The n-Category Café

Durov’s framework has a generalized ring whose modules are sets, and one whose modules are pointed sets. So, his framework generalizes Deitmar’s.

You make it sound like they disagree over the definition of ‘vector spaces over $F_1$’, but I don’t think Deitmar even talks about ‘vector spaces over $F_1$’. I wrote:

A ‘vector space over $F_1$’ is just a set on which this monoid acts via multiplication… but that amounts to just a plain old set. The ‘dimension of such a ‘vector space’ is just its cardinality.

but you shouldn’t really blame Deitmar for this. I was just trying to stretch his idea

$$rings \mapsto monoids$$

to include

$$modules of rings \mapsto actions of monoids$$

and noting that a module over the 1-element monoid (which is what he calls $F_1$) is nothing but a set.

But you see, he’s an algebraic geometer in a hurry. He’s not interested in vector spaces over $F_1$ — he’s interested in going straight to schemes defined over $F_1$. And, I think his example of the ‘projective line over $F_1$’ matches what Durov says that should be: a 2-point set.

To be blunt, I think Durov’s approach is more systematic and general. So, he talks a bit about both generalized rings: the one whose modules are pointed sets, and the one whose modules are sets. It’s probably good for algebraic geometers to deploy both, since the obvious ‘throw out the basepoint’ functor

$$[pointed sets] \to [sets]$$

is just what algebraic geometers would call ‘projectivization’ — turning a ‘vector space’ for Durov’s $F_1$ into a ‘projective space’. It just so happens that a ‘projective space’ for Durov’s $F_1$ is a module of some other generalized ring… the one whose modules are just sets.

There’s also the obvious ‘keep the basepoint’ functor

$$[pointed sets] \to [sets]$$

and the obvious ‘throw in a basepoint’ functor

$$[sets] \to [pointed sets]$$

but I think it’s good to have a total of 3 functors around. After all, whenever we have a homomorphism of (generalized) rings $f: R \to S$ we get the obvious pullback

$$f^* : S Mod \to R Mod$$

but also, one might hope, both a left and right adjoint to this — what Lawvere would call ‘free’ and ‘fascist’ functors.

In short, there’s room for plenty of fun and confusion here, but it should all make nice sense in the end. If I seem a bit casual about pointed sets versus sets, it’s because of my faith in that.

What if one tried to categorify a generalized field? Perhaps after doing the same for a generalized ring.

What do we know about algebraic 2-theories?

No doubt TWF won’t let us down. Ah yes, Week 170. So Noson Yanofsky gives us a definition of algebraic 2-theories in Coherence, Homotopy and 2-Theories.

What kind of 2-theory corresponds to a commutative theory? And what to the terminal theory?

Maybe somebody should think harder about what fields really are.

(And maybe somebody is already doing that. But then maybe somebody should think harder about what the first somebody is really doing.)

Lol, this is brilliant!

Someone’s thinking my Lord, come to him,

Someone’s thinking my Lord, come to him,

Someone’s thinking my Lord, come to him,

Oh lord, kumbaya.

Whoops — thanks for catching that! I’ll fix it.

That cracked me up while reading it as well.

Hmmmm… If we do have liches, then I’d like to dub the maximal lich a lich lord. At last! Mathematics that sounds like an AD&D-game!

If you like Halloween-themed mathematical concepts, you’ll love Dan Christensen’s paper on phantoms, ghosts, and skeleta. Too bad Gavin Wraith didn’t write it.

Jim Dolan called with a few remarks.

I’d written:

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

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

This is correct up to the phrase ‘so we get an equivalence’, which is wrong. For any operad $O$ we have a ‘forgetful’ functor

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

and when $O$ is commutative my argument gives a functor going the other way:

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

but these are not weak inverses of each other!

An abstract way to see this uses the tensor product of operads. This is defined by the property that an $O_1 \otimes O_2$-algebra in $C$ is the same as an $O_1$-algebra in $O_2 alg(C)$, or more precisely:

$$(O_1 \otimes O_2) alg (C) \simeq O_1 alg (O_2 alg(C))$$

Syntactically, we generate the operad $O_1 \otimes O_2$ by throwing in all the operations of $O_1$ and all the operations of $O_2$, and imposing relations saying that they commute.

If we really had

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

we would have

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

But, it’s easy to find counterexamples, even when $O$ is commutative. Why? Simply because $O \otimes O$ is generated by two commuting copies of all the operations in $O$. And that can be a lot more operations than there are in $O$.

In fact, the easiest counterexamples arise when our commutative operad $O$ has just unary operations, so that it’s a commutative monoid in disguise.

For example, suppose $O$ is just $\mathbb{N}$ in disguise, and take $C = Set$. Then an $O$-algebra in $C$ is a set $X$ with a function $f: X \to X$. An $O \otimes O$-algebra in $C$ is a set $X$ with two commuting functions $f,g : X \to X$. So, they’re very different.

So, the following ‘corollary’ of my false claim is also false:

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’.)

The rest of what I wrote seems true.

In particular, it seems correct that the tensor product of operads, restricted to commutative operads, is the coproduct of commutative operads. So, the answer to this:

Does it make the opposite of the category of commutative operads into a cartesian category?

seems to be ‘yes’.

