Claim 1: $H^{2n}(X(\mathbf{C}),\mathbf{C}) \simeq H^n(X_{\mathbf{C}},\Omega^n)$ 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 $X$ over $\overline{\mathbf{Q}}$ (or over $\overline{\mathbf{F}_p}$, it's the same for most $p$) X_{\overline{\mathbf{Q}}}$are all$p^n$(call such representations of the Galois group potentially Tate). Here large "large" means that these are the primes that split in some number field, up to finite nuisance. We refer to Galois representations with the preceding property as being "mixed Tate", in analogy with the situation in Hodge theory. Sketch of proof of Claim 1: If one has the condition that the Galois representation$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$.p^n$, as desired.

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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.

Claim 2: 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 there $V_\ell = H^n_c(X_{\overline{\mathbf{Q}}},\mathbf{Q}_\ell)$ is a mixed Tate Galois representation of the correct weight (where "correct" means the weight predicted by the Hodge structure).

In concrete terms, Claim 2 predicts the existence of a number field $K$ K such that for any prime $p$ splitting completely in $K$, the eigenvalues of $\mathrm{Frob_p}$ Frob_p$-eigenvalues on$V_\ell = H^n_c(X_\mathbf{Q},\mathbf{Q}_\ell)$V_\ell$ are given by corresponding powers of the form $p$, p^{w/2}$where "corresponding" means determined by$w$is a non-zero weight occuring in$H^n_c(X)$. Dualising gives the desired statement for normal cohomology. Recall that a$\mathbf{Q}$-mixed Hodge numbers structure is said to have mixed Tate type if its weight graded pieces are concentrated in even degrees, and in each even degree they are spanned by copies of the Tate Hodge structure$H^n_c(X_\mathbf{C})$. Here's \mathbf{Q}(n)$ 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 sketch 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.

Sketch of an argumentproof of Claim 2: 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 $(k,k)$.

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/theory 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.

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(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?):

Claim: 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 there is a number field $K$ such that for any prime $p$ splitting completely in $K$, the eigenvalues of $\mathrm{Frob_p}$ on $V_\ell = H^n_c(X_\mathbf{Q},\mathbf{Q}_\ell)$ are given by corresponding powers of $p$, where "corresponding" means determined by the Hodge numbers of $H^n_c(X_\mathbf{C})$.

Here's a sketch of an argument:

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 $(k,k)$.

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/theory 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.

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.

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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:

$H^{2n}(X(\mathbf{C}),\mathbf{C}) \simeq H^n(X_{\mathbf{C}},\Omega^n)$ 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 $X$ over $\overline{\mathbf{Q}}$ (or over $\overline{\mathbf{F}_p}$, it's the same for most $p$) are all $p^n$ (call such representations of the Galois group potentially Tate). Here large means that these are the primes that split in some number field, up to finite nuisance.

(Edit: I added a proof sketch and references.references; the proof was discovered jointly with Andrew Snowden when we were unaware of the literature, and we do not claim any originality.)

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 $p^n$.

For 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.

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