What does the existence of self complemented elements tell us about a complete lattice?

Let $L$ be a complete lattice with an involution operation $*$ (a unary operation such that for any $x, y \in L$, $x \leq y$ implies $x^{*} \geq y^{*}$). Now, suppose that there is an element of $L$ that satisfies $x = x^{*}$. How much does that tell us about $L$? Clearly, it means that excluded middle and non-contradiction can fail ($x \vee x^{*} \neq \top$, $x \wedge x^{*} \neq \bot$) when $L \neq \{0, 1\}$, but does it tell us anything interesting about the basic lattice structure (distributivity, existence of implication operations etc)? Sorry if this is a naive question :)

It's hard to imagine what the existence of $x=x^*$ could possibly tell us. To wit, there is a simple way to take any bounded lattice $L$ with involution $*$ and adjoin a new element $x$ with $x^*=x$. In detail, put $L':= L\cup\{x\}$ with ordering such that $0\leq x\leq 1$, but $x$ is not ordered relative to any other element of $L$. Alternatively, extend $\lor$ and $\land$ from $L$ to $L'$ by putting $x\lor y:=1$ and $x\land y:=0$ for any $y\in L\setminus\{0,1\}$. This gives a new lattice $L'$, and you can canonically extend the involution from $L$ to $L'$ by setting $x^*:=x$.

For a concrete example, start with $L=M_2$ being the Boolean lattice with four elements (power set of a 2-set), and you get $L=M_3$.

In conclusion, the existence of a "self-complemented" element doesn't really tell you anything. In particular, it doesn't imply distributivity, let alone the existence of an implication operation.

Edit: as another simple class of examples, start with any lattice $L$ (without involution) and stack the opposite lattice $L^{\mathrm{op}}$ on top of $L$, identifying the top element $1\in L$ with the bottom element of $L^{\mathrm{op}}$. Then the total lattice also has a canonical involution that takes this "middle" element to itself. (By the way, what was the name of this "stacking" construction again?)

• Thanks a lot. That's helpful. More than anything, I was just checking that this condition doesn't render the lattice pathological in some way, since I'm interested in a case where I require the existence of such elements. – King Kong May 25 '16 at 16:18
• I remember Esakia calling this "stacking" (of several lattices) a cascade - but maybe this is not a well-established term. – მამუკა ჯიბლაძე May 26 '16 at 5:26

I claim that a Boolean algebra $B$ has a bijective involution $*$ with some $x\in B$ where $x^{*}=x$ if and only if $B\simeq C\times C$ for some Boolean algebra $C$.

Suppose that $x\in B$ and $*$ is an involution with $x^{*}=x$. Then define a mapping $\phi:B\rightarrow B$ by letting $\phi(x)=(x^{*})'$. Then $\phi$ is an automorphism of $B$. Since $x^{*}=x$, we conclude that $\phi(x)=x'$ and $\phi(x')=x$. Therefore $\phi$ maps $B\upharpoonright x$ isomorphically to $B\upharpoonright x'$. Since $B\upharpoonright x$ is isomorphic to $B\upharpoonright x'$, we conclude that $B\simeq(B\upharpoonright x)\times(B\upharpoonright x')\simeq(B\upharpoonright x)\times(B\upharpoonright x)$.

For the converse, we can define an involution $*$ on $C^{2}$ by letting $(x,y)^{*}=(y',x')$ and $(0,1)^{*}=(0,1).$

Under Stone duality, these Boolean algebras to the compact zero-dimensional spaces $X$ where $X$ cannot be written as a disjoint union $Y\coprod Y$ for some other space $Y$. The countable spaces $X$ that can be written as $Y\coprod Y$ are easy to characterize. The Sierpinski-Mazurkiewicz theorem states that every infinite compact space is of the form $\omega^{\alpha}\cdot n+1$ for some unique $(\alpha,n)$ where $\alpha$ is a non-zero countable ordinal and $n$ is a non-zero natural number. Therefore, the countable compact spaces of the form $Y\coprod Y$ are up-to-isomorphism the spaces $\omega^{\alpha}\cdot 2n+1$

There are complete atomless Boolean algebras $B$ that cannot be written as $C\times C$ such as those mentioned in this paper.