Semi-direct product and linear function

Declaration

The connection between semi-direct product and linear function is observed by my friend 王进一(Jinyi Wang)

If you can read Chinese characters, just read 半直积与一次函数 - 单纯猫的文章 - 知乎 https://zhuanlan.zhihu.com/p/362094211

$H⋊K$$H\rtimes K$ is a little bit confused concept.

I am writing this Blog because I think it is a natural way to consider it ⋊, and I think this idea should be awarded to much more people.

And I also reference 李文威 《代数学方法》卷一

Linear Function and semi-direct product

As a set, $H⋊K$$H\rtimes K$ is isomorphic to $H\times K$$H\times K$

And the operator is $(h_1,k_1)(h_2,k_2)=(h_1\alpha (k_1)\cdot h_2,k_1{k}_2)$$(h_1,k_1)(h_2,k_2)=(h_1\alpha(k_1)\cdot h_2,k_1k_2)$

$\alpha :=K\mapsto \mathrm{Aut}(H)$$\alpha:=K\mapsto \mathrm{Aut}(H)$, Thus $\alpha (k_1)\cdot h_2\in H$$\alpha(k_1)\cdot h_2\in H$

I wouldn't say I like the representation since it is weird.

But, what if we consider $f(x)=B+Ax$$f(x)=B+Ax$?

Let us go back to the linear function $f:=\mathbb{R}\to \mathbb{R}$$f:=\mathbb R\to\mathbb R$ first.

For $f(x)=B+Ax$$f(x)=B+Ax$, $A$$A$ is a automorphism for $(\mathbb{R},+)$$(\mathbb R,+)$

Observe that $(B,A)⟼B+Ax$$(B,A)\longmapsto B+Ax$ is an isomorphism in $\mathrm{S}\mathrm{e}\mathrm{t}$$\rm Set$

And if we have $f(x)=B+Ax,g(x)=D+Cx$$f(x)=B+Ax,g(x)=D+Cx$

$f\circ g(x)=B+(A(D+Cx))=B+(AD+ACx)=(B+AD)+ACx$$f\circ g(x)=B+(A(D+Cx))=B+(AD+ACx)=(B+AD)+ACx$

The identity is $1(x)=x$$1(x)=x$ since $f\circ 1=f=1\circ f$$f\circ 1= f=1\circ f$

$f^{-1}(x)=A^{-1}(-B)+A^{-1}x$$f^{-1}(x)=A^{-1}(-B)+A^{-1}x$ since $f\circ f^{-1}(x)=x=f^{-1}\circ f$$f\circ f^{-1}(x)=x=f^{-1}\circ f$

See, $y=kx+b$$y=kx+b$ is a kind of $\mathbb{R}⋊\mathbb{R}$$\R\rtimes\R$

If we represent the element of $H⋊K$$H\rtimes K$ as $h+k\cdot x$$h+k\cdot x$, $+$$+$ is the operator in $H$$H$, but we do not need $H\in \mathrm{A}\mathrm{b}$$H\in \rm Ab$

then we will feel much more familiar with it!

After understanding the operator in $H⋊K$$H\rtimes K$

We need to consider the relation between $H,K,H⋊K$$H,K,H\rtimes K$

Consider the monomorphism ${\tau }_h:=h⟼h+x,{\rho }_k:=k⟼k\cdot x$$\tau_h:=h\longmapsto h+x,\rho_k:=k\longmapsto k\cdot x$

${\tau }_{h_1}\circ {\tau }_{h_2}=h_1+(h_2+x)=(h_1+h_2)+x={\tau }_{h_1+h_2}$$\tau_{h_1}\circ\tau_{h_2}=h_1+(h_2+x)=(h_1+h_2)+x=\tau_{h_1+h_2}$

${\rho }_{k_1}\circ {\rho }_{k_2}=k_1\cdot (k_2\cdot x)={\rho }_{k_1{k}_2}$$\rho_{k_1}\circ\rho_{k_2}=k_1\cdot(k_2\cdot x)=\rho_{k_1k_2}$

Then we can view $H,K$$H,K$ as subgroups of $H⋊K$$H\rtimes K$

And I guess readers already observe that $⋊$$\rtimes$ looks like $◃$$\triangleleft$

That is because $H◃H⋊K$$H\triangleleft H\rtimes K$

since $(h+x)\circ (b+k\cdot x)\circ (h+x{)}^{-1}=(h+b+k\cdot x)\circ (-h+x)=b+k\cdot x$$(h+x)\circ(b+k\cdot x)\circ(h+x)^{-1}=(h+b+k\cdot x)\circ(-h+x)=b+k\cdot x$,

$\mathrm{E}\mathrm{x}\mathrm{a}\mathrm{m}\mathrm{p}\mathrm{l}\mathrm{e}.$$\rm Example.$ $D_{2n}\cong \mathbb{Z}/n\mathbb{Z}⋊\mathbb{Z}/2\mathbb{Z}$$D_{2n}\cong\mathbb Z/n\mathbb Z\rtimes\mathbb Z/2\mathbb Z$

Define automorphism $\alpha :\mathbb{Z}/2\mathbb{Z}\to \mathrm{Aut}(\mathbb{Z}/n\mathbb{Z})$$\alpha:\mathbb Z/2\Z\to\mathrm{Aut}(\Z/n\Z)$

$0⟼\mathrm{id},1⟼(x\mapsto -x)$$0\longmapsto\mathrm{id},1\longmapsto(x\mapsto-x)$

$\tau \circ r\circ \tau =-r$$\tau\circ r\circ\tau=-r$ since $(\alpha (1)\cdot x)\circ (n+\alpha (0)\cdot x)\circ (\alpha (1)\cdot x)=(-x)\circ (n+x)\circ (-x)=(-n+x)$$(\alpha(1)\cdot x)\circ(n+\alpha(0)\cdot x)\circ(\alpha(1)\cdot x)=(-x)\circ(n+x)\circ(-x)=(-n+x)$

$\mathrm{E}\mathrm{x}\mathrm{a}\mathrm{m}\mathrm{p}\mathrm{l}\mathrm{e}.\mathrm{S}\mathrm{E}(2)$$\rm Example. SE(2)$

The group of rigid motions $\mathrm{S}\mathrm{E}(2)\cong {\mathbb{R}}^2⋊\mathrm{SO}(2)$$\rm SE(2)\cong\R^2\rtimes\mathrm{SO}(2)$

Since the rigid motion can be decomposition into translation and rotation