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This is the converse question to an earlier question. More precisely,

Let $X/K$ be a curve(or variety) over a global field $K$. We consider the Galois representation obtained by the absolute Galois group $G_K$ acting on $H_{et}^i(X_{/\bar K}, \mathbb{Q}_\ell)$.

Do properties of this representation, such as "unramified at a place $v$", semistable, de Rham, crystalline, Hodge-Tate, and so on and so forth, imply some geometric properties about $X_{/K}$? (I must confess that I know the proper definition of only the first property in this list, but I nevertheless put in the whole list from the original question for good measure.)

If so, please give examples.

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    $\begingroup$ The original question was asked by norondion. To him/her: I hope you will forgive me for asking similar question, since imitation is the sincerest form of flattery. $\endgroup$
    – Regenbogen
    Commented Mar 12, 2010 at 20:34

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The converse is false. See the lecture notes by Chandan Singh Dalawat at http://arxiv.org/abs/math/0605326, which give some examples of varieties over finite extensions of $\mathbb{Q}_p$ whose $\ell$-adic cohomology is unramified for $\ell \ne p$ and crystalline at $\ell = p$, but the variety does not have good reduction.

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    $\begingroup$ Neron-Ogg-Schaferevich says that for abelian varieties, the Galois representation on H^1 controls the reduction in the obvious way, but as David says it's too much to hope for in general. $\endgroup$
    – Kevin Buzzard
    Commented Mar 12, 2010 at 23:52
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    $\begingroup$ Some explicit examples of curves have been provided by Qing Liu (mathoverflow.net/questions/10463/…)) One can also write down Châtelet surfaces over $\mathbb{Q}_p$ which have bad reduction and yet the $p$-adic étale cohomology is unramified (mathoverflow.net/questions/416/existence-of-smooth-models/…) $\endgroup$
    – Chandan Singh Dalawat
    Commented Mar 13, 2010 at 3:24
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    $\begingroup$ If $E$ is an elliptic curve with good reduction over $\mathbb Q_p$, and one takes a non-trivial torsor $P$ over $E$, then $P$ will not admit a model with good reduction, yet $H^1(P_{\overline{\mathbb Q}_p})$ and $H^1(E_{\overline{\mathbb Q}_p}, \mathbb Q_{\ell})$ coincide, and hence $H^1(P_{\overline{\mathbb Q}_p},\mathbb Q_{\ell})$ will be unramified (if $\ell \neq p$) or crystalline (if $\ell = p$). (This example is noted in the reference provided by David, but seems fundamental enough to be worth spelling out here.) $\endgroup$
    – Emerton
    Commented Mar 13, 2010 at 4:51
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    $\begingroup$ I should also have noted that "the reference provided by David" is in fact written by Chandan Singh Dalawat! $\endgroup$
    – Emerton
    Commented Mar 13, 2010 at 6:39
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    $\begingroup$ The problem with curves is that $H^1$ represents only the abelian part of $\pi_1$, so you lose some information on the geometry of the curve. There exists however an anabelian analogue by considering the representation in Out$(\pi_1)_{\ell}$. Then one has a good reduction criterium à la Néron-Ogg-Shafarevich. See Theorem 3.2 in T. Oda: A Note on Ramification of the Galois Representation on the Fundamental Group of an Algebraic Curve, II, Journal of Number Theory Volume 53, Issue 2, August 1995, Pages 342-355. $\endgroup$
    – Qing Liu
    Commented Mar 16, 2010 at 9:23

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