Differentiable Manifolds/Product manifolds and Lie groups

The product manifold
Lemma 10.3:

Let $$O \subseteq \mathbb R^{2d}$$ be open. Then for each $$(x, y) \in O$$, there exists $$\epsilon_{(x, y)} > 0$$ such that
 * $$B_{\epsilon_{(x, y)}} (x) \times B_{\epsilon_{(x, y)}} (y) \subset O$$

Proof:

Let $$(x, y) \in O$$ be arbitrary. We choose $$\delta > 0$$ such that $$B_\delta((x, y)) \subseteq O$$. Then we define $$\epsilon_{(x, y)} := \frac{\delta}{2}$$. Then, by the triangle inequality we have for every $$(z, w) \in B_{\epsilon_{(x, y)}} (x) \times B_{\epsilon_{(x, y)}} (y)$$:
 * $$\|(z, w) - (x, y)\| = \|(z, 0) + (0, w) - (x, 0) - (0, y)\| \le \|(z, 0) - (x, 0)\| + \|(0, w) - (0, y)\| = \|z - x\| + \|w - y\| < 2 \epsilon_{(x, y)} = \delta$$

Therefore, $$(z, w) \in B_\delta((x, y)) \subseteq O$$, and since $$(z, w) \in B_{\epsilon_{(x, y)}} (x) \times B_{\epsilon_{(x, y)}} (y)$$ was arbitrary:
 * $$B_{\epsilon_{(x, y)}} (x) \times B_{\epsilon_{(x, y)}} (y) \subseteq O$$

Lemma 10.4:

Let $$O, U \subseteq \mathbb R^d$$ be open sets. Then $$O \times U$$ is open in $$\mathbb R^{2d}$$.

Proof:

Due to the openness of $$O$$ and $$U$$ for each point $$(x, y)$$ in $$O \times U$$, we find $$\epsilon_1, \epsilon_2 \in \mathbb R_{>0}$$ such that $$B_{\epsilon_1}(x) \subseteq O$$ and $$B_{\epsilon_2}(x) \subseteq U$$. If we define $$\epsilon := \min \{\epsilon_1, \epsilon_2\}$$, we have for all $$(z, w) \in B_\epsilon((x, y))$$:
 * $$\|x - z\|^2 = \|(x, y) - (z, w)\|^2 - \|y - w\|^2 \le \|(x, y) - (z, w)\|^2$$

and
 * $$\|y - w\|^2 = \|(x, y) - (z, w)\|^2 - \|x - z\|^2 \le \|(x, y) - (z, w)\|^2$$

and since the function $$\sqrt{ \cdot }$$ is monotonely increasing on $$[0, \infty)$$, it follows
 * $$\|x - z\| < \epsilon$$

and
 * $$\|y - w\| < \epsilon$$

and thus:
 * $$(z, w) \in O \times U$$

Lemma 10.5: Let $$X, Y, Z, W$$ be sets and let $$f: X \to Z$$ and $$g: Y \to W$$ be functions. If $$A \subseteq Z$$ and $$B \subseteq W$$, then
 * $$(f \times g)^{-1}(A \times B) = f^{-1}(A) \times g^{-1}(B)$$

Proof: See exercise 2.

Theorem 10.6: The product manifold of a manifold $$M$$ of class $$\mathcal C^n$$, where $$n \in \mathbb N_0 \cup \{\infty\}$$ with atlas $$\{(O_\upsilon, \phi_\upsilon) | \upsilon \in \Upsilon\}$$ really is a manifold; i. e. $$\{(O_\upsilon \times O_\omicron, \phi_\upsilon \times \phi_\omicron) | \upsilon, \omicron \in \Upsilon\}$$ really is an atlas of class $$\mathcal C^n$$.

Proof:

1. We show that for all $$\upsilon, \omicron \in \Upsilon$$ the set $$(\phi_\upsilon \times \phi_\omicron)(O_\upsilon \times O_\omicron)$$ is open.

We have:
 * $$\begin{align}

(\phi_\upsilon \times \phi_\omicron)(O_\upsilon \times O_\omicron) & = \{(\phi_\upsilon(p), \phi_\omicron(q)) \in \mathbb R^{2d} \big| p \in O_\upsilon, q \in O_\omicron \} \\ & = \phi_\upsilon(O_\upsilon) \times \phi_\omicron(O_\omicron) \\ \end{align}$$ This set is open in $$\mathbb R^{2d}$$ due to lemma 10.4, since it is the cartesian product of two open sets.

