## September 3, 2006

### Doctrines

#### Posted by John Baez

Universal algebra began as an attempt to deal with lots of familiar algebraic gadgets simultaneously: groups, rings, lattices, vector spaces, Lie algebras, and so on. A bunch of the same theorems hold for all of these - so why not prove them all at once and be done with it? It turns out to be possible!

In his famous 1963 thesis, Bill Lawvere showed how universal algebra can be formulated using categories:

He saw that universal algebra was all about algebraic gadgets that make sense in any category with finite products. Shockingly, every such category $C$ can be seen as the “theory” of some such gadget. If $D$ is some other category with finite products, a product-preserving functor $F: C \to D$ then gives such a gadget in $D$. We call this a model of $C$ in $D$.

However, besides categories with finite products, there are also other kinds of categories, which support other kinds of algebraic gadgets. So, around 1969, Lawvere massively generalized universal algebra - so much that he required a precise definition of “a kind of category”! He called such a thing a “doctrine”:

• F. William Lawvere, Ordinal sums and equational doctrines, Springer Lecture Notes in Mathematics No. 80, Springer-Verlag (1969), pp. 141-155.

This is a startling leap in abstraction, typical of Lawvere’s work.

Let’s talk about doctrines here. I’ll start by answering a question Urs Schreiber asked in another thread… and after I get some guesses, I’ll give the precise definition of a doctrine.

Urs writes:

Given any category $C$. Can anyone stop me from addressing it as a syntax?

I can’t stop you - I’m all the way over here in Shanghai!

Given any functor $C \to D$. Do I have the right to call it a semantics for $C$?

Of course you have the “right”. It’s a free country! (Germany, I mean.)

However, to be understood, it will be much better if you call $C$ a “theory”, and $F$ a “model” of that theory in $D$.

More importantly, you are missing the crucial notion of a doctrine. Mike Stay already tried to explain it to you, but let me try again.

Any category can be regarded as a theory of some kind, and any functor can be regarded as a model of some kind.

“Some kind”???

Yes: and a “doctrine” is what picks out a specific kind of category, and a specific kind of functor.

For example, when you mentioned the theory of groups, Th(Grp), you mentioned that this is a category with finite products - that’s a kind of category. But when you said that a model of this theory was a functor

$F: Th(Grp) \to D$

you forgot to note that $F$ must be a functor preserving finite products - that’s a kind of functor.

So, you weren’t clearly specifying the relevant “doctrine” in which Th(Grp) lives - namely, the his the doctrine of algebraic theories:

• categories with finite products,
• functors between these, preserving finite products,
• natural transformations between these.

This doctrine is also called FinProdCat for short. It was Lawvere who first realized that universal algebra was really all about this doctrine. (For an intro to universal algebra, try this.)

For comparison, over on the lambda calculus thread, we’ve been talking about the doctrine of lambda theories. This consists of:

• cartesian closed categories,
• functors between these, preserving finite products
• natural transformations between these

This doctrine is also called CCC for short. It was Lambek who first realized that the lambda calculus was really all about this doctrine. (For an intro to the lambda calculus, try this.)

And on the quantum computation thread, we are talking about various other doctrines, such as the doctrine of PROPs, consisting of

• symmetric monoidal categories,
• symmetric monoidal functors between these
• symmetric monoidal natural transformations between these

This doctrine is also called SymmMonCat. A theory in this doctrine is just a symmetric monoidal category, but Adams and MacLane called it PROP, and people still use that term. Certain special PROPs called “operads” are even more famous. (For an intro to PROPs and operads, try this.)

As an exercise, try to the general concept of “doctrine” more precise. I didn’t want to lay the precise definition on you before you saw these examples! I do want to state the definition, because it’s quite simple and beautiful, but it’s best if people make some guesses first. If people get stuck they can read this:

• A. Kock, G. Reyes, Doctrines in categorical logic, in Handbook of Mathematical Logic, ed. J. Barwise, North Holland 1977.

(Alas, I don’t know an online introduction to doctrines. Does anybody?)

Urs writes:

But assume I feel like it and take any old group $G$, regard it as a category $\Sigma(G)$ with a single object – and then declare that I want to think of $\Sigma(G)$ as a syntax.

First, just to help you sound like an expert, I’d urge you to say theory instead of syntax here, and similarly model instead of semantics. I’ll make those changes below:

But assume I feel like it and take any old group $G$, regard it as a category $\Sigma(G)$ with a single object – and then declare that I want to think of $\Sigma(G)$ as a theory.

Then the first thing I’d ask is “a theory in what doctrine, Urs?”

Next, assume I want to pick any functor

(1)$\Sigma(G) \to \mathrm{Vect}_\mathbb{C}$

and say that this is a model for my theory $\Sigma(G)$?

Then I’d ask “a model in what doctrine, Urs?”

Would you tell me that I am free to do so if I like, or would you point out that $\Sigma(G)$ violates some property that a category must have if we are going to admit that it is a kind of theory?

It all depends on the doctrine! If you want to let any functor count as a model, maybe you want to use the blandest doctrine possible. You can think of a doctrine as an “amount of structure we want our categories to have”, so Mike called this the doctrine of no structure at all. It consists of:

• categories,
• functors between these
• natural transformations between these

This is also called Cat, the 2-category of categories!

If we pick this doctrine, then yes: for any group $G$, the corresponding 1-object category $\Sigma(G)$ is a theory in this doctrine! And, any functor

$F : \Sigma(G) \to D$

is a model. In fact, this example is indeed very important. What do we normally call a model of this sort?

Posted at September 3, 2006 4:14 AM UTC

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### Re: Doctrines

Thanks for this indoctrination!

So instead of admitting that I am talking about a linear representation of some group $G$, I could equivalently say that I am talking about a model in $\mathrm{Vect}$ of the theory $\Sigma(G)$ with respect to the doctrine $\mathrm{Cat}$ of no structure.

As an exercise, try to make the general concept of ‘doctrine’ more precise.

Oh dear. I will fail miserably.

Since $\mathrm{Cat}$ may be addressed as the doctrine of no structure, a first guess would be that a doctrine is a 2-category of categories with certain structure, structure preserving functors between them and natural transformations between these.

Instead of trying to make that more precise, I ask a counter question:

whatever the right answer is, it seems clear that “doctrine” is just the “n=1”-version of a general concept.

What is a 0-doctrine? By my above guess, it would just be any category of sets with structure and structure preserving functions between them.

