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This question has a few bits, and apologies if some questions are phrased poorly since I am not knowledgeable on the language of stacks or deformation theory.

Suppose $\mathscr{X}$ is an algebraic stack over a scheme $S$. What is the correct notion of a tangent space to a point $p \in |\mathscr{X}|$, as a tangent space to the object in the stack and as a tangent space to the set of points?

Whenever we work with the coarse moduli space of semistable sheaves $M$ on a variety $X$, we can identify the tangent space to a point $[\mathcal{E}] \in M$ with the group $\text{Ext}^1(\mathcal{E},\mathcal{E})$, but these groups come from the deformation theory of coherent sheaves. I would like to know how one can work backwards, in a sense, and go from a stack to the deformation theory giving rise to the stack without knowing if this stack represents a moduli functor. I suspect this has something to do with Artin's criteria, but, as mentioned, I am not fluent in deformation theory.

Now, suppose we have a good notion of a tangent space to a point $p \in |\mathscr{X}|$ and $\mathscr{X}$ is irreducible. If $T_p\mathscr{X}$ is trivial, then is $p$ the generic point of $\mathscr{X}$?

The question seems odd since an irreducible variety with vanishing tangent space to a point is a single point, but, again, the motivation comes from semistable sheaves. In some cases, it is more convenient to work with a larger stack of sheaves $\mathscr{X}$ which contains the stack of semistable sheaves $\mathscr{M}$, such as in here. If $\mathscr{X}$ is irreducible and $\text{Ext}^1(\mathcal{E},\mathcal{E})=0$ for some sheaf $[\mathcal{E}] \in |\mathscr{X}|$, then the expectation is that $[\mathcal{E}]$ is the generic point of $|\mathscr{X}|$, but not all of $|\mathscr{X}|$ due to the fact that $\mathscr{X}$ contains a lot more isomorphism classes of sheaves than $\mathscr{M}$. I would like to say that $\text{Ext}^1(\mathcal{E},\mathcal{E})$ is the tangent space to the stack, but I'm not sure if this is correct. For example, what role should the automorphisms and obstruction space $\text{Ext}^2(\mathcal{E},\mathcal{E})$ play in determining the tangent space?

If we rephrase this using morphisms, this is equivalent (I think) to asking whether a morphism $F: \text{Spec}(K) \to \mathscr{X}$ sending $\text{Spec}(K)$ to $p$, where $T_p\mathscr{X}$ is trivial, is a dominant map. In light of this, we can ask more generally:

Is a morphism $F: \mathscr{X} \to \mathscr{Y}$ which induces a surjection on tangent spaces $$F_p: T_p\mathscr{X} \to T_{F(p)}\mathscr{Y}$$ for some point $p$ where $\mathscr{Y}$ is irreducible a dominant map?

The hope is that, with the right definitions, these morphisms behave very much like submersions.

I hope I've made my questions clear, and I'm happy to work through some references if these are very basic questions.

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  • $\begingroup$ Hi Dan. I don't have the answer to your question, but have you looked at the latest version of the stacks project? stacks.math.columbia.edu/tag/0DQT seems relevant to your question. $\endgroup$ – Will Chen Nov 7 '17 at 4:04
  • $\begingroup$ @WillChen Hi Will. I have seen these previously, but for a different reason and not carefully. Now that I look at it, the versal rings section looks especially interesting. Thanks! $\endgroup$ – Daniel Levine Nov 7 '17 at 5:18
  • $\begingroup$ @DanielLevine I think Question 2 and Question 3 have a negative answer. I'm not completely sure, but let me try. Let $X = [\mathbb{A}^1/\mathbb{G}_m]$ over $\mathbb{C}$. Note that $X$ is a finite type smooth irreducible (non-separated) algebraic stack over $\mathbb{C}$ with affine diagonal. It has precisely two $\mathbb{C}$-points: the generic point and the closed point. These two objects correspond to the open orbit (the torus) in $\mathbb{A}^1$ and the origin. Now, the tangent space at both points should be trivial. However, the closed point is not the generic point. (Q1) cont'd.. $\endgroup$ – Ariyan Javanpeykar Nov 7 '17 at 7:17
  • $\begingroup$ @DanielLevine And the inclusion of the closed point into $X$ is not dominant, yet the map on tangent spaces is an isomorphism (because the tangent spaces are trivial). [Caution: I might be wrong about the tangent spaces being trivial here.] $\endgroup$ – Ariyan Javanpeykar Nov 7 '17 at 7:18
  • $\begingroup$ @DanielLevine If your algebraic stack is Deligne-Mumford, then life becomes a bit easier. You can define the tangent space at an object $x$ in $X$ by taking an etale neighbourhood $(U,u)$ with $U$ a scheme and then defining $T_X(x) = T_U(u)$. A reference that might interest you here is arxiv.org/abs/math/9905049 . Also, maybe Section 2.5 of arxiv.org/pdf/1505.02249.pdf could be useful. $\endgroup$ – Ariyan Javanpeykar Nov 7 '17 at 7:19

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