# 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