8
$\begingroup$

As the visuallization of an algebraic stack is virtually impossible I warn about this is a soft question.

I am interested in thinking visually about algebraic stacks (also higher and derived stacks, but let´s start from the beginning) and I find quite suprising the fact that there isn´t a single "picture" of an algebraic stack in the literature. I would expect some analogue to Mumford´s illustration of schemes for some example of algebraic stack but I cannot find it.

To be concrete, take moduli stack of vector bundles $\mathcal{M}$ (of fixed rank and chern class) on a scheme $S$. How would you represent it graphically? The way I usually imagine $\mathcal{M}$ visually is that the fiber over a point of the base scheme is not a single vector bundle but a set of vector bundles related by isomorphims. However the condition that makes the stack to be algebraic (and consequently a geometric space) is roughly (for an Artin stack) that the diagonal is representable, quasi-compact and separated and that it has a smooth cover by an scheme (the atlas). Following T. Gomez "Introduction to Algebraic Stacks" atlas of a stack

I understand from this that if you have a family of isomorphic vector bundles $V_{ki}$ in each point $k$ (where $i=1,2,3,...$ denotes different isomorphic vector bundles over the same point $k$) the scheme $U$ is going to contain at least one $V_{ki}$ for at least one value of $i$ in each point of $U$

enter image description here

It would be great if someone can clarify this and specially to offer some way to imagine visually an algebraic stak (a picture itself would be even better!)

Improve this question
$\endgroup$
13
  • 1
    $\begingroup$ I do not know why do you think (or where it is mentioned) that $\mathcal{M}$ is parametrized by $U$... It only says that there exists a scheme $U$ and a morphism of stacks $U\rightarrow \mathcal{M}$, satisfying some properties... By Yoneda Lemma (I can say more here if you want), this morphism $U\rightarrow \mathcal{M}$ corresponds to the category $\mathcal{M}(U)$, which by definition is the collection of vector bundles over the scheme $U$... $\endgroup$
    – Praphulla Koushik
    Commented Jul 22, 2019 at 13:10
  • 1
    $\begingroup$ I am trying to understand your question... Are you asking why $u:U\rightarrow \mathcal{M}$ is not considered to be surjective? I understand you want to visualize stack.. $\endgroup$
    – Praphulla Koushik
    Commented Jul 22, 2019 at 13:24
  • 1
    $\begingroup$ I understand that the name "surjective" is misleading... Actual name is "representable surjective (submersion)"... Experts use surjective to mean representable surjective (submersion)... It means, the map $u:U\rightarrow \mathcal{M}$ is such that, for any map $v:V\rightarrow \mathcal{M}$, the product $U\times_{\mathcal{M}}V$ is representable by a scheme and that the projection map $U\times_{\mathcal{M}}V\rightarrow V$ is surjective (submersion) of schemes... Does this partially answer your question? $\endgroup$
    – Praphulla Koushik
    Commented Jul 22, 2019 at 13:38
  • 2
    $\begingroup$ I do not have hope for some one saying about picture of $\mathcal{M}$.. I will be more than happy to see if there is any.. :) $\endgroup$
    – Praphulla Koushik
    Commented Jul 22, 2019 at 14:08
  • 3
    $\begingroup$ "Nice" stacks over an algebraically closed field $k$ are locally of the form $[X/G]$ for an affine scheme $X$ and a linearly reductive $k$-group scheme $G$: see arxiv.org/abs/1504.06467. You can picture this by drawing the $G$-orbits on $X$ and decorating points with the stabilizers. $\endgroup$
    – dorebell
    Commented Jul 23, 2019 at 5:46

0

Reset to default

You must log in to answer this question.