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A Lie group $G$ is a smooth manifold that is also a group in the algebraic sense, such that the multiplication map $m: G \times G \to G$ and the inverse $i: G \to G$ are all smooth maps.

Let $g \in G$, we define the left translation $L_g$ and right translation $R_g$ as maps $G \to G$ by $$ L_g(h) = gh, \quad R_g(h) = hg $$

• $GL(n, \R)$, the invertible $n \times n$ matrices with real entries. It's dimension is $n^2$. We can check multiplication is smooth by writing down the formula $C = AB$, then $c_{ij} = \sum_k a_{ik}b_{kj}$. And the inverse is smooth, since we can write $G^{-1} = (\det G)^{-1} G_{adj}$. An open subgroup of $GL(n, \R)$ is $GL_+(n, \R)$.
• $GL(n, \C)$, as a complex manifold.
• If $V$ is a real or complex vector space, we can talk about $GL(V)$, the group of invertible linear maps from $V$ to $V$.
• Translation group $\R^n$ acting.
• The circle group $S^1 \subset \C^*$.
• The n-dimensional torus $\T^n = (\S^1)^n$.
• Important subgroups of $GL(n, \R)$ and $GL(n, \C)$ (later when we know how to produce subgroups)
  • special orthogonal group $SO(n, \R)$,
  • Lorentz group $SO(1,3)$ ,
  • Symplectic group $Sp(2n, \R) \subset GL(2n, \R)$.
  • Unitary group $SU(n) \subset GL(n, \C)$

Let $G, H$ be Lie group, we say $\varphi: G \to H$ is a Lie group homomorphism, if it is a smooth map and also a group homomorphism.

• $(\R, +) \to (\R_+, *)$, $t \mapsto e^t$.
• $S^1 \into \C$
• Let $V, W$ be vector spaces, viewed as Lie group by translation, then any linear map $V \to W$ is a group homomorphism.
• $\det: GL(n, \R) \to \R^*$, since $\det(AB) = \det(A)\det(B)$.
• Let $g \in G$, the conjugation action $Ad_g: G \to G$, $h \mapsto g h g^{-1}$ is a group homomoprhism.

Thm : Group homomorphisms are constant rank maps.

Proof: Let $\varphi: G \to H$ be a Lie group homomorphism. We use left translation to move all the maps on the tangent space $T_g G \to T_{\varphi(g)} H$ back to identity $T_e G \to T_e H$.

A Lie subgroup of $G$ is a subgroup of $G$ endowed with a topology and smooth structure making it into a Lie group and an immersed submanifold of $G$.

Prop 7.11 (Lee) : Let $G$ be a Lie group and $H \subset G$ a subgroup, which is also an embedded submanifold, then $H$ is a Lie subgroup.

Proof: This uses Corollary 5.30, which says, if $F: M \to N$ is a smooth map, $S \subset N$ an embedded submanifold, and image of $F$ is containedin $S$, then $F$ is a smooth map from $M$ to $S$. Of course, one can use slice chart for embedded manifold to prove this corollary directly, but take a look at theorem 5.29 is also useful. Back to this proposition, we just need to check that the multiplication $m: H \times H \to G$ and $i: H \to G$ has image contained in $H$, which is guaranteed by the subgroup condition.

First, we consider just group action. Let $G$ be a group, $M$ be a set. A left group action is a map $$ \rho: G \times M \to M, \quad (g, p) \mapsto g \cdot p$$ such that for all $p \in M$, $g_1, g_2 \in G$, $$ (g_1 g_2) \cdot p) = g_1 \cdot (g_2 \cdot p) $$ and $$ e \cdot p = p. $$

A continuous action, or a smooth action is defined the same way, just imposing the corresponding conditions on $G, M$ and the map $G \times M \to M$.

Right action. And how to translate a right action into a left action.

Lie group usually arises as 'symmetric group of some structure'. For example, if $V$ is a linear space $GL(V)$ is maps from $V$ to $V$ that preserves the linear structure.

Some terminologies: suppose $\theta: G \times M \to M$ is a left action of a group $G$ on a set $M$.

• $\theta_g: M \to M$ is the map $p \mapsto \theta(g,p)$.
• for any $p \in M$, the orbit $G \cdot p$ is the set $\{ g \cdot p \mid g \in G\}$.
• the isotropy group or the stabilizer of $p$ is the subgroup $\{g \in G \mid g \cdot p = p \}$, denoted as $G_p$.
• The action is transitive, if for any two points $p,q \in M$, there exists an element $g \in G$, such that $g \cdot p = q$. In other words the map $G \times M \to M \times M$, $(g,p) \mapsto (p,g \cdot p)$ is surjective.
• The action is free, if the only element of $G$ that fixes some element in $M$ is the identity element, namely if $g \cdot p = p$ for some $p \in M$ then $g = e$. Equivalently, all the isotropy groups are trivial.

Examples:

• Lie group acts by conjugation on itself.
• Lie group acts by left translation on itself.
• $GL(n, \R)$ acts on $\R^n$, it is transitive. What is the isotropy group of $(1,0,\cdots, 0)$?

Suppose $M$ and $N$ are two manifolds where the Lie group $G$ acts on the left. We say a smooth map $F: M \to N$ is $G$-equivariant, if $$ F(g \cdot p) = g \cdot F(p). $$

Thm (constant rank theorem) (7.25) : If $M, N$ are smooth manifold with left $G$-action. Suppose $G$ acts on $M$ transitively. Then any equivariant map $F: M \to N$ is constant rank. In particular, if $F$ is a surjection, then it is a submersion, $F$ is an injectiion then it is a an immersion; finally if $F$ is a bijection, then it is a diffeomorphism. '

This follows immediately from the global rank theorem, which says, if $F: M \to N$ is constant rank, then $F$ is a surjection implies $F$ is a smooth submersion; $F$ is an injection implies $F$ is a smooth immersion; $F$ is a bijection implies $F$ is a diffeomorphism. Morally, it allows one to upgrade a set-wise statement to a smooth manifold statement with control on the differential of $F$.