Difference between revisions of "Aufgaben:Problem 1"
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− | + | ==Task== | |
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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 \(\left(A,x\right)\circ\left(B,y\right):=\left(AB,Ay+x\right)\). | 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 \(\left(A,x\right)\circ\left(B,y\right):=\left(AB,Ay+x\right)\). | ||
<ol style="list-style-type:lower-latin"> | <ol style="list-style-type:lower-latin"> | ||
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<li>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 \(\mathbb{R}^{3}\)</li> | <li>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 \(\mathbb{R}^{3}\)</li> | ||
</ol> | </ol> | ||
− | + | ==Solution== | |
− | + | ===Subtask a)=== | |
The identity element is \(\left(\mathbb{I},0\right)\) and the inverse element is \(\left(A^{-1},-A^{-1}x\right)\). We check the associativity of the group multiplication: | The identity element is \(\left(\mathbb{I},0\right)\) and the inverse element is \(\left(A^{-1},-A^{-1}x\right)\). We check the associativity of the group multiplication: | ||
$$\left(\left(A,x\right)\circ\left(B,y\right)\right)\circ\left(C,z\right)=\left(AB,Ay+x\right)\circ\left(C,z\right)=\left(\left(AB\right)C,ABz+Ay+x\right)$$ | $$\left(\left(A,x\right)\circ\left(B,y\right)\right)\circ\left(C,z\right)=\left(AB,Ay+x\right)\circ\left(C,z\right)=\left(\left(AB\right)C,ABz+Ay+x\right)$$ | ||
$$=\left(A\left(BC\right),A\left(Bz+y\right)+x\right)=\left(A,x\right)\circ\left(BC,Bz+y\right)=\left(A,x\right)\circ\left(\left(B,y\right)\circ\left(C,z\right)\right)$$ | $$=\left(A\left(BC\right),A\left(Bz+y\right)+x\right)=\left(A,x\right)\circ\left(BC,Bz+y\right)=\left(A,x\right)\circ\left(\left(B,y\right)\circ\left(C,z\right)\right)$$ | ||
− | + | ===Subtask b)=== | |
We check that it is indeed a homomorphism: | We check that it is indeed a homomorphism: | ||
$$\left(\rho\left(\left(A,y\right)\circ\left(B,z\right)\right)f\right)\left(x\right)$$ | $$\left(\rho\left(\left(A,y\right)\circ\left(B,z\right)\right)f\right)\left(x\right)$$ |
Revision as of 10:20, 10 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)
The identity element is \(\left(\mathbb{I},0\right)\) and the inverse element is \(\left(A^{-1},-A^{-1}x\right)\). We check the associativity of the group multiplication: $$\left(\left(A,x\right)\circ\left(B,y\right)\right)\circ\left(C,z\right)=\left(AB,Ay+x\right)\circ\left(C,z\right)=\left(\left(AB\right)C,ABz+Ay+x\right)$$ $$=\left(A\left(BC\right),A\left(Bz+y\right)+x\right)=\left(A,x\right)\circ\left(BC,Bz+y\right)=\left(A,x\right)\circ\left(\left(B,y\right)\circ\left(C,z\right)\right)$$
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)