# A characterisation of Boolean algebras

Let $M$ be a meet-semilattice with a least element $0$. Suppose there is an order-reversing involution $a \mapsto -a$ on $M$ such that for all $a, b \in M$, $a \wedge b = 0$ if and only if $b \le -a$. Then $M$ is a Boolean algebra.

Is there a paper/book I can cite for this claim? The proof I have is just ploughing through with symbolic manipulation to show it is a distributive complemented lattice, but it's probably not something people want to read in a group theory paper. Alternatively, does anyone know a more conceptual or slick proof?

-
I haven't thought about this carefully, but are you sure that the noncrossing partition lattice of rank 3 doesn't also satisfy all your conditions? –  Patricia Hersh Aug 30 '12 at 0:20
Todd Trimble has pointed out to me that this does not give a counterexample. You can find a picture of this example at en.wikipedia.org/wiki/Noncrossing_partition. Good luck with your work! –  Patricia Hersh Aug 30 '12 at 11:50

I have a pretty simple proof. As Tom has already observed, we have a bounded lattice. Now one characterization of Boolean algebra is a (bounded) lattice such that for every element $a$ there is an element $-a$ such that

$$a \wedge b \leq c \qquad iff \qquad a \leq -b \vee c$$

In category-speak, notice that the poset map $(-) \wedge b$ has a right adjoint $-b \vee (-)$. This immediately implies that $(-) \wedge b$ distributes over any joins that exist.

So let's check that this iff condition holds. Under the hypotheses, the left side is equivalent to saying

$$a \wedge b \wedge -c = 0.$$

Similarly, the right-hand side says $a \wedge -(-b \vee c) = 0$. But according to the characterization of joins, we can rewrite this as $a \wedge b \wedge -c = 0$, and we are done.

Edit: One reference is the nLab. Alternatively, notice that combining Tom's observation with mine shows that we have a complemented distributive lattice.

-
If it helps at all, the iff condition says precisely that a Boolean algebra is a Heyting algebra for which $p \to q = \neq p \vee q$. I'm sure this must be in thousands of places, but see for example: en.wikipedia.org/wiki/Heyting_algebra#Examples –  Todd Trimble Aug 30 '12 at 2:06
Sorry, $\neq$ should be replaced by $\neg$. –  Todd Trimble Aug 30 '12 at 2:06
Yes, this is exactly the sort of proof I was looking for. –  Colin Reid Aug 30 '12 at 5:33
+1 for taking the time to figure out what was wrong with my suggestion, in addition to giving a proof. –  Patricia Hersh Aug 30 '12 at 13:47
@Patricia: no problem! –  Todd Trimble Aug 30 '12 at 14:08

Edit: Patricia Hersh points out in the comments that I didn't prove it was a Boolean algebra: I only showed that it was a complemented lattice. But maybe this fragment is useful, so I'll leave it here.

Write $M^{\mathrm{op}}$ for $M$ with the reverse ordering. Since the negation map $a \mapsto -a$ is an order-reversing involution, it defines an isomorphism $M^\mathrm{op} \to M$. But $M$ is a meet-semilattice with least element, so $M$ is also a join-semilattice with greatest element. Hence $M$ is a lattice (bounded, if that's not already in your definition of "lattice"). Moreover, the isomorphism $a \mapsto -a$ interchanges joins and meets, i.e. the de Morgan laws hold.

Taking $b = -a$ in your condition, we have $a \wedge (-a) = 0$ for all $a$. But by the de Morgan laws, we also have the dual of your condition: $a \vee b = 1$ if and only if $-b \leq a$. Taking $a = -b$ in this dual condition gives $(-b) \vee b = 1$ for all $b$, as required.

-
Tom, how does this imply $M$ is a Boolean algebra? It seems like e.g. the noncrossing partition lattice of rank 3 also meets the conditions you've proven. Thanks for your help. –  Patricia Hersh Aug 30 '12 at 0:51
Oh, oops. I just forgot that Boolean algebras had to be distributive. Thanks! I don't know what the noncrossing partition lattice of rank 3 looks like - could you describe it in an answer? –  Tom Leinster Aug 30 '12 at 1:01
If $\langle S;\wedge,-\rangle$ is a pseudo-complemented meet-semilattice, then the subset $B(S):=${$-x:x\in S$} with the order inherited from $S$ is a boolean algebra.( A proof of this result can also be found in Grätzer's General Lattice Theory.)
As $-$ is a pseudo-complement on $S$ we have by definition $a\wedge b=0$ iff $b\leq -a$. This implies $a\leq b\rightarrow -a\geq -b$ and $---a=-a$. Therefore $a\mapsto -a$ is an order-reversing involution on $B(S)$.
In your case we have $B(M)=M$ since $--a=a$ for all $a\in M$. Therefore, $M$ is a boolean algebra.