So in mathematics, one can iterate operation of mapping a group to its automorphism group transfinitely to obtain the automorphism tower of a group. And the automorphism tower always terminates. The next natural follow up question to this result is to ask about the congruence lattice tower where one begins with a lattice.
Now, if $L$ is a lattice, then the mapping $e:L\rightarrow\mathrm{Con}(L)$ defined by letting $(x,y)\in e(a)$ if and only if $x\vee a=y\vee a$ is a lattice homomorphism precisely when the lattice $L$ is distributive. Since we want $L$ to embed in $\mathrm{Con}(L)$, we want to restrict our attention to distributive lattices. Furthermore, if $L$ is a lattice, then $\mathrm{Con}(L)$ is always a distributive lattice, so we have no choice but to only consider distributive lattices in the congruence lattice tower problem. It is a not too difficult exercise to show that the mapping $e:L\rightarrow\mathrm{Con}(L)$ is a lattice isomorphism if and only if $L$ is a finite Boolean algebra. Therefore, the congruence tower problem is quite easy and boring for the variety of all distributive lattices. The congruence tower problem for frames instead of distributive lattices is a difficult open question of mathematical significance.
A frame is a complete lattice $L$ that satisfies the infinite distributivity law $x\wedge\bigvee R=\bigvee_{r\in R}(x\wedge r)$. The frames $L$ are precisely the complete lattices which happen to be complete Heyting algebras. If $L$ is a frame, then an equivalence relation $\simeq$ on $L$ is said to be a frame congruence if $\bigvee_{i\in I}x_{i}\simeq\bigvee_{i\in I}y_{i}$ whenever $x_{i}\simeq y_{i}$ for $i\in I$ and $r\wedge s\simeq t\wedge u$ whenever $r\simeq t,s\simeq u$.
If $L$ is a frame, then let $\mathfrak{C}(L)$ denote the collection of all congruences of the frame $L$. Then $\mathfrak{C}(L)$ is itself a frame, and there is a mapping $e:L\rightarrow\mathfrak{C}(L)$ where $(x,y)\in e(a)$ if and only if $x\vee a=y\vee a$ which happens to be a frame homomorphism.
One can define $\mathfrak{C}^{\alpha}(L)$ for all ordinals $\alpha$ by letting $\mathfrak{C}^{\alpha+1}(L)=\mathfrak{C}(\mathfrak{C}^{\alpha}(L))$ for all $\alpha$ and where for limit ordinals $\gamma$, we have $\mathfrak{C}^{\gamma}(L)=\varinjlim_{\alpha<\gamma}\mathfrak{C}^{\alpha}(L)$ (direct limits are taken in the category of frames).
If $e:L\rightarrow\mathfrak{C}^{\alpha}(L)$ is the canonical mapping, then for each complete Boolean algebra $B$, the mapping $\mathrm{Hom}(\mathfrak{C}^{\alpha}(L),B)\rightarrow\mathrm{Hom}(L,B)$ in the category of frames where $\phi\mapsto\phi\circ e$ is a bijection (this result should convince you that $\mathfrak{C}^{\alpha}(L)$ is important).
So there are frames $L$ where the tower $(\mathfrak{C}^{\alpha}(L))_{\alpha}$ never stops growing. If $B$ is a complete Boolean algebra, then the canonical mapping $e:B\rightarrow\mathfrak{C}(B)$ is an isomorphism, and if the tower $(\mathfrak{C}^{\alpha}(L))_{\alpha}$ stops growing, then $\mathfrak{C}^{\alpha}(L)$ is a complete Boolean algebra for large enough $\alpha$.
Now, in all known examples, either $(\mathfrak{C}^{\alpha}(L))_{\alpha}$ never stops growing or the tower $(\mathfrak{C}^{\alpha}(L))_{\alpha}$ stops growing at or before around the 4th step. Suppose that $\alpha$ is an ordinal. Then is there a frame $L$ where $\alpha$ is the least ordinal with $\mathfrak{C}^{\alpha}(L)=\mathfrak{C}^{\beta}(L)$ for $\beta\geq\alpha$ in the sense that the canonical frame homomorphism between these frames is an isomorphism?