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Are there any irreducible representations of $\operatorname{Gal}(\bar{\mathbb Q}/\mathbb Q)$ over $\mathbb Q_l$ that are unramified away from $l$ and crystalline at $l$ but are not known to arise from the etale cohomology of smooth and proper stacks over $\operatorname{Spec}\mathbb Z$?

Thee simplest examples of unramified crystalline Galois representations known, other than powers of the cyclotimc character, are the two-dimensional Galois representations arising from level $1$, weight $n+1$ modular forms. These live inside the etale cohomology of the stack $\overline{M}_{1,n}$, which is indeed smooth and proper. More examples come from other $\overline{M}_{g,n}$, but these are the only examples I am aware of!

The situation for schemes is very different where, as far as I can tell, no Galois representation other than powers of the cyclotomic character is known to arise from smooth and proper schemes over $\mathbb Z$.

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  • $\begingroup$ You probably want to restrict to irreducible representations, since otherwise there are simply too many Gal reps to come from any kind of geometry, just for cardinality reasons. $\endgroup$ Commented Feb 16, 2013 at 20:09
  • $\begingroup$ Alright, fixed. $\endgroup$
    – Will Sawin
    Commented Feb 16, 2013 at 20:26
  • $\begingroup$ When I think of Galois representations that are not known to be motivic, I think of those automorphic ones constructed as $p$-adic limits. The fact that you want "no" ramification is imposing a "high" weight condition on the automorphic representations involved. So, maybe one could find some automorphic representations for GL(n) (n>2). $\endgroup$
    – Rob Harron
    Commented Feb 17, 2013 at 0:13

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Here's a class of examples which isn't in your list as far as I know. Take an even unimodular lattice, e.g. the $E_8$ lattice. The corresponding orthogonal group is a reductive group over $\mathbf{Q}$ which is split at every finite place and compact at $\infty$ (see Gross, "Groups over $\mathbf{Z}$", Inventiones 124 (1996)). There will be lots of algebraic automorphic representations of this group $G$ if you take the weight large enough, and these will give you automorphic representations of $GL(n)$ which are self-dual and unramified at all finite places. It's known that these have Galois representations attached, which will be crystalline at $\ell$ and unramified everywhere else; but I don't think these are known to come from smooth proper stacks over $\mathbf{Z}$.

There is much more on level 1 automorphic representations in the recent preprint of Chenevier and Renard, http://www.math.polytechnique.fr/~chenevier/articles/dimform.pdf.

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  • $\begingroup$ Dear David, how do we construct those representations if not from the cohomology of some algebraic variety? By a p-adic limit argument? $\endgroup$
    – Joël
    Commented Feb 17, 2013 at 19:22
  • $\begingroup$ Sure, these representations are known to be motivic except in some small weight cases, but the original poster asked if the representations came from a very specific class of geometric objects and I don't think that's known here. $\endgroup$ Commented Feb 17, 2013 at 19:34
  • $\begingroup$ What sorts of motives does the construction produce? $\endgroup$
    – Will Sawin
    Commented Feb 17, 2013 at 20:49
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    $\begingroup$ So, David, are you saying that the situation is that we know the representations come from geometry and we know that they are good everywhere, but the reason we know the latter is not because we know the geometric objects are good everywhere? Rather we use some other technique to show this. (I haven't thought about this for a while, so really it's a bit unclear to me whether we know the representations themselves come from geometry, rather their restrictions to imag. quad. extensions are motivic. And the representations are patched together from these restrictions. ?) $\endgroup$
    – Rob Harron
    Commented Feb 17, 2013 at 20:58
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    $\begingroup$ (at the moment the existence of these galois representations is still conditional on the stabilisation of the twisted trace formula, which is in progress, because this is what Arthur needs to transfer to a general linear group.) $\endgroup$
    – fherzig
    Commented Feb 17, 2013 at 22:47

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