A characterisation of Boolean algebras

Question

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?

Answer 1

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.

Answer 2

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.

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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.

Answer 3

The follwing theorem proven by O. Frink in Pseudo-complements in semi-lattices, Duke Math. J. 1962, Vol. 29, may help:

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.