2. We prove that for all $$\upsilon, \omicron \in \Upsilon$$, the function $$\phi_\upsilon \times \phi_\omicron$$ is a homeomorphism.

2.1. For bijectivity, see exercise 1.

2.2. We prove continuity.

Let $$O \subseteq (\phi_\upsilon \times \phi_\omicron)(O_\upsilon \times O_\omicron)$$ be open. Due to the definition of the subspace topology,

Lemma 10.4 implies that we have
 * $$O = \bigcup_{(x, y) \in O} B_{\epsilon_{(x, y)}} (x) \times B_{\epsilon_{(x, y)}} (y)$$

(this equation can be proven by showing '$$\subseteq$$' and '$$\supseteq$$') and thus follows with lemma 10.5, that:
 * $$\begin{align}

(\phi_\upsilon \times \phi_\omicron)^{-1}(O) & = (\phi_\upsilon \times \phi_\omicron)^{-1} \left( \bigcup_{(x, y) \in O} B_{\epsilon_{(x, y)}} (x) \times B_{\epsilon_{(x, y)}} (y) \right) \\ & = \bigcup_{(x, y) \in O} (\phi_\upsilon \times \phi_\omicron)^{-1} (B_{\epsilon_{(x, y)}} (x) \times B_{\epsilon_{(x, y)}} (y)) \\ & = \bigcup_{(x, y) \in O} \phi_\upsilon^{-1} (B_{\epsilon_{(x, y)}} (x)) \times (\phi_\omicron)^{-1} (B_{\epsilon_{(x, y)}} (y)) \end{align}$$ , which is open as the union of open sets (since we had equipped $$M \times M$$ with the product topology).

2.3. We prove continuity of the inverse.

Let $$O \subseteq M \times M$$ be open.

3. We prove that

Lie groups
Theorem 10.11: Let $$G$$ be a $$d$$-dimensional Lie group of class $$\mathcal C^n$$. Then for each $$g \in G$$ the respective left multiplication function and the respective right multiplication are diffeomorphisms of class $$\mathcal C^n$$ from $$G$$ to itself.

Proof:

We only prove the claim for the left multiplication. The proof for the right multiplication goes the same way.

In this proof, the group operation of $$G$$ is denoted by $$*$$.

1. We show that $$L_g$$ is differentiable of class $$\mathcal C^n$$.

Let $$g \in G$$ be arbitrary. Since $$G$$ is a Lie group, the function
 * $$\psi_*: G \times G \to G, \psi_*(g, h) = g * h$$

is differentiable of class $$\mathcal C^n$$, where $$G \times G$$ is equipped with the product manifold structure.

Let now $$(O, \phi)$$ and $$(U, \theta)$$ be two arbitrary elements in the atlas of $$G$$. We choose $$(V, \chi)$$ in the atlas of $$G$$ such that $$g \in V$$. As $$\psi_*$$ is differentiable of class $$\mathcal C^n$$, the function
 * $$\phi|_{*(\psi_*^{-1}(O) \cap V \times U)} \circ \psi_*|_{\psi_*^{-1}(O) \cap V \times U} \circ (\chi \times \theta)|_{\psi_*^{-1}(O) \cap V \times U}^{-1}$$

is contained in $$\mathcal C^n(\mathbb R^{2d}, \mathbb R^d)$$. Therefore, also the function
 * $$f: \mathbb R^d \to \mathbb R^d, f(x) = \phi|_{\psi_*(\psi_*^{-1}(O) \cap V \times U)} \circ \psi_*|_{\psi_*^{-1}(O) \cap V \times U} \circ (\chi \times \theta)|_{\psi_*^{-1}(O) \cap V \times U}^{-1} (\chi^{-1}(g), x)$$

is contained in $$\mathcal C^n(\mathbb R^{2d}, \mathbb R^d)$$; the partial derivatives exist and are equal to the partial derivatives of the last $$d$$ variables of the function
 * $$\phi|_{\psi_*(\psi_*^{-1}(O) \cap V \times U)} \circ \psi_*|_{\psi_*^{-1}(O) \cap V \times U} \circ (\chi \times \theta)|_{\psi_*^{-1}(O) \cap V \times U}^{-1}$$

. But we have for all $$x \in \theta(L_g^{-1}(O) \cap U)$$:
 * $$f(x) = (\phi|_{L_g(|_{L_g^{-1}(O) \cap U})} \circ L_g|_{L_g^{-1}(O) \cap U} \circ \theta|_{L_g^{-1}(O) \cap U}^{-1})(x)$$

and therefore the function
 * $$\phi|_{L_g(|_{L_g^{-1}(O) \cap U})} \circ L_g|_{L_g^{-1}(O) \cap U} \circ \theta|_{L_g^{-1}(O) \cap U}^{-1}$$

is contained in $$\mathcal C^n(\mathbb R^d, \mathbb R^d$$, which means the definition of differentiability of class $$\mathcal C^n$$ is fulfilled.

