The n-Category Café

Here’s the grammar of it: a functor $G: \mathbf{B} \to \mathbf{A}$ may or may not have a codensity monad. If it does, it’s a monad on $\mathbf{A}$, and I’ll call it $T^G$. In the event that $G$ has a left adjoint, $F$, the codensity monad does exist, and it’s simply $G F$.

I’ll state the definition of codensity monad in four ways, using varying amounts of categorical vocabulary.

1. The slickest definition: the codensity monad of $G$ is the right Kan extension of $G$ along itself. (This may or may not exist.) The monad structure comes from the universal property of Kan extensions.

2. Slightly more explicitly, the codensity monad $T^G$ can be defined as an end: for $A \in \mathbf{A}$,

$$T^G(A) = \int_B [\mathbf{A}(A, G B), G B].$$

Here $[\mathbf{A}(A, G B), G B]$ is a power in $\mathbf{A}$: that is, it’s the product of lots of copies of the object $G B$, one for each element of the set $\mathbf{A}(A, G B)$. The end might or might not exist.

3. If you’re not comfortable with ends, maybe you’ll prefer the same formula expressed as a limit. Let $A \in \mathbf{A}$. Let $A/G$ be the comma category whose objects are maps $f\colon A \to G B$ (or more exactly, pairs $(B, f)$ where $B \in \mathbf{B}$ and $f \in \mathbf{A}(A, G B)$). There’s a functor $A/G \to \mathbf{A}$ sending this object to $G B$, and $T^G(A)$ is defined to be its limit (if it exists).

4. Here’s a fourth and final way to say it. For this, let’s assume that $\mathbf{B}$ is small and $\mathbf{A}$ has small limits. Then, à la Kan, the functor $G$ induces an adjunction

$$Hom(-, G) : \mathbf{A} \stackrel{\longleftarrow}{\rightarrow} [\mathbf{B}, \mathbf{Set}]^{op} : Hom(-, G)$$

where I’m calling both functors $Hom(-, G)$, though they’re not the same. The left adjoint $Hom(-, G): \mathbf{A} \to [\mathbf{B}, Set]^{op}$ is defined by $(Hom(A, G))(B) = \mathbf{A}(A, G B)$. I’ll skip the description of the right adjoint — see the paper if you want to know. Of course, the right adjoint is determined by the fact that it’s a right adjoint: explicitly,

$$Hom_{\mathbf{A}}(A, Hom(Y, G)) \cong Hom_{[\mathbf{B}, \mathbf{Set}]}(Y, Hom(A, G))$$

whenever $A \in \mathbf{A}$ and $Y \in [\mathbf{B}, \mathbf{Set}]$. Anyway, being an adjunction, it induces a monad, and this is the codensity monad of $G$.

Example  Suppose $G$ has a left adjoint, $F$. Using any of the descriptions above, you can show that the codensity monad $T^G$ is just the ordinary monad $G F$ (Proposition 6.1 of my paper, but known for donkeys’ years).

Example  Every set $S$ has a monoid of endomorphisms $\mathbf{End}(S)$, and an action of a monoid $M$ on $S$ can be described as a map $M \to \mathbf{End}(S)$ of monoids. Modules over rings, representations of groups, algebras for operads, etc., can all be described in a similar way.

What seems less well known is that algebras for monads can also be described in this way. Let $A$ be an object of a category $\mathbf{A}$. Under mild hypotheses, $A$ has an endomorphism monad $\mathbf{End}(A)$, whose crucial feature is that for any other monad $S$ on $\mathbf{A}$, an $S$-algebra structure on $A$ can be described as a map $S \to \mathbf{End}(A)$ of monads. It is, in fact, the codensity monad of the functor $\mathbf{1} \to \mathbf{A}$ that picks out the single object $A$.

Explicitly, this monad $\mathbf{End}(A)$ sends an object $X$ of $\mathbf{A}$ to the power $[\mathbf{A}(X, A), A]$. (That’s a special case of the end or limit formula above.) Why is it the case that an $S$-algebra structure on $A$ amounts to a monad map $S \to \mathbf{End}(a)$? Keep reading to find out.

Example  Here’s something a bit more substantial. Take the category $\mathbf{Ring}$ of commutative rings, its full subcategory $\mathbf{Field}$, and the inclusion functor $G\colon \mathbf{Field} \hookrightarrow \mathbf{Ring}$.

There’s probably no canonical way of turning a ring into a field, and as a matter of fact, $G$ doesn’t have a left adjoint. Why not? Here’s one of many possible arguments. The category $\mathbf{Ring}$ has an initial object, $\mathbb{Z}$, and any left adjoint would send $\mathbb{Z}$ to an initial object of $\mathbf{Field}$. But $\mathbf{Field}$ has no initial object, since there are no homomorphisms between fields of different characteristics.

