Given a complex projective variety $X$, let's define singular cohomology of $K(X)$ (its function field) as the direct limit of cohomology groups over all of its Zariski open subsets. Similarly let's define the singular cohomology of algebraic closure $\overline{K(X)}$ by taking the direct limit of singular cohomology of all finite field extensions of $K(X)$. Now looking at $H\_i(\overline{K(X)}, \mathbb{Q})$ is there a way to tell which classes come from smooth projective varieties? Is it possible to give an example of a cohomology class that doesn't come from a smooth projective variety?


$\textbf{Tl;dr}:$ For $X$ fixed and smooth complex projective varieties $Y$ such that $\overline{K(X)}\cong \overline{K(Y)}$ what is the image of the following map? Is it surjective?


$$\bigoplus\_{Ys \text{ with the property explained above}}H^i(Y, \mathbb{Q})\rightarrow H^i(\overline{K(X)}, \mathbb{Q})$$



Which cohomology classes come from smooth projective varieties?
Donu Arapura's comment contains the right idea: this fails for Hodge-theoretic reasons. (I will also give a more elementary argument below.)


Indeed, each $H^i(U,\mathbf Q)$ carries a mixed Hodge structure [PS08, Thm. 4.2] that is well-defined and functorial in $U$ [PS08, Prop. 4.18]. Thus, $H^i\big(\overline{K(X)},\mathbf Q\big)$ carries an ind-mixed Hodge structure [PS08, Def. 8.13], and all maps $H^i(U,\mathbf Q) \to H^i\big(\overline{K(X)},\mathbf Q\big)$ are morphisms of (ind) mixed Hodge structures.


But the classes in $H^i\big(\overline{K(X)},\mathbf Q\big)$ coming from smooth projective varieties are pure of weight $i$ by classical Hodge theory. On the other hand, on open varieties $U \subseteq X$ you get many classes of different weights. For instance, $H^1(\mathbf C\!\setminus\!\{0\},\mathbf Q)$ is pure of weight $2$, so these classes can never come from smooth projective varieties.


In this case, there is an elementary argument as well. You have to show that there does not exist a smooth curve $Y$ with a nonempty open subscheme $i \colon U \hookrightarrow Y$ and a morphism $f \colon U \to \mathbf C\!\setminus\!\{0\}$ such that $f^\* \eta \in i^\* H^1(Y,\mathbf Q)$ for the generator $\eta \in H^1(\mathbf C\!\setminus\!\{0\},\mathbf Q)$. The morphism $f \colon U \to \mathbf C\!\setminus\!\{0\}$ extends to a morphism $f \colon Y \to \mathbf P^1$, and there exists an open $V \subseteq \mathbf C\!\setminus\!\{0\}$ such that $f^{-1}(V) \to V$ is finite. Replacing $U$ by $f^{-1}(V)$ if necessary, we may assume $f \colon U \to V$ is finite. Then the Gysin sequences give a commutative diagram with exact rows
$$\begin{array}{ccccccccc}\cdots & \to & H^1(\mathbf P^1,\mathbf Q) & \to & H^1(V,\mathbf Q) & \to & H^0(\mathbf P^1\!\setminus\! V,\mathbf Q) & \to & \cdots \\ & & \downarrow & & \downarrow & & \downarrow & & \\ \cdots & \to & H^1(Y,\mathbf Q) & \to & H^1(U,\mathbf Q) & \to & H^0(Y\!\setminus\! U,\mathbf Q) & \to & \cdots.\!\!\! \end{array}$$
The right vertical map is injective since $Y\!\setminus\!U \to\mathbf P^1\!\setminus\!V$ is a surjection of zero-dimensional (reduced) schemes. Thus the vanishing of $H^1(\mathbf P^1,\mathbf Q)$ shows that a class in the image of $f^\* \colon H^1(V,\mathbf Q) \to H^1(U,\mathbf Q)$ can never come from $H^1(Y,\mathbf Q)$.


In étale cohomology, we would probably say that the class $f^\* \eta$ is 'ramified' at the points $p\_i$ of $Y$ above $0 \in \mathbf P^1$, in the sense that it does not extend to a class on an open containing some $p\_i$. (Like how a nontrivial étale cover of $\mathbf C\!\setminus\!\{0\}$ is ramified above $0$ in that it does not extend to an étale cover of $\mathbf C$.)


(If you want the above diagram to be a commutative diagram of mixed Hodge structures, you should probably add some twists to the $H^0$ terms.)




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**References.**


[PS08] C. A. M. Peters, J. H. M. Steenbrink, [*Mixed Hodge structures*](https://doi.org/10.1007/978-3-540-77017-6). Ergebnisse der Mathematik und ihrer Grenzgebiete. 3. Folge **52**. Springer, 2008. [ZBL1138.14002](https://zbmath.org/?q=an:1138.14002).


