# Is forming the Albanese variety a monad?

I'm trying to understand the idea of an Albanese variety. It reminds me of something simpler:

Given a set $X$ with a chosen point $x \in X$, we can form the free abelian group on the pointed set $(X,x)$, which is just the free abelian group on $X$ modulo a relation saying $x = 1$.

If we call this group $A(X,x)$, it has a nice universal property: any map of pointed sets $f$ from $X$ into an abelian group $A$ factors uniquely as the obvious inclusion

$$i_X \colon X \to A(X,x)$$

followed by some homomorphism

$$\overline{f} \colon A(X,x) \to A.$$

So:

$$f = \overline{f} \circ i_X$$

The process of taking the free abelian group on a pointed set defines a functor

$$A \colon \mathrm{Set}_* \to \mathrm{AbGp}$$

$$U\colon \mathrm{AbGp} \to \mathrm{Set}_*$$

sending any abelian group $A$ to its underlying pointed set $(A,1)$. The composite

$$U A \colon \mathrm{Set}_* \to \mathrm{Set}_*$$

is thus a monad, and if I'm not mistaken, the algebras of this monad are just abelian groups.

The idea of an Albanese variety seems to be similar, but working with projective algebraic varieties instead of sets. Namely:

Given any such variety $X$ with a chosen point $x$ there is an abelian variety called the Albanese variety $A(X,x)$, apparently defined by the following universal property: there is a map of varieties

$$i_X \colon X \to A(X,x)$$

such that any map of pointed varieties $f$ from $X$ into an abelian variety $A$ factors uniquely as $i_X$ followed by some map of abelian varieties

$$\overline{f} \colon A(X,x) \to A.$$

So:

$$f = \overline{f} \circ i_X$$

(The Wikipedia article on Albanese varieties doesn't clearly require that $\overline{f}$ be a map of abelian varieties, but Ravi Vakil's lecture notes do, so I'm going with that.)

So, naturally, I'm wondering: does taking the Albanese variety define a functor from pointed varieties to abelian varieties, which has a right adjoint, which together define a monad on the category of pointed varieties whose algebras are the abelian varieties?

The so-called Albanese map $i_X \colon X \to A(X,x)$ would then be the unit of this monad, while the multiplication of the monad would be the map of abelian varieties $\overline{1} \colon A(A(X,x),x) \to A(X,x)$ obtained from the identity map of varieties $1 \colon A(X,x) \to A(X,x)$.

• In this particular case, since Abelian varieties are a full subcategory of pointed varieties, I think you can apply part 5 of Definition 2.1 here ncatlab.org/nlab/show/idempotent+monad to show that indeed taking the Albanese variety is monadic. – Dmitry Vaintrob Aug 9 '16 at 7:23
• Perhaps it is useful to recall that the Albanese variety is the dual of the Picard scheme. – Leo Alonso Aug 9 '16 at 8:46
• Are abelian varieties a full subcategory of pointed varieties??? That would mean any basepoint-preserving map between pointed varieties that happen to be abelian necessarily preserves the abelian group structure. That would be shocking to me, since it's like saying a basepoint-preserving map between abelian groups is a homomorphism, which is false... but there are some ways in which algebraic varieties are shockingly "rigid", so who knows? – John Baez Aug 9 '16 at 11:59
• @JohnBaez: yes, somewhat amazingly, every map between abelian varieties that maps the identity to the identity is a homomorphism. Reference: Shafarevich Volume 1, Theorem III.4.3. – potentially dense Aug 9 '16 at 12:03
• @John: perhaps not so shocking; an analogous result in topology is that every basepoint-preserving map between tori is homotopic to a homomorphism. (Certainly no statement of this form is going to be true in a setting where the basepoint is disconnected from everything else, which is why looking at (discrete) abelian groups is misleading.) – Qiaochu Yuan Aug 9 '16 at 17:09

I'll go ahead and turn my comment into an answer. It does form a monad, but (probably) not a very interesting one. Namely, first note that any pair of adjoint functors $L:\mathcal{C}\leftrightarrows \mathcal{D}:R$ is associated to a monad $RL$ on $\mathcal{C}$. Multiplication is given by the map $RLRL\overset{LR\to 1}{\to} RL$ and unit is $1\to RL$. Here the natural transformations $1\to RL$ and $LR\to 1$ (with $1$ denoting the identity functor) are the unit and counit of the adjunction, adjoint to the identity maps $L\to L, \text{ } R\to R$, respectively. Further, given an object $A$ of $\mathcal{D}$, we get an algebra $R(A)$ over the monad $RL$ with action morphism $RLR(A)\overset{LR\to 1}{\to} R(A)$ induced from the counit. This gives us a functor $\alpha:\mathcal{D}\to Alg(RL)$ to algebras over the monad $RL$. In "most real-life situations" where the functor $R$ is "forgetful" and $L$ is "free", the functor $\alpha$ to algebras is in fact an equivalence of categories. This can be checked formally, using something called the Bar-Beck monoidicity theorem (in a sense this is similar to the condition for an abelian category $\mathcal{A}$ to be the catgeory of modules over an algebra: in fact, modules over an algebra are a special case of algebras over a monad).
The monad you are considering is associated to the adjunction $A:Sp_*\leftrightarrows Ab:F$, where $F:Ab\to Sp_*$ is the forgetful functor from abelian varieties to pointed spaces and $A:Sp_*\to Ab$ is the Albanese functor. The monad is then $FA:Sp_*\to Sp_*$. The reason this monad is not very interesting is that the forgetful functor $F$ is fully faithful (this result in itself is very interesting, and encapsulates the rigidity of abelian varieties in algebraic geometry). This implies that the composition $FA:Sp_*\to Sp_*$ is an idempotent functor. In particular, here we can apply the theory of idempotent monads, see e.g. https://ncatlab.org/nlab/show/idempotent+monad. Part 6 of definition 2.1 in the above notes then gives us for free that the adjunction $A:Sp_*\leftrightarrows Ab:F$ is monadic, i.e. the category $Ab$ of Abelian varieties is indeed equivalent to the category of algebras over the monad $FA$.
• That's a more familiar fact -- consider the adjoint action, mapping $G\to GL(\mathfrak{g})$. This has connected projective source and affine target, so must be constant. – Allen Knutson Aug 11 '16 at 15:13