nLab
identity component

Definition

If $G$ is a topological group, then the identity component is the connected component of the identity element $e$ in $G$.

Basic results

Proposition

The identity component $G_0$ is a closed normal subgroup of $G$.

Proof

It is clearly closed (indeed, any connected component is closed). If $g,h\in G_0$, then $gh$ is in the same connected component as $g$ (since $h$ is in the same connected component as $e$ and left multiplication by $g$ is a homeomorphism), which in turn is in the same connected component as $e$. Using similar reasoning, if $g$ is in the connected component as $e$, then $e$ is in the same connected component as $g^{-1}$. Hence $G_0$ is a subgroup.

If $\varphi$ is any automorphism of $G$, then $\varphi (G_0)=G_0$. (Indeed, $\varphi (G_0)$ is a connected set containing $e$ and therefore $\varphi (G_0)\subseteq G_0$. Replacing $\varphi$ by its inverse ${\varphi }^{-1}$, we similarly have ${\varphi }^{-1}(G_0)\subseteq G_0$ and therefore $G_0\subseteq \varphi (G_0)$.) Applying this to inner automorphisms $\varphi$, we conclude that $G_0$ is a normal subgroup of $G$.

• Remark: $G_0$ need not be open in $G$; for example, for the group of $p$-adic integers, $G_0$ is the (non-open) singleton $\{e\}$. However, if $G$ is locally connected, for example if $G$ is a Lie group, then $G_0$ is open (and therefore also clopen. In this case, $G/G_0$ is discrete (because $G\to G/G_0$ is an open map, implying that the identity and therefore every point in $G/G_0$ is open).

Proposition

The group $G/G_0$, equipped with the quotient space topology, is a Hausdorff topological group.

Proof

Given the fact that $p:G\to G/G_0$ is an open surjection, the product $p\times p:G\times G\to G/G_0\times G/G_0$ is also an open surjection and therefore a quotient map. It follows easily from the universal property of quotient maps that the multiplication $G\times G\to G$ therefore descends to a continuous multiplication $G/G_0\times G/G_0\to G/G_0$, so that $G/G_0$ is a topological group.

Because a topological group is a uniform space, the Hausdorff condition follows from a weaker separation axiom such as $T_1$ (points are closed). It suffices that the identity of $G/G_0$ be closed. Its complement $C$ is the image under $p$ of the complement of $G_0$ in $G$ (just by examining coset decompositions), which is open. Since $p$ is an open map, it follows that $C$ is open, so that $\{e\}$ is closed, as desired.