Is it known that $AC^1 \subseteq L$?
Matrix powering shows that directed reachability is in $\mathsf{AC^1}$, and so $\mathsf{L} \subseteq \mathsf{NL} \subseteq \mathsf{AC^1}$. In more detail, suppose that $A$ is the adjacency matrix of a directed graph $G$, with self-loops on all vertices. The matrix $B_{ij} = \bigvee_k A_{ik} \land A_{kj}$, which can be computed in $\mathsf{AC^0}$, is the adjacency matrix of $G^2$. In other words, $B_{ij} = 1$ iff there is a path of length at most 2 from $i$ to $j$ in $G$. Iterating this construction $\log n$ times, we obtain an $\mathsf{AC^1}$ circuit computing the adjacency matrix of $G$, that is, which vertex is reachable from which vertex.



What goes wrong when using DFS to evaluate $\mathsf{AC^1}$ circuits? The same approach works, only it uses $\log^2 n$ space rather than $\log n$ space. The reason is that at each level, you need $\log n$ space to store which child you are currently in. Multiplied by the number of levels, we obtain $\log^2 n$.