## 78.5 Groupoid of sections

Suppose we have a groupoid $(\text{Ob}, \text{Arrows}, s, t, c, e, i)$. Then we can construct a monoid $\Gamma$ whose elements are maps $\delta : \text{Ob} \to \text{Arrows}$ with $s \circ \delta = \text{id}_{\text{Ob}}$ and composition given by

$\delta _1 \circ \delta _2 = c(\delta _1 \circ t \circ \delta _2, \delta _2)$

In other words, an element of $\Gamma$ is a rule $\delta$ which prescribes an arrow emanating from every object and composition is the natural thing. For example

$\vcenter { \xymatrix{ & \bullet \ar[dl] \\ \bullet \ar@(ul, dl)[] & \bullet \ar[d] \\ & \bullet \ar[lu] } } \quad \circ \quad \vcenter { \xymatrix{ & \bullet \ar[d] \\ \bullet \ar[ru] & \bullet \ar[d] \\ & \bullet \ar[lu] } } \quad = \quad \quad \vcenter { \xymatrix{ & \bullet \ar@/^/[dd] \\ \bullet \ar@(ul, dl)[] & \bullet \ar[l] \\ & \bullet \ar[lu] } }$

with obvious notation

The same procedure can be applied to a groupoid in algebraic spaces $(U, R, s, t, c, e, i)$ over a scheme $S$. Namely, as elements of $\Gamma$ we take the set

$\Gamma = \{ \delta : U \to R \mid s \circ \delta = \text{id}_ U\}$

and composition $\circ : \Gamma \times \Gamma \to \Gamma$ is given by the rule above

78.5.0.1
\begin{equation} \label{spaces-more-groupoids-equation-composition} \delta _1 \circ \delta _2 = c(\delta _1 \circ t \circ \delta _2, \delta _2) \end{equation}

The identity is given by $e \in \Gamma$. The groupoid $\Gamma$ is not a group in general because there may be elements $\delta \in \Gamma$ which do not have an inverse. Namely, it is clear that $\delta \in \Gamma$ will have an inverse if and only if $t \circ \delta$ is an automorphism of $U$ and in this case $\delta ^{-1} = i \circ \delta \circ (t \circ \delta )^{-1}$.

For later use we discuss what happens with the subgroupoid $\Gamma _0$ of $\Gamma$ of sections which are infinitesimally close to the identity $e$. More precisely, suppose given an $R$-invariant closed subspace $U_0 \subset U$ such that $U$ is a first order thickening of $U_0$. Denote $R_0 = s^{-1}(U_0) = t^{-1}(U_0)$ and let $(U_0, R_0, s_0, t_0, c_0, e_0, i_0)$ be the corresponding groupoid in algebraic spaces. Set

$\Gamma _0 = \{ \delta \in \Gamma \mid \delta |_{U_0} = e_0\}$

If $s$ and $t$ are flat, then every element in $\Gamma _0$ is invertible. This follows because $t \circ \delta$ will be a morphism $U \to U$ inducing the identity on $\mathcal{O}_{U_0}$ and on $\mathcal{C}_{U_0/U}$ (Lemma 78.5.1) and we conclude because we have a short exact sequence $0 \to \mathcal{C}_{U_0/U} \to \mathcal{O}_ U \to \mathcal{O}_{U_0} \to 0$.

Lemma 78.5.1. In the situation discussed in this section, let $\delta \in \Gamma _0$ and $f = t \circ \delta : U \to U$. If $s, t$ are flat, then the canonical map $\mathcal{C}_{U_0/U} \to \mathcal{C}_{U_0/U}$ induced by $f$ (More on Morphisms of Spaces, Lemma 75.5.3) is the identity map.

Proof. To see this we extend the bottom of the diagram (78.3.0.2) as follows

$\xymatrix{ Y \ar[r] \ar[d] & R \times _{s, U, t} R \ar@<1ex>[r]^-c \ar@<-1ex>[r]_-{\text{pr}_0} \ar[d]_{\text{pr}_1} & R \ar[r]^ t \ar[d]^ s & U \\ U \ar[r]_\delta & R \ar@<1ex>[r]^ s \ar@<-1ex>[r]_ t & U }$

where the left square is cartesian and this is our definition of $Y$; we will not need to know more about $Y$. There is a similar diagram with similar properties obtained by base change to $U_0$ everywhere. We are trying to show that $\text{id}_ U = s \circ \delta$ and $f = t \circ \delta$ induce the same maps on conormal sheaves. Since $s$ is flat and surjective, it suffices to prove the same thing for the two compositions $a, b : Y \to R$ along the top row. Observe that $a_0 = b_0$ and that one of $a$ and $b$ is an isomorphism as we know that $s \circ \delta$ is an isomorphism. Therefore the two morphisms $a, b : Y \to R$ are morphisms between algebraic spaces flat over $U$ (via the morphism $t : R \to U$ and the morphism $t \circ a = t \circ b : Y \to U$). This implies what we want. Namely, by the compatibility with compositions in More on Morphisms of Spaces, Lemma 75.5.4 we conclude that both maps $a_0^*\mathcal{C}_{R_0/R} \to \mathcal{C}_{Y_0/Y}$ fit into a commutative diagram

