Exponential map and smooth structure on qoutient space of Lie group

Question

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?

Answer 1

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 $0∈U⊆\mathfrak f$ and $0∈V⊆\mathfrak h$ 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.

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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*}

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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 $e^U H$.

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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.