Posted by: urs on September 4, 2006 6:04 PM | Permalink | Reply to this

### Re: Doctrines

Urs wrote:

So instead of admitting that I am talking about a linear representation of some group G, I could equivalently say that I am talking about a model in Vect of the theory $\Sigma(G)$ with respect to the doctrine Cat of no structure.

Yes, but you didn’t really answer my question. A “model in $D$ of the theory theory $\Sigma(G)$ with respect to the doctrine Cat of no structure” is highly esoteric way of talking about a functor

$F: \Sigma(G) \to D.$

But my question was, what do we normally call such a functor?

As you note, when $D$ is $\Vect$ we call it a representation of the group $G$. But what about for general $D$?

I’m trying to get you to answer this so we can work our way up to seeing precisely which class of algebraic gadgets can be described using the doctrine of “categories with no structure”.

This is the first case of a much larger project that people have carried out, namely to take lots of doctrines and see precisely which class of algebraic gadgets can be described in each one. If you meet a sophisticated categorical logician, you can hand him any sort of algebraic gadget - a Hopf algebra or a field or a local ring or a Boolean algebra or a Heyting algebra - and they’ll you which doctrine this concept lives in. That means they know which kind of category you can define this gadget in!

For example, groups - or “group objects” - can be defined in any category with finite products. A Lie group is a group object in the category of smooth manifolds, a strict 2-group is a group object in the category of categories, and so on. But monoids - or “monoid objects” - are far more robust, since their axioms don’t require duplication or deletion of variables. They make sense in any monoidal category! For example, an algebra is a monoid object in $(Vect, \otimes)$, a ring is a monoid object in $(Ab, \otimes)$, and so on.

Concepts that can be defined in any category are the most robust of all. This is why the “doctrine of categories with no extra structure” is sort of important.

On to another question…

As an exercise, try to make the general concept of ‘doctrine’ more precise.

Urs sighed:

Oh dear. I will fail miserably.

Since Cat may be addressed as the doctrine of no structure, a first guess would be that a doctrine is a 2-category of categories with certain structure, structure preserving functors between them and natural transformations between these.

Yes! So, a doctrine involves a 2-category $D$, and the question becomes: how can we be more precise in saying that $D$ is a 2-category of categories equipped with extra structure?

Urs wrote:

Instead of trying to make that more precise…

Well, you took a good step forwards! You’ve told us that a doctrine consists, first of all, of a 2-category $D$. And I bet you can say something about what it means to say this $D$ consists of categories with extra structure.

Maybe you can’t say everything about what this means… all I want is something.

Urs wrote:

… I ask a counter question:

whatever the right answer is, it seems clear that “doctrine” is just the “n=1”-version of a general concept.

What is a 0-doctrine? By my above guess, it would just be any category of sets with structure and structure preserving functions between them.

Right! Excellent!

Since this is the n-Category Café, as soon as we understand doctrines we’re going to $n$-categorify them. But Toby has taught us the power of negative thinking: before we try to categorify a concept, we should try to do decategorify it, to understand it better. Before we get too far on understanding 1-categories with extra structure, maybe we should understand 0-categories with extra structure: that is, sets with extra structure.

So, I could also ask you this: what does it mean to say a category $D$ is a “category of sets with extra structure”?

(By the way, just for the experts in the crowd, I must admit that a doctrine is not just a 2-category of categories with extra structure. It’s a 2-category $D$ of categories with extra properties, structure, and stuff.

We haven’t been emphasizing the role of $n$-stuff in this discussion - we’ve been loosely calling it “structure”. But if we want to get technical, we must admit that while FinProdCat only has more structure than Cat, SymmMonCat has more stuff than Cat.

Why? Well, the morphisms in FinProdCat are functors with extra properties - they’re product-preserving functors. But, the morphisms in SymmMonCat are functors equipped with extra structure - they’re symmetric monoidal functors. Equipping the morphisms with extra structure means we’re equipping the objects with extra stuff.

We probably shouldn’t talk about this much now; it’s sort of distracting, but I had to mention it for the sake of honesty!)

Posted by: John Baez on September 5, 2006 3:53 AM | Permalink | Reply to this

### Re: Doctrines

[…] a functor

(1)$F : \Sigma(G) \to D \,.$

But my question was, what do we normally call such a functor?

As you note, when $D$ is $\mathrm{Vect}$ we call it a representation of the group $G$. But what about for general $D$?

For general $D$, we call such a functor an action of $G$ on an object in $D$.

For instance, if $D$ is the category of manifolds, a functor $F : \Sigma(G) \to D$ is a manifold $X = F(\bullet)$, together with a diffeomorphism $X \stackrel{F(g)}{\to} X$ for each $g \in G$ such that - well - such that these diffeomorphisms represent an action of the group $G$ on $X$.

Or let $D$ be the category of some kind of bundle on some base space $X$. Then $F : \Sigma(G) \to D$ is a $G$-equivariant bundle.

Now let $X$ be a point and let the “kind of bundles” be “vector bundles”. Then the above example reduces to an ordinary linear representation of the group $G$.

This little trick allows us, for instance, to address linear reps $F : \Sigma(G_2) \to D$ of a 2-group $G_2$ as an equivariant version of 2-bundles over a point. This is a big deal in the recent work by Ganter and Kapranov.

Posted by: urs on September 5, 2006 10:51 AM | Permalink | Reply to this

### Re: Doctrines

John wrote approximately:

What do we normally call a functor $F : \Sigma(G) \to D ?$ As you note, when $D$ is $Vect$ we call it a representation of the group $G$. But what about for general $D$?

Urs wrote:

For general $D$, we call such a functor an action of $G$ on an object in $D$.

Yeah!

And the point is, we can describe such actions within the simplest doctrine in the world: the doctrine of categories with no extra structure. That’s because the operations involved in a group action all have just one input and one output. To get multi-input, multi-output operations we need to go to some doctrine like that of monoidal categories, or maybe some especially nice monoidal categories.

But, we can describe more than just group actions while staying within the doctrine of categories with no extra structure. We can describe actions of monoids - dropping the requirement that our operations be invertible. We can even describe actions of categories! Any functor

$F : C \to D$

can be considered an action of the category $C$, generalizing the notion of group action.

If we prefer to act like logicians, we can call $C$ a “theory” in the doctrine Cat. Then the objects of $C$ should be called types, while the morphisms should be called operations.

