Complexity of sorting
Suppose you have $n$ numbers $x_1,\ldots,x_n$ of width $m \geq \log n$. Without loss of generality, all numbers are different (add an extra $\log n$ lower-order bits). Two numbers can be compared in AC0, so in AC0 we can compute, for each $x_i$, a binary vector $v$, defined by $v_j = 1$ if $x_j \geq x_i$. In TC0 we can sort $v$ to $w$, and then locate (in AC0) the position $k$ such that $w_k = 1$ while $w_{k-1} = 0$ (where $w_0=0$, $w_{n+1}=1$). Given these values $k$ for each $i \in [n]$, we can compute the sorted array in AC0. In total, we get a TC0 circuit.