I am trying to understand a proof of the following fact. Let $G$ be a Lie group and $H$ a closed subgroup of $G.$ Then $G/H$ can be endowed with a smooth structure such that the action $(g,g_1H)\mapsto gg_1H$ is smooth. Denote $\pi:G\to G/H$ to be the quotient map.
First denote $\mathfrak h$ to be the Lie algebra corresponding to $\mathfrak g.$ Let $\mathfrak f$ be the complemented subspace of $\mathfrak h.$ Therefore, we have an orthogonal decomposition $\mathfrak g=\mathfrak f \oplus \mathfrak h$. Define $\phi:\mathfrak f \oplus \mathfrak h\to G$ as $\phi(X,Y)=exp(X)exp(Y).$ One can use inverse mapping theorem etc to conclude that there exists two open neighbourhoods $0\in U\subseteq \mathfrak f$ and $0\in V\subseteq \mathfrak h$ such that $\phi:U\times V\to \phi(U\times V)$ is a smooth diffeomorphism and the map $\alpha _{H}:exp(X)\mapsto \pi(exp(X))$ is a homeomorphism from $U$ to a neighborhood of $eH\in G/H$. Denote it by $\widetilde{U}_{H}.$ Define $\psi_H:(\alpha_H)^{-1}.$ For any $gH\in G/H$ one defines $\widetilde{U}_{gH}$ to be $g\widetilde{U}.$ Also define the map $\psi_{gH}:\widetilde{U}_{gH}\to U$ as $g\tilde{g}H\mapsto \psi_H(gH).$ Now in the book from where I am reading the proof (S.Kumaresan) one defines the atlas on $G/H$ as $\{(\widetilde{U}_{gH},\psi_{gH})\}.$ Though the book has a proof but it seems little mysterious. Can any one show me how to prove the given atlas is smooth?
Exponential map and smooth structure on qoutient space of Lie group
differential-geometrylie-groupsmanifoldssmooth-manifolds
Related Solutions
Consider the composition $G\rightarrow G\times G\rightarrow G$ where the first map sends $g$ to $(g,g^{-1})$, and the second map is $m$: it sends $(g,h)$ to $gh$.
This composition is constant, so its differential must be $0$. On the other hand, if you compute the differential via the chain rule, you'll get the result you want.
Here is a better (more elegant) proof for both questions using Cartan-Von Neumann's theorem on closed subgroups.
Let $H$ be the graph of $\gamma$, then $H$ is closed (and thus a Lie subgroup of $G\times \mathbb{R} \cong \mathbb{R}\times G$, so $H\hookrightarrow \mathbb{R}\times G$ is smooth) The projection from $H$ to $\mathbb{R}$ is a bijective lie group morphism, and therefore its inverse is smooth. Now just have a look at the following diagram
As for the second one, if $\varphi : G\to H$ is continuous, let $\Gamma_\varphi : G\to G\times H$ be the "graph map" which sends $g$ to $\big(g, \varphi(g) \big)$. Note that $\Gamma_\varphi $ is a group homeomorphism from $G$ onto its image $\mathrm{Graph}(\varphi)$ whose inverse is the restriction of the projection $G\times H \to G$. Now, $\mathrm{Graph}(\varphi)$ is clearly closed (therefore a Lie subgroup), and the inverse of $\Gamma_\varphi$ is smooth with constant rank, and thus $\Gamma_\varphi $ is a diffeomorphism. If $q: G\times H \to H$ is the second projection (smooth), you just have to note that $\varphi=q\circ \Gamma_\varphi $ and we are done :)
Best Answer
I will make a little adaptation to your notation. I will initially consider $\phi_1:\mathfrak f\oplus \mathfrak h = G $ as the map $\phi_1 (X, Y) = e^X e^Y$. As you told us there exist two open neighborhoods $0āUā\mathfrak f$ and $0āVā\mathfrak h$ such that $\phi_1: U_1\times V_1\to W_1 = e^{U_1} e^{V_1}$ is a diffeomorphism and $e^{U_1} \cap H = \{1\}$ and $e^{V_1} \cap H = V_1$ . It is easy to see that we can find $U\subset U_1$, $V\subset V_1$, $W\subset W_1$, such that, $W^2\subset W_1$, $W^{-1}W\subset W_1$, and $\phi:U\times V\to W$, $\phi:=\left.\phi_1\right|_{U\times V}$ is a diffeomorfism. This will be useful during the proof.
