## September 2, 2006

### Ringoids

#### Posted by John Baez

Just as a “group with many objects” is a groupoid, a “ring with many objects” is called a ringoid.

Gregory Muller has emailed me a question about ringoids. I don’t want to get into the habit of posting emailed questions on this blog, because I already burnt myself out years ago helping moderate a newsgroup. But just this once, I will.

(Famous last words…)

Gregory Muller writes:

I was excited recently when I learned how to categorify the notion of
rings. I’ve long thought of groups as categories, and up to a few days ago,
it had bothered me that I lacked a parallel notion for rings, especially
given the significance of category-theoretic notions in algebraic geometry.
However, my current method of categorifying rings is ugly; specifically, it
relies on a definition that uses underlying sets and elements, which is
obviously distasteful. What is further frustrating is that this method
generalizes in a way that allows Lie groups and other common mathematical
objects to be categorified.

The method I am using is as follows. Let X be a category with a forgetful
functor F to Set. Then define an “X-valued category” to be a category in
the usual sense, except that the Hom(A,B) is an object in X instead of a
set, and satisfying the following compatibility with composition:

The composition function is

o: F(Hom(A,B)) × F(Hom(B,C)) → F(Hom(A,C)),

with the map from F(Hom(B,C)) → F(Hom(A,C)) given by ‘plugging in’
is the image of a morphism in X, and the similar statement for plugging in
on the right.

This is a useful definition, since:

1) A ring (with unit) is the same as an AbGrp-valued category with one
object.

2) A k-algebra (associative, with unit) is the same as a k-Vect-valued
category with one element.

3) An R-algebra (associative, with unit) is the same as an R-Mod-valued
category with one element.

4) A Lie Group is the same as a SmoothManifold-valued groupoid with one
element.

…and likely others that I haven’t noticed yet.

The problem is the use of sets and elements in the defintion. I have tried
to clean things up and use a category-theoretic definition, but (at least in
the case of AbGrp-valued categories) it seems to be related to the problem
of expressing the formula “ab+cd” as a product of sums in an arbitrary ring,
which I think is about as hard and convoluted a question as any you are
likely to find in math.

Another question is how to philosophically reconcile this notion of a
“category with arrows in category X”, with the other notion of such a
category coming from n-categories. Specifically, a 2- category can be
thought of as a category with Hom(A,B) taking its values in Cat, instead of
Set, subject to some very different notions of “compatibility with
compositions”.

I would be very thankful for any insight into this stuff,
Greg

Your concept of an “X-valued category” is usually called an X-enriched category, or X-category for short. The idea is to fix a category X and define a X-category to have a set of objects and, for any pair of objects a and b, not a set but an object of X called hom(a,b). We can write down the whole definition of category this way as long as X is a monoidal category, that is, a category with tensor product. This allows us to say that for any objects a,b,c in our X-category, composition is a morphism in X:

o: hom(a,b) ⊗ hom(b,c) → hom(a,c)

By this means we can avoid referring to “elements” of hom(a,b).

Enriched categories were invented by Max Kelly, and you will enjoy reading his book, since he gives a very clean treatment:

Kelly was the first to bring category theory to Australia, and enriched category theory has been a mainstay of Australian category theory ever since he invented it. In addition to X-enriched categories, he defined X-enriched functors and X-enriched natural transformations. He then went ahead and redid all of category theory - well, lots of it anyway! - in this X-enriched setting. The basics are straightforward; things get more tricky when you reach the theory of limits and colimits.

One of Kelly’s most famous students is Ross Street, and you can read about the history of enriched category theory near the beginning of Street’s Australian conspectus of higher categories.

People usually denote the category of abelian groups by Ab instead of AbGrp. With the usual tensor product of abelian groups, Ab becomes a monoidal category, and an Ab-category is sometimes called a ringoid.

As you note, a one-object ringoid is a ring, just as a one-object groupoid is a group. These are not “categorifications” of the concept of ring and group, not in the technical sense anyway. A group is already a category; when we go to groupoids we are just letting it have more objects. Similarly for rings and ringoids. So, instead of categorification, one should call this process many-object-ification, or maybe oidization.

To categorify the concepts of group and ring, we need to go up to 2-categories. The resulting concepts are called “ring categories” and “categorical groups” (or “2-groups”). Ring categories were introduced by Kelly and Laplaza.

