So I've been working with moduli stacks in algebraic geometry for a while now, with no formal training in the technicalities of the theory of algebraic stacks (ie, I've read a few articles and I learn what I need, without having spent much time doing exercises or working through examples/counterexamples).

One of the difficulties I've experienced is that while discussing morphisms of algebraic stacks, some authors only require certain morphisms to be representable by algebraic spaces, while others require them to be representable (by schemes).

Of course I've always found the latter to be more comprehensible (since it saved me from having to learn too much about alg. spaces on the way to working with alg. stacks). On the other hand, all the really comprehensive references on the subject usually fall into the former category.

For example, the stacks project's definition of an algebraic stack only requires the diagonal to be representable by alg spaces, while Gomez's article https://arxiv.org/pdf/math/9911199v1.pdf requires that it be representable (by schemes).

So far, whenever I read "representable by alg spaces", I sort of "pretend" that it says "representable by schemes" and proceed with that assumption.

My questions are:

  1. What is the difference between requiring the diagonal of an algebraic stack to be representable by spaces vs representable by schemes? When are the two notions equivalent? What pitfalls are there to avoid?

  2. In general, what are some good illustrative examples of morphisms of alg stacks which are only representable by spaces, but not by schemes? When are the two notions equivalent? What pitfalls are there to avoid?

  • $\begingroup$ One pitfall is that being representable by schemes is not fpqc-local. I think this is one of the basic reasons "representable by algebraic spaces" is much more common. For example, if G is a (smooth) algebraic group and X → Y is a G-equivariant morphism, the induced map [X/G] → [Y/G] is not necessarily representable by schemes. $\endgroup$ Sep 13, 2016 at 2:29
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    $\begingroup$ @MarcHoyois: The property of an fpqc sheaf being an algebraic space is not visibly fpqc-local on the base either, so perhaps it is more apt to note that being representable in schemes is not generally etale-local on the base (whereas being an algebraic space is etale-local in the base, since etale-descent is effective for algebraic spaces). Of greater relevance to how proofs work is that a condition (on the relative diagonal) with algebraic spaces is amenable to verification by Artin's abstract criteria (applied to Isom-functors) rather than being something to be checked "by hand". $\endgroup$
    – nfdc23
    Sep 13, 2016 at 5:34
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    $\begingroup$ It would be a valuable investment of your time to learn some basic things about algebraic spaces given that you are using stacks, since algebraic spaces are a much simpler notion (corresponding to set-valued functors rather the more general fibered categories). Even if you were to only consider situations with diagonal representable in schemes, it still makes good sense psychologically to spend some time understanding algebraic spaces as a way of acclimating to the use of stacks, just as it is a good idea to spend some time with classical varieties before diving into scheme-land. $\endgroup$
    – nfdc23
    Sep 13, 2016 at 5:38

2 Answers 2


To answer question 2, the best example I know is $\mathscr{M}_1$, the stack of (proper smooth geom. connected) curves of genus 1. Indeed, Raynaud has contructed an elliptic curve $E\to S$ over a scheme $S$ and an $E$-torsor $X\to S$ which is (an algebraic space but) not a scheme.

This implies two things. First, in order to define $\mathscr{M}_1$ we are forced to take "curve" to mean "algebraic space in curves": if we insist that curves must be schemes, the resulting $\mathscr{M}_1$ will not be a stack for the flat topology, because the above $X$ is locally a (projective) scheme for the flat (even étale) topology on $S$.

Concerning the diagonal, put $I:={\underline{\mathrm {Isom}}}_{\mathscr{M}_1}(E,X)$. There is a monomorphism $X\to I$ (in fact, a closed immersion) identifying $X$ with the subsheaf of $E$-torsor isomorphisms. So, $I$ is not a scheme because $X$ isn't. Viewed as a morphism $S\to \mathscr{M}_1$, $I$ is the pullback of the diagonal under the morphism $S\to \mathscr{M}_1\times\mathscr{M}_1$ given by $(E,X)$. Hence the diagonal is not representable in the scheme sense.