So what’s a commutative (or do we want some other term, like symmetric?) algebraic 2-theory? Presumably a 2-theory made up of things like $n$-ary Vect-linear combinations would count.

Though Todd was pipped by Jim, Todd’s post is probably easier to understand, since it’s more detailed. So, it’s all for the best in this best of all possible worlds.

Todd wrote:

Are you unhappy with the standard way of defining the notion of commutative operad?

I was at the time, mainly because it seemed too ‘syntactical’, as opposed to ‘semantic’. I was trying to figure out why Durov prefers to do algebraic geometry with commutative operads (or really, algebraic theories). And, it seemed to me that one of the the main things he does with an operad $O$ is play around with their categories of algebras $O alg(C)$. (Not having read all of his 568-page time, I’m not sure how true this really is, but anyway, this is one of the main things everyone always does with operads, so I was willing to guess it’s true for him too.) So, I was seeking a characterization of commutative operads solely in terms of their categories of algebras!

And so, I was slowly blundering my way towards a realization that you and Jim had more clearly.

Namely: while $Op$ (the category of operads) has a tensor product, this becomes the coproduct when restricted to $CommOp$ (the category of commutative operads). A tensor product is a coproduct when it has a codiagonal

$$\nabla: O \otimes O \to O$$

and a unit

$$!: 1 \to O$$

satisfying some nice properties. In paarticular, they make $O$ into a commutative monoid in $Op$. So, Jim also hazarded a speculation that perhaps a commutative operad is the same as a commutative monoid in $Op$. Is anyone up to figuring this out?

Anyway, in terms of categories of models, our operad being commutative gives us

$$\Delta^* : O alg(C) \to (O \otimes O) alg(C) \simeq O alg(O alg (C))$$

as we have been discussing. It also gives us

$$!^*: 1 alg(C) \to O alg(C)$$

Hmm. Since the algebras of the terminal operad $1$ are just commutative monoids, $1 alg (C)$ is the category of commutative monoids in $C$. So, we’re saying any commutative monoid in $C$ becomes an algebra of $O$ in $C$, which is quite familiar to me — just a turned-around, semantic way of saying $1$ is the terminal operad.

Anyway, trying to assemble these observations into something a bit more cogent, I’d like to say that $O$ being commutative makes it a commmutative monoid in $Op$, so that $O alg(C)$ gets a kind of ‘cocommutative comonoid’ structure. But, I’m not quite sure how to make that precise, because what symmetric monoidal 2-category is $O alg(C)$ an object in, exactly?

To be frank, I’m not sure where all this stuff is going. It now seems to me that something else is more important. We’re trying to generalize algebraic geometry. Algebraic geometers are called ‘geometers’ because $CommRing^{op}$ (the category of affine schemes) has many of the features typical of a category of ‘spaces’. For starters, it’s cartesian! But $CommRing^{op}$ lacks some nice features: namely, we can’t always glue together affine schemes to get other affine schemes. So, algebraic geometers embed $CommRing^{op}$ in a larger category, the category of schemes, which has more colimits. (All colimits?)

Now we’re seeing that $CommOp$ is a bit similar to $CommRing$. In particular, $CommOp^{op}$ — the punster in me wants to call this the category of commutative ‘erads’ — is cartesian, just like the category of affine schemes.

I’d like to know exactly how far the similarity goes.

And then, I’d like to see Deitmar and Durov’s tricks for generalizing the concept of ‘scheme’ as very general tricks, applicable whenever we have a cartesian category with some extra features, and want to be able to from new objects by ‘gluing together’ objects form this cartesian category.

(We could take presheaves on our cartesian category, to get arbitrary colimits, or we could do something weirder, analogous to the textbook definition of ‘scheme’.)

When I’m done, I’ll probably discover that this stuff is already well-known among experts in topos theory. But that’s okay.

Kea wrote:

Could Lambda-rings help?

I like $\lambda$-rings, and I keep meaning to talk about them in This Week’s Finds. But, I don’t know what a ‘module’ of a $\lambda$-ring should be.

However, it’s easy to believe that just as a ring $R$ can be thought of as an algebraic theory whose $n$-ary operations are ‘$R$-linear combinations’

$$(x_1, \dots, x_n) \mapsto r_1 x_1 + \cdots r_n x_n$$

so that an $R$-module is just a model of this algebraic theory, so too the modules of a $\lambda$-ring are models of some algebraic theory. This would mean that $\lambda$-rings are generalized rings in Durov’s sense.

That would be nice.

Sadly, googling under ‘lambda-scheme’ turns up stuff that seems utterly irrelevant.

Thanks for reminding me, David! I’ll add Haran’s paper as an official Addendum to week259, since I want that to be my main location for references to $F_1$, at least until I get my math wiki going.

The abstract of a course by Haran suggests relations between the field with 1 element, the diagonal in $Spec(\mathbb{Z}) \times_{Spec(F_1)} Spec(\mathbb{Z})$, and the Riemann hypothesis — exactly along the lines Minhyong was asking for!

I don’t understand this stuff at all, but it sounds cool, so I should give it a try. Any heuristic help would be greatly appreciated.

We ought to form a list

Yes, here.