Relativized world where $L^A=NP^A$

Question

I wonder¹ whether there is a known relativization barrier against proving $L\neq NP$. Hence I'm looking for a language $A$ for which $L^A=NP^A$.

My first idea was to try $A:=SAT$, but then I thought that $L^A\subset P^A = \Delta_2^P$ and $NP^A=\Sigma_2^P$. This seems to disqualify not only $A:=SAT$, but any complete problem $A$ from the polynomial hierarchy for my purpose.

My next idea was to try $A:=TQBF$, where $TQBF$ is the $PSPACE$-complete problem to decide true quantified Boolean formulas. But $P^A=NP^A$ is well known, and $L^A=NP^A$ is a stronger statement, so $L^A=NP^A$ would be well known, if anybody had proved it.

My question is just whether there is a known relativization barrier. We certainly can't prove that such a language $A$ can't exist, because otherwise we would get $L\neq NP$ as corollary.

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_{1. I know that logarithmically space bounded TMs with stack recognize exactly the languages from $P$, independent of whether the TM is deterministic or not. I don't know whether this result relativizes, but I guess it does. So I wonder whether there can be a language $A$ with $NL^A=P^A$. But asking for $L^A=NP^A$ instead seems easier, because of the connection to the polynomial hierarchy.}

Answer 1

My question is just whether there is a known relativization barrier.

Yes, there is a known relativization barrier. It's given by $A:=TQBF$, because Emil Jeřábek (see comments) is right: the statement that $TQBF$ is $PSPACE$-complete under logspace many-one reductions is well known.

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Emil Jeřábek's remark "(or even uniform $\mathbf{AC}^0$)" seems less relevant, because $\mathbf{AC}^0$ is easily separated from $L$ (or even uniform $\mathbf{NC}^1$). Because the resulting formula actually depends only very locally on the input for the $PSPACE$-Turing machine, one could even claim that $TQBF$ is $PSPACE$-complete under $L$-uniform $\mathbf{NC}^0$ (instead of uniform $\mathbf{AC}^0$). But I find this misleading, because the circuit doesn't really do any non-trivial work at all, it basically just writes down a template.

(The reduction writes down $n^k$ times $\exists c_{1.5}\forall (c_3, c_4) \in \{(c_1, c_{1.5}), (c_{1.5}, c_2)\}$ with systematically varying indices, where each $c_i$ are actually $n^k$ Boolean variables. This entire part only depends on the length of the input, hence the corresponding output variables are hardwired to constants. The final formula could be written as $\forall z. (z_1=i_i\land\dots\land z_n=i_n)\land \phi(x,y,z)$ such that only $i_i,\dots,i_n$ depends on the actual input, and such that this is the only place where $i_i,\dots,i_n$ appear. So this is nearly trivially an $\mathbf{NC}^0$, the entire work has been done by the $L$-algorithm writing down the circuit, whose outputs are nearly all directly hardwired to constant.)