The Moral: If we’re trying to describe some algebraic gadget with a bunch of operations satisfying equations, as long as these operations all have one input and one output, we can describe it within the simplest doctrine in the world: Cat. In other words, such gadgets, living in a category $D$, are nothing but functors $F : C \to D$ for some specific category $C$ - the “theory” of these gadgets.

Posted by: John Baez on September 6, 2006 11:21 AM | Permalink | Reply to this

### Re: Doctrines

And I bet you can say something about what it means to say this $D$ consists of categories with extra structure.

Yes, yes, I can. I have heard somebody (was it you? ;-) ) talk about the yoga of stuff, structure and property once in a while.

Let me just make a quick search through the available web documents… hm, lots of email discussion,… lots of usenet discussion,… some notes by Toby Bartels,… Not sure which the canonical reference is at the moment. I think I’ll stick to section 4.4 (pp. 27-29) of Jeffrey Morton’s paper math.QA/0601458 for the moment.

So, let’s see. Given an $n$-functor, we can measure how forgetful it is by checking if it is more or less surjective on $p$-morphisms.

Restricting to 1-functors for the moment, we have.

1) Functors that are surjective on objects (up to isomorphism), on morphisms and on 2-morphisms. These are called essentially, surjective, full and faithful. By a standard theorem, such a functor is an equivalence of categories, hence does not forget anything.

2) Some functors may not hit every isomorphism class of objects in the codomain but are still surjective on the remaining 1- and 2-morphisms, hence full and faithful. We say (or the Baez-ian school says) that such functors forget property (and only property). That’s because this makes sense if you look at examples, like the functor from abelian groups to all groups.

3) Some functors may not hit every isomorphism class of objects and not even every morphism in the remaining Hom-sets. They are not full. We say these also forget structure. If they are still faithful, though, we say they forget only property and structure.

4) Finally, if the functor is neither (essentially) surjective on 0-morphisms, 1-morphisms nor 2-morphisms, it is said to forget not just properties and structures, but even stuff.

(People should check the examples in Jeffrey’s text.)

Good. So as a warmup, I will now try to say what a 0-doctrine should be.

A 0-doctrine should be a category of sets with extra structure and structure preserving morphisms between them.

I use the above yoga to say this in a more precise fashion.

A 0-doctrine is any category $C$ with a functor

(1)$F : C \to \mathrm{Set}$

such that $F$ is faithful.

To borrow Jeffrey’s example 10, I’d say there is for instance a 0-doctrine of total orders, namely the category of all totally ordered sets and order-preserving functors. When we inject this into all sets, i.e. into $\mathrm{Set}$, we clearly don’t hit every morphism between sets, just those that were order preserving as long as we remembered a total order on our sets. Still, it’s injective on morphisms, hence faithful. So this functor forgets the structure of being totally ordered. It still remembers all the available stuff, though, being injective on morphisms.

So now I have to lift all of the above from 1-functors to 2-functors. Once I manage to do that, I will be able to say what a (1-)doctrine is. It should read

A doctrine is any 2-category $D$ together with a 2-functor

(2)$F : D \to \mathrm{Cat}$

such that $F$ forgets at most structure.

And this should mean that either

1) that $F$ is surjective on 3-morphisms (i.e. injective on 2-morphisms) and not required to be anything else

2) or that $F$ is surjective on 3 and on 2-morphisms.

In order to decide this, I should look at some example.

But I will do that later, since have to do something else right now.

Posted by: urs on September 5, 2006 11:58 AM | Permalink | Reply to this

### the syntax of quantum mechanics

Maybe I should remind our physically inclined $n$-Café readers that my questions cited by John above were part of an effort to understand what it means to say that

The syntax of quantum mechanics is quantum $\lambda$-calculus. #

This is a pretty cool statement!

In my attempts to decode this statement I first caught a mutant strain # of the true concept; and the $n$-category doctor diagnosed conceptual sickness.

(I don’t know if it’s related, but over the week-end I also must have cought a more ordinary virus and became sick in the more traditional sense. Poor me.)

Given what John said here maybe I can phrase things like this:

A particular quantum theory is this:

1) a collection of Hilbert spaces $\{H_i\}$, which are the spaces of $i$-states.

For instance for the ordinary quantum mechanics of a single nonrelativistic particle there is just a single such Hilbert space, which is the space of states of precisely that particle.

As another example, in the quantum theory of a linearly extended thing like a string, we may have in general at least two different Hilbert spaces, $H_c$ of closed and $H_o$ of open string states.

Usually we want to admit several copies of the “single” quantum object we are describing, and hence we also consider all the tensor powers of the Hilbert spaces $\{H_i\}$. For instance $H_c\otimes H_c$ is the space of states of two closed strings.

It is a little easier to illustrate the following concepts in the string example than in the particle example, so I’ll stick to the string. If you don’t like strings you are invited to replace “string” by “1-dimensional spacetime” without loss of anything substantial here.

2) A quantum object in a given state may usually evolve into some other state. There are in general different “channels” along which to involve. Each such channel is characterized by the space of states of objects coming into the channel and the space of states of objects coming out of the channel on the other side.

So we are tempted to regard a “channel” (it’s really a common term in particle physics) as a morphism

(1)$H_{\mathrm{in}} \stackrel{\text{channel}}{\to} H_\mathrm{out} \,.$

For instance, a cylinder (usually called a cylinder diagram in this context) can be regarded as a channel

(2)$H_c \stackrel{\text{cylinder}}{\to} H_c$

which describes a closed string coming in, just propagatinng without interaction and coming out again.

Slightly more interestingly, a sphere with three disks cut out (usually called the pair-of-pants diagram) may be regarded as a channel

(3)$H_c \otimes H_c \stackrel{\text{p.o.p.}}{\to} H_c \,.$

In order to emphasize the relation of this business to the theory of (quantum) computation, we simply go ahead and rename everything encountered so far.

Instead of spaces of states we’ll talk about types. The elementary spaces of states (like $H_c$ and $H_o$ above) we call generating types.

Instead of channels we now say operations.

Composition of different channels usually obeys certain rules. For instance in the theory of “topological strings” we want a rule that the two ways to compose the pair-of-pants channel with itself lead to the same result. Such conditions we now call equations between functions.

Finally, we want to be able to consider different theories of topological strings, say, each of which involves the same sort of channels. In a lucky coincidence of terminology, different such theories are often called different models of the given quantum theory (like the “A-model” or the “B-model” of the topological string).

This “model”-terminology now becomes precise as follows.

We realize that all such theories can be regarded as functors

(4)$2\mathrm{Cob} \to \mathrm{Hilb}$

from 2-dimensional (open/closed) cobordisms to Hilbert spaces.