As you note, we can also many-object-ify the concept of algebra. The category of R-modules is called R-Mod, and it’s a monoidal category with the usual tensor product of R-modules whenever R is commutative. An R-Mod-category is called an R-algebroid or simply an algebroid. As you note, a one-object algebroid is an associative algebra with unit.

Note that when R = Z, R-Mod is just Ab, so a Z-algebroid is just a ringoid.

Similarly, as you note, a Cat-category is a 2-category.

Lately we’ve been talking about symmetric monoidal closed categories, for example cartesian closed categories. Any such category is enriched over itself! (Being symmetric is actually irrelevant here.)

There’s been a lot of work on all these subjects.

But don’t feel bad that you’re reinventing the wheel a bit here - it’s a very good wheel, and you can roll quite far with it.

One thing fans of category theory enjoy is how sufficiently general concepts can bend back, bite their own tails, and swallow themselves. This happens in the case of ringoids. Whenever R is any ring, R-Mod is a ringoid. But, this is also true when R is a ringoid! We define a module of a ringoid R to be an Ab-enriched functor

F: R → Ab

and define a homomorphism between these to be an Ab-enriched natural transformation. These notions reduce to the standard ones when R is a ring.

So, we get a category R-Mod of modules for any ringoid R… and R-Mod is again a ringoid!

For a very practical text on algebroids try this:

• P. Gabriel and A. V. Roiter, Representations of Finite-Dimensional Algebras, Enc. of Math. Sci., 73, Algebra VIII, Springer, Berlin 1992.

The terminology is a bit quirky, but there’s some amazing stuff in here.

Posted at September 2, 2006 2:35 AM UTC

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### enrichment

Gregory Muller described the concept of composition in $X$-enriched categories this way:

define an “$X$-valued category” to be a category in the usual sense, except that the $\mathrm{Hom}(A,B)$ is an object in $X$ instead of a set, and satisfying the following compatibility with composition:

The composition function is $\circ : F(\mathrm{Hom}(A,B)) \times F(\mathrm{Hom}(B,C)) \to F(\mathrm{Hom}(A,C))\,,$ with the map from $F(\mathrm{Hom}(B,C)) \to F(\mathrm{Hom}(A,C))$ given by ‘plugging in’ is the image of a morphism in $X$, and the similar statement for plugging in on the right.

He noticed that various familiar algebraic concepts can hence nicely be understod as $X$-enriched categories of various sorts. In this context he also remarks that

Specifically, a 2- category can be thought of as a category with $\mathrm{Hom}(A,B)$ taking its values in $\mathrm{Cat}$, instead of $\mathrm{Set}$, subject to some very different notions of “compatibility with compositions”.

I am not fully sure what this last comment on a “different notion” of compatibility is addressing. But I’ll say this:

Abstractly, the “compatibility of composition” is always the same, namely always given by the monoidal structure of the category $X$ that we enrich over. The only subtlety to be aware of is that there may be quite different monoidal structures on one and the same category $X$.

There is a standard monoidal structure $\times$ on $\mathrm{Cat}$. For categories enriched over $\mathrm{Cat}$ this implies that composition

(1)$\array{ \mathrm{Hom}(A,B) \times \mathrm{Hom}(B,C) &\stackrel{F}{\to}& \mathrm{Hom}(A,C) }$

is a functor from a product category, which implies that

(2)$\array{ x && x' \\ f_1\downarrow\;\; && \;\f'_1\downarrow \\ y && y' \\ f_2\downarrow\;\; && \;\f'_2\downarrow \\ z && z' } \; \mapsto \; \array{ F(x,x') \\ F(f_1,f'_1)\downarrow\;\;\;\; \\ F(y,y') \\ F(f_2)\downarrow\;\;\;\; \\ F(z,z') }$

does not depend on whether we first apply $F$ on $f_1, f'_1$ and on $f_2,f'_2$ seperately, and then compose the result “vertically”, or if we first compose vertically and apply $F$ (the “horizontal composition”) to the result.

This compatibility condition is called the exchange law in 2-categories. It is implied by the standard monoidal structure on $\mathrm{Cat}$.

I am assuming that what Gregory had in mind is that this looks different from the “distributive” compatibility condition which we have for $\mathrm{Ab}$-categories.

But the reason for that is just the choice of monoidal product in $\mathrm{Ab}$.

One choice would be the cartesian product of abelian groups. Using that for enriching over $\mathrm{Ab}$ does not produce the expected distributivity of composition over addition.