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    $\begingroup$ So in your second paragraph, are you saying that $\mathcal{M}_1$ is not a stack because of the failure of descent? (Ie, $X/S$ is a "descent" of $E$, which doesn't exist in $\mathcal{M}_1$)? $\endgroup$ Sep 13, 2016 at 15:29
  • $\begingroup$ In your last paragraph, I don't see how $I\cong X$. For example, if $X$ were just $E$ itself, then $\underline{Isom}_{\mathscr M_1}(E,E) = \underline{Aut}_{\mathscr M_1}(E)$, which for most elliptic curves is $\{\pm 1\}\times E\not\cong E$. $\endgroup$ Sep 13, 2016 at 15:57
  • $\begingroup$ Instead of "algebraic space in curves" did you actually mean "(relative) curve in (the category of) algebraic spaces"? $\endgroup$
    – Qfwfq
    Sep 13, 2016 at 19:00
  • $\begingroup$ @Qfwfq: yes, that's what I mean. $\endgroup$ Sep 13, 2016 at 21:18
  • $\begingroup$ @rtz: about the failure of descent, this is indeed what I mean. $\endgroup$ Sep 14, 2016 at 5:48

This may not directly answer your question. However, one reason to introduce the notion of an algebraic space is that sometimes it is easier to show that the diagonal is representable by an algebraic space than by a scheme. Consider the following example. Let $G$ be a smooth algebraic group over a field $k$. Consider the classifying stack BG that parametrizes principal bundles that are locally trivial in the fpqc topology. Since $G$ is smooth over a point, and smooth surjective morphisms \'{e}tale locally admit a section, we note that these bundles are a fortiori trivial in the \'{e}tale topology.

Let $T$ be a scheme and suppose we have a morphism $$T \stackrel{(P_1, P_2)}{\longrightarrow} BG \times BG.$$ Here the $P_i$'s are two principal bundles over the scheme $T$. The fiber product $$T \times_{BG \times BG} BG \cong \underline{\operatorname{Isom}}_T(P_1, P_2)$$ where the right hand side is the functor of isomorphisms between $P_1$ and $P_2$. Now choose an \'{e}tale cover $T' \to T$ that trivializes both the $P_i$'s (for instance if $T_i \to T$ trivializes $P_i$ then we can just take $T' := T_1 \times_{T} T_2$). Hence $$\underline{\operatorname{Isom}}_T(P_1, P_2) \times_T T' \cong \underline{\operatorname{Isom}}_{T'}(G_{T'}, G_{T'}) \cong h_{G \times T'}.$$ This shows that the base change of the Isom functor by an \'{e}tale cover is a scheme, from which we conclude that Isom is an algebraic space.

  • $\begingroup$ If, as you assume, $G$ is affine, then the diagonal of $\mathrm{B}G$ is an affine morphism, hence representable by (affine) schemes. This follow from fpqc descent of quasicoherent sheaves. $\endgroup$ Sep 13, 2016 at 11:06
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    $\begingroup$ If $G$ is not smooth then the $P_i$'s generally are split only by an fppf cover of $T$ (giving "just" an fppf cover of the diagonal by a scheme, but it is a deep theorem of Artin proved in his work on stacks -- with alg. space diagonal! -- that a quotient of a finitely presented scheme by an fppf equivalence relation is an algebraic space). Perhaps it depends what hypotheses are implicit in your phrase "affine algebraic group" (unclear if you intended smoothness; it is best to make such hypotheses explicit). Taking $G$ to be a nonzero abelian variety fixes this error too. $\endgroup$
    – nfdc23
    Sep 13, 2016 at 12:08
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    $\begingroup$ An illustration of the phenomenon pointed out by Laurent Moret-Bailly (and of the theorem of Artin that I mentioned in my comment above) is that if $G \rightarrow S$ is an fppf affine commutative (say) group scheme then classes in the fppf cohomology group ${\rm{H}}^1(S,G)$ are represented by schemes but if we drop the affineness assumption then all one can expect is that such $G$-torsors as sheaf functors are algebraic spaces rather than schemes (but at least it gives some "geometric structure" to those cohomology classes). $\endgroup$
    – nfdc23
    Sep 13, 2016 at 12:16
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    $\begingroup$ That principal bundles for smooth groups are split by an etale cover shouldn't be considered as a definition, but rather as a theorem via the Zariski-local description of smooth morphisms. It is better to use the uniform definition that a "principal bundle" for a flat group scheme $G \rightarrow S$ is defined to be locally trivial for the fpqc topology (and then one gets more tangible splitting properties as one puts nicer hypotheses on $G$). $\endgroup$
    – nfdc23
    Sep 13, 2016 at 14:45
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    $\begingroup$ Yes, that is much better. Note that the base being a field is never used, and from that perspective this is very close in spirit to Moret-Bailly's example, insofar as a genus-1 algebraic space curve $X \rightarrow S$ is a torsor for its Jacobian $P_X := {\rm{Pic}}^0_{X/S}$ that is a-priori just a smooth proper algebraic space group but is actually a scheme (that an abelian algebraic space is necessarily a scheme is a hard theorem of Deligne and Raynaud, but in relative dimension 1 it is easy via descent theory for schemes since the identity section yields a relatively ample line bundle). $\endgroup$
    – nfdc23
    Sep 13, 2016 at 15:58

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