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So let $R$ be a discrete valuation ring and let $X$ be a scheme which is proper and flat over $R$. Let $X_s$ denote the special fiber of $X$.

So intuitively, when somebody says that a curve $X$ is semistable I kind of equate this in my mind with the property that $X_s$ has only ordinary double points as singularities.

Q1: So in general (i.e. in higher dimension) what is the geometrical meaning for a scheme to be semistable?

On the Galois representation side we have a very precise definition of what semistable means using Fontaine's ring $\mathbf{B}_{st}$.

Q2 If there is a precise answer to Q1, is there a good reference (more on the intuitive side than on the technical side) where one shows the equivalence (under suitable assumptions) that the geometrical definition coincides with the galois representation one?

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In your setting $X$ is semi-stable means that its special fiber $X_s$ is a reduced divisor with normal crossings on $X$.

The link with Galois representations is very deep. In fact in general only one implication is known, namely if $X$ is semi-stable then its associated Galois representation is semi-stable. This was known as (a consequence of) the conjecture $C_{\mathrm{st}}$ of Fontaine and Jannsen. There are now at least three different proofs of this conjecture. One was given by the Japanese school (Hyodo, Kato, Tsuji), see Tsuji's survey in Astérisque 279. Another was given by Faltings using his theory of almost étale extensions. Recently Niziol gave another proof using $K$-theory.

The converse implication seems very difficult in general. For abelian varieties this was proved by Coleman-Iovita (Duke Math. 1999) and Breuil (Annals of Math. 2000). For curves this follows from Deligne-Mumford's theorem that a curve is semi-stable if and only if its Jacobian is semi-stable (Publ. Math. IHÉS 1969) (see Mathieu Romagny's article).

Faltings also has a result that the Galois representations associated to proper schemes over $K = \operatorname{Frac}(R)$ are de Rham (and thus potentially semi-stable). So if we knew the converse implication in general, then we would deduce that every scheme is potentially semi-stable (in the sense that it acquires semi-stable reduction after a finite extension), but this is not known in general.

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  • $\begingroup$ Thanks Francois. So is there some kind of relationship between Hironaka's desingularisation over a field $K=Frac(R)$ of characteristic 0 (the existence of a smooth compactification $X'$ of $X$ which is isomorphic to $X$ away from a normal crossing divisor) and the conjecture which predicts that every scheme of $R$ is potentially semi-stable. $\endgroup$
    – Hugo Chapdelaine
    Commented Nov 21, 2011 at 18:29
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    $\begingroup$ A few days ago, Alexander Beilinson, On the crystalline period map (arxiv.org/abs/1111.3316) gave "a new proof of the Fontaine-Jannsen conjecture based on a crystalline version of the $p$-adic Poincar'e lemma (different proofs were found earlier by Faltings, Niziol and Tsuji)". $\endgroup$
    – Chandan Singh Dalawat
    Commented Nov 22, 2011 at 3:07
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    $\begingroup$ @Ariyan : I forgot to mention that in the definition of semi-stability, one also asks the scheme $X$ to be regular, so the generic fiber $X_K$ is normal. It is also safe to put some assumption on the residue field of $R$ (perfect of char. $p>0$). @Hugo : I don't know a precise relationship, but finding a good model over $R$ of a $d$-dim variety over $K$ can be seen as desingularizing a $(d+1)$-dim variety over $R$ (which is harder than over $K$). @Chandan : Thanks for the reference. I guess I should have said "there are now at least four proofs"... $\endgroup$
    – François Brunault
    Commented Nov 22, 2011 at 10:06
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Just to elaborate on Francois Brunault's answer: it is not true in general that having semistable etale cohomology implies having semistable reduction (just as it is not true that having crystalline etale cohomology implies having good reduction). So the implication only goes in one direction.

Added in response to comment below: E.g. if $E$ is an elliptic curve over $\mathbb Q_p$ with good reduction, and $P$ is an $E$-torsor with no $\mathbb Q_p$-rational point, then $P$ will not have good reduction. It will not have semi-stable reduction either, since it has potentially good reduction (it obtains a rational point over some extension of $\mathbb Q_p$, and hence becomes isomorphic to $E$ over that same extension). The etale cohomology ($\ell$-adic or $p$-adic) of $P$ coincides with that of $E$, and so $P$ is an example of a variety over $\mathbb Q_p$ with semistable (indeed crystalline) Galois action on its $p$-adic etale cohomology, which does not have semistable reduction.

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  • $\begingroup$ Hi Mat, could you elaborate a little bit on the sentence "it is not true in general that having semistable etale cohomology implies having semistable reduction". I'd like to understand your comment. And what is the meaning of your comment on the geometrical side? $\endgroup$
    – Hugo Chapdelaine
    Commented Nov 21, 2011 at 18:44
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    $\begingroup$ Dear Hugo, On the geometrical side, it simply means that Galois action on cohomology can't capture everything about singularities. As for examples, see this answer and the comments to it: mathoverflow.net/questions/18006/… (Note that it deals with the case when the etale cohomology is crystalline but the variety has bad reduction, but similar examples will exist with crystalline replaced by semistable and bad reduction replaced by non-semistable reduction.) Regards, Matt $\endgroup$
    – Emerton
    Commented Nov 21, 2011 at 20:39
  • $\begingroup$ Thanks a lot Mat, this is a very nice example. With this example in mind it also gives the motivation why one should base change to a larger field in order to acquire semi-stable reduction. $\endgroup$
    – Hugo Chapdelaine
    Commented Nov 21, 2011 at 23:58
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    $\begingroup$ For an example of a surface over ${\mathbf Q}_p$ whose $l$-adic cohomology is unramified (including at $l=p$) but which still fails to have good reduction, see mathoverflow.net/questions/416/existence-of-smooth-models/…. $\endgroup$
    – Chandan Singh Dalawat
    Commented Nov 22, 2011 at 4:06

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