What is the logical complexity of the Hodge conjecture?

The Hodge conjecture seems to me the most mysterious among the Millennium problems (and many others). In particular, I am not sure about its logical complexity. It is not difficult to see that the conjecture is equivalent to a $\Pi^0_2$ statement if restricted to varieties over $\overline{\mathbb{Q}}$. (Basically, it is because these varieties form a recursive set, and the statement sounds like "for any Hodge class there is an algebraic cycle".) However the conjecture may be true over $\overline{\mathbb{Q}}$ but false in general ; if I am not mistaken, this possibility has not being ruled out yet. My guess is that the Hodge conjecture may be formulated as a $\Pi^0_3$ statement but, at present, not as $\Pi^0_2$. Is it true?

Some remarks. As is well known, the Hodge conjecture may be separated into two parts:

I. Hodge classes are absolute,

II. Absolute Hodge classes are algebraic.

The second part is $\Pi^0_2$. (It follows from results of Voisin, "Hodge loci and absolute Hodge classes", Compositio Mathematica, Vol. 143 Part 4, 945-958, 2007. ) So, the logically hard part seems to be the first one. In view of Proposition 0.5 of the Voisin's paper, it boils down to whether Hodge loci are defined over $\overline{\mathbb{Q}}$ or not. I have some ideas about how to translate this into $\Pi^0_3$ but nothing close to $\Pi^0_2$. Naturally, if it is true that the Hodge conjecture can not be formulated as a $\Pi^0_2$ statement for now, there is no way one can actually prove this fact. In case it is not unlikely to be so, I am only asking for an expert opinion. Also, if possible, I would appreciate a slick and sound proof of the "upper bound" $\Pi^0_3$ (although I have some ideas about it, I do not really like them).

• what are $\Pi_{2}^0$ and $\Pi_{3}^0$? – Libli Jan 2 '18 at 14:01
• – David E Speyer Jan 2 '18 at 14:28