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.)



Lie group operation and tangent vectors
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:


1. The "pair of paths" map $(\alpha, \beta): (-\epsilon, \epsilon) \times (-\epsilon, \epsilon) \to G \times G$.
2. The multiplication map $m: G \times G \to G$.
3. The diagonal $\Delta: (-\epsilon, \epsilon) \to (-\epsilon, \epsilon) \times (-\epsilon, \epsilon)$.
4. (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 $i\_2$.


