Comparing cohomology over ${\mathbb C}$ and over ${\mathbb F}_q$ - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T17:11:32Z http://mathoverflow.net/feeds/question/40486 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/40486/comparing-cohomology-over-mathbb-c-and-over-mathbb-f-q Comparing cohomology over ${\mathbb C}$ and over ${\mathbb F}_q$ Alexander Braverman 2010-09-29T15:50:56Z 2010-10-18T02:57:30Z <p>I have the following (probably well-known) question: let $X$ be a regular scheme over $\mathbb Z$. Let $p$ be a prime and Let us denote the reduction of $X$ mod $p$ by $X_p$. Let also $X_{\mathbb C}$ be the corresponding complex variety. Suppose that I know everything about cohomology of $X_{\mathbb C}$ (i.e. I know the mixed Hodge structure). Is there a situation where I can use that to describe the $\ell$-adic cohomology of $X_p$ with the action of Frobenius? More precisely, are there conditions that would guarantee that all the eigen-values of Frobenius are powers of $q$ and that those powers are exactly those predicted by the Hodge structure on the cohomology of $X_{\mathbb C}$?</p> <p>UPDATE: I would like to emphasize that the scheme $X$ I am talking about is not proper and its cohomology is not pure. However, it is not very bad either: roughly speaking nothing worse than ${\mathbb G}_m$ can appear there. In my specific situation conjecturally (I don't know how to prove this) there is a ${\mathbb G}_m$-action on $X$ which contracts it to a subscheme $Y$ (i.e. there exists a morphism ${\mathbb A}^1\times X\to X$ extending the ${\mathbb G}_m$-action such that ${ 0 } \times X$ goes to $Y$) where $Y$ is a disjoint union of locally closed subschemes of the form ${\mathbb A}^k\times {\mathbb G}_m^l$. Note that in this case the cohomology of $X$ is the same as cohomology of $Y$, so this should give you an idea how complicated the cohomology of $X$ is.</p> http://mathoverflow.net/questions/40486/comparing-cohomology-over-mathbb-c-and-over-mathbb-f-q/40490#40490 Answer by Bhargav for Comparing cohomology over ${\mathbb C}$ and over ${\mathbb F}_q$ Bhargav 2010-09-29T16:15:32Z 2010-10-18T02:57:30Z <p>Assume $f:X \to \mathrm{Spec}(\mathbf{Z})$ is proper with $X_{\mathbf{Q}}$ smooth and geometrically connected, and fix an integer $n$. Then I think the following is true:</p> <p><strong>Claim 1</strong>: $H^{2n}(X(\mathbf{C}),\mathbf{C}) \simeq H^n(X_{\mathbf{C}},\Omega^n)$ if and only if there exists a large number of primes $p$ such that the eigenvalues of Frobenius at $p$ acting on the $2n$-th $\ell$-adic cohomology group $V$ of $X_{\overline{\mathbf{Q}}}$ are all $p^n$ </p> <p>Here "large" means that these are the primes that split in some number field, up to finite nuisance. We refer to Galois representations with the preceding property as being "mixed Tate", in analogy with the situation in Hodge theory.</p> <p>(Edit: I added a proof <em>sketch</em> and references; the proof was discovered jointly with Andrew Snowden when we were unaware of the literature, and we do not claim any originality.)</p> <p><em>Sketch of proof of Claim 1</em>: If one has the condition that the Galois representation $V$ is potentially Tate, then, as Emerton points out, one can use $p$-adic Hodge theory to show that all (2n)-classes have type (n,n). For the converse direction, the easiest way I know to see this is by using Mazur's theorem that the Newton polygon lies or or above the Hodge polygon for a smooth projective scheme over $\mathbf{Z}_p$. This theorem implies that the traces of the Frobenius eigenvalues on $V$ are divisible by $p^n$ for almost all primes $p$. Hence, the traces on $V(n)$ are integral (doing a Tate twist amounts to dividing the eigenvalues by p). On the other hand, by the Weil conjectures, these traces have bounded absolute values (for weight reasons). Hence, the trace function associated to $V(n)$ takes on finitely many values at all Frobenius elements. By Chebotarev, it follows that the trace function associated to $V(n)$ is finitely valued. Since it is also continuous, one can find an extension $K/Q$ such that the Galois representation $V(n)$ restricted to $K$ has constant trace function. This implies that the semisimplifed Galois representation underlying $V(n)$ is constant over $K$. Hence, for any prime $p$ that splits completely in $K$, the eigenvalues of the Frobenius at $p$ on $V \simeq V(n)(-n)$ are given by $p^n$, as desired.</p> <p>--</p> <p>For a more general statement, please see Kisin-Wortmann (see Emerton's answer) who prove a much better version of the statement above. Also, Bloch-Esnault (http://arxiv.org/abs/math/0212256) prove related results.