📂Representation Theory

Connected Lie Group

Definition¹

Identity component

For the matrix Lie group $G$, the identity component of $G$ is the set satisfying the following.

$$ G_{0} := \left\{ A \in G : \text{$A$ has a continuous path $A(t)$, $a \le t \le b$, } \right. \\ \qquad \qquad \qquad \qquad \qquad \left. \text{lying in $G$ with $A(a) = I$ and $A(b) = A$} \right\} $$

Explanation

The identity component is the set of matrices for which there exists a continuous path from the identity element to the matrix itself.

By the theorem below, the identity component $G_{0}$ is a subgroup of the matrix Lie group $G$, and since it is a closed set it is in fact a matrix Lie group.

Types

The following matrix Lie groups are connected.

• General linear group $\operatorname{GL}(n, \mathbb{C})$
• Special linear group $\operatorname{SL}(n, \mathbb{C})$
• Unitary group $\operatorname{U}(n)$
• Special unitary group $\operatorname{SU}(n)$
• Special orthogonal group $\operatorname{SO}(n)$

Theorem

If $G$ is a matrix Lie group, then the identity component $G_{0}$ of $G$ is a normal subgroup of $G$.

Proof

Let $G_{0}$ denote the identity component of the matrix Lie group $G$.

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Step 1: $G_{0}$ is a subgroup of $G$

Let $A, B \in G_{0}$. Then in $I$ there exist continuous paths $A(t)$ and $B(t)$ connecting $A$ and $B$, respectively.

Subgroup test

For a nonempty subset $H$ of a group $G$, if the following two conditions are satisfied, then $H$ is a subgroup of $G$.

1. $a$, $b \in H \implies ab \in H$
2. $a \in H \implies a^{-1} \in H$

By the subgroup test, it suffices to show the existence of a continuous path in $I$ joining $I$ to $AB$ and a continuous path in $I$ joining $I$ to $A^{-1}$. $AB(t) = A(t)B(t)$ is a product of continuous functions, hence still continuous, and it connects $I$ to $AB$ within $G$. Therefore $AB \in G_{0}$. Also, $A^{-1}(t)$ is a continuous path in $G$ connecting $I$ to $A^{-1}$. Hence $A^{-1} \in G_{0}$, and $G_{0}$ is a subgroup of $G$.

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Step 2: $G_{0}$ is a normal subgroup of $G$

Let $A \in G_{0}$ and $B \in G$. Then there exists a continuous path $A(t)$ in $G$ connecting $I$ to $A$. Then $BA(t)B^{-1}$ is a continuous path in $G$ connecting $I$ to $BAB^{-1}$. Therefore $BAB^{-1} \in G_{0}$, and $G_{0}$ is a normal subgroup of $G$.

Normal subgroup test

$$ \forall g \in G, \forall h \in H,\quad ghg^{-1} \in H \implies H \triangleleft G $$

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