Tannakian theory for Lie algebras Let $G$ be a reductive (just in case) linear algebraic group over $\mathbb{C}$ and let $\mathfrak{g}$ be the Lie algebra of $G$. Consider the category $\operatorname{Rep}(G)$ of finite dimensional representations of $G$. Let $F\colon \operatorname{Rep}(G) \rightarrow \operatorname{Vect}_{\mathbb{C}}$ be the forgetfull (fiber) functor. According to the Tannkian formalism, group $G$ coincides with the group of tensor authomorphisms of the functor $F$. 
Is it true that the Lie algebra $\mathfrak{g}$ is nothing else but the Lie algebra of endomorphisms $E\colon F \rightarrow F $ such that for any two representations $V,W \in \operatorname{Rep}(G)$ the following map
$ F(V) \otimes F(W) \xrightarrow{E(V)\otimes \operatorname{Id} + \operatorname{Id} \otimes E(W)} F(V)\otimes F(W)
$
coincides with the map $F(V\otimes W) \xrightarrow{E(V\otimes W)} F(V\otimes W)$
with respect to the identification $F(V)\otimes F(W) \simeq F(V\otimes W)$?
 A: Yes, the statement is true for general $G$ (not just reductive).
Write $A$ for the ring of regular functions on $G$, which is a commutative Hopf algebra. For each representation $V$, there is a comodule structure $\Delta_V:F(V)\to F(V)\otimes A$. The multiplication $m:A\otimes A\to A$ is compatible with the tensor product of representations in the sense that the following two maps are equal:
$$
F(V)\otimes F(W)\xrightarrow{\Delta_V\otimes\Delta_W} (F(V)\otimes A)\otimes (F(W)\otimes A)\to F(V)\otimes F(W)\otimes A\otimes A\xrightarrow{\mathrm{Id}\otimes m}F(V)\otimes F(W)\otimes A,
$$
$$
F(V)\otimes F(W)=F(V\otimes W)\xrightarrow{\Delta_{V\otimes W}} F(V\otimes W)\otimes A=F(V)\otimes F(W)\otimes A.
$$
There is a bijection between the functorial endomorphisms of $F$ (ignoring the tensor structure) and the linear maps $\varphi:A\to \mathbb{C}$: the endomorphism corresponding to $\varphi$ is 
$$
F(V)\xrightarrow{\Delta_V} F(V)\otimes A\xrightarrow{\mathrm{Id}\otimes\varphi} F(V).
$$
Write $\epsilon:A\to\mathbb{C}$ for the counit, which corresponds to the identity endomorphism ($\epsilon$ takes a regular function to its value at the identity element of $G$). The condition that $\varphi$ corresponds to an endomorphism satisfying your condition is that the map
$$
F(V)\otimes F(W)\xrightarrow{\Delta_V\otimes\Delta_W} (F(V)\otimes A)\otimes (F(W)\otimes A)\to F(V)\otimes F(W)\otimes A\otimes A\xrightarrow{\mathrm{Id}\otimes(\varphi\otimes \epsilon+\epsilon\otimes\varphi)}F(V)\otimes F(W)
$$
coincides with the map
$$
F(V)\otimes F(W)\xrightarrow{\Delta_{V\otimes W}}F(V\otimes W)\otimes A\xrightarrow{\sim}F(V)\otimes F(W)\otimes A\xrightarrow{\mathrm{Id}\otimes\varphi}F(V)\otimes F(W).
$$
Combining this with the compatibility of the multiplication on $A$ and the tensor product, this condition is equivalent to the condition that $\varphi\circ m=\varphi\otimes \epsilon+\epsilon\otimes\varphi$. This is the same as
$$
\varphi(xy)=\varphi(x)\epsilon(y)+\epsilon(x)\varphi(y),
$$
i.e., that $\varphi$ is a derivation at the identity on $G$. The Lie algebra of a group is identified with the derivations at the identity, so we conclude that endomorphisms $\varphi$ satisfying your condition correspond to elements of the Lie algebra of $G$.
