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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.

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    $\begingroup$ 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". $\endgroup$ – Jason Starr May 14 '14 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. $\endgroup$ – Martin Brandenburg May 14 '14 at 12:33
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    $\begingroup$ @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! $\endgroup$ – Steven Gubkin May 14 '14 at 18:37
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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

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

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  • $\begingroup$ I haven't asked what $\Omega^1$ is in homotopical algebra. $\endgroup$ – Martin Brandenburg May 14 '14 at 20:27
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    $\begingroup$ 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. $\endgroup$ – bananastack May 14 '14 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 '15 at 13:15
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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.

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  • $\begingroup$ 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 $X=B \mathbb{G}_m$? $\endgroup$ – Martin Brandenburg May 14 '14 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 '14 at 3:00
  • $\begingroup$ Ok, thank you! I think that Jack Hall has obtained more general algebraicity results for Hom stacks. $\endgroup$ – Martin Brandenburg May 16 '14 at 20:20
  • $\begingroup$ The tangent bundle of $BGL_n$ is trivial because locally every extension of vector bundles splits? $\endgroup$ – Martin Brandenburg May 16 '14 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 '14 at 4:23

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