Aufgaben:Problem 1

Task

We endow the Cartesian product \(G=SO\left(3\right)\times\mathbb{R}^{3}\) with the map \(\circ:G\times G\rightarrow G\) defined by \(\left(A,x\right)\circ\left(B,y\right):=\left(AB,Ay+x\right)\).

1. Check that the map \(\circ\) turns \(G\) into a group. Find explicit formulas for the neutral element and inverse.
2. Verify that the formula \(\left(\rho\left(A,y\right)f\right)\left(x\right):=f\left(A^{-1}\left(x-y\right)\right)\) defines a representation of \(G\) on the space of real valued functions \(f\) on \(\mathbb{R}^{3}\)

Solution

Subtask a)

Check that \((A,x) \circ (B,y) \in G\): \(SO(3)\) is a group under matrix multiplication, so \(AB \in SO(3)\). \(Ay, x \in \mathbb{R}^3\), and since \(\mathbb{R}^3\) is a vector space, \(Ay + x \in \mathbb{R}^3\). So, the statement is proven.

We check the associativity of the group multiplication: $$ \begin{align} \left(\left(A,x\right)\circ\left(B,y\right)\right)\circ\left(C,z\right)&=\left(AB,Ay+x\right)\circ\left(C,z\right)\\ &=\left(\left(AB\right)C,ABz+Ay+x\right)\\ &=\left(A\left(BC\right),A\left(Bz+y\right)+x\right)\\ &=\left(A,x\right)\circ\left(BC,Bz+y\right)\\ &=\left(A,x\right)\circ\left(\left(B,y\right)\circ\left(C,z\right)\right)\\ \end{align} $$ The identity element is \(\left(\mathbb{I},0\right)\) and the inverse element is \(\left(A^{-1},-A^{-1}x\right)\) because $$(A,x) \circ (\mathbb{I},0) = (A\mathbb{I},A0 + x) = (A,x) = (\mathbb{I}A, \mathbb{I}x + 0) = (\mathbb{I},0) \circ (A,x)$$ and $$(A,x) \circ (A^{-1},-A^{-1}x) = (AA^{-1},-AA^{-1}x + x) = (\mathbb{I},0) =(A^{-1}A,A^{-1}x-A^{-1}x)=(A^{-1},-A^{-1}x) \circ (A,x)$$

Subtask b)

We check that it is indeed a homomorphism: $$ \begin{align} \left(\rho [ \left(A,y\right)\circ\left(B,z\right) ] f\right)\left(x\right) &= \left(\rho [ AB,Az+y ] f\right)\left(x\right) \\ &= f\left(\left(AB\right)^{-1}\left(x-Az-y\right)\right) \\ &= f\left(B^{-1}A^{-1}\left(x-Az-y\right)\right) \\ &= f\left(B^{-1}\left(A^{-1}\left(x-y\right)-z\right)\right) \\ &= \left(\rho\left(A,y\right)\left(f\left(B^{-1}\left(\cdot-z\right)\right)\right)\right)\left(x\right) \\ &= \left(\rho\left(A,y\right)\rho\left(B,z\right)f\right)\left(x\right) \\ \end{align}$$ (Here the dot indicates where the argument of the function goes)

It remains to show that \(\rho(g^{-1}) = \rho(g)^{-1}.\) We plug the formula of the neutral element into the definition of \(\rho\). We then see \((\rho(Id)f)(x) = f(x)\). It follows: $$(Id \circ f)(x) = ((\rho(Id)f)(x) = (\rho(g^{-1} \circ g)f)(x) = (\rho(g^{-1}) \circ \rho(g)) \circ f(x)$$

This is the same as:

$$\rho(g^{-1}) \circ \rho(g) = Id$$

and therefore:

$$\rho(g^{-1}) = \rho^{-1}(g)$$

Alternative solution for \(\rho(g^{-1}) = \rho(g)^{-1}.\)

(It´s basically the same thing as above, but for some people it might be a bit easier to remember it this way.)

We want to show that

$$ \rho ( (A,x)^{-1} ) \circ \rho ( (A,x) ) = Id $$

because this implies that \(\rho((A,x)^{-1}) = \rho((A,x))^{-1}.\)

$$ \rho ( (A,x)^{-1} ) \circ \rho (A,x) \stackrel{\rho \ homomorphism}{=} \rho ( (A,x)^{-1} \circ (A,x) ) \stackrel{a)}{=} \rho ( \mathbb{I},0 ) $$

and we see, that indeed:

$$ ( \rho ( \mathbb{I},0) f ) (x) = f (\mathbb{I}x - 0) = f(x) $$ $$ \Rightarrow \rho (\mathbb{I},0) = Id $$