I hope you are doing well. Recently I've been self studying a course on Differential Geometry from this website. I've been dealing with Lie-Groups for the last few days and I've been stuck on one of the Problems in this Sheet. I've done some work but I am lost in thinking about derivatives of $C^\infty$ and other definitions. First of all, here's the problem:
- In the problem, by $r_{\gamma(t)}$ the author means the right multiplication by $\gamma(t)$, i.e. the map $\left(g\mapsto g\gamma(t)\right)$.
- Here, $\Phi^\xi_t$ denotes the unique global flow that corresponds to the left-invariant vector field $X_\xi$ with initial condition $X_\xi(e) = (e, \xi)$
- In general, by $\left(\left.\dfrac{\partial}{\partial x^i}\right|_p\right)$ the author means the $i^{th}$ member of the standard basis for $T_pM$ given by the chart $x: U\subseteq M \rightarrow \mathcal{O}\subseteq \mathbb{R}^n$.
- Also by a one-parameter subgroup he means for $\gamma:\mathbb{R} \rightarrow G$ to be a Lie-Group homomorphism.
This is the main theorem that the author proved in his lecture notes (I add this so you can see the convention they used. For example $X:M\rightarrow TM$ and by (9.4) the author means the second component of RHS is equal to LHS.)
Here's my progress (only for the first part):
Suppose we know that $\gamma$ is a one-parameter subgroup. Define $\Phi(t, g):= g\gamma(t)$. We need to show that $\Phi$ is the corresponding flow of $X_\xi$. In other words, one has to show that $\left((t, g), D\Phi(t, g)\left(\left.\dfrac{\partial}{\partial t}\right|_{(t, g)}\right)\right) = X_\xi(\Phi(t, g))$. To show this let $f$ be a smooth map on $G$. Then $D\Phi(t, g)\left(\left.\dfrac{\partial}{\partial t}\right|_{(t, g)}\right) (f) = \left(\left.\dfrac{\partial}{\partial t}\right|_{(t, g)}\right)(f\circ \Phi) = \dfrac{\partial}{\partial t} \left( f\circ \Phi \circ \left(\underset{(id, x)}{\underbrace{ (t, a) \mapsto(t, x^{-1}(a) }}) \right) \right)((t, x(g)))$
where in here $x$ is a chart about $g$ and now $\dfrac{\partial }{\partial t}$ is the usual partial derivative in Euclidean spaces.
But this last equality gives us $\dfrac{\partial}{\partial t} \Bigg( f\circ \bigg( (t, a) \mapsto x^{-1}(a)\gamma(t)) \bigg) \Bigg)\Big((t, x(g)\Big)$.
To finish off I understand that I have to use the homomorphism property to show that $\pi_2\left(X_\xi(t, g)\right)$ acts the same way on $f$. But here I got stuck (because of the heavy notations I guess. I believe the problem shouldn't be hard at all). I appreciate your help. (And if you have any advice of how I should think about these problems while not getting lost in the definitions and at the same time, not making up sentences that lack mathematical precision).
Sorry for the long post. And thanks in advance for your help.
Update: I guess I have finally figured it out. I have poste my answer below. Please feel free to tell me if there was a simpler solution. Thanks again! 🙂
Best Answer
I talked with one of my friends and he gave me a hint for why I was lost. In short the main thing that clicked in my mind was when he told me maybe it's easier to first deal with the equality in the origin $e$ and then try to prove it in other points! I will now post my solution.
First suppose that $\gamma$ is a one-parameter subgroup. Let $\Phi$ be as it was defined in the post. Now by definition, letting $g:= e$ we get
$$ D\Phi(t, e)\left(\left.\dfrac{\partial}{\partial t}\right|_{(t, e)}\right) = \dfrac{\partial}{\partial t} \Bigg( f\circ \bigg( (t, a) \mapsto e\gamma(t)) \bigg) \Bigg)\Big((t, x(e)\Big) = (f\circ \gamma)'(t) = \dot{\gamma}(t)(f) = Dl_{\gamma(t)}(e)(\dot{\gamma}(0)) = Dl_{\gamma(t)}(e)(\xi) = \text{(by definition)} \ X_\xi(\gamma(t)) $$
Note that we used the fact that $\dot{\gamma}(t) = Dl_{\gamma(t)}(e)(\dot{\gamma}(0))$. This is true since $$ Dl_{\gamma(t)}(e)(\dot{\gamma}(0))(f) = \dot{\gamma(0)}(f\circ l_{\gamma(t)}) = (f\circ l_{\gamma(t)}\circ\gamma)'(0)$$ $$ \text{(and this is where we really use the homomorphism property)}$$ $$ = \lim_{s\rightarrow 0}\dfrac{f\Big(l_{\gamma(t)}(\gamma(s))\Big) - f\Big(l_{\gamma(t)}(\gamma(0))\Big)}{s} = \lim_{s\rightarrow 0}\dfrac{f\Big(\gamma(t)\gamma(s)\Big) - f\Big(\gamma(t)\gamma(0)\Big)}{s} = \lim_{s\rightarrow 0}\dfrac{f\Big(\gamma(t + s)\Big) - f\Big(\gamma(t)e\Big)}{s} = \lim_{s\rightarrow 0}\dfrac{f\Big(\gamma(t + s)\Big) - f\Big(\gamma(t)\Big)}{s} = (f\circ \gamma)'(t) = \dot{\gamma}(t)(f)$$
This proves that $\gamma(t)$ is an integral curve of $X_\xi$ passing $e$ at time $0$. Thus $\Phi(t, e) = \gamma(t) = \Phi^\xi(t, e)$. The rest is really easy now. We continue from our computation before: $$ D\Phi(t, g)\left(\left.\dfrac{\partial }{\partial t} \right|_{(t, g)}\right)(f) = \dfrac{\partial}{\partial t} \Bigg( f\circ \bigg( (t, a) \mapsto x^{-1}(a)\gamma(t)) \bigg) \Bigg)\Big((t, x(g)\Big) = \dfrac{d}{dt}(f\circ l_{g} \circ \gamma)(t) = \dot{\gamma}(f\circ l_{g}) = \Big(Dl_g(e)(\dot{\gamma}(t))\Big)(f) $$ which by what we have proved is the same as $$ Dl_g(e)(Dl_{\gamma(t)}(e)(\underset{\xi}{\underbrace{\dot{\gamma}(0)}}))(f) = \Big(D(l_{g}\circ l_{\gamma(t)})(e)(\xi)\Big)(f) = \Big(Dl_{g\gamma(t)}(e)(\xi)\Big)(f) $$ and by definition this is $X_{\xi}(g\gamma(t))(f)$. We have thus, finally proved that $D\Phi(t, e)\left(\left.\dfrac{\partial}{\partial t}\right|_{(t, g)}\right) = X_\xi(g\gamma(t))$. This concludes the proof.