[Proposition 3.15, BookAV]

Lemma 39.6.3. Let $(G, m, e, i)$ be a group scheme over the scheme $S$. Denote $f : G \to S$ the structure morphism. Then there exist canonical isomorphisms

$\Omega _{G/S} \cong f^*\mathcal{C}_{S/G} \cong f^*e^*\Omega _{G/S}$

where $\mathcal{C}_{S/G}$ denotes the conormal sheaf of the immersion $e$. In particular, if $S$ is the spectrum of a field, then $\Omega _{G/S}$ is a free $\mathcal{O}_ G$-module.

Proof. By Morphisms, Lemma 29.32.10 we have

$\Omega _{G \times _ S G/G} = \text{pr}_0^*\Omega _{G/S}$

where on the left hand side we view $G \times _ S G$ as a scheme over $G$ using $\text{pr}_1$. Let $\tau : G \times _ S G \to G \times _ S G$ be the “shearing map” given by $(g, h) \mapsto (m(g, h), h)$ on points. This map is an automorphism of $G \times _ S G$ viewed as a scheme over $G$ via the projection $\text{pr}_1$. Combining these two remarks we obtain an isomorphism

$\tau ^*\text{pr}_0^*\Omega _{G/S} \to \text{pr}_0^*\Omega _{G/S}$

Since $\text{pr}_0 \circ \tau = m$ this can be rewritten as an isomorphism

$m^*\Omega _{G/S} \to \text{pr}_0^*\Omega _{G/S}$

Pulling back this isomorphism by $(e \circ f, \text{id}_ G) : G \to G \times _ S G$ and using that $m \circ (e \circ f, \text{id}_ G) = \text{id}_ G$ and $\text{pr}_0 \circ (e \circ f, \text{id}_ G) = e \circ f$ we obtain an isomorphism

$\Omega _{G/S} \to f^*e^*\Omega _{G/S}$

as desired. By Morphisms, Lemma 29.32.16 we have $\mathcal{C}_{S/G} \cong e^*\Omega _{G/S}$. If $S$ is the spectrum of a field, then any $\mathcal{O}_ S$-module on $S$ is free and the final statement follows. $\square$

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