The invariant differentials of a group scheme

Notation and the shearing automorphism. Put $W=G{\times }_{k}G$ with projections $p_0,p_1$ and consider the shearing map $\tau :W\to W$, $\tau (g,h)=(m(g,h),h)$. Then $p_1\circ \tau =p_1$, and $\tau$ is an isomorphism over $G$ with inverse ${\tau }^{-1}(u,h)=(m(u,i(h)),h)$: indeed $\tau ({\tau }^{-1}(u,h))=(m(m(u,i(h)),h),h)=(u,h)$ and ${\tau }^{-1}(\tau (g,h))=(m(m(g,h),i(h)),h)=(g,h)$ by associativity and the inverse laws of [F7]. Moreover $m=p_0\circ \tau$. The map $s=(e\circ f,{\mathrm{id}}_G):G\to W$ satisfies $m\circ s={\mathrm{id}}_G$ and $p_0\circ s=e\circ f$ by the identity law of [F7].

Base change along the structure morphism. Apply [F1] to the Cartesian square with $X=G$, $S=\mathrm{Spec}k$, $S^{\mathrm{\prime }}=G$ and $S^{\mathrm{\prime }}\to S=f$. Its fibre product is $W=G{\times }_{k}G$, the projection to $X$ is $p_0$, and the structure map to $S^{\mathrm{\prime }}$ is $p_1$. Thus $p_0^{\ast }{\mathrm{\Omega }}_{G\mathrm{/}k}\cong {\mathrm{\Omega }}_{W\mathrm{/}G}$ canonically, where the relative differentials on the right are taken for $p_1$.

Differentials of automorphisms are isomorphisms. Let $\phi :X\to Y$ be an isomorphism of $G$-schemes with inverse $\psi$; for the application below $X=Y=W$ over $G$ and $\phi =\tau$ with $\psi ={\tau }^{-1}$ by step 1.1. By the identity clause of [F2], $\mathrm{d}({\mathrm{id}}_X)$ is the canonical identification ${\mathrm{id}}_X^{\ast }{\mathrm{\Omega }}_{X\mathrm{/}G}\cong {\mathrm{\Omega }}_{X\mathrm{/}G}$, and by the chain-rule clause applied to $X\stackrel{\phi }{\to }Y\stackrel{\psi }{\to }X$ the composite $\mathrm{d}\phi \circ ({\phi }^{\ast }\mathrm{d}\psi )$, formed with the canonical identification ${\phi }^{\ast }{\psi }^{\ast }\cong (\psi \circ \phi {)}^{\ast }$, equals $\mathrm{d}(\psi \circ \phi )=\mathrm{d}({\mathrm{id}}_X)$; applying the chain rule to the reversed composite gives $\mathrm{d}\psi \circ ({\psi }^{\ast }\mathrm{d}\phi )=\mathrm{d}({\mathrm{id}}_Y)$. Hence $\mathrm{d}\phi$ and $\mathrm{d}\psi$ are mutually inverse under the canonical pullback identifications, so $\mathrm{d}\phi$ is an isomorphism.

Comparison of $m$ with the first projection. The canonical identification ${\tau }^{\ast }{p}_0^{\ast }\cong (p_0\circ \tau {)}^{\ast }$ of [F2], together with $m=p_0\circ \tau$ from step 1.1, identifies $m^{\ast }{\mathrm{\Omega }}_{G\mathrm{/}k}\cong {\tau }^{\ast }{p}_0^{\ast }{\mathrm{\Omega }}_{G\mathrm{/}k}$; pulling the isomorphism of step 1.2 back along $\tau$ gives ${\tau }^{\ast }{p}_0^{\ast }{\mathrm{\Omega }}_{G\mathrm{/}k}\cong {\tau }^{\ast }{\mathrm{\Omega }}_{W\mathrm{/}G}$, and step 2.1 applied to $\phi =\tau$ gives ${\tau }^{\ast }{\mathrm{\Omega }}_{W\mathrm{/}G}\cong {\mathrm{\Omega }}_{W\mathrm{/}G}$. Combining with step 1.2 a second time yields the canonical isomorphism $m^{\ast }{\mathrm{\Omega }}_{G\mathrm{/}k}\cong p_0^{\ast }{\mathrm{\Omega }}_{G\mathrm{/}k}$.

