If $k$ is a field of characteristic zero and $X$ is a smooth irreducible projective variety over $k$, then $X$ satisfy Hodge symmetry, meaning that $$\dim H^p(X, \Omega_{X/k}^q) = \dim H^q(X, \Omega_{X/k}^p).$$ The only proof I know of this fact goes by reducing to the case of $k=\mathbb{C}$ (since $X$ can always be defined over a finitely generated extension of $\mathbb{Q}$, which can then be embedded inside $\mathbb{C}$), and then using classical Hodge theory for complex Kahler manifolds, which is proved using analytic techniques.
This proof has something extremely unsatisfying: if the statement holds for any field of characteristic zero, it seems reasonable that there should be proof that works uniformly for all such fields. Hence my question:
Is there an algebraic proof of Hodge symmetry for all fields of characteristic zero that does not involve a reduction to $\mathbb{C}$?
