Lie group operation and tangent vectors

Question

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.)

Answer 1

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 $i_2$.

Answer 2

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 $m_*(\alpha'(0),\beta'(0)) = \alpha'(0) + \beta'(0)$.

Answer 3

If $\alpha,\beta : \mathbb{R} \to G$, then $\alpha \beta : \mathbb{R} \to G$ is the composition $\mu \circ (\alpha,\beta)$, where $\mu : G \times G \to G$ is the multiplication. Identify $T_{(e,e)} (G \times G)$ with $T_e G \oplus T_e G$. By the chain rule, $(\alpha \beta )_* = \mu_* \circ (\alpha _{*},\beta _{*})$. Since the differential of multiplication is addition in the Lie algebra, you get $\alpha_{*} + \beta_{*}$.

Answer 4

EDIT: Wrong.

Forgive me my naiveté (and my spelling), but I think that on a small enough chart around $e$, we can actually subtract elements of $G$, and we have

$\displaystyle \left(\alpha\beta\right)^{\prime}\left(0\right)=\lim_{t\to 0}\underbrace{\frac{\alpha\left(t\right)\beta\left(t\right)-e}{t}}_{=\alpha\left(t\right)\cdot\frac{\beta\left(t\right)-e}{t}+\frac{\alpha\left(t\right)-e}{t}}$

$\displaystyle =\lim_{t\to 0}\left(\alpha\left(t\right)\cdot\frac{\beta\left(t\right)-e}{t}+\frac{\alpha\left(t\right)-e}{t}\right)$

$\displaystyle =\underbrace{\left(\lim_{t\to 0}\alpha\left(t\right)\right)}_{=\alpha\left(0\right)=e} \cdot \underbrace{\left(\lim_{t\to 0}\frac{\beta\left(t\right)-e}{t}\right)}_{=\beta^{\prime}\left(0\right)} + \underbrace{\left(\lim_{t\to 0} \frac{\alpha\left(t\right)-e}{t}\right)}_{=\alpha^{\prime}\left(0\right)}$

$\displaystyle =\beta^{\prime}\left(0\right)+\alpha^{\prime}\left(0\right) = \alpha^{\prime}\left(0\right)+\beta^{\prime}\left(0\right)$.

Note that I have used the product rule for limits ($\displaystyle \lim_{t\to 0}\left(U\left(t\right)\cdot V\left(t\right)\right)=\left(\lim_{t\to 0}U\left(t\right)\right)\cdot\left(\lim_{t\to 0}V\left(t\right)\right)$). This is okay because the multiplication on $G$ is continuous.

This looks a bit awkward, but as far as I know there is no other way to define the addition of tangent vectors than to use a chart, at least if tangent vectors are defined as stalks of curves. Thus, we cannot hope for a chart-less proof. I'd like to be proven wrong!

Answer 5

The first equation in Darij's answer needs a justification, but this follows from the Baker-Campbell-Hausdorff formula. For $X,Y \in \mathfrak{g}$ close enough to the origin, there is an identity

$$ exp (X) exp(Y) = exp(X+Y+ R(X,Y)) $$

holds, where the remainder $R(X,Y)$ is a power series in iterated commutators of $X$ and $Y$ has vanishing derivative at $(X,Y)=0$.