Difference between revisions of "Aufgaben:Problem 1"
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(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 \(\rho(g^-1) = \rho(g)^{-1}.\) | + | 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: | We plug the formula of the neutral element into the definition of \(\rho\). We then see \((\rho(Id)f)(x) = f(x)\). It follows: | ||
$$(Id \circ f)(x) = ((\rho(Id)f)(x) = (\rho(g^{-1} \circ g)f)(x) = (\rho(g^{-1}) \circ \rho(g)) \circ f(x)$$ | $$(Id \circ f)(x) = ((\rho(Id)f)(x) = (\rho(g^{-1} \circ g)f)(x) = (\rho(g^{-1}) \circ \rho(g)) \circ f(x)$$ | ||
Revision as of 08:32, 15 June 2015
Contents
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 \(\left(A,x\right)\circ\left(B,y\right):=\left(AB,Ay+x\right)\).
- Check that the map \(\circ\) turns \(G\) into a group. Find explicit formulas for the neutral element and inverse.
- 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)
We check the associativity of the group multiplication: $$\left(\left(A,x\right)\circ\left(B,y\right)\right)\circ\left(C,z\right)=\left(AB,Ay+x\right)\circ\left(C,z\right)=\left(\left(AB\right)C,ABz+Ay+x\right)$$ $$=\left(A\left(BC\right),A\left(Bz+y\right)+x\right)=\left(A,x\right)\circ\left(BC,Bz+y\right)=\left(A,x\right)\circ\left(\left(B,y\right)\circ\left(C,z\right)\right)$$ 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: $$\left(\rho\left(\left(A,y\right)\circ\left(B,z\right)\right)f\right)\left(x\right)$$ $$=\left(\rho\left(AB,Az+y\right)f\right)\left(x\right)$$ $$=f\left(\left(AB\right)^{-1}\left(x-Az-y\right)\right)$$ $$=f\left(B^{-1}A^{-1}\left(x-Az-y\right)\right)$$ $$=f\left(B^{-1}\left(A^{-1}\left(x-y\right)-z\right)\right)$$ $$=\left(\rho\left(A,y\right)\left(f\left(B^{-1}\left(\cdot-z\right)\right)\right)\right)\left(x\right)$$ $$=\left(\rho\left(A,y\right)\rho\left(B,z\right)f\right)\left(x\right)$$ (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:
$$\rho(g^{-1}) = \rho^{-1}(g)$$
