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Stream: learning: questions

Topic: complete atomic boolean algebras

John Baez (Jul 12 2021 at 00:05):

I'm a bit confused by this old comment of @Mike Shulman's on the n-Cafe:

John Baez (Jul 12 2021 at 00:05):

Let me just discuss the case of $\mathrm{Set}$, using classical logic.

John Baez (Jul 12 2021 at 00:06):

$\mathrm{Set}^{\rm op}$ is equivalent to $\mathrm{CABA}$, the category of complete atomic boolean algebras.

John Baez (Jul 12 2021 at 00:07):

So, if $P \colon \mathrm{Set}^{\rm op} \to \mathrm{Set}$ is monadic, this implies $\mathrm{CABA}$ is monadic over $\mathrm{Set}$.

John Baez (Jul 12 2021 at 00:11):

I would like to understand this monad rather explicitly. Mike says the adjoint of $P$ is just " $P$ itself", or a bit more cautiously it's the functor I'll call

$Q : \mathrm{Set} \to \mathrm{Set}^{\rm op}$

which is just another way of thinking about $P$.

John Baez (Jul 12 2021 at 00:12):

So the monad is $PQ : \mathrm{Set} \to \mathrm{Set}$.

John Baez (Jul 12 2021 at 00:13):

I want to keep on talking until I get to my actual questions.

John Baez (Jul 12 2021 at 00:14):

This monad $PQ$ is a bit scary since it's "without rank", i.e. it's not a $\kappa$-ary monad for any cardinal $\kappa$. So we have to be a bit careful with it.

John Baez (Jul 12 2021 at 00:17):

If I have a $\kappa$-ary monad $T$ we can associate to it a $\kappa$-ary algebraic theory whose algebras are the same as algebras of the monad. For any cardinal $\alpha \lt \kappa$ we can define the $\alpha$- ary operations of this algebraic theory to be the set $T(\alpha)$ (which makes sense since $\alpha$ is a set).

John Baez (Jul 12 2021 at 00:18):

I will try doing this for the monad $PQ$, with a bit of trepidation since it's a monad without rank.

John Baez (Jul 12 2021 at 00:19):

If this works, the $\alpha$-ary operations for complete atomic boolean algebras form the set

$P(Q(\alpha)) = 2^{2^\alpha}$

John Baez (Jul 12 2021 at 00:21):

Now, this seems to make sense. For example, when $\alpha$ is finite, we are used to the fact that boolean algebras have $2^{2^\alpha}$ different $\alpha$ -ary operations. We usually draw these as truth tables! Here are the $2^{2^2}$ different binary operations for boolean algebras:

binary operations for boolean algebras

John Baez (Jul 12 2021 at 00:27):

But we are letting $\alpha$ be any cardinal.

So my question is: are complete atomic boolean algebras precisely the algebras of some algebraic theory without rank, whose $\alpha$-ary operations form the set $P(Q(\alpha)) = 2^{2^\alpha}$?

John Baez (Jul 12 2021 at 00:28):

And if so, what I want to understand is this: why are the algebras of this monad precisely the complete atomic boolean algebras?

John Baez (Jul 12 2021 at 00:29):

It's easy for me to imagine that any complete boolean algebra will be an algebra of some algebraic theory whose $\alpha$-ary operations form the set $P(Q(\alpha))$.

John Baez (Jul 12 2021 at 00:32):

But this may be false.

How does the atomicity of a complete atomic boolean algebra get captured by the operations in an algebraic theory, and the laws obeyed by these operations?

John Baez (Jul 12 2021 at 00:33):

For some related subtleties see Section 6 here:

• Complete Boolean algebra.

John Baez (Jul 12 2021 at 00:35):

This section says complete boolean algebras are algebras of some algebraic theory, but that there is no corresponding monad: there is no free complete boolean algebra on an infinite set.

John Baez (Jul 12 2021 at 00:36):

That's about complete boolean algebras... but my question is about complete atomic boolean algebras.

John Baez (Jul 12 2021 at 00:37):

How does atomicity change things?

Fawzi Hreiki (Jul 12 2021 at 00:39):

This probably won’t answer your question but there is a syntactic presentation of this ‘double power set theory’ in Ernest Manes’ book ‘Algebraic Theories’ (which is actually a book about monads)

Fawzi Hreiki (Jul 12 2021 at 00:40):

And a more detailed analysis of this monad and it’s algebras.

Zhen Lin Low (Jul 12 2021 at 00:46):

Martin Brandenburg remarks here (Example 6.8) that atomicity = complete distributivity, which is an algebraic condition.
https://arxiv.org/pdf/2106.11115.pdf

Zhen Lin Low (Jul 12 2021 at 00:49):

I think the paper also answers your implicit question about axiomatisability of monads without rank.

John Baez (Jul 12 2021 at 02:07):

Thanks, Zhen! If atomiticity is equivalent to complete distributivity, that answers my question. I hadn't suspected that.

And thanks, Fawzi: I'd really like to see a syntactic presentation of this theory. I haven't looked at Manes' book.

John Baez (Jul 12 2021 at 02:37):

Brandenburg writes:

Now Definition 5.16 allows us to define CABAs internal to any complete category C. This might be surprising, since a priori atomic is not an algebraic condition, but actually it is equivalent to completely distributive by [31, Proposition VII.1.16], and this is an algebraic condition.

Here [31] is

• P. T. Johnstone, Stone Spaces, Cambridge Studies in Advanced Mathematics Vol. 3, Cambridge University Press, 1982.

so that's where I need to look!

John Baez (Jul 12 2021 at 02:39):

Brandenburg also writes:

Manes explains in [47, Example 3.46] directly why $\tilde{P}(P(X))$ is the free CABA on a set X.

John Baez (Jul 12 2021 at 02:44):

Here his $\tilde{P}(P(X))$ is my $P(Q(X))$, and [47] is

• E. Manes, Monads of sets, Handbook of algebra. Vol. 3. North-Holland, 2003, 67-153

Mike Shulman (Jul 12 2021 at 03:36):

I don't think I've ever worked out explicitly how this happens. But it should also be possible to just $\beta$-reduce the proof of monadicity of $P$ and then apply the known equivalence between powersets and CABAs.