Whereas at the beginning I was thinking of quantum theories in all generality, right now I am making some sort of restriction to what are usually called “topological theories”.

As far as I understand, the crucuial point of this restriction for our current purpose is that the category $n\mathrm{Cob}$ is cartesian closed. For instance, the internal $\mathrm{hom}$-object of morphisms from the circle to the circle is just

(5)$S^1 \sqcup \bar S^1 \,,$

where the overbar denotes orientation reversal.

This now finally is getting close to answering the question what it means to say that quantum $\lambda$-calculus is the syntax of quantum mechanics.

Namely, $n\mathrm{Cob}$ is a CCC (a cartesian closed category), and, using the above language of types, functions and equations, a functor (preserving the relevant structure) from any $CCC$ $C$ to $\mathrm{Hilb}$ we would call a model of the $\lambda$-theory $C$.

For the special case that $C = n\mathrm{Cob}$ we know that we may address such a functor equivalently as a model of topological “$n$-particles (= $(n-1)$-branes). For other CCCs we may still imagine addressing the functor as a quantum theory, though it might be an exotic sort of quantum theory that no physicist has ever dreamed of, I guess.

That’s currently roughly my understanding of what John and Mike are trying to tell me.

One thing I don’t understand yet is this:

is it reasonable to restrict the notion of “models in quantum mechanics” to functors whose domain is cartesian closed?

Doesn’t that exclude any “non-topological” theory?

For instance in ordinary non-relativistic QM of a single point particle we actually need 1-dimensional Riemannian cobordisms as domain.

Or the non-relativistic string. It needs 2-dimensional conformal cobordisms.

But these categories of cobordisms with extra structure are not cartesian closed, are they?

Is it therefore maybe better to say that “Quantum $\lambda$-calculus is the syntax of topological quantum mechanics”?

Posted by: urs on September 4, 2006 7:51 PM | Permalink | Reply to this

### Re: the syntax of quantum mechanics

Urs writes:

The syntax of quantum mechanics is quantum $\lambda$-calculus.

This is a pretty cool statement!

Yeah! I hope the folks at the NSF think so too.

In my attempts to decode this statement I first caught a mutant strain of the true concept; and the n-category doctor diagnosed conceptual sickness.

(I don’t know if it’s related, but over the week-end I also must have caught a more ordinary virus and became sick in the more traditional sense. Poor me.)

Oh dear - I hope you get better soon. It probably is related: as David’s book shows, there’s a strong relation between mental health and bodily health - so if you get infected by mutant doctrines, you’re bound to catch a cold too.

Your example of open-closed topological string theories is great! Thanks for bringing physics back into the game - doctrines can seem like very rarified abstract notions until we bring them down to earth and show how they apply to something concrete and practical like… umm, string theory.

Seriously, open-closed topological strings are great for illustrating the idea of objects as types - we have a type “open string”, and a type “closed string”, and various operations involving these types.

I just have one complaint - probably you’re not completely recovered from that mutant strain. You say:

… the category $n$Cob is cartesian closed.

But it’s not! It’s a symmetric monoidal category, but the tensor product - disjoint union of $(n-1)$-manifolds - is not cartesian: it’s not the product in the category-theoretic sense. This is one reason $n$Cob is “quantum” in nature: we can’t duplicate and delete information.

And, $n$Cob is closed, but far from being cartesian closed it has duals for objects: the internal hom is given by

$\mathrm{hom}(a,b) = a^* \otimes b$

where $*$ is orientation-reversal and $\otimes$ is disjoint union. This is completely different from a cartesian closed category like Set: in Set we have

$\mathrm{hom}(a,b) = b^a \ne a^* \times b$

Here $b^a$ is the set of functions from $a$ to $b$, and there’s no way to define a “dual” of a set $a$ that gives $b^a = a^* \times b$!

is it reasonable to restrict the notion of “models in quantum mechanics” to functors whose domain is cartesian closed?

I have to unask the question. The ordinary lambda calculus concerns cartesian closed categories. The quantum lambda calculus concerns symmetric monoidal categories with duals. The latter is what you should be talking about here.

(If anyone reading this is getting bogged down with all the jargon, they might try the list of definitions here.)

By the way, Lauda and Pfeiffer develop open-closed topological string theory very beautifully along the lines we’re discussing here:

They show the category of open-closed topological strings is the free symmetric monoidal category on a “knowledgeable Frobenius algebra”. This turns out to be a symmetric monoidal category with duals.

Posted by: John Baez on September 5, 2006 5:02 AM | Permalink | Reply to this

### Re: the syntax of quantum mechanics

the category $n\mathrm{Cob}$ is cartesian closed.

But it’s not!

Ah, of course. What I meant to say is just that it’s closed. That’s why I continued in the next sentence with describing the internal hom-objects.

I have to unask the question.

Is it reasonable to restrict the notion of “models in quantum mechanics” to functors whose domain is symmetric monoidal with duals? Doesn’t this restrict us to what are usually called “topological” quantum theories.

For instance, ordinary non-relativistic 1-particle quantum mechanics is a functor with domain 1-dimensional Riemannian cobordisms. That doesn’t have duals.

Posted by: urs on September 5, 2006 11:05 AM | Permalink | Reply to this

### Re: the syntax of quantum mechanics

John wrote:

I have to unask the question.

Urs wrote:

Is it reasonable to restrict the notion of “models in quantum mechanics” to functors whose domain is symmetric monoidal with duals? Doesn’t this restrict us to what are usually called “topological” quantum theories?

For instance, ordinary non-relativistic 1-particle quantum mechanics is a functor with domain 1-dimensional Riemannian cobordisms. That doesn’t have duals.

I tend to think of nonrelativistic quantum mechanics this way: we have a Hilbert space $U$, and a self-adjoint Hamiltonian on this Hilbert space. This generates a 1-parameter family of unitary operators

$U(t) : H \to H$

describing time evolution, where $t \in \mathbb{R}$. They form a 1-parameter group, meaning

$U(t+s) = U(t)U(s)$

In other words, we have a unitary representation of $\mathbb{R}$ on $H$.

Now, how do we think of this as a model of some theory in the doctrine of symmetric monoidal categories with duals?

It’s a model of a certain symmetric monoidal category with duals which you would probably call $\Sigma(\mathbb{R})$. As a category, this has one object. The morphisms from this object to itself are real numbers, and composition is addition.

