Question about proof that a closed Lie subgroup is embedded in John Lee's Smooth Manifolds: constructing a slice chart

Question

Below is part of the proof that for a Lie group $G$ and a Lie subgroup $H\subset G$, if $H$ is closed in $G$ then it is embedded (Theorem 7.21 of John Lee's Introduction to Smooth Manifolds).

The proof assumes that the dimension of $H$ is less than the dimension of $G$. The goal is to show that for some $h_1 \in H$, there is a neighborhood $U_1$ of $h_1$ in $G$ such that $H \cap U_1$ is an embedded submanifold of $U_1$.

We first construct a slice chart $(U,\phi)$ for $V$ in $G$ where $V$ is a neighborhood of $e$ in $H$. Then we can assume that $U \cap V$ is the slice $(x^1, \dots, x^m,0,\dots, 0)$. Also we can define $S \subset U$ to be the set of points with coordinates $(0,\dots, 0, x^{m+1}, \dots, x^n)$, which is the slice perpendicular to $U\cap V$.

Then we define the map $\psi:V\times S \to G$ as the group multiplication map and find a neighborhood $V_0 \times S_0$, where $V_0$ and $S_0$ are neighborhoods of $e$ in $V$ and $S$ respectively, and $U_0$ a neighborhood of $e$ in $G$ such that $\psi: W_0 \to U_0$ is a diffeomorphism.

Given these, in the final line of the proof, it states that $U_1 = \psi(V_0 \times S_1)$ is a neighborhood of $h_1$ in $G$ with the property that $U_1 \cap H$ is the slice $\psi(V_0 \times \{h_1\})$ in $U_1$.

My question is why is $\psi(V_0 \times \{h_1\})$ a slice in $U_1$? I can't figure out how to show that this gives a slice chart. I would greatly appreciate some help about this proof.

Answer 1

We need to show that $$\psi\left(V_0 \times\left\{h_1\right\}\right)=U_1 \cap H.$$ Let $h\in \psi\left(V_0 \times\left\{h_1\right\}\right)$ then exists $v\in V_0$ such that $h=v\cdot h_1$. So, $h\in H\cap U_1$.

Now let $h^\prime $ be an element of $U_1 \cap H.$ Thus exists $s \in S_1$ and $v\in V_0$ such that $v\cdot s=h^\prime$. This implies that $s=v^{-1}\cdot h^{\prime}$ is an element of $H$. However $S_1\cap H=\{h_1\}$, hence $h^\prime \in \psi\left(V_0 \times\left\{h_1\right\}\right)$.

Edit: This is indeed a slice chart. $\psi_{\restriction V_0 \times S_1}$ is a diffeomorphism as $V_0 \times S_1\subset W_0$. We can now map $V_0 \times S_1\to R^{2n}$ by $(x,y)\mapsto (\phi(x),\phi(y)).$ By definition of $S$ and $V$ this map will be given in coordinates by $$(x^1\ldots,x^m,0,\ldots,0,x^{m+1},\ldots,x^n).$$ Then, forgetting the $0,\ldots,0$ in the middle, we get the desired slice chart and $$\psi\left(V_0 \times\left\{h_1\right\}\right)\cong\{(x^1,\ldots,x^m,\phi(h_1) )|(x_1,\ldots,x^m) \text{ are the coordinate given for } V_0 \}$$

Answer 2

Let me give this a try.

So we have that the group multiplication $\psi: V_0\times S_0\to U_0$ is a diffeomorphism. Here $V_0\subset V$ has coordinates $(x^1, \dots, x^m)$ and $S_0\subset S$ has coordinates $(x^{m+1}, \dots, x^n)$, induced from their slice structure in the original $U$.

Here is the main point, to me. Since $\psi$ is a diffeomorphism, we use new coordinates on $U_0$ coming from this diffeomorphism. That is, the coordinates of $g\in U_0$ are the above coordinates of $\pi_{V_0}(\psi^{-1}(g))$ and $\pi_{S_0}(\psi^{-1}(g))$ put together.

This is a different coordinate system from the original one on $U$, since here we use the group multiplication. It is with this new coordinate sysmten that you can see the slice structure.

Now we find $h_1\in S_0\cap H$ with a neighborhood $S_1\subset S_0$ such that $S_1\cap H = \{h\}$. Then consider the restricted diffeomorphism $\psi: V_0\times S_1\to U_1$, where $U_1=\psi(V_0\times S_1)\subset U_0$ with the above new coordinates.

Now $U_1\cap H = \psi(V_0\times \{h_1\})$, and in the new coordinates on $U_1$, the points in $\psi(V_0\times \{h_1\})$ have coordinates $$ (x^1,\dots, x^m, c^{m+1},\dots, c^n) $$ where $(c^{m+1},\dots,c^n)$ are the coordinates of $h_1\in S_1\subset S_0$.

Hence $\psi(V_0\times \{h_1\})$ is a slice in $U_1$, just at some other constant values. See p. 101 of Lee's book for this natural generalization.