$\xymatrix{ a_0^*\mathcal{C}_{R_0/R} \ar[rr] & & \mathcal{C}_{Y_0/Y} \\ a_0^*t_0^*\mathcal{C}_{U_0/U} \ar[u] \ar@{=}[rr] & & (t_0 \circ a_0)^*\mathcal{C}_{U_0/U} \ar[u] }$

whose vertical arrows are isomorphisms by More on Morphisms of Spaces, Lemma 75.18.1. Thus the lemma holds. $\square$

Let us identify the group $\Gamma _0$. Applying the discussion in More on Morphisms of Spaces, Remarks 75.17.3 and 75.17.7 to the diagram

$\xymatrix{ (U_0 \subset U) \ar@{..>}[rr]_{(e_0, \delta )} \ar[rd]_{(\text{id}_{U_0}, \text{id}_ U)} & & (R_0 \subset R) \ar[ld]^{(s_0, s)} \\ & (U_0 \subset U) }$

we see that $\delta = \theta \cdot e$ for a unique $\mathcal{O}_{U_0}$-linear map $\theta : e_0^*\Omega _{R_0/U_0} \to \mathcal{C}_{U_0/U}$. Thus we get a bijection

78.5.1.1
\begin{equation} \label{spaces-more-groupoids-equation-isomorphism} \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_{U_0}}(e_0^*\Omega _{R_0/U_0}, \mathcal{C}_{U_0/U}) \longrightarrow \Gamma _0 \end{equation}

by applying More on Morphisms of Spaces, Lemma 75.17.5.

Proof. Let $\delta _1, \delta _2 \in \Gamma _0$ correspond to $\theta _1, \theta _2$ as above and the composition $\delta = \delta _1 \circ \delta _2$ in $\Gamma _0$ correspond to $\theta$. We have to show that $\theta = \theta _1 + \theta _2$. Recall (More on Morphisms of Spaces, Lemma 75.17.2) that $\theta _1, \theta _2, \theta$ correspond to derivations $D_1, D_2, D : e_0^{-1}\mathcal{O}_{R_0} \to \mathcal{C}_{U_0/U}$ given by $D_1 = \theta _1 \circ \text{d}_{R_0/U_0}$ and so on. It suffices to check that $D = D_1 + D_2$.

We may check equality on stalks. Let $\overline{u}$ be a geometric point of $U$ and let us use the local rings $A, B, C$ introduced in Section 78.4. The morphisms $\delta _ i$ correspond to ring maps $\delta _ i : B \to A$. Let $K \subset A$ be the ideal of square zero such that $A/K = \mathcal{O}_{U_0, \overline{u}}$. In other words, $K$ is the stalk of $\mathcal{C}_{U_0/U}$ at $\overline{u}$. The fact that $\delta _ i \in \Gamma _0$ means exactly that $\delta _ i(I) \subset K$. The derivation $D_ i$ is just the map $\delta _ i - e : B \to A$. Since $B = s(A) \oplus I$ we see that $D_ i$ is determined by its restriction to $I$ and that this is just given by $\delta _ i|_ I$. Moreover $D_ i$ and hence $\delta _ i$ annihilates $I^2$ because $I = \mathop{\mathrm{Ker}}(I)$.

To finish the proof we observe that $\delta$ corresponds to the composition

$B \to C = (B \otimes _{s, A, t} B)^ h_{\mathfrak m_ B \otimes B + B \otimes \mathfrak m_ B} \to A$

where the first arrow is $c$ and the second arrow is determined by the rule $b_1 \otimes b_2 \mapsto \delta _2(t(\delta _1(b_1))) \delta _2(b_2)$ as follows from (78.5.0.1). By Lemma 78.4.1 we see that an element $\zeta$ of $I$ maps to $\zeta \otimes 1 + 1 \otimes \zeta$ plus higher order terms. Hence we conclude that

$D(\zeta ) = (\delta _2 \circ t)\left(D_1(\zeta )\right) + D_2(\zeta )$

However, by Lemma 78.5.1 the action of $\delta _2 \circ t$ on $K = \mathcal{C}_{U_0/U, \overline{u}}$ is the identity and we win. $\square$

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