In other words: morphisms in a category always describe processes, and morphisms in this category describe processes of “time evolution”!

This category becomes a symmetric monoidal category with duals in a thoroughly trivial way, except that the dual of the morphism $t \in \R$ is $-t$.

Then, a model of this symmetric monoidal category with duals in $\mathrm{Hilb}$ - that is, a symmetric monoidal functor preserving duals

$U: \Sigma(\mathbb{R}) \to \mathrm{Hilb}$

is nothing other than a unitary representation of $\mathbb{R}$ on some Hilbert space!

Note that the “symmetric monoidal” business was irrelevant here. That’s because we’re not talking about processes like this:

where several types of things come in and several go out. We just have one thing coming in and one going out. So, we don’t need tensor products in our doctrine to describe what’s going on!

We do, however, need the duals for morphisms, if we want to insist that time evolution is unitary.

It’s easy to generalize all this from $\mathbb{R}$ to whatever group you like, for example the Poincaré group.

You know, all this stuff might better fit into the discussion of Quantum computation and symmetric monoidal categories, because that’s where I’m trying to advertise the virtues of certain doctrines like “Symmetric Monoidal Categories with Duals”. But oh well… everything is related.

Posted by: John Baez on September 6, 2006 11:00 AM | Permalink | Reply to this

### Re: the syntax of quantum mechanics

You know, all this stuff might better fit into the discussion of Quantum computation and symmetric monoidal categories, because that’s where I’m trying to advertise the virtues of certain doctrines like “Symmetric Monoidal Categories with Duals”.

Okay. I have now moved the discussion back to that entry.

Posted by: urs on September 6, 2006 2:17 PM | Permalink | Reply to this

### Re: Doctrines

This is interesting. I had the idea that a doctrine is a 2-monad on Cat, but that is not consistent with your referring to various 2-categories as doctrines.

So perhaps a doctrine is the 2-category of (pseudo-)algebras for some 2-monad on Cat. But that isn’t right, because you mentioned the doctrine of CCCs.

Could it be that a doctrine is the 2-category of (pseudo-)algebras for some 2-monad on Catg, the 2-category of categories, functors and natural isomorphisms? That can’t be right either, because in some of your examples the 1-cells preserve all the relevant structure, but in others they preserve only some of it (e.g. the maps of CCCs preserve finite products up to isomoprhism, but not the exponentials).

So I’m guessing that the answer is something much simpler, that ignores distinctions of this sort. Perhaps a doctrine is a 2-category with finite products, together with a finite-product-preserving functor to Cat. That seems to fit your examples, but I’m sure it’s not the right answer. I look forward to learning what the right answer is. (Or else I’ll cheat and look in the Big Yellow Book tomorrow, as you suggested.)

Posted by: Robin on September 5, 2006 8:30 PM | Permalink | Reply to this

### Re: Doctrines

Robin wrote:

I look forward to learning what the right answer is.

It’s quite possible that in some of my posts I am stretching Lawvere’s original definition of “doctrine” in some dangerous ways. This would raise some interesting problems.

However, I don’t want to tackle these issues until Urs has guessed a bit more about Lawvere’s definition. It shouldn’t take long.

Posted by: John Baez on September 6, 2006 11:27 AM | Permalink | Reply to this

### Re: Doctrines

Okay, Robin - now I think we should talk about your objection, namely that the lambda-calculus doesn’t seem to fit into the “doctrine” framework, since we don’t require that a “model”

$F : X \to Y$

of the CCC $X$ in the CCC $Y$ be a functor that preserves internal homs. (Preserving the products seems to be all we want.)

Surely someone must have addressed this issue somewhere. No?

There’s no way to save the day?

Posted by: John Baez on September 7, 2006 3:14 PM | Permalink | Reply to this

### Re: Doctrines

I think I’ll keep posting the “main line” of my Socratic dialog with Urs as comments that will appear on the very bottom of this page, instead of commenting on a comment on a comment….

John wrote approximately:

Can you say anything about what it means to say $D$ is a doctrine, or 1-doctrine - that is, a 2-category whose objects are “categories with extra structure”?

Urs wrote:

Yes, yes, I can. I have heard somebody (was it you? ;-) ) talk about the yoga of stuff, structure and property once in a while.

Yes - most recently, in this thread I wrote:

I was trying to tell you not to bring the yoga of stuff, structure and property into this game! It’s relevant, but only in a subtle way. It’s not the royal road to the official “right answer” - that is, Lawvere’s definition of “doctrine”.

However, even if we don’t march down the royal road, we can still get there pretty quick by sneaking down back alleys. I will try not to let you get too lost.

So - you wisely decategorified my question and tackled this warmup question first:

What does it mean to say $C$ is a 0-doctrine - that is, a category whose objects are “sets with extra structure”?

It is any category $C$ with a functor $U : C \to \mathrm{Set}$ such that $U$ is faithful.

Great! This is the correct answer if you follow the yoga of stuff, structure and properties. Here $U$ is a functor that sends any object of $C$ to its underlying set - I’m using $U$ here because that’s the traditional letter: $U$ for “underlying”. You used $F$, but $F$ stands for something else that’s about to be very important.

You are right that $U$ forgets at most structure iff $U$ is faithful - in other words, the objects of $C$ don’t have any extra “stuff”, they’re just a set with extra structure and properties.

However, the official right answer is much better, because it’s extremely elegant yet allows us to prove lots more theorems. It relies on noticing something that lots of examples have in common.

Here are a few classic examples:

• Groups: $U : \mathrm{Grp} \to \mathrm{Set}$
• Rings: $U : \mathrm{Ring} \to \mathrm{Set}$
• Complex vector spaces: $U : \mathrm{Vect}_{\mathbb{C}} \to \mathrm{Set}$

There’s something very nice and simple you can do in all these cases that you can’t do in these cases:

• Let $C$ be the category of 5-element sets and functions between them, and let $U : C \to \mathrm{Set}$ be the obvious functor.
• Let $\mathrm{Set}_0$ be the group of sets and bijections betwen them, and let $U : \mathrm{Set}_0 \to \mathrm{Set}$ be the obvious functor.

In both these cases $U$ is faithful - but there’s something that doesn’t work for these that works for the previous ones. And, I hope you’ll agree that intuitively, these cases feel different than the previous ones!

So, I’ve given you a clue, but I’m not sure you’ll get it. If you don’t get it maybe someone else can chime in - or else I can say two magic words that will instantly make it all clear.

