How to use the chain rule to prove this couple of differentiations

differential-geometrygroup-theorylie-algebraslie-groupsmanifolds

Consider the following from van den Ban’s Lie Groups (webspace.science.uu.nl/~ban00101/lie2019/lie2010.pdf), pp. 18-20:

enter image description here
enter image description here
enter image description here
enter image description here

  1. I am trying to prove the last equation

$$\DeclareMathOperator{\ad}{ad} \DeclareMathOperator{\Ad}{Ad} \ad
(X)=\left.\frac{d}{dt}\right|_{t=0}\Ad
(\exp(tX)) \tag{*}$$

How do I use the chain rule to prove it?

My solution is the following but I don’t think I am using the chain rule (or I maybe I am using without being aware of it):

I write $X$ as the derivative at $t=0$ of a curve $\gamma$ living in $G$, that would be the exponential map $\exp(tX)$:

$$\ad
(X)=T_e(\Ad
)(X)=T_e(\Ad
) \left.\frac{d}{dt}\right|_{t=0}\gamma(t).$$

Now I use the definition of the action of a tangent map (differential) on a vector of the tangent space, meaning that it maps velocities into velocities (tangent vectors into tangent vectors). If $\left.\frac{d}{dt}\right|_{t=0}\alpha(t)$ is a tangent vector this translates into $T_e(f)\left.\frac{d}{dt}\right|_{t=0}\alpha(t)= \left.\frac{d}{dt}\right|_{t=0}(f(\alpha(t))$, with $f$ some smooth function. In my problem:

$$ \ad(X)=T_e(\Ad)(X)=T_e(\Ad) \left.\frac{d}{dt}\right|_{t=0}\gamma(t)=\left.\frac{d}{dt}\right|_{t=0} \Ad
(\exp(tX)).$$

Is this a correct procedure? Is using the chain rule an alternative approach? How do I use it then?

  1. If this is correct, where is that $T_0(\exp)$ coming out from in the following differentiation (p.20 in the notes), for $Z \in T_eG$:

$$\left.\frac{d}{ds}\right|_{s=0}\Ad(\exp(sZ))Z = \ad(Z)T_0(\exp)Z ?$$

I would just substitute equation $(*)$ like this:

$$ \left. \frac{d}{ds}\right|_{s=0} \Ad(\exp(sZ))Z= \ad(Z)Z \tag{**}$$

and the result is the same because actually $T_0(\exp)$ can be shown to be the identity map, but still what is going on here?

I am not very familiar with differential geometry manipulations so please be detailed in your use of the chain rule.

Edit Let me do (2) in detail as suggested using the chain rule:
calling $\alpha: s \mapsto \alpha(s)=sZ$

$$\frac{d}{ds}|_{s=0} Ad(\exp(sZ)))Z = T_0 Ad(exp(\alpha(s)))Z = T_0 (Ad \circ \exp \circ \alpha(s) Z) $$
$$= T_{exp(\alpha(0))}Ad\circ T_{\alpha(0)}\exp\circ T_{0}\alpha(s) Z $$
$$= T_{e}Ad \circ T_{0} \exp \circ T_{0}\alpha(s) Z =ad \circ T_{0} \exp \circ Z Z$$ $$=ad(Z)Z$$ Makes sense? However since $Z \in T_e G $ and $\alpha(s) $ is the curve whose derivative at $0$ is $Z$, shouldn't I have $\alpha(0)=e$? , instead of that I am using $\alpha(0)=0Z=0 $ And something else how's that Z at the end of the expression is not affected by anything the whole time?

Best Answer

You are right in 1. and in 2. it seems a more explicit composition of functions is being considered.

For $X\in T_eG$, the adjoint operator $\text{ad}_X$ is defined as the derivative of the function $\mathbb{R}\to \text{GL}(T_eG)$, $t\mapsto \text{Ad}(\exp(tX))$ evaluated at $t=0$. Writing this function out explicitly as a composition we have

enter image description here

(Here $f:(A,a)\to(B,b)$ means $f$ is a function from $A$ to $B$ and $f(a)=b$.)

You are using the chain rule with this composition in the last equality in the chain

$$\text{ad}(X)=T_e(\text{Ad})(X)=T_e(\text{Ad}) \left.\frac{d}{dt}\right|_{t=0}\gamma(t)=\left.\frac{d}{dt}\right|_{t=0} \text{Ad}(\exp(tX))$$

On the other hand (for 2.) one can also consider the composition as

enter image description here

in which case there would be a factor of $T_0(\exp)$, which like you said is identity and can be ignored. Note that the derivative of the first map at $0$ is $X$.

Added: As per request, here is a comparison of the tangent functor notation and the Leibniz notation:

$$\left.\dfrac{d}{dt}\right|_{t=0} \gamma(t) = \left.\dfrac{d}{dt}\right|_{t=0} \exp(\bullet X) = T_0 \exp(\bullet X) \left( \left.\dfrac{d}{dt}\right|_{t=0}\right) = T_0 \gamma \left( \left.\dfrac{d}{dt}\right|_{t=0}\right).$$

Here in the last expression $\left.\dfrac{d}{dt}\right|_{t=0}\in T_0\mathbb{R}$ stands for the basis vector normalized by way of the standard Riemannian (Euclidean) metric of $\mathbb{R}$. The functorial chain rule does not see this choice of an element in $T_0\mathbb{R}$, hence would give the equality of two linear operators:

$$T_0 [\text{Ad}(\exp(\bullet X))] = T_e \text{Ad} \circ T_0\exp(\bullet X): \mathbb{R}\cong T_0\mathbb{R}\to T_I (\text{GL}(T_e G))\cong \text{End}(T_e G).$$

Related Question