### Algebras as 2-Categories and its Effect on Algebraic Geometry

#### Posted by Urs Schreiber

Here is another question from Bruce Bartlett which really deserves to be a post of its own (see also his last question of this type)… it’s about categories of algebras, algebras as categories and the possible implications for *non-commutative* algebraic geometry.

As adequate for a deep question, this one extends over three seperate emails.

Hi Urs,

Today at Sheffield we had a colloquium by Alexey Bondal (Steklov Mathematics Institute) on “Noncommutative deformations of algebraic varieties and Poisson brackets”.

In the beginning of the program he outlined the grand scheme (excuse the pun) of “noncommutative algebraic geometry”. Namely, (1) first set up ordinary commutative algebraic geometry in a nice categorical framework, and then (2) apply this to noncommutative algebras.

To do (1), it seems the best way is to set up ordinary algebraic geometry from a “functor of points” perspective. (At this point, I am going to become very vague and factually incorrect, since I am still learning the basics of algebraic geometry. But you could probably fix up mentally what I’m saying.)

I believe that in this perspective, you simply

definethe category of affine schemes to be the opposite of the category of commutative algebras. Then you put a Grothendieck topology on it - which allows you to glue things together, amongst other things - and then you play the game of sheaves and stacks over this “site”. In particlar, at some point one considers “prestacks” on this category which are simply weak 2-functors : $F : \mathrm{CommAlg}^\mathrm{op} \to \mathrm{Groupoids}$ Here we are considering $\mathrm{CommAlg}$ as a 2-category withtrivial2-morphisms. (By the way, I get this “stacky” stuff from notes from a “chromatic homotopy” seminar going on at Sheffield - go to “An introduction to stacks”).Now it seems easy to perform step (2) : simply cross out the word “commutative” in the above paragraph. So the category of “noncommutative schemes” is the opposite of the category $\mathrm{Alg}$.

The trouble is, as he explained, is that its quite difficult to work with this - there are all sorts of technical problems that arise, and one is forced to use a kind of work-around by using derived categories of coherent sheaves, etc.

Anyway, here is what made me think:

Simon [

Willerton, -urs] taught me a month ago that the category of algebras is really a 2-category. An $n$-category cafe regular such as yourself will agree that an algebra is best defined as a kind of one object category. Thus morphisms are functors, and 2-morphisms… are natural transformations. What are these? Well work it out : if $A$ and $B$ are algebras, and $f$ and $g$ are homomorphisms $A \to B$, then a 2-morphism $\phi : f \Rightarrow g$ is an element $b$ of $B$ such that $b f(a) = g(a) b$ for all $a \in A$.Okay… so if Alg is really a 2-category, then how does this affect algebraic geometry? Well, one thing all kindergarten algebraic geometers (like me) know is that the category of integral domains is equivalent to the opposite of the category of affine varieties… give or take some technicalities.

How does the notion “Alg is really a 2-category” affect this? Well, suppose $A$ and $B$ are integral domains, with $f,g : A \to B$. What is a 2-morphism? Well we look at the formula above, and use the fact that the algebra is commutative to get : $b[ f(a) - g(a) ] = 0.$ Now we use the fact that it is an integral domain, to conclude that $f(a) - g(a) = 0$… that is, $f$ and $g$ must be the same map, and there is no condition on $b$ at all! In other words, the 2-morphisms are

trivialprecisely for integral domains - there is no information in them at all!Thus we elegantly arrive at the fact that the theory of “ordinary” affine varieties does not need to care about the 2-morphisms… they are trivial… precisely when the algebras are integral domains! It is reassuring to see how the notion “integral domain” kind of “emerges” from a 2-category point of view.

Here is my point : when we change our attention to non-integral domains, or more seriously, noncommutative algebras, then the 2-morphisms become important! We must take $\mathrm{Alg}$ seriously as a 2-category. This will change the way we set up sheaves, stacks, etc. This might fix up a lot of the fundamental problems that we are experiencing.

For example, a prestack on $\mathrm{Alg}^\mathrm{op}$ will now be a weak 2-functor: $F : \mathrm{Alg}^\mathrm{op} \to \mathrm{Groupoids}$ but where we are taking seriously the 2-morphisms inside $\mathrm{Alg}^\mathrm{op}$ - we haven’t just set them to be trivial. Thus this definition of a stack is a bit stricter and also richer.

What do you think?

Regards Bruce

P.S. To me, a lot of the problems in going from commutative to noncommutative geometry seems to be linked to the following. To an “old-fashioned” algebraist, there is no difference in principle between a noncommutative algebra and a commutative one - the latter just obeys an extra equation.

To an $n$-category café patron, there is a

world of difference: a (non)commutative algebra is a 1-object category, while a commutative algebra is a 1-object, 1-morphism 2-category… and this affects our viewpoint quite a lot.

$\;$

Hi Urs,

In my last mail I wrote a P.S…. which made me rethink a lot of what I said.

Namely, a commutative algebra is really a one-object, one-morphism 2-category… which means that $\mathrm{CommAlg}$ is really a 3-category : objects are commutative algebras, morphisms are weak 2-functors, 2-morphisms are transformations, and 3-morphisms are modifications.

Its a bit late… but I seem to have worked out that, give or take some stuff:

A 2-functor in this sense is just an ordinary algebra homomorphism. (I should check this… being a weak 2-functor might introduce extra data).