Pulling back along the identity section. The composite $m\circ s={\mathrm{id}}_G$ and the chain-rule identification $s^{\ast }{m}^{\ast }\cong (m\circ s{)}^{\ast }$ of [F2], followed by the identity clause for ${\mathrm{id}}_G$, identify $s^{\ast }{m}^{\ast }{\mathrm{\Omega }}_{G\mathrm{/}k}\cong {\mathrm{\Omega }}_{G\mathrm{/}k}$; likewise $p_0\circ s=e\circ f$ identifies $s^{\ast }{p}_0^{\ast }{\mathrm{\Omega }}_{G\mathrm{/}k}\cong (e\circ f{)}^{\ast }{\mathrm{\Omega }}_{G\mathrm{/}k}=f^{\ast }{e}^{\ast }{\mathrm{\Omega }}_{G\mathrm{/}k}$. Applying $s^{\ast }$ to step 3.1 therefore yields the canonical isomorphism ${\mathrm{\Omega }}_{G\mathrm{/}k}\cong f^{\ast }{e}^{\ast }{\mathrm{\Omega }}_{G\mathrm{/}k}$ of invariant differentials. Now $e^{\ast }{\mathrm{\Omega }}_{G\mathrm{/}k}\cong {\mathfrak{m}}_e\mathrm{/}{\mathfrak{m}}_e^2$ by [F3], so by [F4] the module ${\mathrm{\Omega }}_{G\mathrm{/}k}\cong f^{\ast }{e}^{\ast }{\mathrm{\Omega }}_{G\mathrm{/}k}\cong {\mathcal{O}}_G{\otimes }_k({\mathfrak{m}}_e\mathrm{/}{\mathfrak{m}}_e^2)$ is free. Its rank is ${\dim }_k({\mathfrak{m}}_e\mathrm{/}{\mathfrak{m}}_e^2)$, which equals ${\dim }_k\mathrm{Lie}(G)$: the cotangent space ${\mathfrak{m}}_e\mathrm{/}{\mathfrak{m}}_e^2$ is finite-dimensional, its dual $\mathrm{Lie}(G)=T_{G\mathrm{/}k,e}$ is the intrinsic tangent space $T_{e}G$ at the $k$-rational point $e$ and hence has the same finite dimension, by [F5].

Left translations. Let $x\in G(k)$ and let ${\mathrm{\ell }}_x:G\to G$, ${\mathrm{\ell }}_x(g)=m(x,g)$, be left translation by $x$; then ${\mathrm{\ell }}_x$ is an isomorphism with inverse ${\mathrm{\ell }}_{x^{-1}}$, and ${\mathrm{\ell }}_x\circ e=x$ by the identity law of [F7]. By step 2.1 applied to $\phi ={\mathrm{\ell }}_x$ (an isomorphism of $k$-schemes, the base being $\mathrm{Spec}k$), the differential $\mathrm{d}{\mathrm{\ell }}_x:{\mathrm{\ell }}_x^{\ast }{\mathrm{\Omega }}_{G\mathrm{/}k}\to {\mathrm{\Omega }}_{G\mathrm{/}k}$ is an isomorphism; taking the induced map on the fibre at $e$ in the sense of the fibre clause of [F2], and using that ${\mathrm{\ell }}_x(e)=x$ so that the source fibre is ${\mathrm{\Omega }}_{G\mathrm{/}k}{\otimes }_{{\mathcal{O}}_{G,x}}\kappa (x)$, it identifies ${\mathrm{\Omega }}_{G\mathrm{/}k}{\otimes }_{{\mathcal{O}}_{G,x}}\kappa (x)$ with its target ${\mathrm{\Omega }}_{G\mathrm{/}k}{\otimes }_{{\mathcal{O}}_{G,e}}\kappa (e)={\mathfrak{m}}_e\mathrm{/}{\mathfrak{m}}_e^2$, the identification of the target with the cotangent space at the identity being step 4.1.