Instead, this can in fact be understood as a special case of the above “exchange law” in 2-categories, namely if we think of an abelian group as a special case of a category (with a single object) with addition being the composition of morphisms.

But there is another monoidal structure on $\mathrm{Ab}$, namely the tensor product obtained by regarding abelian groups as $\mathbb{Z}$-modules. Using this monoidal structure when enriching produces the expected distributive compatibility condition.

This is more or less obvious, but maybe it doesn’t hurt saying it. In fact, the only reason why I am making this comment is that I was myself mixed up about this at one point.

Posted by: urs on September 2, 2006 1:04 PM | Permalink | Reply to this

### Re: Ringoids

John wrote:

An $R-\mathrm{Mod}$-category is called an $R$-algebroid or simply an algebroid.

What is puzzling, though, is that nobody seems to be aware of an equally nice characterization of the concept of Lie algebroid.

(Or is it maybe the same, and I just don’t see it?)

A Lie group has a Lie algebra. A Lie groupoid has a Lie algebroid.

But while groups and groupoids have nice arrow-theoretic definitions, the definition of a Lie algebroid ($\to$) is a mess, comparatively.

Posted by: urs on September 2, 2006 1:10 PM | Permalink | Reply to this

### Re: Ringoids

Urs writes:

But while groups and groupoids have nice arrow-theoretic definitions, the definition of a Lie algebroid ($\to$) is a mess, comparatively.

Indeed! On page 43 here I show a hypothesized “periodic table” of Lie n-groupoids, and on page 44 a corresponding periodic table of Lie n-algebroids. The first row of the periodic table of Lie n-algebroids is funny, because they have a smooth space of objects instead of a vector space.

It’s possible that someone good at making things elegant can polish up the existing theory of Lie algebroids… but there seems to be something we don’t understand about this stuff, which may only become clear when we study the whole periodic table of Lie n-algebroids.

Posted by: John Baez on September 3, 2006 1:33 AM | Permalink | Reply to this

### Re: Ringoids

It’s possible that someone good at making things elegant can polish up the existing theory of Lie algebroids…

Maybe it would be helpful to understand the semi-discrete case first.

Assume our groupoid has just a set of objects (instead of a manifold of them). Just a finite set, say. But assume that the automorphism group of each object is a Lie group $G$.

Can one associate a nice algebroid $A$ to such a groupoid? $\mathrm{Hom}_A(x,x)$ should probably be the enveloping algebra of the Lie algebra of $G$. What is $\mathrm{Hom}_A(x,y)$ for $x\neq y$?

Posted by: urs on September 19, 2006 11:59 AM | Permalink | Reply to this

### Re: Ringoids

John wrote:

It’s possible that someone good at making things elegant can polish up the existing theory of Lie algebroids…

Urs wrote:

Maybe it would be helpful to understand the semi-discrete case first. Assume our groupoid has just a set of objects (instead of a manifold of them). Just a finite set, say. But assume that the automorphism group of each object is a Lie group $G$.

Can one associate a nice algebroid $A$ to such a groupoid? $\mathrm{Hom}_A(x,x)$ should probably be the enveloping algebra of the Lie algebra of $G$. What is $\mathrm{Hom}_A(x,y)$ for $x\neq y$?

I’ve always been a bit confused when you say “algebroid” to mean “Lie algebroid”. To me they are different things, which reduce to “algebras” and “Lie algebras”, respectively, in the one-object case.

(For me an algebra with no adjectives in front means an associative unital algebra. An algebroid with no adjectives in front is a $\mathrm{Vect}$-enriched category. A one-object algebroid is then an algebra.)

But now I’m even more unhappy because you seem to be trying to construct an honest algebroid, rather than a Lie algebroid, from a Lie groupoid!

Anyway, let’s follow your suggestion and take a Lie groupoid with a discrete set (= 0-manifold) of objects. This is just a disjoint union (= coproduct) of Lie groups

$\coprod_i G_i .$

And, its Lie algebroid will be just the disjoint union of Lie algebras

$\coprod_i \mathrm{Lie}(G_i).$

And you’re right that in this case, we can go a further step and form a “universal enveloping algebroid” of our Lie algebroid, namely just the disjoint union of universal enveloping algebras:

$\coprod_i \mathrm{U}(\mathrm{Lie}(G_i)).$

But, does anyone ever try to form a “universal enveloping algebroid” for a general Lie algebroid???

