In many references (Toen, Higher and derived stacks: a global overview, Toen, Vezzosi, Homotopical algebraic geometry II, and so on), the definition of $n$-geometric stack appears.

In the non-derived case, the definition starts by declaring affine schemes as $(-1)$-geometric stacks and inductively defines $n$-geometric stacks by some procedure.

Toen-Vezzosi, HAG II, Remark says that algebraic spaces and schemes are automatically 1-geometric stacks. I can check that schemes are 1-geometric. (It does not depend on whether a scheme is separated or not.) But I can't check easily that algebraic spaces are 1-geometric stacks.


1 Answer 1


Always check the definitions being used in the reference - there are even significant differences between different arXiv versions of HAG2. Once you know you have an epimorphism from a union of affines, saying that $X$ is $n$-geometric in the HAG2 v7 sense basically amounts to saying that the higher diagonal $$ X \to \mathrm{map}(S^{n},X) $$ is affine. Thus $0$-geometric is equivalent to semi-separated, and any algebraic space $X$ is $1$-geometric because $\mathrm{map}(S^1,X)\cong X$.

EDIT: in response to keaton's comment, the condition isn't quite equivalent to $n$-geometricity, as there are some epimorphism conditions to check, but arises inductively because $$ X\times^h_{\mathrm{map}(S^{n},X)}X \cong \mathrm{map}(S^{n+1},X). $$

FURTHER EDIT with more details now I have time:

If you take affines $U,V$ etale over $X$, you want to show that $U\times_XV$ is $0$-geometric, and you already know that it is a scheme. Since the map $U\times_XV \to U \times V$ is a pullback of the diagonal $X \to X \times X$ of $X$, the relative diagonal $U\times_XV \to (U\times_XV)\times^h_{U \times V}(U\times_XV)$ is a pullback of $X \to \mathrm{map}(S^1,X)$, so is an isomorphism. The (absolute) diagonal of $U\times_XV$ is then a pullback of the diagonal of $U\times V$, hence affine.

  • 2
    $\begingroup$ Thank you. I think you know something and have intuition. Since I'm a just beginner of derived algebraic geometry, I cannot understand that n-geometric is equivalent to that higher diagonal $X \to map(S^n, X)$ is affine. May I ask you the reason why? Is there a theorem saying this in HAG2? $\endgroup$
    – keaton
    Feb 20, 2018 at 10:19
  • $\begingroup$ Sorry I'm little confusing. What is a definition of $\mathrm{map}(S^n,X)$? I think $X$ lives in the category of functors from $\mathrm{Ho}(sComm)^{op}$ to $\mathrm{SSet}$. Is it make sence to consider a map space from $S^n$ to $X$? $\endgroup$
    – keaton
    Feb 21, 2018 at 4:29
  • $\begingroup$ On the other hand, why $\mathrm{map}(S^1,X)$ is isomorphic to $X$? $\endgroup$
    – keaton
    Feb 21, 2018 at 4:40
  • $\begingroup$ By $\mathrm{map}(S^n,X)$, I just mean the functor $A \mapsto \mathrm{map}(S^n,X(A))$. For set-valued functors $X$, you then have $\mathrm{map}(S^n,X) \simeq X$. $\endgroup$ Feb 21, 2018 at 7:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.