TITLE: Lie group operation and tangent vectors
QUESTION [6 upvotes]: Suppose we have two differentiable paths $\alpha$ and $\beta$ thru the identity of a Lie group $G$, $\alpha(0)=\beta(0)=e$ the identity element. Denote $\alpha\beta$ the path given by $\alpha\beta(t)=\alpha(t).\beta(t)$ where the dot denotes the group operation. We also have $\alpha\beta(0)=e$.
The paths are differentiable, we can take the derivative, giving tangent vectors $\alpha'(0), \beta'(0), (\alpha\beta)'(0),$ which are all elements of $T_e G =\mathfrak{g},$ the lie algebra. Question: Is is true that $(\alpha\beta)'(0)=\alpha'(0)+\beta'(0)$? If not in general, then under what additional condition is it true? 
I know that it's true if $\alpha,\beta$ are 1-parameter subgroups commuting with each other, since in that case $\exp(u+v)=\exp(u)\exp(v)$. 
(This question arises when I try to study the tangent space of the space of representations from a fundamental group of a surface into a lie group.)

REPLY [11 votes]: Here's another way to look at the problem.  The derivative of a differentiable map at any point is a linear map of tangent spaces.  We have five differentiable maps in play:

The "pair of paths" map $(\alpha, \beta): (-\epsilon, \epsilon) \times (-\epsilon, \epsilon) \to G \times G$.
The multiplication map $m: G \times G \to G$.
The diagonal $\Delta: (-\epsilon, \epsilon) \to (-\epsilon, \epsilon) \times (-\epsilon, \epsilon)$.
(and 5.) The coordinate inclusions $i_1, i_2: (-\epsilon, \epsilon) \to (-\epsilon, \epsilon) \times (-\epsilon, \epsilon)$

We want to say that the derivative of $m \circ (\alpha, \beta) \circ \Delta$ is equal to the derivative of $m \circ (\alpha, \beta) \circ i_1$ plus the derivative of $m \circ (\alpha, \beta) \circ i_2$ (where the derivatives are evaluated at $0 \in (-\epsilon, \epsilon)$).  By the chain rule, this follows from the fact that the derivative of $\Delta$ is the sum of derivatives of $i_1$ and $i_2$.

REPLY [6 votes]: Heres a pretty clean proof.  Let $m:G \times G \to G$ denote the multiplication map.  Then we have (identifying $T_{e,e} G\times G$ with $T_e G \oplus T_e G$)
$$
m_*(\alpha'(0), 0) = \frac{d}{dt}\vert_{t=0} m(\alpha, e) = \alpha'(0).
$$
The same thing shows that $m*(0,\beta'(0)) = \beta'(0)$.  By linearity we get $m_*(\alpha'(0),\beta'(0)) = \alpha'(0) + \beta'(0)$.