# Open and Dense Substack

I am looking for a definition of open and dense substack of a Deligne-Mumford stack $\mathcal X$. I have encountered this notion many times, but I am not able to find any references in which dense substacks are defined.

The only thing I have found is the one on nlab - dense subtopos, but, as I don't know much about topos theory, I am wondering whether there is a definition which does not use the characterisation of DM-stacks as ringed topoi locally equivalent to the étale spectrum of a commutative ring.

Furthermore, is this notion stable under 2-pullback in the 2-category of stacks?

• Are you working with algebraic stacks? Let $X$ be an algebraic stack. Let $U$ be a substack of $X$. The stack-theoretic closure of $U$, denoted $\bar{U}$, is a closed substack of $X$ containing $U$ with the following universal property. If $Z$ is a closed substack of $X$ containing $U$, then $Z$ contains $\bar{U}$. Now, define $U$ to be dense if $\bar{U} =X$. – Ariyan Javanpeykar Apr 11 '17 at 9:41
• To complement Ariyan: every "smooth local" property of morphisms has a well-defined extension to algebraic stacks. A property $P$ of a morphism $f:X'\to X$ is "smooth local" if for every quasi-compact, surjective, smooth morphism $Y\to X$, the property $P$ holds for $f$ if and only if the property holds for the base change $f':Y'\to Y$. In that case, for every algebraic stack $X$, since there exists a quasi-compact, surjective, smooth $1$-morphism $Y\to X$ with $Y$ a scheme, we define $P$ to hold for $f$ if and only if $P$ holds for $f'$. Now apply this to "open immersion with dense image". – Jason Starr Apr 11 '17 at 9:46
• In the above, the $1$-morphism $f$ should be representable. If the property $P$ is simultaneously smooth local on the target and on the source, then we can extend $P$ to "not-necessarily-representable" $1$-morphisms between algebraic stacks. Anyway, every morphism that is smooth locally an open immersion is representable. – Jason Starr Apr 11 '17 at 9:56
• Regarding stability under pullback, that is not even true in the category of schemes. If $i:U\to X$ is a dense open immersion, then for the complementary closed subspace $j:C\to X$, the pullback of $i$ by $j$ is the empty scheme. However, this is stable under pullback by smooth morphisms to $X$, and even by flat morphisms to $X$, both in the category of schemes and in the $2$-category of algebraic stacks. – Jason Starr Apr 11 '17 at 10:50
• @JasonStarr could you suggest any references in which stability under pull-back (by smooth morphism) is discussed? – Symòn Apr 17 '17 at 10:23

## 1 Answer

Given an algebraic stack $\mathcal{X}$ there is a canonically associated topological space $|\mathcal{X}|$ of points of $\mathcal{X}$. A point is an equivalence class of morphisms $\text{Spec}(k) \to \mathcal{X}$, where $k$ is a field. Two morphisms $\text{Spec}(k) \to \mathcal{X}$ and $\text{Spec}(k') \to \mathcal{X}$ are equivalent if $\text{Spec}(k) \times_\mathcal{X} \text{Spec}(k')$ is nonempty (as an algebraic stack, i.e., there is a point). Given a morphism $f : \mathcal{X} \to \mathcal{Y}$ of algebraic stacks, the obvious map $|\mathcal{X}| \to |\mathcal{Y}|$ is continuous and if $f$ is smooth then it is open. This together with the requirement that $|\mathcal{X}|$ should be the usual topological space when $\mathcal{X}$ is a scheme, uniquely determines the topology.

Definition: An open substack $\mathcal{U} \subset \mathcal{X}$ is dense if $|\mathcal{U}| \subset |\mathcal{X}|$ is dense.

IMHO opinion this is the correct definition. More generally, topological properties should be, as much as possible, defined in terms of $|\mathcal{X}|$. For example a morphism $f : \mathcal{X} \to \mathcal{Y}$ of algebraic stacks is universally closed if for all morphisms $\mathcal{Z} \to \mathcal{Y}$ the map $|\mathcal{Z} \times_\mathcal{Y} \mathcal{X}| \to |\mathcal{Z}|$ is closed.

• That sounds fine, but how are you going to define "schematically dense"? – Jason Starr Apr 11 '17 at 11:42
• That wasn't the question? Of course I can give you the definition if you want, but that's ridiculous as you know the definition. – Darx Apr 12 '17 at 3:08
• My question was, indeed, rhetorical. – Jason Starr Apr 12 '17 at 3:18