By the way, here is a paper that we should read:

It talks about getting Lie n-groupoids from Lie n-algebroids, and it cites our paper on 2-groups from loop groups. Alissa Crans pointed it out to me.

Posted by: John Baez on September 21, 2006 2:22 AM | Permalink | Reply to this

### Re: Ringoids

I’ve always been a bit confused when you say “algebroid” to mean “Lie algebroid”.

I’ll stop doing that. Bad habit.

you seem to be trying to construct an honest algebroid, rather than a Lie algebroid, from a Lie groupoid!

Yes, I was trying to get a handle on the question what a nice arrow-theoretic description of Lie-algebroid would be by passing from Lie algebras to their enveloping algebras. Maybe it’s not fruitful. It was just a suggestion for how to possibly make progress.

take a Lie groupoid with a discrete set (= 0-manifold) of objects. This is just a disjoint union (= coproduct) of Lie groups

Wait, not necessarily. That case is not interesting enough to be of value here.

We can have a discrete set of objects and still have morphisms $x \to y$ for $x \neq y$.

Consider a group $G_x = \mathrm{Hom}(x,x)$ and a group $G_y = \mathrm{Hom}(y,y)$. Then morphisms $x \stackrel{f}{\to} y$ would be labeled by group isomorphisms $G_y \to G_x$, because

(1)$x \stackrel{f}{\to} y \stackrel{g_y}{\to } y \stackrel{f^{-1}}{\to} x := x \stackrel{g_x = f(g_y)}{\to} x \,.$

So, I thought, since I know how to associate an algebra $\mathrm{Hom}_A(x,x) = U(\mathrm{Lie}(G_x))$ to $G_x$ and analogously for $G_y$, can I maybe consistently find a vector space $\mathrm{Hom}_A(x,y)$ such that $A$ becomes an algebroid?

Posted by: urs on September 21, 2006 12:58 PM | Permalink | Reply to this

### Re: Ringoids

Urs wrote:

We can have a discrete set of objects and still have morphisms $x \to y$ for $x \ne y$.

Whoops - can I blame my mistake on jet lag? I think I got confused because a category is said to be discrete when there are no morphisms of the sort you mention… but you clearly were talking about the other sort of discreteness, namely the objects forming a discrete space.

So, I thought, since I know how to associate an algebra $\mathrm{Hom}_A(x,x) = U(\mathrm{Lie}(G_x))$ to $G_x$ and analogously for $G_y$, can I maybe consistently find a vector space $\mathrm{Hom}_A(x,y)$ such that $A$ becomes an algebroid?

Yes, and you can just use $U(\mathrm{Lie}(G_x))$, with composition of morphisms being multiplication.

The reason is that if a Lie groupoid $C$ has a discrete space of objects, it’s smoothly equivalent to any skeleton $\mathrm{Sk}(C)$, and the latter is just a coproduct of Lie groups. This means we can just do stuff in the trivial case I mentioned and transport it over to your case using the smooth equivalence. The result will be as I said, if $x$ lies in the skeleton.

A couple of words of explanation in case any lurking readers want them:

A topological category is a category internal to $\mathrm{Top}$. These are different than categories in certain ways: in particular, they aren’t always equivalent to a skeletal subcategory.

Given a category $C$ we can form a skeletal subcategory $\mathrm{Sk}(C)$ by taking one representative of each isomorphism class of objects, and all the morphisms between these. This skeleton will then be equivalent to $C$.

But, this doesn’t hold internal to $\mathrm{Top}$, since constructing the equivalence uses the axiom of choice, and the axiom of choice fails in $\mathrm{Top}$.

In other words, a skeleton automatically comes with an inclusion

$\mathrm{Sk}(C) \to C$

but finding a map going the other way:

$C \to \mathrm{Sk}(C)$

requires picking for each object of $C$ an object in the skeleton, and this choice can’t usually be done in a continuous way.

However, this choice can be done continuously when the space of objects of $C$ is discrete - since then continuity becomes vacuous.

All this is also true for “smooth categories”, i.e. categories internal to your favorite category of smooth spaces, e.g. smooth manifolds.

Posted by: John Baez on September 21, 2006 4:09 PM | Permalink | Reply to this

### Re: Ringoids

This means we can just do stuff in the trivial case I mentioned and transport it over to your case using the smooth equivalence.

Oh, of course, sure.

