Finite Reflection Groups in Two Dimensions - R2

1. Apr 22, 2012

Math Amateur

I am seeking to understand reflection groups and am reading Grove and Benson: Finite Reflection Groups

On page 6 (see attachment - pages 5 -6 Grove and Benson) we find the following statement:

"It is easy to verify (Exercise 2.1) that the vector $x_1 = (cos \ \theta /2, sin \ \theta /2 )$ is an eigenvector having eigenvalue 1 for T, so that the line
$L = \{ \lambda x_1 : \lambda \in \mathbb{R} \}$ is left pointwise fixed by T."

I am struggling to se why it follows that L above is left pointwise fixed by T (whatever that means exactly).

Can someone please help - I am hoping to be able to formally and explicitly justify the statement.

The preamble to the above statement is given in the attachment, including the definition of T

Notes (see attachment)

1. T belongs to the group of all orthogonal transformations, $O ( \mathbb{R} )$.

2. Det T = -1

For other details see attachment

Peter

Attached Files:

• Grove and Benson - Finite Reflection Groups -Pages 5 -6.pdf
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Last edited: Apr 22, 2012
2. Apr 22, 2012

DonAntonio

"$\,x_1$ is an eigenvector of $\,T$ with eigenvalue $\,1\,\,$" means $\,Tx_1=x_1$

To leave a subspace L fixed pointwise means $\,\,Tv=v\,,\,\,\forall v\in L$

Can you now apply the above to see the statement in that book is trivial?

DonAntonio

3. Apr 22, 2012

Math Amateur

Thanks for the help - thinking that over

What is meant by (or what is the significance of) 'left' pointwise fixed

Peter

4. Apr 22, 2012

DonAntonio

In this case "left" is the past participle of "leave", and NOT the left of "side", as opposite to "right"...:P)

DonAntonio

5. Apr 22, 2012

Math Amateur

Oh ... right ... completely misunderstood the text :-(

Thanks

Peter