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Stream: learning: questions

Topic: extensive categories

John Baez (Dec 28 2021 at 01:28):

When is the opposite of the category of algebras of a Lawvere theory an [[extensive category]]?

Any sort of necessary or sufficient conditions would be interesting.

John Baez (Dec 28 2021 at 01:31):

Here's why: an extensive category acts to some extent like a category of "spaces". So, this question asks when a category of "algebraic gadgets" has an opposite that acts like a category of spaces. This is a basic idea behind algebraic geometry.

John Baez (Dec 28 2021 at 01:33):

The classic example is the category of commutative rings, whose opposite is the category of [[affine schemes]].

John Baez (Dec 28 2021 at 01:34):

Another example is the category of commutative $R$-algebras for any commutative ring $R$.

John Baez (Dec 28 2021 at 01:35):

But I think there are also other examples that show up in modern generalizations of algebraic geometry.

Nathanael Arkor (Dec 28 2021 at 01:42):

I suspect that the answer may be extracted with a little work from Lack–Vitale's When do completion processes give rise to extensive categories?. In particular, the filtered-cocompletion of a category C is extensive if and only if C is (Proposition 6.4). Categories of algebras for algebraic theories are locally finitely presentable categories, and so this proposition should apply.

John Baez (Dec 28 2021 at 01:49):

Great! Can someone please do the "little work" required to extract an answer to my question from Lack-Vitale's paper? I don't see how.

Todd Trimble (Dec 28 2021 at 02:03):

But John said the opposite of the category of algebras, not the category of algebras itself, which is the filtered cocompletion of the finitely presentable algebras.

Nathanael Arkor (Dec 28 2021 at 02:04):

Ah good point, I somehow skipped over that.

Todd Trimble (Dec 28 2021 at 02:12):

So to try to get the ball rolling, let me sketch how extensivity (lextensitvity) is proven for the classical case of commutative rings:

• The category $\mathrm{CRing}$ of commutative rings is of course finitely cocomplete.

• Product projections $\pi_1: R \times S \to R$ and $\pi_2: R \times S \to S$ are epic, and their pushout is terminal.

• The pushout in $\mathrm{CRing}$ of two ring maps $R \to S$, $R \to T$ is given by the tensor product $S \otimes_R T$, with the evident commutative ring structure.

• The pushing-out functor along a map $R \to S$, namely $S \otimes_R -: \mathrm{CAlg}_R \to \mathrm{CAlg}_S$, preserves finite products (since finite products are reflected and preserved by the forgetful functor to $\mathrm{Mod}_R$, where they are biproducts and therefore coproducts that are preserved by tensoring).

Todd Trimble (Dec 28 2021 at 02:14):

So this does look very specialized. In how many algebraic categories do we have coproducts distributing over products?

John Baez (Dec 28 2021 at 02:17):

Thanks, Todd!

Zhen Lin Low (Dec 28 2021 at 02:40):

The arguments can be applied verbatim for commutative rigs, however.

Zhen Lin Low (Dec 28 2021 at 02:46):

All the examples of co-extensive categories of algebras I can think of are categories of commutative rigs: commutative rings, of course, but also $\mathscr{C}^\infty$-rings, distributive lattices, frames, boolean algebras, ...

John Baez (Dec 28 2021 at 02:53):

I would be delighted to see the commutative rig axioms show up in the answer to my question, maybe with the aid of some extra assumptions. This would, in my mind, "justify" the idea that algebraic geometry is about commutative rigs.

John Baez (Dec 28 2021 at 02:54):

I decided to post a version of my question on MathOverflow.

Zhen Lin Low (Dec 28 2021 at 02:57):

Tim points out that commutative monoids in semiadditive idempotent-complete symmetric monoidal categories also form a co-extensive category, which seems to be another generalisation of commutative rigs. (I don't think $\mathscr{C}^\infty$-rings form such a category, though?)

John Baez (Dec 28 2021 at 02:59):

I added some of your examples to my question... maybe I should have credited you, but I'll just credit you here: THANKS!

Zhen Lin Low (Dec 28 2021 at 04:40):

I recall reading somewhere, I think in a paper referenced by @Nathanael Arkor, that an algebraic theory whose category of algebras has a strict terminal object must have at least two distinct constant symbols. I don't remember the argument at all, though.

Zhen Lin Low (Dec 28 2021 at 05:03):

Oh, silly me, it's actually easy. If there is exactly one constant then the initial algebra is a singleton, so if there are any algebras with more than one element then the singleton is not a strict terminal object. So either there are no constants or there are at least two constants. I'm not sure the former case can be excluded...

Fawzi Hreiki (Dec 28 2021 at 08:26):

@John Baez, you may want to take a look at http://www.tac.mta.ca/tac/volumes/20/14/20-14abs.html

Todd Trimble (Dec 28 2021 at 10:39):

If there are no constants, then the initial algebra is empty as a set. But then the two product projections $0 \times 0 \to 0$ are both isomorphisms, as we can check at the underlying set level, so their pushout would be $0$ and not $1$.

Graham Manuell (Dec 28 2021 at 14:41):

This has been solved here.

John Baez (Dec 28 2021 at 16:39):

Nice! It's also worth looking at part 1 of that paper by David Neal Broodryk.

He gives necessary and sufficient conditions for a variety ($\simeq$ the category of algebras of a Lawvere theory) to be coextensive. Unfortunately the conditions are a complicated mess of syntax and he doesn't give any verbal explanation to help me understand what it means. He also doesn't give any examples other than CommRing. So I would need to spend an hour thinking about his paper to extract some useful information from it, and right now I'm not in the mood.

John Baez (Dec 28 2021 at 16:41):

Can anyone understand what his conditions "really mean"?

John Baez (Dec 28 2021 at 16:44):

Yes, this is a lazy request.