Re: [math-fun] Bott periodicity
Gareth wrote: << [I wrote:]
Er, guys, would one of you be so kind as to state what theorem is being discussed here?
The answer: Any isometry of R^n that fixes a k-dimensional subspace is the product of at most n-k reflections.
OK, I'll give it a shot: The proposition is equivalent to saying that any isometry of R^p with no nontrivial fixed subspace is the product of at most p reflections. Any isometry of R^n preserves mutually orthogonal subspaces: of dim = 2, and another one of dim = 1 (if n is odd). (This follows readily from complexification.) The cases of 1 dimension, and one reflection in 2D, are trivial. And a rotation of R^2 of angle theta is the product of a reflection about the x-axis, and one about the line at angle theta/2. QED. --Dan
[about this theorem:]
Any isometry of R^n that fixes a k-dimensional subspace is the product of at most n-k reflections.
[Dan's proof:]
The proposition is equivalent to saying that any isometry of R^p with no nontrivial fixed subspace is the product of at most p reflections.
Any isometry of R^n preserves mutually orthogonal subspaces: of dim = 2, and another one of dim = 1 (if n is odd). (This follows readily from complexification.)
The cases of 1 dimension, and one reflection in 2D, are trivial.
And a rotation of R^2 of angle theta is the product of a reflection about the x-axis, and one about the line at angle theta/2. QED.
Looks OK to me, but Fred's question wasn't "can someone give me a proof of this?" but "can someone explain the proof of this found in Conway&Smith's book". :-) -- g
On 1/9/06, Gareth McCaughan <gareth.mccaughan@pobox.com> wrote:
[about this theorem:]
Any isometry of R^n that fixes a k-dimensional subspace is the product of at most n-k reflections.
I think the isometries of S^n rather than R^n are intended --- the presence of translations would complicate the argument. Also Lemma 3 carefully says "can be written as", since the conclusion is false unless the reflection hyperplanes are chosen with linearly independent coefficient vectors.
[Dan's proof:] Looks OK to me, but Fred's question wasn't "can someone give me a proof of this?" but "can someone explain the proof of this found in Conway&Smith's book". :-)
Quite so. However, if you're going to use induction [Gareth's proof] --- which C&S doesn't mention --- why not just say that the intersection of n-k independent hyperplanes has dimension exactly k, and a point is fixed just when it lies in this set? [I'm assuming that when Lemma 3 says "fixes", it means pointwise, and implies only that the axis includes the k-flat.] In fairness, this glitch is unimportant and seems to be isolated --- but rather unfortunate that it occurs so early in the book! Fred Lunnon
participants (3)
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Daniel Asimov -
Fred lunnon -
Gareth McCaughan