspondence.


As you've spotted though, things aren't quite the same with $\rho$ and $\pi$ because the symmetry is broken. $\pi$ is a representation over the complexes, and $\rho$ is a representation over the $p$-adics. So the first thing you have to do is to choose embeddings of $\overline{\mathbf{Q}}$ into $\mathbf{C}$ and $\overline{\mathbf{Q}}\_p$. If you were working with a general number field you would now have set up some sort of weird correspondence between the places of the field above $p$ and the places above infinity, and there is some interplay in the local-global story with these places. 


You have base field $\mathbf{Q}$ so the interplay is between $p$ and $\infty$. Basically, the behaviour of $\pi$ at both $p$ and $\infty$ should be determined by the behaviour of $\rho$ at both $p$ and $\infty$, but as you have spotted it is *not* the case that the behaviour of one of these things at $p$ is determined by the other one, and the same for infinity; you have to treat both places at once.


The answer to your question is that the Langlands parameter for $\pi\_\infty$ can be read off from the Hodge--Tate weights of $\rho$ plus the behaviour of complex conjugation. The infinitesimal character of $\pi\_\infty$ is the Hodge--Tate cocharacter of $\rho$, for example. 


All the details can be found in


<http://arxiv.org/abs/1009.0785>


