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Topological Group 📂Abstract Algebra

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

  1. Every group becomes a topological group when endowed with the discrete topology.
  2. The unit circle in the complex plane $\braket{S^{1}(\mathbb{C}), \cdot}$ is a topological group. Here $\cdot$ is complex multiplication.
  3. 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.

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 $\mathbb{C}$.


  1. 박대희·안승호. 위상수학 (5/E, 2022), p648 ↩︎