### Re: This Week’s Finds in Mathematical Physics (Week 259)

I was trying to figure out why Durov likes *commutative* generalized rings so much better than generalized rings in general. Of course this should be related to why algebraic geometers like commutative rings better than rings in general.

And, this morning while I was waiting for my pupils to dilate at the optometrist’s, I came up with some ideas. I’m hoping Tom Leinster or Todd Trimble or someone could say if these are new or not.

First, let’s stop calling them ‘generalized rings’ and call them what they are: ‘algebraic theories’.

Second, let’s start by seeing how much we can do using operads.

There’s a down-to-earth definition of what it means for an operad to be ‘commutative’. But the main thing this definition accomplishes, I think, is this:

Given an operad $O$ we can talk about its algebras in any symmetric monoidal category $C$. We get a category of $O$-algebras in $C$, which I’ll call $O alg(C)$.

If $O$ is commutative, all its operations act on any algebra $A$ not just as morphisms

$A^{\otimes n} \to A$

in $C$, but *as $O$-algebra homomorphisms!*

So, for $O$ commutative, an $O$-algebra in $C$ automatically becomes an $O$-algebra in $O alg(C)$! We also have a forgetful functor going the other way, so we get an equivalence

$O alg (C) \simeq O alg (O alg(C))$

There’s something very Eckmann–Hiltonesque about this. And in fact it seems to be Eckmann–Hiltonesque in two ways — two ways related by a curious ‘level shift’.

First, consider an operad with only unary operations. This is just a monoid in disguise. Further, the operad is commutative iff the monoid is commutative. We don’t usually talk about ‘algebras’ of a monoid, though — we call them ‘actions’. The above story, restricted to this special case, thus says ‘if $M$ is a commutative monoid, an action of $M$ in $C$ is the same as an action of $M$ in the category of actions of $M$ in $C$’.

(And, if $C = Set$, this statement is an ‘if and only if’.)

Second, consider the operad $O$ whose algebras are commutative monoids. This $O$ is a commutative operad! So, we get

$O alg (C) \simeq O alg (O alg(C))$

or in other words, ‘a commutative monoid in $C$ is the same as a commutative monoid in the category of commutative monoids in $C$’.

Anyway, while all this is cute, I don’t yet see how to use it to give a slick *definition* of ‘commutative operad’. Or maybe I do: maybe we can say $O$ is commutative if the forgetful functor

$O alg(O alg(C)) \to O alg (C)$

is an equivalence of categories! Does that seem right?

Also, I don’t see why this makes commutative operads better than general operads as a method of generalizing algebraic geometry. For *that*, I think I want something more like this:

There’s a tensor product of rings, but for commutative rings this tensor product is more interesting, since it makes $Comm Ring^{op}$ into a *cartesian* category. This is a fundamental sense in which affine schemes act like spaces.

There’s also a tensor product of operads. Is this more interesting for commutative operads? Does it make the opposite of the category of commutative operads into a cartesian category?

I thought of some other stuff too, but this is already more than anyone wants to read.

## Nebulae and Meadows; Re: This Week’s Finds in Mathematical Physics (Week 259)

That was great!

The astrophotography, vision of the hot dusty electron-atmospheric future (don’t bother Al Gore, though, as he’s busy today in Oslo), and Field ponderings.

Anything to this?:

arXiv:0712.0917

Title: Some properties of finite meadows

Authors: Inge Bethke, Piet Rodenburg

Comments: 8 pages, 1 table

Subjects: Symbolic Computation (cs.SC)