TQBF consists of alternating quantifiers, so does $\Sigma^2_n$ for fixed $n$. So given a formula in TQBF, shouldn't there be a level of the polynomial hierarchy that solves it?
I think this is incorrect because TQBF consists of any number of quantifiers, but each level in the PH has fixed number of quantifiers, so there's no level of the PH that can solve TQBF. Does it have to do with the fact that the number of quantifiers is part of the input in TQBF, whereas in PH only the formula is the input?
So this could also be worded as, if there exists a PH-complete problem, does TQBF reduce to that problem?