# Tangent space to lie group at identity.

I'm supposed to show that for a Lie group G, $T_{(e,e)}G\times G \simeq T_eG\oplus T_eG$ and that $T_{(e,e)}m$ is given by $(X,Y)\mapsto X+Y$. I'm having trouble proving this. I'm not exactly clear what this means or how one would go about proving this.

To show $T_{(e,e)} G \times G \cong T_e G \oplus T_e G$, let $v$ be a tangent vector to $G \times G$ at $(e,e)$, viewed as $X = \gamma'(0)$ for some curve $\gamma : (-\epsilon, \epsilon) \rightarrow G \times G$ with $\gamma(0) = e$. Then, letting $pr_1, pr_2: G \times G \rightarrow G$ denote the two natural projections, the isomorphism in question sends $X$ to $$((pr_1 \circ \gamma)'(0), (pr_2 \circ \gamma)'(0)) \in T_e G \oplus T_e G.$$ This map is simply $d_{(e,e)} pr_1 \times d_{(e,e)} pr_2$. It is not hard to show the map is a well-defined bijective linear transformation.
Then I think you want to study the differential of the multiplication map $m : G \times G \rightarrow G$, $$d_{(e,e)}m : T_{(e,e)} G \times G \rightarrow T_e G.$$ Via the above isomorphism, this becomes $$d_{(e,e)}m : T_e G \oplus T_e G \rightarrow T_e G.$$ Given a tangent vector $(X,Y) = (\gamma'(0), \rho'(0))$ for smooth curves $\gamma, \rho : (-\epsilon, \epsilon) \rightarrow G$ with $\gamma(0) = \rho(0) = e$. Then $$d_{(e,e)}m (X,Y) = (m \circ (\gamma, \rho))'(0) = (\gamma \rho)'(0) = \gamma'(0) \rho(0) + \gamma(0) \rho'(0) = X + Y.$$
• Why is $(\gamma \rho )'(0)=\gamma'(0)\rho(0) +\gamma(0)\rho '(0)$? Oct 22 '14 at 20:00
• That's the product rule. Here I am identifying $G$ with a closed subgroup of $GL_n$ and doing all my calculations in the space of $n \times n$ matrices. Oct 22 '14 at 20:10