# Why are monadicity and descent related?

This question is probably too vague for experts, but I really don't know how to avoid it.

I've read in several places that under mild conditions, a morphism is an effective descent morphism iff the base-change functor it induces is monadic.

Now, I don't really know anything about descent and I'm trying to probe around and figure out how to start learning about it. I think of monads in terms of algebraic theories. Descent on the other is supposedly of geometric nature; a formalism which generalizes familiar gluing over open subsets to a more general setting without a spatial topology. I can't imagine how or why these two notions should be related.

What lies at the core of the relationship between monadicity - a seemingly algebraic and concrete notion - and (effective) descent (morphisms)?

• Monadicity is an expression of associativity; gluability, as expressed by the usual cocycle conditions, is also a form of associativity. – Mariano Suárez-Álvarez Dec 6 '15 at 13:14
• @MarianoSuárez-Alvarez I see.. Would you say I should delete this question as too basic? – Arrow Dec 6 '15 at 13:17
• I think it's a good question, and something I as confused by for a long time, for what it's worth. – Jonathan Beardsley Dec 6 '15 at 13:56
• Want to see just how shortly this may be formulated. Descent datum is very close to being an algebraic structure (in many cases it is). Monads are gadgets to describe maximally general algebraic structures. In other words, forgetting descent data is very close to being a monadic functor, and in many cases it actually is monadic. – მამუკა ჯიბლაძე Dec 6 '15 at 22:40
• You might like the paper Descent in triangulated categories by Paul Balmer (even if you aren't interested in triangulated categories). It has a very readable introduction in which the basic setup of monadic descent theory is recalled. – Ingo Blechschmidt Mar 28 '16 at 11:39

Probably the reason "monadicity" gets connected with descent (and the associated terminology of descent theory) is because of its relevance to the question of descent for rings.

If you're talking about a morphism of rings $\phi:A\to B$ there is a functor $-\otimes_AB:Mod_A\to Mod_B$. Then you can ask the question: Can I recover the category $Mod_A$ from $Mod_B$? This question is answered by knowing whether or not $-\otimes_AB$ is comonadic. In particular, this tells us that $Mod_A$ is equivalent to the category of $\phi^\ast\circ(-\otimes_AB)$-comodules in $Mod_B$. It takes a little bit of work, but as an exercise for really understanding descent, it may be useful to prove that in this case, the category of comodules for this comonad is equivalent to the category of $B\otimes_AB$-comodules, where $B\otimes_AB$ is a $B$-coring with comultiplication $$B\otimes_AB\cong B\otimes_AA\otimes_AB\to B\otimes_AB\otimes_AB\cong B\otimes_AB\otimes_B B\otimes_AB.$$

This object is also sometimes called the descent coring, and for reasons that will hopefully become clear in the next paragraph, its comodules are often referred to as descent data.

With a bit of finagling one can prove that a $B\otimes_AB$-comodule structure on a $B$-module $N$ (so in the case that $-\otimes_AB$ is comonadic we know that $N\cong M\otimes_AB$ for a unique $A$-module $M$) is the same thing as a "cocycle condition" satisfying isomorphism $$M\otimes_AB \cong M\otimes_B B\otimes_AB\overset{\sim}\to B\otimes_AB\otimes_B M\cong B\otimes_AM.$$ The basic idea is to apply the adjunction $$Hom_B(M,B\otimes_AM)\cong Hom_{B\otimes_AB}(M\otimes_AB,B\otimes_AM).$$

And so this is the thing that lets us take a geometric perspective. If we consider the equivalent problem of whether or not we can recover the category of quasicoherent sheaves on $Spec(A)$ from the category of quasicoherent sheaves on $Spec(B)$, with adjunction $\phi^\ast:QC_{Spec(A)}\rightleftarrows QC_{Spec(B)}:\phi_*$ then the descent data (as defined above for modules) for a sheaf $\mathcal{F}$ on $Spec(B)$ is an isomorphism between the two ways of pulling back $\mathcal{F}$ to $Spec(B)\times_{Spec(A)}Spec(B).$ This, as you may recall, is precisely the way that we typically state "gluing" conditions. In other words, if a sheaf supports such an isomorphism, it lives in the coequalizer (in categories, so necessarily the coequalizer computed as a 2-colimit) of the diagram $$QC_{Spec(B)}\leftleftarrows QC_{Spec(B)\times_{Spec(A)} Spec(B)}\Lleftarrow QC_{Spec(B)\times_{Spec(A)} Spec(B)\times_{Spec(A)} Spec(B)}.$$

The only further thing maybe to mention is that usually geometric-type descent is stated in terms of covers $X\overset{f_i}\leftarrow\{U_i\}$ but we can do everything above by just considering $X\overset{\coprod f_i}\leftarrow\coprod U_i.$

So now whenever you've got any kind of monad or comonad, you can ask about "descent" for that monad, which is really just a question of whether or not you can recover some category from some other category of (co)monadic (co)modules.

• Thanks for the great answer. I will probably have questions when I start studying it. For now, I'll just link to this related question for completeness. – Arrow Dec 6 '15 at 14:23
• @Arrow yeah that question was really useful to me in working some of this stuff out. – Jonathan Beardsley Dec 7 '15 at 0:15
• I should add that there are many books and papers in which these concepts are discussed, for instance Knus and Ojanguren's "Descent Theory and Azumaya Algebras" as well as Brzezinski and Wisbauer's "Corings and Comodules." There are lots of other things I'm not thinking of right now as well. – Jonathan Beardsley Dec 7 '15 at 0:18

I think of monads in terms of algebraic theories.

Monads are substantially more general than this intuition suggests! Here is a better intuition: monads are categorified idempotents.

