[itex]\renewcommand{\vec}[1]{\mathbf{#1}}[/itex]
Here is an excerpt from the text:
"[...]Theorem 12.5 The only finite symmetry groups in [itex]ℝ^2[/itex] are [itex] \mathbb{Z}_n[/itex] and [itex]D_n.[/itex]
PROOF. Any finite symmetry group [itex]G[/itex] in [itex]\mathbb{R}^2[/itex] must be a finite subgroup of [itex]O(2)[/itex]; otherwise, [itex]G[/itex] would have an element in [itex]E(2)[/itex] of the form [itex](A,\vec{x})[/itex], where [itex]\vec{x} ≠ \vec{0}[/itex]. Such an element must have infinite order. [...]"
But if I understand this correctly, the last sentence is false. As a counterexample, I can give the element [itex](-I, \vec{x})[/itex] which will have order 2 for any [itex]\vec{x}[/itex].
So is this an error, or did I misunderstand something?
Here is an excerpt from the text:
"[...]Theorem 12.5 The only finite symmetry groups in [itex]ℝ^2[/itex] are [itex] \mathbb{Z}_n[/itex] and [itex]D_n.[/itex]
PROOF. Any finite symmetry group [itex]G[/itex] in [itex]\mathbb{R}^2[/itex] must be a finite subgroup of [itex]O(2)[/itex]; otherwise, [itex]G[/itex] would have an element in [itex]E(2)[/itex] of the form [itex](A,\vec{x})[/itex], where [itex]\vec{x} ≠ \vec{0}[/itex]. Such an element must have infinite order. [...]"
But if I understand this correctly, the last sentence is false. As a counterexample, I can give the element [itex](-I, \vec{x})[/itex] which will have order 2 for any [itex]\vec{x}[/itex].
So is this an error, or did I misunderstand something?
via Physics Forums RSS Feed http://www.physicsforums.com/showthread.php?t=703866&goto=newpost
0 commentaires:
Enregistrer un commentaire