Timeline for Is forming the Albanese variety a monad?
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 10, 2016 at 6:27 | comment | added | Lazzaro Campeotti | @JohnBaez: I haven't checked the foundational material, so you can take this comment with a shovelful of salt. But I would be surprised if the implication you cite in the previous comment is a "real" implication. In other words, I would guess that to prove some construction of the Albanese has the desired universal properties, you need to use the fact about regular maps and homomorphisms (which is much more elementary, in any case). | |
| Aug 10, 2016 at 0:06 | comment | added | John Baez | I just read that the Albanese variety of an abelian variety is itself; this implies that every map between abelian varieties that maps the identity to the identity is a homomorphism. This also says we have an "idempotent monad": the Albanese variety of the Albanese variety is the Albanese variety. You get idempotent monads from adjunctions where the right adjoint merely forgets a property, not a structure: for example, the right adjoint of "abelianization" the forgetful functor from groups to abelian groups, and the abelianization of the abelianization is the abelianization. | |
| Aug 9, 2016 at 18:49 | answer | added | Dmitry Vaintrob | timeline score: 12 | |
| Aug 9, 2016 at 17:09 | comment | added | Qiaochu Yuan | @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.) | |
| Aug 9, 2016 at 12:04 | history | edited | John Baez | CC BY-SA 3.0 |
added 23 characters in body
|
| Aug 9, 2016 at 12:03 | comment | added | Lazzaro Campeotti | @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. | |
| Aug 9, 2016 at 11:59 | comment | added | John Baez | 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? | |
| Aug 9, 2016 at 8:46 | comment | added | Leo Alonso | Perhaps it is useful to recall that the Albanese variety is the dual of the Picard scheme. | |
| Aug 9, 2016 at 7:23 | comment | added | Dmitry Vaintrob | 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. | |
| Aug 9, 2016 at 7:11 | comment | added | Dmitry Vaintrob | Note that any pair of adjoint functors $L:C\leftrightarrows D :R$ give rise to a monad via the natural transformations $RLRL\overset{LR\to 1}{\to} RL$ and $1\to RL$ given by the unit/counit maps of the adjunction. Any object of $D$ then gives an algebra over this monad via the transformation $RLR\overset{LR\to 1}{\to} R$. A pair is said to be "monadic" if in fact $D$ is equivalent to the category of algebras over $C$. This condition can be checked formally (and holds for most "free-like" functors): see en.wikipedia.org/wiki/Beck%27s_monadicity_theorem. | |
| Aug 9, 2016 at 6:29 | history | edited | John Baez | CC BY-SA 3.0 |
added 19 characters in body
|
| Aug 9, 2016 at 6:19 | history | edited | John Baez | CC BY-SA 3.0 |
added 32 characters in body
|
| Aug 9, 2016 at 6:13 | history | asked | John Baez | CC BY-SA 3.0 |