Difference between revisions of "Aufgaben:Problem 1"
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− | == | + | ==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 \(\left(A,x\right)\circ\left(B,y\right):=\left(AB,Ay+x\right)\). | ||
+ | <ol style="list-style-type:lower-latin"> | ||
+ | <li>Check that the map \(\circ\) turns \(G\) into a group. Find explicit formulas for the neutral element and inverse.</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> | ||
+ | ==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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Latest revision as of 10:35, 30 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)
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 \(\mathbb{R}^3\)