Topological Group
Definition1
A group $\braket{G, \cdot}$ that is a topological space and satisfies the following is called a topological group.
- The group multiplication $\cdot : G \times G \to G$, $\quad (g, h) \mapsto g \cdot h$ is continuous.
- The map to inverses $i : G \to G$, $\quad g \mapsto g^{-1}$ is continuous.
From the topological viewpoint
A Hausdorff space $G$ is called a topological group if it forms a group with respect to the following two continuous functions.
$$ m : G \times G \to G,\qquad (g, h) \mapsto g \cdot h \\ i : G \to G,\qquad g \mapsto g^{-1} $$
Explanation
It is a blend of a group and a topological space. If the condition that the product map and inversion map be continuous is replaced by “[differentiable]”(../3116), one obtains a Lie group.
$$ \text{Lie group} \implies \text{topological group} $$
Examples
- Every group becomes a topological group when endowed with the discrete topology.
- The unit circle in the complex plane $\braket{S^{1}(\mathbb{C}), \cdot}$ is a topological group. Here $\cdot$ is complex multiplication.
- The group formed by Euclidean space and addition $\braket{\mathbb{R}^{n}, +}$ is a topological group. Here $+$ is vector addition.
Proof
1.
If we give the group $\braket{G, \cdot}$ the discrete topology $\mathcal{P}(G)$, then every subset of $G$ is open.
Equivalence condition for continuous functions:
- $f : X \to Y$ is a continuous function.
- For every open set $V \subset Y$, $f^{-1}(V)$ is an open set.
Since every subset of $G$ is open, by the above condition every $f : G \to G$ is continuous. Hence the inversion $i : G \to G$ is continuous. By the same reasoning, if we give $G \times G$ the discrete topology then the multiplication $\cdot : G \times G \to G$ is also continuous. Therefore the group $\braket{G, \cdot}$ is a topological group.
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2.
The unit circle in the complex plane is the set satisfying the following.
$$ S^{1}(\mathbb{C}) = \{ z \in \mathbb{C} : |z| = 1 \} $$
$\braket{S^{1}(\mathbb{C}), \cdot}$ is a group.
- Closed: If $z_{1}, z_{2} \in S^{1}(\mathbb{C})$, then |$z_{1} z_{2}| = |z_{1}||z_{2}| = 1$, hence $z_{1} z_{2} \in S^{1}(\mathbb{C})$.
- Associativity: If $z_{1}=e^{i\theta_{1}}, z_{2}=e^{i\theta_{2}}, z_{3}=e^{i\theta_{3}}$, then $z_{1} (z_{2} z_{3}) = e^{i(\theta_{1} + \theta_{2} + \theta_{3})} = (z_{1} z_{2}) z_{3}$.
- Identity: $(1,0) \in S^{1}(\mathbb{C})$.
- Inverse: If $z \in S^{1}(\mathbb{C})$ then $z^{-1} = \frac{1}{z} \in S^{1}(\mathbb{C})$.
$S^{1}(\mathbb{C})$ is a topological space.
Equip $S^{1}(\mathbb{C})$ with the subspace topology induced from the usual topology on the complex plane. Then $S^{1}(\mathbb{C})$ is a subspace of $\mathbb{C}$.
박대희·안승호. 위상수학 (5/E, 2022), p648 ↩︎