A transformation is just an ordinary natural transformation in the previous email sense; i.e. if $f, g : A \to B$ then a natural transformation from $f$ to $g$ is an element $b$ of $B$ such that $b [ f(a) - g(a)] = 0 \; \forall\; a \in A.$ If $b, b' : f \Rightarrow g$ is a transformation then a 3-morphism from $b$ to $b'$ is an element $c$ of $B$ such that $c [ b - b'] = 0.$ Anyhow… you get the picture. It is an interesting thing to think about - in the Baez/Dolan stabilization hypothesis, it is normally stated that , e.g. “1-object 1-morphism 2-categories are commutative algebras”. But we should check this for the arrows as well : do we introduce any interesting new data into the game?

For instance, it is supposed to stabilize, so that a “one object, one 1-morphism, one 2-morphism 3-category” is also just a commutative algebra. But then we can consider 4-morphisms… what will they turn out to be?

In other words -> do the

morphismsalso stabilize? I hope they do, but my calculation above seems to say otherwise, as it suggests that modifications bring new data into the game.Perhaps its true that somehow, if we think of commutative algebras as “one object, one 1-morphism, one 2-morphism, one 3-morphism, …., one 9-morphism 10 categories” then this will introduce all sorts of extra morphisms inside $\mathrm{CommAlg}$ $\to$ it will become a 10-category! And the game can be continued…

What is going on here?

Regards, Bruce

$\;$

Hi Urs,

I’m not so sure anymore about integral domains forcing the 2-morphisms to be

totallyuninteresting… after all, having around $B$’s worth of 2-morphisms from $f \Rightarrow f$ for every $f: A\to B$ might be significant - what do they correspond to one the affine varieties side? I kind of worked that out though, but how to motivate it geometrically?What seems to be somehow true though is that,

ifconsidering commutative algebras as a “one object, one 1-morphism, …, n-category” really does throw in new morphisms into the game (which I’m not certain of at all), then they seem to be all of the form $c [ a - b]$ so they’re kind-of-trivial precisely when you have an integral domain. Said more succinctly : $\mathrm{Alg}$ collapses to a 1-category precisely when you have commutative integral domains. Otherwise it seems to be some kind of infinity-category. (Gulp! Can that be?? Think I’m making a blunder here :-))Regards, Bruce

## Re: Algebras as 2-Categories and its Effect on Algebraic Geometry

Hi Bruce,

I’ll comment on one aspect of your remarks.

One of your points is, paraphrased, that whenever we use an algebra or some other monoid in some context, we might want to ask ourselves:

“How would our perception of that context change, if we systematically used the fact that a monoid is really a certain kind of category? For instance: would we think differently about the natural notions of morphisms that we are dealing with?”

Specifically, the context of interest here is, vaguely speaking, the op-description of spaces in terms of their representations, and the notions of generalized spaces derived from that.

Here I say “representations” instead of “functions”, in order to conjure up the picture which we expect # to be the one that categorifies nicely.

Using this point of view as a guiding principle, it might be fruitful to consider the following aspect of your remark:

A complex function on a space $X$

maps points of $X$ to elements of $\mathbb{C}$. Obviously.

But $\mathbb{C}$ is not just any old set. It is in fact a monoid. Following your remark, we want to be very sophisticated and instead think of $\Sigma(\mathbb{C})$: the category with a single object and $\mathbb{C}$-worth of morphisms, with composition rule of morphisms given by the product in $\mathbb{C}$.

From this point of view, $f$ is a actually a map to morphisms of a category:

A triviality.

On top of this triviality, I’ll now demand that we do with the domain what we did with the codomain and think of $f$ as a map

where $\mathrm{Disc}(X)$ denotes the discrete category on the set $X$.

This reformulation is utterly trivial and might seem void.

But your remark suggests that we take it serious and try to figure out where this leads us.

Okay, so what can a map from objects of one category to morphisms of another possibly be, if we demand everything to have nice category-theoretic interpretation?

Answer: it must be a natural transformation of functors.

That’s what it means to take a complex-valued function and be serious about the fact that complex numbers are really to be thought of as morphisms.

Now, I claim, the above trivialities that I went through lead to something a little more interesting:

there is not much of a choice concerning the functors that $f$ could be a natural transformation for.

For instance, let

be a trivial line bundle with trivial connecton

Then $f$ would be a morphism

While still not really non-trivial, this is somewhat pleasing. Because this now has lead us automagically from

- replacing the monoid $\mathbb{C}$ by the category $\Sigma(\mathbb{C})$

to

- replacing the monoid $[X,\mathbb{C}]$ by the category $\simeq \Sigma([X,\mathbb{C}])$, whose single object we regard as $\mathrm{tra}^0$ and whose morphisms are functions $f$, regarded as natural transformations

The product in the algebra of functions on $X$ is now nothing but composition of natural transformations

Notice that this picture nicely categorifies: from the categorified Gelfand-Naimark theorem # we know that the right categorification of the algebra of complex functions on spaces $X$ is the 2-algebra of Hermitean vector bundles with connection on $X$.

But the 2-algebra of vector bundles with connections is nothing but the monoid of endomorphisms of the trivial (bundle-)gerbe on $X$!

(There is a nice diagrammatic way to say this, using pseudonatural transformations of trivial 2-functors, but I’ll refrain from reproducing this right now.)

Now I would claim that this way of looking at things helps us decide which answers to the problems you mentioned are more natural than others.

But I have to end here for the moment.