# Semi-rigid boolean algebras

A forcing construction I'm trying to do seems to require a complete atomless boolean algebra (used as a forcing poset) that is "semi-rigid" in the sense defined below. I'm wondering if anyone has studied this property, maybe using a different term, and shown it to be realizable or not realizable.

A rigid boolean algebra is one that has no nontrivial automorphisms. Call a boolean algebra "semi-rigid" if it does have at least one nontrivial automorphism, but none of these has any fixed points other than 0 and 1 (the algebra's least and greatest elements). The justification for calling this "semi-rigidity" is that although there are nontrivial automorphisms, knowing where an automorphism puts one element -- any element, other than 0 and 1 -- completely determines where it will put every element; there is no further flexibility. (For if $$\phi$$ and $$\phi'$$ were distinct automorphisms that mapped some $$b$$ ($$\neq 0,1$$) to the same element, then $$\phi^{-1} \circ \phi'$$ would be a nontrivial automorphism with fixed point $$b$$.) The four-element algebra $$\{0, b, \neg b, 1\}$$ is a simple example of semi-rigidity.

Preliminary question: Are there semi-rigid complete atomless boolean algebras (CABA's)? I suspect you can get one by gluing together copies of a rigid CABA $$C$$ (such $$C$$'s are well known to exist): let $$B$$ be a CABA with an element $$b$$ such that $$B \upharpoonright b$$ and $$B \upharpoonright \neg b$$ (i.e. the principal ideals with greatest elements $$b$$ and $$\neg b$$ respectively) are isomorphic copies of $$C$$. I suspect this $$B$$ will have exactly one nontrivial automorphism, which interchanges $$B \upharpoonright b$$ and $$B \upharpoonright \neg b$$.

Even if this is right and such a $$B$$ can be proved semi-rigid, though, it would be an uninteresting example because its semi-rigidity would reduce to rigidity (of principal ideals). So my main question is:

Question: Are there semi-rigid CABA's none of whose principal ideals are rigid?