I am currently studying Lie groups and I cannot solve the following exercise, which I think is vital to my understanding.

The Lie bracket is defined as $[X,Y]=\text{ad}(X)Y$. Let the group commutator be defined by $c(x,y)=xyx^{-1}y^{-1}$. Show that $$[X,Y]=\frac{d}{ds}\frac{d}{dt}c(\exp sX,\exp tY) \text{ at } s=0,t=0.$$

If I write it out, I get $$[X,Y]=\text{ad}(X)(Y)=\frac{d}{ds}\text{Ad}(\exp sX)(Y)=\frac{d}{ds}\frac{d}{dt}C_{\exp sX}\exp tY=\frac{d}{ds}\frac{d}{dt}\exp sX \exp tY (\exp sX)^{-1}$$

What am I doing wrong?

  • 2
    $\begingroup$ See the comments here, why this is false. $\endgroup$ Apr 24, 2015 at 15:00
  • $\begingroup$ So my line of thought was right? Thanks! $\endgroup$ Apr 24, 2015 at 20:39

1 Answer 1


The formula is correct. Your approach will work and it should become clear what to do after putting in a cleverly placed "$-0$", namely consider that

$$\frac{d}{ds}\text{Ad}(\exp sX)(Y)|_{s=0}=\frac{d}{ds}(\text{Ad}(\exp sX)(Y)-Y)|_{s=0}$$

Now use the fact that $$\frac{d}{dt}f(t,t)|_{t=0}=\frac{d}{dt}f(t,0)|_{t=0}+\frac{d}{dt}f(0,t)|_{t=0}$$ for every smooth map $f:U\rightarrow M$ from a neigborhood $U$ of $0$ in $\mathbb{R}^2$ to a smooth manifold $M$.

Here it is shown that the statement $d_{(e,e)}c(X,Y)=[X,Y]$ is false. Note however that $$d_{(e,e)}c(X,Y)=\frac{d}{dt}\exp(tX)\exp(tY)\exp(-tX)\exp(-tY)|_{t=0}$$ which is not the same as the formula you wrote. I think a source for what you are after is given in the comments at that post though.

Naming the Lie group $G$, the difference comes from either viewing $c$ as a smooth map from $G\times G$ to $G$ or as a smooth map from $G$ to $G$ parametrized by $G$. The latter is what you want and it yields first (differentiation w.r.t. $t$) a smooth map from $G$ to $End(T_eG)$ given by $$x\mapsto Ad(x)-Id$$ and then (differentiation w.r.t. $s$) a linear map from $T_eG$ to $T_0End(T_eG)=End(T_eG)$ given by $$X\mapsto ad(X).$$


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