Posted by: John Baez on September 6, 2006 10:33 AM | Permalink | Reply to this

### Re: Doctrines

Posted by: David Corfield on September 6, 2006 11:22 AM | Permalink | Reply to this

### Re: Doctrines

David has spoken the magic words…

Posted by: John Baez on September 6, 2006 11:29 AM | Permalink | Reply to this

### Re: Doctrines

I need to run to get some lunch. But here is a quick guess what you all are trying to tell me:

A $0$-doctrine is a category with a faithful functor to $\mathrm{Set}$ such that it has an adjoint functor (the free construction).

Posted by: urs on September 6, 2006 12:55 PM | Permalink | Reply to this

### Re: Doctrines

Hm, now I am not sure what precisely I am supposed to say.

Having another look at John’s lecture on universal algebra, I see that, as Robin indicates, too, we are really talking about $n$-monads.

So from an adjunction we get a monad, and we can always assume the context to be such that every monad comes from an adjunction (p. 21 of the above lectures).

So maybe I should just say that a 0-doctrine is a monad on $\mathrm{Set}$.

Does that imply that the adjunction

(1)$\mathrm{Set} \stackrel{L}{\to} D \stackrel{R}{\to} \mathrm{Set}$

corresponding to that monad has faithful $R$? Or is that an extra condition? (I have never thought about this, but maybe faithfulness is in fact a necessary condition for there being an adjoint?)

In any case, this should maybe show how what I have been trying to say in terms of SSP yoga and what Robin was thinking (and what John was trying to make us figure out) merges.

Maybe an $n$-doctrine is just an $(n+1)$-monad, which, if we realize it in terms of an $(n+1)$-adjunction, involves an $(n+1)$-functor $R$ that forgets at most $(n+1)$-structure.

Or something like that.

Posted by: urs on September 6, 2006 1:52 PM | Permalink | Reply to this
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Tracked: September 6, 2006 12:23 PM

### Re: Doctrines

Hm, now I am not sure what precisely I am supposed to say.

Having another look at John’s lecture on universal algebra, I see that, as Robin indicates, too, we are really talking about $n$-monads.

So from an adjunction we get a monad, and we can always assume the context to be such that every monad comes from an adjunction (p. 21 of the above lectures).

So maybe I should just say that a 0-doctrine is a monad on $\mathrm{Set}$.

Does that imply that the adjunction

(1)$\mathrm{Set} \stackrel{L}{\to} D \stackrel{R}{\to} \mathrm{Set}$

corresponding to that monad has faithful $R$? Or is that an extra condition? (I have never thought about this, but maybe faithfulness is in fact a necessary condition for there being an adjoint?)

No, but… you’ve got the right idea, Urs!

A 0-doctrine is just a monad on $\mathrm{Set}$! This implies that the forgetful functor from the category of “algebras” of this monad to $\mathrm{Set}$ is faithful - but it’s much more powerful.

Let me sketch how this beautiful story goes - for details, people can read my universal algebra notes or else get serious and look at Toposes, Triples and Theories. (A “triple” is the same as a monad.)

As explained in “week89” of This Week’s Finds, a monad on a category $C$ is just a monoid $T$ in the monoidal category of functors from $C$ to itself. This monoidal category acts on $C$, so we can say what it means for $T$ to act on an object in $C$, and such an action is called an algebra of $T$.

Don’t worry if this seems mysterious. The first time someone told me this, here’s what I looked like:

Let me unpack what I just said. A monad is a functor

$T : C \to C$

with a multiplication

$m: T^2 \Rightarrow T$

and unit

$i: 1 \Rightarrow T$

satisfying the usual laws of a monoid: associativity and the left/right unit laws. A great example is the functor

$T: \mathrm{Set} \to \mathrm{Set}$

sending any set $S$ to the underlying set of the free group on $S$. Here we are building $T$ as the “free” functor (left adjoint)

$F : \mathrm{Set} \to \mathrm{Grp}$

followed by the “underlying” functor (right adjoint)

$U : \mathrm{Grp} \to \mathrm{Set}$

The properties of adjoint functors make $T$ into a monad. Indeed, any pair of adjoint functors gives a monad, and there’s an utterly simple picture proof of this using string diagrams, given in “week92”.

Continuing to unpack what I said, an algebra of a monad is an object $c \in C$ together with a morphism

$a: T c \to c$

satisfying the usual laws for an “action” of a monoid: the associative law and the left unit law.

It’s a great exercise to check that the algebras of the monad $T = U F$ built from

$F : \mathrm{Set} \to \mathrm{Grp}$

and

$U : \mathrm{Grp} \to \mathrm{Set}$

are precisely groups. There’s no possible way to truly understand what I’m saying until one does this exercise.

For any monad $T: C \to C$ we can define a category of algebras, often called $C^T$ for some silly reason. This is just like the usual category of actions of a monoid. But it’s a cool idea: in the example above it’s the category of groups!

There’s an obvious forgetful functor from algebras of our monad to the category we started with, say

$U': C^T \to C$

This is always faithful, since being an algebra of a monad is just extra structure on an object of $C$! The proof is not too hard - just axiom-juggling.

See, Urs? We get the faithfulness for free, from a very elegant set of ideas - we don’t need to stick it in by hand!

Furthermore, the functor $U'$ always has a left adjoint

$F': C \to C^T$

sending each object $c \in C$ to the “free” algebra $T c$.

$F' : C \to C^T$

and

$U' : C^T \to C$

and $U'$ is always faithful.

You’ll note we went around a kind of loop. If you give me a pair of adjoint functors

$F: C \to D$

$U: D \to C$

I can cook up a monad

$T = U F: C \to C$

Then I can cook up the category of algebras $C^T$ of this monad, and then I get an adjunction

$F' : C \to C^T$

and

$U' : C^T \to C$

Do I get back the adjoint functors I started with? Not always! But I do in the example of groups and sets, and all the examples we’re really interested in now. So, that’s the case we want to focus on. We call such adjoint functors monadic.

We can also run around the loop starting with a monad! Start with a monad $C$, build its category of algebras $C^T$, get a pair of adjoint functors between $C$ and $C^T$, and then use this to define a monad!

In this case we always do get back where we started. So this is the nicest way to play the game. Say we take $C = \mathrm{Set}$ to be specific:

We define a 0-doctrine to be a monad

$T: \mathrm{Set} \to \mathrm{Set}$

This allows us to define a certain “category of sets with extra structure”, namely namely the category of algebras of $T$, or $C^T$ for short. Then we get a pair of adjoint functors

$F' : C \to C^T$

and

$U' : C^T \to C$

and $U'$ is always faithful.

