# Lie group operation and tangent vectors

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.)

• About some of following answers: Excuse me, but is not the content of the question exactly to prove that $T_{e,e}\mu(\xi,\eta)=\xi+\eta$ for any $\xi,\eta\in T_eG$? So we should not appeal to it in a proof. But we should point out that this expression is just a consequence of the canonical identification of $T(G\times G)$ with the direct product $TG\times TG$.
– agt
Commented Apr 25, 2011 at 11:47
• I'm sorry but it's not quite clear to me how the problem follows from the identification $T(G\times G)=TG\times TG$. Commented Apr 25, 2011 at 12:28
• This is a good question, but probably not a great fit for this forum --- it easily arises (and is worded as if it has arisen) as a homework problem in a first course on Lie theory. Since it has been answered and accepted, it's not a big deal. Commented Apr 25, 2011 at 21:37

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 $i_2$.

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

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 $\alpha_{*} + \beta_{*}$.

• "Since the differential of multiplication is addition in the Lie algebra": that is more or less the original question, isn't it? Commented Apr 25, 2011 at 10:56

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!

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 $(X,Y)=0$.

• Which "first equation"? Commented Apr 25, 2011 at 11:00
• Ah, you mean distributivity. Commented Apr 25, 2011 at 11:01
• I don't believe we can use BCH here, as the proof of BCH (the geometric one) requires lots of assertions of the same kind (but harder) than the OP's question. Commented Apr 25, 2011 at 11:06
• The equation $(\alpha\beta)'(0)=\lim_{t\to 0}\underbrace{\frac{\alpha(t)\beta(t)-e}{t}}$ does not make sense without additional information, because you cannot subtract. Either you embed the group in $GL_n$, or you pick a chart in which you can take differences, for example exponential charts. How the multiplication relates to addition in exponential charts is the CBH formula. Commented Apr 26, 2011 at 6:00