First of all, consider the function \begin{align*}\Psi: U\times H&\to G\\ (Y,h)&\mapsto e^Y h, \end{align*} where $U$ is the open neighborhood from $0$ in $\mathfrak f$ defined in your question. We wil prove that $\Psi$ is differentiable. Let $A\in \mathfrak f$ and $B^*\in T_hH$, such that $B^*= \mathrm{d}(L_h)_1 B$, $B\in \mathfrak h$
\begin{align*} \text{d}\Psi_{Y,h}(A,B^*) &= \left.\frac{\mathrm d}{\mathrm dt}\Psi(Y+tA,e^{tB}h)\right|_{t=0}\\ &= \frac{\mathrm d}{\mathrm dt} \left.R_h(e^{Y+tA}e^{tB})\right|_{t=0}\\ &= \mathrm d \left(R_h\right)_{e^Y} \psi_{(Y,0)}(A,B)\\ &= \mathrm d \left(R_h\circ\psi\right)_{(Y,0)}(A,B)\\ &= \mathrm d \left(R_h\circ\psi\right)_{(Y,0)}\left(A,\mathrm{d}\left(L_{h^{-1}}\right)_h B^* \right).\quad (*) \end{align*}
Lemma 1. $\Psi: U\times H\to G$ is a diffeomorphism under its image $e^U H$.
Proof: Let us see that $\Psi$ is an injection. If $e^{Y_1}h_1 = e^{Y_2}h_2$, then $e^{-Y_2} e^{Y_1} = h_2h_1^{-1}.$ Note that $e^{-Y_2} e^{Y_1}\in W^2\subset W_1$, on the other hand $e^{-Y_2} e^{Y_1}\in H$. Therefore $e^{-Y_2} e^{Y_1}\in W_1\cap H = e^{V_1}$, therefore $e^{Y_1} = e^{Y_2}e^{X}$, for some $X\in V_1$. This implies that $\phi_1(Y_1,0) = \phi_1(Y_2,X)$, since $\phi_1$ is a diffeomorphism $Y_1 = Y_2$ and $X=0$. Thus $h_1=h_2$ and $\Psi$ is an injective map. And by $(*)$ $\mathrm{d}\Psi_{(Y,h)}$ is an isomorphism. Therefore $\Psi$ is a diffemorphism between $U\times H$ and $e^U H$.
Let us prove that the atlas $\left\{\left(\widetilde U_{gH},\psi_{gH}\right)\right\}_{gH\in G/H}$ is smooth. So, consider $g_1H$ and $g_2H$ such that $\widetilde U_{g_1H}\cap \widetilde U_{g_2H}\neq \emptyset$.
Adapting a bit you notation, we can define the maps $\alpha_g: U \to g_1\tilde U$, $\alpha_g = \pi\left(g e^Y\right) = g \pi\left(e^Y\right)$. Note that $\alpha_g = (\psi_{gH})^{-1}$. And by our construction $\alpha_g = g\pi \circ \left.\Psi\right|_{V\times\{1\}}$.
So it is sufficient to prove that $\alpha_{g_2}^{-1}\circ \alpha_{g_1}$ is smooth. However,
\begin{align*} \alpha_{g_2}^{-1}\circ \alpha_{g_1}(Y) &= p_1\left(\Psi^{-1}\left(g_2^{-1}g_1\Psi(Y,1)\right)\right)\\ &= p_1\circ \Psi^{-1} \left(g_2^{-1}g_1e^Y\right)\\ &= p_1\circ \Psi^{-1} \circ E_{g_2^{-1}g_1}\circ \exp(Y) . \end{align*} where $p_1: U\times H\to U$, $p_1 (Y,h)= Y$. Since we were able to write $\alpha_{g_2}^{-1}\circ \alpha_{g_1}$ as composition a of smooth maps, $\alpha_{g_2}^{-1}\circ \alpha_{g_1}$ is a smooth map, which implies that $\psi_{g_2H}\circ (\psi_{g_1H})^{-1}$ is smooth. Therefore the atlas the atlas $\left\{\left(\widetilde U_{gH},\psi_{gH}\right)\right\}_{gH\in G/H}$ is smooth.