So, we don’t get an induced monad on $\mathbf{Ring}$ by the standard monad-from-adjunction method.

But we do get a codensity monad, $T^G$. It does its best to turn rings into fields. What it actually does is turn a ring into a product of fields: for a ring $A$,

$$T^G(A) = \prod_{p \in Spec(A)} Frac(A/p)$$

where $Spec(A)$ is the set of prime ideals of $A$ and $Frac(A/p)$ is the field of fractions of the quotient ring $A/p$.

You might wonder whether the unit map $A \to T^G(A)$ embeds our original ring $A$ into a product of fields (in other words, whether it’s injective). An obvious necessary condition is that $A$ contains no nilpotent elements other than zero. In fact, some basic commutative algebra implies that this is also a sufficient condition (Example 5.1 of my paper).

For instance,

$$T^G(\mathbb{Z}) = \mathbb{Q} \times (\mathbb{Z}/2\mathbb{Z}) \times (\mathbb{Z}/3\mathbb{Z}) \times (\mathbb{Z}/5\mathbb{Z}) \times \cdots$$

which is the product of one copy each of the prime fields. (It contains $\mathbb{Z}$ as a subring.) Also, for $n \geq 1$,

$$T^G(\mathbb{Z}/n\mathbb{Z}) = \prod_{primes  p | n} \mathbb{Z}/p\mathbb{Z}.$$

You can rewrite this as

$$T^G(\mathbb{Z}/n\mathbb{Z}) = \mathbb{Z}/rad(n)\mathbb{Z}$$

where $rad(n)$ is the radical of $n$. If you’ve been reading about the ABC conjecture recently, you’ll have met the radical: it’s the product of the distinct prime factors of $n$.

I want to finish by convincing you that codensity monads really do deserve to be thought of as substitutes for monads induced by adjunctions, usable in situations where no adjoint exists.

The first example above was one piece of evidence for this point of view: that if $G$ does have a left adjoint, then the codensity monad is the same thing as the monad induced by the adjunction. So, the two constructions agree where they’re both defined, but codensity monads are defined more often.

Here’s some stronger evidence that we’re doing the categorically right thing. Take, as before, a functor $G\colon \mathbf{B} \to \mathbf{A}$, and suppose that the codensity monad $T^G$ exists. Then it’s a theorem that given any monad $S$ on $\mathbf{A}$, there’s a natural one-to-one correspondence between:

• maps of monads $S \to T^G$, and
• functors $\mathbf{B} \to \mathbf{Alg}(S)$ over $\mathbf{A}$.

Here “over $\mathbf{A}$” refers to the fact that both $\mathbf{B}$ and $\mathbf{Alg}(S)$ come equipped with functors into $\mathbf{A}$ (namely, $G$ and the forgetful functor); the functor $\mathbf{B} \to \mathbf{Alg}(S)$ is required to make the obvious triangle commute. If you want a proof, you can find this as Proposition 6.2 of my paper.

(Back in the example of endomorphism monads, maybe you wondered why maps $S \to \mathbf{End}(A)$ were the same thing as $S$-algebra structures on $A$. It’s because of the result just mentioned: just take $G\colon \mathbf{1} \to \mathbf{A}$ to be the functor picking out the object $A$.)

So: the process of constructing codensity monads is adjoint to the process of constructing the category of algebras for a monad. This can be put precisely in a couple of ways, which you can find in Section 6 of my paper. I’ll just mention one.

Fix a category $\mathbf{A}$. Not every functor into $\mathbf{A}$ has a codensity monad, and certainly not every functor into $\mathbf{A}$ is monadic. Write:

• $\mathbf{CAT}/\mathbf{A}$ for the slice category of $\mathbf{CAT}$ over $\mathbf{A}$
• $(\mathbf{CAT}/\mathbf{A})_{CM}$ for the full subcategory consisting of those functors that have a codensity monad
• $(\mathbf{CAT}/\mathbf{A})_{mndc}$ for the full subcategory of monadic functors.

These are successively smaller, since any monadic functor has a left adjoint and therefore (by our very first example) a codensity monad. So we have

$$(\mathbf{CAT}/\mathbf{A})_{mndc}   \subseteq  (\mathbf{CAT}/\mathbf{A})_{CM}.$$

The theorem now is that it’s a reflective subcategory. That is, the inclusion functor $(\mathbf{CAT}/\mathbf{A})_{mndc} \hookrightarrow (\mathbf{CAT}/\mathbf{A})_{CM}$ has a left adjoint. The adjoint takes a functor $G\colon \mathbf{B} \to \mathbf{A}$ and turns it into the forgetful functor $\mathbf{Alg}(T^G) \to \mathbf{A}$.

In other words, the canonical way to force a functor to be monadic is to first take its codensity monad, then form the corresponding monadic functor.

In future installments: ultrafilters, ultraproducts, and: what could it sensibly mean for a vector space to be compact?