Semi-direct products

Posted on 4th February 2019

Motivated by some reading about quantum groups, I want to sketch how a semi-direct product of (topological) groups is the same as having an idempotent group homomorphism.

Firstly, let's remember what the (external) semi-direct product of groups is. I will follow the notation of the book of Kaniuth and Taylor. Let $$N,H$$ be (topological) groups, and denote by $$\newcommand{\aut}{\operatorname{Aut}}\aut(N)$$ the collection of continuous group automorphisms of $$N$$. Suppose we have a group homomorphism $$\alpha:H\rightarrow\aut(N)$$, written as $$h\mapsto \alpha_h$$, which is continuous in the sense that $$N\times H\rightarrow N; (n,h)\mapsto \alpha_h(n)$$ is continuous.

Let $$N \rtimes H$$, the semi-direct product, be the topological space $$N\times H$$ equipped with the group product $(n,h)(m,g) = (n \alpha_h(m), hg).$ It is easy to check that this is associate, that $$(e,e)$$ is the unit, and that $$(n,h)^{-1} = (\alpha_{h^{-1}}(n^{-1}), h^{-1})$$. Similarly, it is easy to check that the product and inverse are continuous, so that $$N \rtimes H$$ becomes a topological group.

Let $$\tilde N = \{(n,e) : n\in N\}$$ and $$\tilde H = \{(e,h):h\in H\}$$. These can be checked to be closed subgroups of $$N \rtimes H$$ such that $$\tilde N \cap \tilde H = \{e\}$$ and $$\tilde N\tilde H = N \rtimes H$$. For $$(m,e)\in \tilde N$$ and $$(n,h)\in N \rtimes H$$ we have that $(n,h)(m,e)(n,h)^{-1} = (n\alpha_h(m), h)(\alpha_{h^{-1}}(n^{-1}), h^{-1}) = (n\alpha_h(m)n^{-1}, e),$ which shows that $$\tilde N$$ is normal in $$N \rtimes H$$, and that $$(\alpha_h(m),e) = (e,h)(m,e)(e,h)^{-1}$$, so that conjugation by $$\tilde H$$ in $$N \rtimes H$$ implements $$\alpha$$.

Conversely, suppose that $$G$$ is a topological group, with closed subgroups $$N,H$$, with $$N$$ normal, $$N\cap H=\{e\}$$ and $$G = NH$$. Define $$\alpha:H\rightarrow\aut(N)$$ by $$\alpha_h(n) = hnh^{-1}$$ for $$h\in H, n\in N$$. As $$N$$ is normal, it follows that $$\alpha$$ makes sense, and is a (continuous) group homomorphism. Consequently, we can define $$N \rtimes H$$.

Define $$\theta:N \rtimes H\rightarrow G$$ by $$\theta(n,h)=nh$$. Then $\theta(n,h)\theta(m,g) = nhmg = nhmh^{-1}hg = \theta(n\alpha_h(m), hg),$ showing that $$\theta$$ is a group homomorphism. By assumption, $$\theta$$ is onto, and if $$nh=e$$ then $$n = h^{-1} \in N\cap H$$ so $$n=h=e$$, so $$\theta$$ is a bijection. Thus $$G$$ is isomorphic to $$N \rtimes H$$.

We now show the (I think less common) link with idempotent group homomorphisms. With $$G,N,H$$ as before, for $$g\in G$$ we can write $$g=nh$$. This decomposition is unique, either by the above map $$\theta$$, or by observing that if $$nh=mg$$ then $$m^{-1}n = gh^{-1} \in N\cap H$$ so $$m^{-1}n=e$$ and $$gh^{-1}=e$$ so $$n=m, g=h$$. Define $$\phi : N \backslash G \rightarrow H$$ by $$\phi(Ng) = h$$ when $$g=nh$$ for some $$n\in N$$. Notice that $$g=nh$$ for some $$n\in N$$ is equivalent to $$Ng = Nh$$. Then $$\phi(Nh) = h$$ for $$h\in H$$ so $$\phi$$ is onto, and if $$\phi(gN) = e$$ then $$g\in N$$ so $$gN=N$$. As $$N$$ is normal, of course $$Ng_1 Ng_2 = N g_1 g_1^{-1}Ng_1 g_2 = N g_1g_2$$. Let $$\phi(Ng_1)=h_1$$ and $$\phi(Ng_2)=h_2$$, so that $$Ng_1=Nh_1$$ and $$Ng_2=Nh_2$$, so $$Ng_1g_2 = Ng_1 Ng_2 = Nh_1 Nh_2 = Nh_1h_2$$. Thus $$\phi$$ is a group isomorphism. To argue that $$\phi$$ is continuous seems a little tricky, but it follows rather easily if you view $$G$$ as being isomorphic to $$N \rtimes H$$.

Now let $$p$$ be the composition $G \twoheadrightarrow N \backslash G \cong H \hookrightarrow G$ which is a (continuous) group homomorphism. Thus $$p(g)=h\in H$$ if and only if $$h = \phi(Ng)$$, that is, $$Nh=Ng$$ or equivalently, $$gh^{-1} \in N$$. Thus $$p(h)=h$$ for any $$h\in H$$, so the image of $$p$$ is $$H$$, and the kernel of $$p$$ is $$N$$. In particular, $$p\circ p=p$$.

Conversely, let $$G$$ be a topological group, and let $$p$$ be a continuous group homomorphism from $$G$$ to $$G$$, with $$p\circ p =p$$. Let $$N=\ker p$$, a closed normal subgroup of $$G$$, and let $$H$$ be the image of $$p$$, a subgroup. As $$p\circ p=p$$ and $$p$$ is continuous, it follows that $$H$$ is closed. If $$g\in N\cap H$$ then $$p(g)=e$$ and $$p(g)=g$$, so $$N\cap H=\{e\}$$. For $$g\in G$$, let $$n = g p(g)^{-1} = gp(g^{-1})$$ so $$p(n) = p(g) p(g^{-1}) = p(g)p(g)^{-1} = e$$. Thus $$g = n p(g) \in NH$$.

Thus the semi-direct product construction is exactly captured by the notion of a group with an idempotent group homomorphism.