# Group action of linear algebraic group $G$ on itself induces a representaion of $G$ on $Lie(G)$

Let us be given a linear algebraic group $G$ over a field $K$ of characterstic zero. This group $G$ is defined as the common zeroes of a finite set of polynomials $\{f_1, \ldots ,f_r\}$ $\in K [T_1,\ldots,T_n]$with some additional properties like having the structure of a group along with 2 morphisms of varieties.

Now, the tangent space of $G$ at a point $x$ can be thought of as common zeroes of $\{d_xf_1,\ldots,d_xf_r\}$ where $d_xf= \sum_{i=1}^{n} \frac{\partial f}{\partial T_i}(x) (T_i - x_i)$. Let us think of the $Lie(G)$ as the tangent space at $e$ defined this way.

Now, suppose we are given a group action of $G$ on $G$, say $$\phi : G \times G \to G$$ sending $$(g,x) \mapsto gxg^{-1}$$

How does this map induce a represention :

$$\psi : G \to GL_{Lie(G)}$$

Please help me understand in terms of the common zeroes as how the common zeroes are mapped by an element of $G$. I have just started to read algebraic geometry. So, an intutive explanation will be really helpful.

• For any $g$, the map $x\mapsto gxg^{-1}$ has a differential $dg:T_eG\to T_eG$. That's your map. You need to read on what the differential of a map is. – Mariano Suárez-Álvarez Feb 3 '16 at 19:50
• @MarianoSuárez-Alvarez Please let me know where should I read about the differential map so that I can get an intutive feeling as well. Thanks for the help and wherever I have read the differential, Lie(G) is either thought of as $M/M^2$ dual or the derivations on the coordinate ring commuting with the translation map. Is there any definition which explictly tells in terms of these common zeroes ? – Jagdeep Singh Feb 3 '16 at 19:52
• I suggest you pick a good calculus book treating functions of many variables to get an intuition. The differential in algebraic geometry is just the one from calculus done algebraically. – Mariano Suárez-Álvarez Feb 3 '16 at 19:53