</p> <p>--</p> <p>(Second edit 10/17/10): It seems to us that the above proof works in the mixed case as well, to show the following (is this implicit in Kisin-Wortmann?):</p> <p><strong>Claim 2</strong>: If $X/\mathbf{Q}$ is a variety such that $H^n_c(X_\mathbf{C})$ has a Hodge structure of mixed Tate type for some n, then $V_\ell = H^n_c(X_{\overline{\mathbf{Q}}},\mathbf{Q}_\ell)$ is a mixed Tate Galois representation of the correct weight (where "correct" means the weight predicted by the Hodge structure). </p> <p>In concrete terms, Claim 2 predicts the existence of a number field K such that for any prime $p$ splitting completely in $K$, the $Frob_p$-eigenvalues on $V_\ell$ are of the form $p^{w/2}$ where $w$ is a non-zero weight occuring in $H^n_c(X)$. Dualising gives the desired statement for normal cohomology.</p> <p>Recall that a $\mathbf{Q}$-mixed Hodge structure is said to have mixed Tate type if its weight graded pieces are concentrated in even degrees, and in each even degree they are spanned by copies of the Tate Hodge structure $\mathbf{Q}(n)$ where $-2n$ is the weight. In particular, if $Y$ is smooth projective, then $H^{2m}(Y)$ is mixed Tate if and only if all classes in $H^{2m}(Y)$ have Hodge-type $(m,m)$. Hence, Claim 2 recovers Claim 1. The standard examples of varieties carrying a mixed Tate Hodge structure are those whose cohomology is generated by algebraic cycles, eg, G/B for a reductive group G with Borel B; one can get more/singular examples by taking quotients by finite/reductive group actions.</p> <p><em>Sketch of proof of Claim 2</em>: The main idea is to use the Hodge and weight filtrations on $V$, and their compatibility with the $p$-adic comparison isomorphisms. More specifically, let $W_\cdot(V)$ denote the weight filtration on $V$. By the Weil conjectures, the weights are all $\leq n$. The assumption that $X$ has mixed Tate type over $\mathbf{C}$ gives that $gr^W_k(V) = 0$ for $k$ odd, and that $H_{2k} = gr^W_{2k}(V)$ is pure of weight $2k$ with the Hodge structure entirely of type $(k,k)$.</p> <p>Now fix a prime p such that all the $p$-adic Galois representations in sight are crystalline; most primes will work. Applying Fontaine's functor to $V_p$ gives a weakly admissible module $D(V_p)$, i.e, a filtered $\mathbf{Q}_p$-vector space with a Frobenius action satisfying the condition that the Newton polygon (defined using the $p$-adic valuations of the eigenvalues of Frobenius) lies above Hodge polygon (defined using the dimensions of the filtered pieces). The comparison theorems identify the filtered module underlying $D(V_p)$ with the Hodge filtered de Rham cohomology of $X$ (base changed to $\mathbf{Q}_p$); the Frobenius action comes from crystalline cohomology.</p> <p>Since the comparison isomorphisms respect the weight filtration, the filtered module underlying the weakly admissible module $D(H_{2k})$ is given by the $2k$-th weight graded piece of the de Rham cohomology of $X$. Thus, $D(H_{2k})$ has a Hodge filtration that is entirely of type $(k,k)$. It follows from weak admissibility that the Frobenius eigenvalues on $D(H_{2k})$ have $p$-adic valuation at least $k$, and so the same is true for $H_{2k}$. The same argument as in the pure case (do a Tate twist, argue some character is finitely valued using purity, etc) now finishes the proof.</p> http://mathoverflow.net/questions/40486/comparing-cohomology-over-mathbb-c-and-over-mathbb-f-q/40531#40531 Answer by Emerton for Comparing cohomology over ${\mathbb C}$ and over ${\mathbb F}_q$ Emerton 2010-09-29T20:17:13Z 2010-09-29T20:17:13Z <p>You have to be a little careful because of the following sort of phenomenon: the pair of conjugate points $x^2 + 1 = 0$ has Frobenius eigenvalues equal to $1$ if $p$ is 1 mod 4, but equal to $1$ and $-1$ if $p$ is $-1$ mod 4. But its Hodge structure is no different to the pair of rational points $x^2 - 1$, for which the Frobenius eigenvalues are always 1. (If you want an example where the scheme is a variety, rather than geometrically disconnected, instead consider the plane blown up at these two points.)</p> <p>The condition you want is that the cohomology over $\mathbb C$ be generated by Hodge classes (i.e. that the cohomology be all in even degree, and that in degree $2i$ it all be of type $(i,i)$). This will give that the Frobenius eigenvalues are powers of $q$ for almost all $p$, up to multiplication by a root of unity. (This is the Artin--Tate statement of Bhargav's answer.) The above example shows that these roots of unity can appear.</p> <p>A paper which discusses this kind of question pretty comprehensively, and where the preceding statement is proved, is the linked <a href="http://www.math.harvard.edu/~kisin/dvifiles/hodge.dvi" rel="nofollow">paper of Kisin and Lehrer</a>. (See Corollary 2.3.) See also <a href="http://www.math.harvard.edu~kisin/dvifiles/artin.dvi" rel="nofollow">this paper of Kisin and Wortmann</a>, which provides key results for the argument of Kisin and Lehrer. </p> <p>The basic technique that you can use to connect the Hodge theory at infinity to the Frobenius elements mod primes is $p$-adic Hodge theory, and this is what Kisin--Lehrer and Kisin--Wortmann use.</p>