Actually, nobody ever talks about 0-doctrines. But this is the idea that Lawvere categorified to define doctrines, or what we would call “1-doctrines”. In brief:

A doctrine is a monad on $\mathrm{Cat}$.

We can play the whole game as before, defining algebras of these and so on.

There’s a lot more to say, but it’s late - and this is probably far more than anyone can digest who doesn’t sort of know this stuff already!

Posted by: John Baez on September 6, 2006 4:00 PM | Permalink | Reply to this

### Re: Doctrines

Thanks for the detailed explanation!

Okay, great.

I had cursorily looked at this universal algebra stuff before, but never thought about it seriously.

My little detour through SSP yoga was instructive, though. The faithfulness property you mention does not really make a prominent appearance in the usual discussion of monads etc,. but now I see why it is important conceptually.

In fact, our “socratic dialogue” has made explicit what might be for practical purposes the most sensible algorithm for producing new doctrines, or for choosing the doctrine one needs.

Namely: start with some 2-category of categories with extract structure, find the corresponding faithful 2-functor to $\mathrm{Cat}$ and check if it has an adjoint. If not, generalize your original 2-category and try again.

One thing that is irritating when learning this stuff is the use of the word “algebra”. I am more used to thinking of monads as algebras and of what are called algebras here as modules for these algebras. But all right.

Another thing that is bugging me:

A couple of prominent “algebras for monads on Set” are actually categories with a single object. Monoidal sets, groups, abelian groups, etc.

On the other hand, a category $C$ (in $\mathrm{Set}$) itself is a monad in $\mathrm{Span}(\mathrm{Set})$, where the span is

(1)$\array{ \mathrm{Mor}(C) & \overset{t}{\to} & \mathrm{Obj}(C) \\ s\downarrow\;\; \\ \mathrm{Obj}(C) } \,.$

This must be related to how a category with a single object (a monoidal set) is an algebra for a monad on $\mathrm{Set}$. Somehow.

Posted by: urs on September 6, 2006 4:39 PM | Permalink | Reply to this

### Re: Doctrines

Now that Dr. Doctrine has cured Urs of his conceptual virus, I hope his cold goes away too!

But right now, he still seems to be suffering from a minor irritation of the cerebral cortext:

One thing that is irritating when learning this stuff is the use of the word “algebra”. I am more used to thinking of monads as algebras and of what are called algebras here as modules for these algebras. But all right.

Yes, that used to irritate me too. I first ran into this nuisance in the theory of operads: everyone uses the term “algebra” for what I’d like to call a “representation”, “module” or “action” of an operad.

The reason is historical, going back to universal algebra, where any “kind” of “algebraic gadget” is called a “variety” of “algebra”. An example is an algebra in the traditional sense - an associative unital algebra, that is. Both monads and operads can describe this example. In other words, there’s both a monad and an operad on $\mathrm{Set}$ whose algebras are just algebras in this traditional sense!

But, the modern viewpoint is deeper: monads and operads are both just monoids $m$ in certain monoidal categories $M$, and their “algebras” are really just actions of these monoids on objects $x$ in categories $X$! In both cases, we have a big monoidal category

$\otimes: M \times M \to M$

containing a little monoid

$\cdot: m \otimes m \to m$

and this big monoidal category acts on another category:

$A: M \times X \to X$

allowing our monoid to act on an object in that category:

$a: A(m,x) \to x$

The little guys “ride” the big guys and copy their behavior. This funny pattern is what led James Dolan and I to enunciate the Microcosm Principle (see page 10 here).

But I digress… Urs has also caught another bug:

Another thing that is bugging me:

A couple of prominent “algebras for monads on Set” are actually categories with a single object. Monoidal sets, groups, abelian groups, etc.

It’s true. Part of this is just the mysterious prevalence of monoids: something about the universe or the human brain seems to delight in this simple structure, so lots of “different” mathematical entities are all just monoids in different monoidal categories - or more generally, $n$-monoids in monoidal $n$-categories.

For example: the notion of “monoid” makes sense in any monoidal category. But, a monoidal category is just a 2-monoid in the monoidal 2-category Cat. And a category, as you point out, is a monoid in Span. So, the concept of “monoid” makes sense in any 2-monoid in the monoidal 2-category of monoids in Span! Yaaaaah! One begins to fear the whole universe is fundamentally the elaborate gibberings of some crazed, monomaniacal monoid-lover.

Maybe that’s why they call it monotheism!

But, let’s not be fooled: monads are also great for describing lots of algebraic gadgets that aren’t monoids - like sets equipped with 3 different 17-ary operations satisfying some complicated equations.

But anyway: you ask for a relation between monads and the fact that categories are monoids. I don’t know one, except the mysterious prevalence of monoids.

Posted by: John Baez on September 7, 2006 4:22 AM | Permalink | Reply to this

### Re: Doctrines

a monad on a category C is just a monoid T in the monoidal category of functors from C to itself

I’m having some trouble reconciling what you said in week89 with what you said in this thread, so I’m going to try to unpack this, too.

Is this a correct restatement of what you said above?

$HOM(C, C)$ is the image of a product-preserving functor $Th(Mon) \rightarrow Cat$ and a monad is the image of a product-preserving functor $Th(Mon) \rightarrow HOM(C, C)$.

$HOM(C, C)$ is the monoidal category with endofunctors as objects, natural transformations between them as morphisms, composition being the multiplication, and the identity endofunctor as the unit.

I don’t see how $HOM(C, C)$ has finite products, so I don’t know how to get a model of $Th(Mon)$ in it in the usual way.

If we use the monoidal product instead, then the type $M$ maps to an endofunctor $T:C \to C$; the type 1 maps to “no compositions of $T$”, or the endofunctor $1_C$. The unit $i:1\to M$ maps to an “element of $T$”, a natural transformation $\iota:1_C \Rightarrow T$. The multiplication $m:M \times M\to M$ is a natural transformation $\mu:T^2 \Rightarrow T$, where $T^2$ is $T$ composed with itself. It would be nicer if this were a map from a pair of elements to a new element, but that doesn’t seem to be what’s going on. The monoid axioms, of course, hold for $\iota$ and $\mu$.

