Computing a differential on a derivation Let $\varphi:G\to G'$ be a morphism of algebraic groups over an algebraically closed field $k$, so that $d\varphi:\mathscr{L}(G)\to\mathscr{L}(G')$ is a morphism of Lie algebras.  Here I view $\mathscr{L}(G)$ as the set of all left-invariant derivations $\delta:k[G]\to k[G]$.
Given a left-invariant derivation $\delta$ of $k[G]$, how can we compute the left-invariant derivation $d\varphi(\delta)$ of $k[G']$?  In other words, if $f\in k[G']$, what is $d\varphi(\delta)(f)\in k[G']$?
 A: A homomorphism $\varphi : G \to G'$ induces a map $(d\varphi)_1 : T_1(G) \to T_1(G')$ of the tangent spaces at the identiy elements of the two groups, but does not induce a map $\operatorname{Der}_k(k[G]) \to \operatorname{Der}_k(k[G'])$, unless $\varphi$ is an isomorphism. Therefore, given $\delta \in \mathscr{L}(G)$, we define the left-invariant derivation $d\varphi(\delta) \in \mathscr{L}(G')$ by translation of the tangent vector $(d\varphi)_1(\delta_1) \in T_1(G')$ to the tangent spaces at all $k$-points of $G'$ (here $\delta_1 \in T_1(G)$ denotes the $k$-valued derivation of the local ring $\mathscr{O}_{G,1}$ induced by $\delta$). Since the coordinate ring $k[G']$ is reduced and $k$ is algebraically closed, an element $h \in k[G']$ can be uniquely recovered from the map $G'(k) \to k$ it induces. Thus, if for $x \in G'$, we denote by $\ell_x$ the ring automorphism of $k[G']$ defined via the rule $\ell_x(f)(y) := f(x^{-1}y)$, and by $\varphi^\sharp$ the homomorphism $k[G']\to k[G]$ induced by $\varphi$, we have
$$ d\varphi(\delta)(f)(x) = \delta(\varphi^\sharp\circ\ell_{x^{-1}}(f))(1).$$
One could show that, as a function of $x \in G'$, the right hand side defines an element of $k[G']$, as follows (cf. e.g. the proof of Theorem 3.4 from Chapter I, $\S 3$ in the second edition of Linear algebraic groups by A. Borel). Let $\mu : k[G'] \to k[G'] \otimes_k k[G']$ denote the co-multiplication map induced by the multiplication $G' \times_k G' \to G'$, and write $\mu(f) = \sum g_j \otimes h_j$ with $g_j, h_j \in k[G']$. It follows that $\ell_{x^{-1}}(f) = \sum g_j(x)\cdot h_j$, and hence
$$ d\varphi(\delta)(f)(x) = \sum g_j(x)\cdot\delta(\varphi^\sharp h_j)(1).$$
Thus, $d\varphi(\delta)$ maps $k[G']$ into itself. The proof that this is in fact a left-invariant derivation of $k[G']$ is routine.
