# Is the cotangent complexes of groupoids bounded above by degree $1$?

Let $\mathcal{X}$ be a stack given by a groupoid $X_1\rightrightarrows X_0$, where $X_0$ and $X_1$ are smooth $k$-varieties. Let $\mathbb{L}_{\mathcal{X}/k}$ be the cotangent complex of $\mathcal{X}$.

I've heard the fact that the amplitude of $\mathbb{L}_{\mathcal{X}/k}$ is bounded above by degree $1$, i.e. the cohomology of this complex vanishes for degree $>1$. I think this fact is well-known to experts but I cannot figure out a proof by myself.

$\bf{My question}$ is: how to prove the above property? Is there any written-down proofs in the literature?

• The standard references for cotangent complexes of stacks are Laumon - Moret-Bailly and Olsson. – Jason Starr Sep 9 '14 at 11:17
• @JasonStarr Thank you very much! Is there any illustrative description in easy cases? – Zhaoting Wei Sep 9 '14 at 19:55

## 1 Answer

Here's a proof assuming basic properties of the cotangent complexes of stack. At least for the proof to work I need to assume that $$X_0$$ is smooth over $$k$$ and the source/target maps $$s, t: X_1 \to X_0$$ of the groupoid structure are smooth (implying that $$X_1$$ is also smooth over $$k$$).

Note: when I say "concentrated in degree 0" I mean "has cohomology only in degree 0."

There's a cartesian diagram of the form $$\begin{equation} \begin{matrix} X_1 & \xrightarrow{t} & X_0 \\ s \downarrow & & \downarrow \pi \\ X_0 & \xrightarrow{\pi} &\mathcal{X} \\ \end{matrix} \end{equation}$$

By the construction of $$\mathcal{X}$$, $$\pi$$ is a smooth cover. It follows that $$\begin{equation} t^* L_{\pi} = L_{s} \end{equation}$$ Since $$s$$ is smooth by hypothesis, $$L_s \simeq \Omega_{X_1/X}$$. So in particular $$L_s$$ is concentrated in degree 0. Since $$\pi$$ (and hence $$t$$) are smooth covers, by descent $$L_\pi$$ is concentrated in degree 0.

Now consider the composition $$\begin{equation} X_0 \xrightarrow{\pi} \mathcal{X} \to \mathrm{Spec} \, k \end{equation}$$ and the resulting distinguished triangle $$\pi^* L_{\mathcal{X}/k} \to L_{X_0/k} \to L_{\pi} \to \pi^* L_{\mathcal{X}/k}$$ in $$\mathrm{D}(X_0)$$. Since $$X_0$$ is smooth over $$k$$, $$L_{X_0/k}$$ is concentrated in degree 0, and we just showed $$L_{\pi}$$ is concentrated in degree 0. Taking the long exact sequence of cohomology sheaves completes the proof.

• Thanks @DavidRoberts for fixing the diagrams. – cgodfrey Mar 12 '19 at 20:14