Posted by: Mike on September 6, 2006 9:04 PM | Permalink | Reply to this

### Re: Doctrines

Now I want to get a monoid as a special case of a monad. A monad on $C$ is a monoid object in $HOM(C,C)$, and a monoid is a monoid object in Set. So Set has to be the functor category $HOM(C, C)$ for some $C$. It looks way too complicated for me to figure out what $C$ is, since there has to be one way to assign objects to objects and morphisms to morphisms for each set. If we simply posit its existence, then we get the 2-category of week89 by letting $C$ be the one object, the endofunctors (sets) be the morphisms, and the natural transformations (functions) be the 2-morphisms.

Posted by: Mike on September 6, 2006 9:53 PM | Permalink | Reply to this

### Re: Doctrines

Mike wrote:

Now I want to get a monoid as a special case of a monad. A monad on $C$ is a monoid object in $HOM(C,C)$, and a monoid is a monoid object in Set. So Set has to be the functor category $HOM(C, C)$ for some $C$. It looks way too complicated for me to figure out what $C$ is, since there has to be one way to assign objects to objects and morphisms to morphisms for each set. If we simply posit its existence…

I’m not sure such a $C$ exists.

Here’s an easier way to see a monoid $m \in \mathrm{Set}$ as a special sort of monad. We define a endofunctor

$T : \Set \to \Set$

by

$T: s \to m \times s$

Then if $m$ is a monoid, we can make $T$ into a monad!

Cool, huh?

This works not just for $\Set$ but for any monoidal category, replacing $\times$ by $\otimes$.

In short: making an object into a monoid makes multiplying by that object into a monad!

(I said “multiplying” instead of “tensoring” to make it even more of a tongue-twister - and to make it clearer how self-referential this whole business is!)

Posted by: John Baez on September 7, 2006 6:38 AM | Permalink | Reply to this
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Tracked: September 6, 2006 6:57 PM

### Re: Doctrines

Older patrons of the n-Category Café may recall this 70’s hit song - we used to play it all the time here:

Doctor Doctrine, give me the news!
I’ve got a hard case of logic blues!

Anyway, Mike Stay seems to have a case of logic blues here, so Dr. Doctrine had better try to cure it:

John wrote:

A monad on a category $C$ is just a monoid $T$ in the monoidal category of functors from $C$ to itself.

Mike wrote:

I’m having some trouble reconciling what you said in week89 with what you said in this thread, so I’m going to try to unpack this, too.

Okay… but you seem intent on “unpacking” each phrase by treating it as a special case of something more general, creating a truly frightening slag-heap of abstraction:

$\mathrm{hom}(C,C)$ is the image of a product-preserving functor $\mathrm{Th}(\mathrm{Mon}) \to \mathrm{Cat}$ and a monad is the image of a product-preserving functor $\mathrm{Th}(\mathrm{Mon}) \to \mathrm{hom}(C,C)$.

Zounds! I’ve created a monster! It’s not clear that taking every concept and expressing it in terms of algebraic theories actually helps here. I’m going to have to break this down into tiny shards to understand it.

Anyway, the first half is correct:

$\mathrm{hom}(C,C)$ is the image of a product-preserving functor $\mathrm{Th}(\mathrm{Mon}) \to \mathrm{hom}(C,C)$.

Translating into slightly less scary language, this says

$\mathrm{hom}(C,C)$ is a model in Cat of the algebraic theory of monoids.

Here you are using the fact that Cat has finite products to describe monoids in Cat using the algebraic theory of monoids. But you’re really just saying this:

$\mathrm{hom}(C,C)$ is a monoid in Cat.

A monoid in Cat is usually called a strict monoidal category: it has a tensor product

$\otimes : C \times C \to C$

and unit

$1 \in C$

for which the monoid laws (associativity, right/left unit laws) hold “strictly”, as equations. So, you’re saying:

$\mathrm{hom}(C,C)$ is a strict monoidal category.

And indeed, this is true if we take composition

$\circ : \mathrm{hom}(C,C) \times \mathrm{hom}(C,C) \to \mathrm{hom}(C,C)$

as the tensor product, and the identity functor

$1 \in \mathrm{hom}(C,C)$

as the unit object. So, you’re really just saying this:

Composition in $\mathrm{hom}(C,C)$ is associative, and $1: C \to C$ serves as a left and right unit for composition.

Whew! Now it doesn’t sound so scary.

The second part, alas, is wrong:

A monad is the image of a product-preserving functor $\mathrm{Th}(\mathrm{Mon}) \to \mathrm{hom}(C,C)$.

And, you explain why: $\mathrm{hom}(C,C)$ doesn’t have finite products! It’s a monoidal category - as we’ve just seen - but it’s tensor product is not the “product” in the categorical sense.

It’s not a big problem. You’re just using the wrong doctrine to describe monoids in $\mathrm{hom}(C,C)$: you’re trying to use categories with finite products (= algebraic theories), but you should be using monoidal categories (= PROPs). There’s an algebraic theory for monoids, but there’s also a PROP for monoids, and we should use that. So, you should have said:

A monad is the image of a monoidal functor $\mathrm{PROP}(\mathrm{Mon}) \to \mathrm{hom}(C,C)$.

but this is just a fancy way of saying

A monad is a monoid in $\mathrm{hom}(C,C)$.

A monoid in $\mathrm{hom}(C,C)$ would be, as usual, an object $T$ together with a multiplication

$m : T T \Rightarrow T$

(note the tensor product in $\mathrm{hom}(C,C)$ is composition, and morphisms are natural transformations) and a unit

$i: 1 \Rightarrow T$

(note the unit object in $\mathrm{hom}(C,C)$ is $1: C \to C$), satisfying associativity and the left/right unit laws. So, you’re saying

$T : C \to C$

together with natural transformations

$m : T^2 \Rightarrow T$

$i: 1 \Rightarrow T$

satisfying the monoid laws.

I could unpack further and show what the monoid laws look like here - but they’re hard to draw, and you can find them in Tom Leinster’s notes or any decent book on category theory.

For the present purposes, “decent books on category theory” include Categories for the Working Mathematician and Toposes, Triples and Theories. The latter calls monads “triples”, and it’s tough going for the beginner, but it’s free, and it’s a great place to read if you’re interested in both monads and algebraic theories as ways of describing algebraic gadgets.

There are lots of nice relations between monads and algebraic theories, which we haven’t gotten to yet…

(By the way, Mike: I know most of what I said in this post you already figured out yourself in yours. I just thought it would be good to go through everything in detail, so everyone could get a nice workout!)

Posted by: John Baez on September 7, 2006 5:31 AM | Permalink | Reply to this
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