Timeline for Pro-algebraic fundamental groups
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 8 at 21:34 | comment | added | Jon Pridham | For the point on the pro-unipotent fundamental group, Morgan established formality over $\mathbb{Q}$, so the Malcev completion of $\pi_1$ is just the pro-nilpotent Lie algebra generated by $H_1$ with relations induced by the dual of the cup product on $H_2$. Degeneration of the Hodge-de Rham spectral sequence allows you to identify that with infinitesimal Higgs bundles. [For the other point, $\mathbb{G}_m^2$ is the space of reductive $1$-dimensional reps of $\pi_1=\mathbb{Z}^2$ of an elliptic curve.] | |
| Nov 8 at 21:08 | comment | added | Antoine Labelle | That's fair, I can believe that this isomorphism is something really special to $K=\mathbb{C}$. I'm still curious about the pro-unipotent quotients though. | |
| Nov 8 at 20:06 | comment | added | Will Sawin | One can prove the existence of a bijection (even a bijection that is a restriction of a continuous map on $\mathbb C$-points) but this will just be an arbitrary choice, there surely won't be a canonical bijection. | |
| Nov 8 at 20:05 | comment | added | Will Sawin | Such an isomorphism would have to be pretty exotic. As you note, an isomorphism of pro-algebraic groups arising from Tannakian categories whose objects have moduli spaces need not give an isomorphism of the moduli spaces. But it gives a bijection between the $K$-points of one moduli space and the $K$-points of the other. For $K \subset \mathbb C$ there is a natural complex-analytic bijection between the $K$-points of one and the $K$-points of the other, but it is transcendental and hence usually won't send $K$-points to $K$-points. | |
| Nov 8 at 19:58 | comment | added | Antoine Labelle | I'm also a bit confused about where the $\mathbb{G}_m^2$ comes from. The moduli space of $1$-dimensional vector bundles with flat connection on an elliptic curve $X$ is an affine space bundle over $\operatorname{Jac}(X)=X$, not $\mathbb{G}_m^2$. | |
| Nov 8 at 19:55 | comment | added | Antoine Labelle | Can you say a bit more on how formality of the de Rham algebra yields an isomorphism of the pro-unipotent quotients? Regarding your second point, I'm not sure why that forbids an isomorphism of the pro-algebraic groups; in general the set of 1d representations of a pro-algebraic group doesn't come with a natural moduli space structure. | |
| Nov 8 at 17:33 | comment | added | Jon Pridham | For the pro-unipotent quotients, you can deduce this from Morgan's "The algebraic topology of smooth algebraic varieties"; it follows from formality of the de Rham algebra over $K$. For the Tannakian groups, you can't expect an isomorphism: just consider the case where $X$ is an elliptic curve and look at the respective spaces of $1$-dimensional representations; you know that $\check{X} \times \mathbb{A}^1$ isn't isomorphic as a scheme to $\mathbb{G}_m^2$. | |
| Nov 8 at 16:31 | history | asked | Antoine Labelle | CC BY-SA 4.0 |