2. We show that $$L_g$$ is bijective.

We do so by noticing that an inverse function of $$L_g$$ is given by $$L_{g^{-1}}$$: For arbitrary $$h \in G$$, we have:
 * $$L_{g^{-1}}(L_g(h)) = g * (g^{-1} * h) = (g * g^{-1}) * h = e * h = h$$

and
 * $$L_g(L_{g^{-1}}(h)) = g^{-1} * (g * h) = (g^{-1} * g) * h = e * h = h$$

3. We note that the inverse function is differentiable of class $$\mathcal C^n$$:

We use 1. with $$g^{-1}$$; $$g^{-1}$$ also is an element of $$G$$, and 1. proved that the left multiplication function is differentiable for every element of $$G$$, including $$g^{-1}$$.

Left invariant vector fields
Let us repeat the definition of a Lie subalgebra:

Definition 6.2:

Let $$L$$ with $$[ \cdot, \cdot ]$$ be a Lie algebra. A subset of $$L$$ which is a Lie algebra with the restriction of $$[ \cdot, \cdot ]$$ on that subset is called a Lie subalgebra.

Proof:

Let $$\mathbf V, \mathbf W \in \mathfrak g$$. It suffices to show that $$[\mathbf V, \mathbf W] \in \mathfrak g$$, because then $$\mathfrak g$$ is a Lie algebra with the restriction of the vector field Lie bracket.

Indeed, we have for all $$g, h \in G$$ and $$\varphi \in \mathcal C^n(G)$$:
 * $$\begin{align}

(d L_g)_h ([\mathbf V, \mathbf W](h))(\varphi) & = [\mathbf V, \mathbf W](h)(L_g^* \varphi) \\ & = \mathbf V (h) (\mathbf W (L_g^* \varphi)) - \mathbf W(h) (\mathbf V (L_g^* \varphi)) \\ & = \mathbf V (h) ((d L_g)_{\cdot} (\mathbf W \varphi) ) - \mathbf W (h) ((d L_g)_{\cdot} (\mathbf V \varphi) ) \\ & = \mathbf V(h) (\mathbf W(g \cdot)(\varphi)) - \mathbf W(h) (\mathbf V(g \cdot)(\varphi)) \\ & = \mathbf V(h) (\mathbf W \varphi \circ L_g) - \mathbf W(h) (\mathbf V \varphi \circ L_g) \\ & = \mathbf V(h) (L_g^*(\mathbf W \varphi)) - \mathbf W(h) (L_g^*(\mathbf V \varphi \circ L_g)) \\ & = (d L_g)_h (\mathbf V(h)) (\mathbf W \varphi) - (d L_g)_h (\mathbf W(h))(\mathbf V \varphi) \\ & = \mathbf V(gh) (\mathbf W \varphi) - \mathbf W(gh) (\mathbf V \varphi) \\ & = [\mathbf V, \mathbf W](gh)(\varphi) \end{align}$$ , where by $$(d L_g)_{\cdot} (\mathbf W \varphi)$$ the function
 * $$p \mapsto (d L_g)_p (\mathbf W \varphi)$$

is meant and by $$\mathbf W(g \cdot)(\varphi)$$ the function
 * $$p \mapsto \mathbf W(g p)(\varphi)$$

is meant (as both are equal to $$\mathbf W (L_g^* \varphi)$$, both are differentiable of class $$\mathcal C^n$$).

Proof:

We choose the function
 * $$\Theta: \mathfrak g \to T_e G, \Theta(\mathbf V) := \mathbf V(e)$$

We will now show that this function is the desired isomorphism.

