7
$\begingroup$

Let $S$ be a base scheme. For which algebraic stacks $X$ over $S$ can we define a sheaf of differentials $\Omega^1_{X/S}$ (classifying derivations)? Probably it works when $X$ is Deligne Mumford because $\Omega^1$ is stable under pullbacks of étale morphisms? But for example for the classifying stack $B \mathbb{G}_m$ there is no $\Omega^1$ because deformations of invertible sheaves have non-trivial automorphisms, right?

Even if there is no $\Omega^1_{X/S}$ in general, I would like to know if there is a "tangent bundle" $T(X/S)$ which satisfies the adjunction $\hom_S(Y[\varepsilon]/\varepsilon^2,X) \simeq \hom_S(Y,T(X/S))$ for algebraic stacks $Y$ over $S$. If $\Omega^1_{X/S}$ exists then one may take $T(X/S) = \mathrm{Spec} \mathrm{Sym} \Omega^1_{X/S}$, but perhaps this definition is "too discrete" for algebraic stacks. Perhaps one can encode the cotangent complex into $T(X/S)$?

Edit: At this moment I am not interested in derived stacks. Algebraic stacks mean Artin stacks in the usual sense.

$\endgroup$
3
  • 7
    $\begingroup$ Didn't you already ask this question before? In fact, it seems to me, that a number of your questions disappear shortly after you ask them. I do not think that is "best practice". $\endgroup$ May 14, 2014 at 10:27
  • $\begingroup$ Sorry I've deleted my previous question because it made no sense. But I can guarantee that this here is a different question. Before I asked what $\Omega^1_{B \mathbb{G}_m}$ is and you explained to me why it doesn't exist at all. $\endgroup$ May 14, 2014 at 12:33
  • 12
    $\begingroup$ @MartinBrandenburg Questions that do not make sense are still valuable! You are a smart guy, so if you ask a question which does not make sense, it is probably easy to make the same mistake. Having a record of "false questions" and answers explaining why they are wrong is a great service of this site: such things are usually not contained in the literature, and must be gleaned by word of mouth. It would be great if there was a record of why $\Omega^1_{BG_n}$ does not exist! $\endgroup$ May 14, 2014 at 18:37

2 Answers 2

9
$\begingroup$

I might be "too derived" for the kind of framework you're looking for, but in general, if you think of a stack as a functor from CDGAs to spaces, the replacement for $\Omega^1$ is the cotangent complex for the stack. As you point out, you have to prove the existence for such a thing, but when a stack $X$ is n-geometric, a cotangent complex exists. In particular, it exists for $BG$. In that case, a sheaf on $BG$ is the same thing as giving a representation of $G$, and the cotangent complex can be written as

$$ \mathfrak{g}^\vee[1] $$

i.e., the dual of the Lie algebra, shifted in degree. The representation is the dual of the adjoint representation.

In this framework, the adjunction property you seek won't hold, because $Y[\epsilon]/\epsilon^2$ isn't semi-free. (Semi-free resolutions are the usual ways you get cofibrant objects in CDGAs.)

Regardless, the cotangent complex still "classifies" derivations in the following sense:

If $A$ is an affine scheme (i.e., a CDGA), then for every $A$-point

$$f: Spec(A) \to X$$

there exists an $A$-module $\mathbb{L}_{X,f}$ such that

\begin{equation} (*) \qquad Map_{AMod}(\mathbb{L}_{X,f},M) \simeq hofib(X(A\oplus M) \to X(A)) \end{equation}

for every $A$-module $M$ with cohomology concentrated in non-positive degree. (Note that $A \oplus M$ is an algebra as the square-zero extension.)

This must be functorial in the sense that for all commutative diagrams

$$ Spec(B) \to Spec(A) \to X $$

you have that pullback preserves the cotangent complex---i.e.,

$$ \mathbb{L}_{X,f_A} \otimes_A^{ho} B \simeq \mathbb{L}_{F,f_B}. $$

It might help to note that (*) is just a homotopical way to write a universal property from the non-derived setting: Whenever you have a map of rings $f: B \to A$, then

$$ Hom_A(f^* \Omega_B, M) \cong \{ \text{maps $B \to A \oplus M$ factoring the map $B \to A$} \}. $$

$\endgroup$
3
  • $\begingroup$ I haven't asked what $\Omega^1$ is in homotopical algebra. $\endgroup$ May 14, 2014 at 20:27
  • 2
    $\begingroup$ in which category is the dual $\mathfrak{g}^\vee[1]$ taken? If it's just the dual vector space then this provides an answer to Martin's underived question: whatever theory of Kahler differentials you use for underived stacks should be equal to a truncation of the full derived theory. As $H^0(\mathfrak{g}^\vee[1]) = H^{-1}(\mathfrak{g}^\vee) = 0$ this shows that BG has no Kahler differentials. $\endgroup$ May 14, 2014 at 23:12
  • $\begingroup$ The twist in $\mathfrak{g}^\vee[1]$ seems wrong to me, I think $\mathfrak{g}^\vee[-1]$ is correct. $\endgroup$
    – Niels
    May 25, 2015 at 13:15
2
$\begingroup$

There is a first-order jet stack, given by the Hom stack construction $\underline{\operatorname{Hom}}_S(S \times \operatorname{Spec} \mathbb{Z}[x]/(x^2), X)$. When $X/S$ is representable, this describes the total space of the relative tangent sheaf.

It does not encode the cotangent complex, and is insensitive to inertia - you would need a derived jet stack to capture more data.

$\endgroup$
7
  • $\begingroup$ This Hom stack is just a reformulation of the definition of $T(X/S)$, but there seems to be a theorem which says that it is algebraic? What does it look like for example when $X=B \mathbb{G}_m$? $\endgroup$ May 14, 2014 at 12:35
  • $\begingroup$ Yes, this is a 2006 theorem of Masao Aoki. The properties we need are that the dual numbers are proper and flat, and the algebraic stack $X$ is separated of finite type. When $X = B\mathbb{G}_m$, you just get $B\mathbb{G}_m$. $\endgroup$
    – S. Carnahan
    May 15, 2014 at 3:00
  • $\begingroup$ Ok, thank you! I think that Jack Hall has obtained more general algebraicity results for Hom stacks. $\endgroup$ May 16, 2014 at 20:20
  • $\begingroup$ The tangent bundle of $BGL_n$ is trivial because locally every extension of vector bundles splits? $\endgroup$ May 16, 2014 at 20:21
  • $\begingroup$ @MartinBrandenburg Thanks for letting me know about Hall's work - I have not been keeping track of recent progress. I see that Anatoly Preygel also has a paper on Hom stacks, and it contains a footnote on page 8 stating that Aoki's paper has some serious errors. Yes, that is one way to explain the triviality of $T(BGL_n)$. $\endgroup$
    – S. Carnahan
    May 17, 2014 at 4:23

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.