The point of idempotents (acting on, say, a module) is to pick out nice subobjects: subobjects that are so nice that they are simultaneously subobjects and quotient objects (say, direct summands of modules) in a compatible way. More formally, every idempotent $m : X \to X$ wants to become a pair of a map $f : X \to Y$ and a map $g : Y \to X$ such that $g \circ f = m$ and $f \circ g = \text{id}_Y$. Similarly, the point of monads is to pick out nice categories which simultaneously map into and out of a category in a compatible way (via an adjunction). More formally, every monad $M : C \to C$ wants to become a pair of a functor $F : C \to D$ and a functor $G : D \to C$ such that $F$ and $G$ are adjoint and $G \circ F \cong M$.

This analogy is quite robust: for example, the analogue of taking the fixed points of an idempotent is taking the category of algebras of a monad. And the analogue of an adjunction being monadic is a submodule being a direct summand.

Descent on the other is supposedly of geometric nature; a formalism which generalizes familiar gluing over open subsets to a more general setting without a spatial topology. I can't imagine how or why these two notions should be related.

Here's a simple toy model. Let $f : X \to Y$ be a map of sets and let $\text{Sh}(X)$ be, for concreteness, the functor assigning a set $X$ the category of sheaves of sets on $X$, which just means the category of assignments, to each $x \in X$, of a set $A_x$. There is a pullback functor

$$f^{\ast} : \text{Sh}(Y) \to \text{Sh}(X).$$

It has a right adjoint $f_{\ast} : \text{Sh}(X) \to \text{Sh}(Y)$ given by taking fiberwise products: that is, if $A$ is a sheaf of sets on $X$, then

$$f_{\ast}(A)_y = \prod_{f(x) = y} A_x.$$

(It also has a left adjoint which we'll ignore.) This adjunction induces a comonad $f^{\ast} f_{\ast} : \text{Sh}(X) \to \text{Sh}(X)$ sending a sheaf $A$ of sets on $X$ to the sheaf

$$f^{\ast} f_{\ast}(A)_x = \prod_{f(x') = f(x)} A_{x'}.$$

Now, what should descent mean in this situation? $f$ should be descent iff it is surjective, and descent should intuitively say that a sheaf on $X$ descends to a sheaf on $Y$ iff for all $y \in Y$, all of the sets $A_x, f(x) = y$ are canonically identified. This reflects the fact that $f$ is surjective iff $Y$ itself is obtained from $X$ by quotienting by the equivalence relation $x \sim x' \Leftrightarrow f(x) = f(x')$. (Which in turn says that surjections in $\text{Set}$ are effective epimorphisms.)

This is encoded by the comonad above as follows. A coalgebra for the above comonad is a sheaf $A$ on $X$ together with a map $A \to f^{\ast} f_{\ast}(A)$ satisfying some compatibilities. What does such a map look like? Stalkwise it looks like a map

$$A_x \to \prod_{f(x') = f(x)} A_{x'}$$

and this map will end up encoding a bunch of isomorphisms $A_x \cong A_{x'}$. These isomorphisms should satisfy a cocycle condition which is encoded by the coalgebra compatibilities.

The decategorified version of this story is that a real-valued function $A : X \to \mathbb{R}$ descends to a function $Y \to \mathbb{R}$ iff $A_x = A_{x'}$ for all $x, x'$ such that $f(x) = f(x')$. Moreover, if $f$ has finite fibers, we can pick out which functions these are as the fixed points of the idempotent

$$m(A)_x = \frac{1}{|f^{-1}(f(x))|} \sum_{f(x') = f(x)} A_{x'}$$

acting on the vector space of functions $X \to \mathbb{R}$.

A more interesting case to work out is the case that $f : X \to X/G$ is a Galois cover with Galois group $G$. In this case descent will say that $\text{Sh}(X/G)$ is the category of homotopy fixed points $\text{Sh}(X)^G$ for the action of $G$ on $\text{Sh}(X)$, and the way in which the comonad $f^{\ast} f_{\ast}$ encodes this fact is a categorification of the fact that if $G$ is a finite group acting on a vector space $V$ (over a field of suitable characteristic), the subspace $V^G$ of fixed points is a direct summand picked out by the idempotent $\frac{1}{|G|} \sum_{g \in G} g$. This is a geometric form of Galois descent.

• I have accepted Jon Beardsley's answer because it clarifies the link between algebraic structure and descent for rings. Still, your answer is very illuminating and I will surely ask more about it in time. – Arrow Dec 7 '15 at 14:18
• In your analogy "monads are categorified idempotents" you say "the analogue of taking the fixed points of an idempotent is taking the category of algebras of a monad." Here "taking fixed points" means splitting the idempotent, forming the equalizer of $m, 1_X \colon X \rightrightarrows X$, but the same object is recovered by instead taking the coequalizer of this pair (because, as you note, a splitting is a "direct summand": both a submodule and a quotient)... [to be continued] – Emily Riehl Dec 16 '15 at 15:32
• "Taking the category of algebras" is a limit construction, more precisely analogous to "fixed points", but is there any reason to prefer this to the dual colimit construction, which yields the Kleisli category? – Emily Riehl Dec 16 '15 at 15:35
• @Emily: yeah, I ran into this issue while writing a blog post fleshing this answer out (qchu.wordpress.com/2015/12/15/monads-are-idempotents) and I don't have a principled answer. The two agree if $m$ is an idempotent monad, so I guess the "monads are idempotents" intuition doesn't have enough resolution to distinguish the Eilenberg-Moore and Kleisli categories. – Qiaochu Yuan Dec 16 '15 at 16:56
• I am not sure if I would agree with "monads are idempotents". Algebras for a monad are not just special objects, they come with new structure. At least idempotent monads behave like idempotents. – HeinrichD Sep 23 '16 at 13:45