Difference between revisions of "Aufgaben:Problem 1"

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(Added proof of trivial second condition for linearity.)
 
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==Solution==
 
==Solution==
 
===Subtask a)===
 
===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:
 
We check the associativity of the group multiplication:
−
$$\left(\left(A,x\right)\circ\left(B,y\right)\right)\circ\left(C,z\right)=\left(AB,Ay+x\right)\circ\left(C,z\right)=\left(\left(AB\right)C,ABz+Ay+x\right)$$
+
$$
−
$$=\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)$$
+
\begin{align}
 +
\left(\left(A,x\right)\circ\left(B,y\right)\right)\circ\left(C,z\right)&=\left(AB,Ay+x\right)\circ\left(C,z\right)\\
 +
&=\left(\left(AB\right)C,ABz+Ay+x\right)\\
 +
&=\left(A\left(BC\right),A\left(Bz+y\right)+x\right)\\
 +
&=\left(A,x\right)\circ\left(BC,Bz+y\right)\\
 +
&=\left(A,x\right)\circ\left(\left(B,y\right)\circ\left(C,z\right)\right)\\
 +
\end{align}
 +
$$
 
The identity element is \(\left(\mathbb{I},0\right)\) and the inverse element is \(\left(A^{-1},-A^{-1}x\right)\) because
 
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
 
$$(A,x) \circ (\mathbb{I},0) = (A\mathbb{I},A0 + x) = (A,x) = (\mathbb{I}A, \mathbb{I}x + 0) = (\mathbb{I},0) \circ (A,x)$$ and
 
$$(A,x) \circ (A^{-1},-A^{-1}x) = (AA^{-1},-AA^{-1}x + x) = (\mathbb{I},0) =(A^{-1}A,A^{-1}x-A^{-1}x)=(A^{-1},-A^{-1}x) \circ (A,x)$$
 
$$(A,x) \circ (A^{-1},-A^{-1}x) = (AA^{-1},-AA^{-1}x + x) = (\mathbb{I},0) =(A^{-1}A,A^{-1}x-A^{-1}x)=(A^{-1},-A^{-1}x) \circ (A,x)$$
 +
 
===Subtask b)===
 
===Subtask b)===
 
We check that it is indeed a homomorphism:
 
We check that it is indeed a homomorphism:
−
$$\left(\rho\left(\left(A,y\right)\circ\left(B,z\right)\right)f\right)\left(x\right)$$
+
$$
−
$$=\left(\rho\left(AB,Az+y\right)f\right)\left(x\right)$$
+
\begin{align}
−
$$=f\left(\left(AB\right)^{-1}\left(x-Az-y\right)\right)$$
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\left(\rho [ \left(A,y\right)\circ\left(B,z\right) ] f\right)\left(x\right) &= \left(\rho [ AB,Az+y ] f\right)\left(x\right) \\
−
$$=f\left(B^{-1}A^{-1}\left(x-Az-y\right)\right)$$
+
&= f\left(\left(AB\right)^{-1}\left(x-Az-y\right)\right) \\
−
$$=f\left(B^{-1}\left(A^{-1}\left(x-y\right)-z\right)\right)$$
+
&= f\left(B^{-1}A^{-1}\left(x-Az-y\right)\right) \\
−
$$=\left(\rho\left(A,y\right)\left(f\left(B^{-1}\left(\cdot-z\right)\right)\right)\right)\left(x\right)$$
+
&= f\left(B^{-1}\left(A^{-1}\left(x-y\right)-z\right)\right) \\
−
$$=\left(\rho\left(A,y\right)\rho\left(B,z\right)f\right)\left(x\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)
 
(Here the dot indicates where the argument of the function goes)
  
−
It remains to show that \(\rho(g^-1) = \rho(g)^{-1}.\)
+
We furthermore show trivially that \(\rho(e) = e\\
−
We plug the formula of the neutral element into the definition of \(\rho\). We then see \((\rho(Id)f)(x) = f(x)\). It follows:
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\rho(e)(f)(x) = \rho((\mathbb{I},0))(f)(x) = f(\mathbb{I}(x-0)) = f(x)\)
−
$$(Id \circ f)(x) = ((\rho(Id)f)(x) = (\rho(g^{-1} \circ g)f)(x) = (\rho(g^{-1}) \circ \rho(g)) \circ f(x)$$
+
−
 
+
−
This is the same as:
+
−
 
+
−
$$\rho(g^{-1}) \circ \rho(g) = Id$$
+
  
−
and therefore:
+
Using the homomorphic property of \(\rho\) and the fact that \(G\) is a group: \(e = \rho(e) = \rho(gg^{-1}) = \rho(g) \rho(g^{-1}) \overset{!}{=} \rho(g) \rho(g)^{-1} \Rightarrow \rho(g) \in GL(V)\)
  
−
$$\rho(g^{-1}) = \rho^{-1}(g)$$
+
\(\Rightarrow \rho\) is a representation of \(G\) on \(V\) the space of all real vaued functions on \(\mathbb{R}^3\)

Latest revision as of 10:35, 30 June 2015

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)

We furthermore show trivially that \(\rho(e) = e\\ \rho(e)(f)(x) = \rho((\mathbb{I},0))(f)(x) = f(\mathbb{I}(x-0)) = f(x)\)

Using the homomorphic property of \(\rho\) and the fact that \(G\) is a group: \(e = \rho(e) = \rho(gg^{-1}) = \rho(g) \rho(g^{-1}) \overset{!}{=} \rho(g) \rho(g)^{-1} \Rightarrow \rho(g) \in GL(V)\)

\(\Rightarrow \rho\) is a representation of \(G\) on \(V\) the space of all real vaued functions on \(\mathbb{R}^3\)

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