This amounts operationally to picking a fixed isomorphism $G_x \to G_y$ and then using that to map all such isomorphisms to elements of $G_x$, say.

Hm, let’s see. Suppose now I have a proper Lie groupoid, with a smooth manifold of objects. I wanto to build an algebroid $A$, such that $\mathrm{Hom}_A(x,x) = U(\mathrm{Lie}(G_x))$ for all objects $x$.

Then I might take $\mathrm{Hom}_A(x,y)$ to be the collection of pairs

(1)$(t,k) \in U(\mathrm{Lie}(G_x)) \times \mathrm{Iso}(G_x,G_y)$

divided out by the equivalence relation

(2)$(t,k) \sim (t',k') \;\; \Leftrightarrow t' = k'^{-1}(k(t)) \,.$

On the right is the induced action of a group automorphism on the Lie algebra.

The linear structure would be

(3)$c_1(t,k) + c_2(t',k) = (c_1 t + c_2 t', k) \,.$

Composition would be defined as

(4)$x \stackrel{(t_1,k_1)}{\to} y \stackrel{(t_2,k_2)}{\to} z = x \stackrel{(t_1 k_1^{-1}(t_2),k_2 \circ k_1)}{\to} z \,.$

Do I get an algebroid this way? Or did I make a mistake?

Posted by: urs on September 21, 2006 5:17 PM | Permalink | Reply to this

### from algebroids to Lie algebroids

Above, I tried to associate to any Lie groupoid $\mathbf{G}$ a smooth algebroid $\mathbf{A}$ in such a way that if $\mathbf{G}$ has a single obect then $\mathbf{A}$ is the universal enveloping algebra of the Lie algebra of $\mathbf{G}$.

I haven’t checked this carefully, but let me assume for a moment a construction along the above lines does work.

Then it seems to be rather straightforward to pass from $\mathbf{A}$ to a Lie algebroid, in a manner that generalizes how we would pass from $U(\mathrm{Lie}(G))$ to $\mathrm{Lie}(G)$ - namely by restricting to generators and taking the Lie product to be the commutator.

It’s obvious what to do in the case that $\mathbf{G}$ is the transport groupoid of a trivial $G$-bundle $P$ (for $G$ some Lie group), i.e.

(1)$\mathbf{G} = P \times P /G \,.$

So let’s look at how it works in this case an then try to reduce the general case to this trivial case.

Since in the trivial case all vertex groups $\mathrm{Hom}_\mathbf{G}(x,x) = G$ are canonically identified, we may forget all the isomorphism gymnastics that I mentioned above and simply identify all $\mathrm{Hom}$-sets as $G$.

Let’s then say that a section of $\mathbf{G}$ is a vector field on the manifold of objects equipped with a smooth assignment of an element of $\mathrm{Lie}(G)$ to each vector of the vector field.

On these sections, we naturally have a Lie bracket operation obtained by a slight adaption of the Lie bracket on vector fields. To compute $[v,w]$ we flow, at each point, a little along $v$, then a little along $w$, then back along $v$ and back along $w$. Only that we now accompany this process by the corresponding trajectory on $G$, which is at each step tangent to the correcponding Lie algebra element associated to our vector field.

I guess this does indeed reproduce the Lie algebroid structure of the Lie algebroid associated to $P \times P/G$.

For the case of a general groupoid, I guess we simply add to the definition of section given above the additional data consisting of, for each object, a small neighbourhood of that object and a choice of isomorphisms of all vertex groups in that neighbourhood. Then we compute the brackets of sections as above, by doing the computation pointwise in one of these neighbourhoods, using the locally chosen isomorphisms to get us back to the trivial case, locally.

Posted by: urs on September 21, 2006 9:01 PM | Permalink | Reply to this

### Re: Ringoids and categories

An earlier, useful textbook with contents and references relevant to this question is also:

– Abelian Categories with Applications to Rings and Modules –

by N. Popescu, Academic Press: New York and London, 1973.

Posted by: I.C. Baianu on September 3, 2006 2:17 AM | Permalink | Reply to this
Read the post Kosmann-Schwarzbach & Weinstein on Lie Algebroid Classes
Weblog: The n-Category Café
Excerpt: Kosmann-Schwarzbach on classes of Lie algebroids generalizing the character of a Lie algebra.
Tracked: October 10, 2006 2:15 PM
This page now linked under “horizontal categorification” in the $n$Lab.