1. We prove linearity: Let $$\mathbf V, \mathbf W \in \mathfrak g$$ and $$c \in \mathbb R$$. We have:
 * $$\begin{align}

\Theta (\mathbf V + c \mathbf W) & = (\mathbf V + c \mathbf W)(e) \\ & = \mathbf V(e) + c \mathbf W(e) \\ & = \Theta (\mathbf V) + c \Theta (\mathbf W) \end{align}$$

2. We prove bijectivity.

2.1. We prove injectivity.

Let $$\Theta(\mathbf V) = \Theta(\mathbf W)$$ (i. e. $$\mathbf V(e) = \mathbf W(e)$$) for $$\mathbf V, \mathbf W \in \mathfrak g$$. Since $$\mathbf V, \mathbf W$$ are left invariant, it follows for all $$g \in G$$ and $$\varphi \in \mathcal C^n$$, that


 * $$\begin{align}

\mathbf V(g) (\varphi) & = \mathbf V(ge)(\varphi) \\ & = (d L_g)_e (\mathbf V) (\varphi) \\ & = \mathbf V(e) (L_g^*(\varphi)) \\ & = \mathbf W(e) (L_g^*(\varphi)) \\ & = (d L_g)_e (\mathbf W) (\varphi) \\ & = \mathbf W(ge)(\varphi) \\ & = \mathbf W(g) (\varphi) \end{align}$$

2.2. We prove surjectivity.

Let $$\mathbf V_e \in T_e G$$ be arbitrary. We define
 * $$\mathbf U_{\mathbf V_e}(g) := (d L_g)_e \mathbf V_e$$

Due to theorem 2.19, this is a vector field. It is also left invariant because for all $$g, h \in G$$ and $$\varphi \in \mathcal C^n(M)$$, we have:
 * $$\begin{align}

(d L_g)_h (\mathbf U_{\mathbf V_e}(h))(\varphi) & = \mathbf U_{\mathbf V_e}(h)(L_g^*(\varphi)) \\ & = (d L_h)_e \mathbf V_e (L_g^*(\varphi)) \\ & = \mathbf V_e (L_h^*(L_g^*(\varphi))) \\ & = \mathbf V_e (\varphi \circ L_g \circ L_h) \\ & = \mathbf V_e (\varphi \circ L_{gh}) \\ & = (d L_{gh})_e \mathbf V_e (\varphi) \\ & = \mathbf U_{\mathbf V_e}(gh)(\varphi) \end{align}$$ Further, we have for all $$\varphi \in \mathcal C^n(M)$$:
 * $$\begin{align}

\Theta(\mathbf U_{\mathbf V_e})(\varphi) = \mathbf U_{\mathbf V_e} (e)(\varphi) = (d L_e)_e \mathbf V_e(\varphi) = \mathbf V_e(\varphi) \end{align}$$

3. We note that the inverse of $$\Theta$$ is linear since the inverse of a linear bijective function is always linear.

The next theorem shows that in a Lie group, all left invariant vector fields are complete.

Lemma 10.15:

Let $$G$$ be a Lie group, let $$\mathbf V \in \mathfrak g$$ and let $$\Phi_{\mathbf V}$$ be the flow of $$\mathbf V$$. Then for all $$g \in G$$ and all $$(x, h)$$ in the domain of $$\Phi{\mathbf V}$$, we have:
 * $$\Phi_{\mathbf V}(x, g * h) = h * \Phi_{\mathbf V}(x, g)$$

Proof:

Let $$h \in G$$ be arbitrary, and let I_h be the unique largest interval such that $$0 \in I_h$$ and there exists a unique integral curve $$\gamma_h: I_h \to M$$ such that
 * $$\gamma_h'$$

Proof:

Let $$g \in G$$ be arbitrary and let $$\gamma_g$$ be an integral curve at $$g$$.

The adjoint function
For the next definition, we recall that the automorphism group of a group was given by the set of group isomorphisms from the group to itself with composition as the group operation. Indeed, this is a group (see exercise 3).

We further recall that for a group $$G$$, the automorphism group is denoted by $$\text{Aut}(G)$$.

Theorem 10.19:

Let $$G$$ be a manifold of class $$\mathcal C^n$$, where $$n \in \mathbb N_0 \cup \{\infty\}$$. For each $$g \in G$$, $$\text{Ad}(g)$$ is of class $$\mathcal C^n$$.

Proof:

We have:
 * $$\text{Ad}(g) = L_g \circ R_{g^{-1}}$$

Therefore, the claim follows from theorems 2.29 and 10.11.

Exercises

 * 1) Let $$X, Y, Z, W$$ be sets $$f: X \to Z$$ and $$g: Y \to W$$ be two bijective functions. Prove that $$f \times g$$ is bijective.
 * 2) Prove lemma 10.5.
 * 3) Let $$G$$ be a group and $$\text{Aut}(G)$$ be the set of group isomorphisms from $$G$$ to $$G$$. Prove that $$\text{Aut}(G)$$ together with the composition as operation is a group.