# Math Help - Finite Reflection Groups in Two Dimensions - R2

1. ## Finite Reflection Groups in Two Dimensions - R2

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 see why it follows that L above is left pointwise fixed by T (whatever that means exactly! - can someone please clarify this matter?).

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 transformatios, $O ( \mathbb{R} )$.

2. Det T = -1

For other details see attachment

Peter

2. ## Re: Finite Reflection Groups in Two Dimensions - R2

saying that x1 is an eigenvector of T with eigenvalue 1 means that Tx1 = x1. verifying this is just a routine computation:

$Tx_1 = \begin{bmatrix}\cos\theta&\sin\theta\\ \sin\theta&-\cos\theta \end{bmatrix} \begin{bmatrix}\cos(\theta/2)\\ \sin(\theta/2) \end{bmatrix} =$

$\begin{bmatrix}\cos\theta\cos(\theta/2) + \sin\theta\sin(\theta/2)\\ \sin\theta\cos(\theta/2) - \sin(\theta/2)\cos\theta \end{bmatrix} =$

$\begin{bmatrix}\cos(\theta - \theta/2)\\ \sin(\theta - \theta/2) \end{bmatrix} = x_1$

now, for any real number λ, we have: T(λx1) = λ(Tx1) (since T is linear)

= λx1, so any vector λx1 is also an eigenvector for T (with eigenvalue 1).

this means that the line L = {λx1: λ in R} is an invariant subspace of T: T(L) ⊂ L (in fact, since T has non-zero determinant, T(L) = L).

geometrically, this means that T does not change the line L at all, in other words (since we know that T2 = I) T reflects the plane through the line L (we know it is a reflection and not a rotation of 180 degrees since det(T) = -1).

personally, i find the use of the trigonometry somewhat distracting. the underlying algebraic fact is this: if R is a matrix with det(R) = 1, and Rn = I, then R is a rotation "1/n-th of the way around a circle". of course, to generalize this entirely, one must allow n to be any real number greater than 1 (which is where the trigonometry comes in, since it becomes difficult to speak of non-integral powers of a matrix). but the important fact is that given a rotation matrix R, then there is a corresponding reflection matrix T, where T = RC, where C is the matrix:

$C = \begin{bmatrix}1&0\\0&-1 \end{bmatrix}$

geometrically, this means that for any rotation of the plane, if we pick the line left invariant by the associated reflection T, as the "new y-axis", and the line perpendicular to that line as the "new x-axis", then in these "new coordinates" T is just reflection about the y-axis, and in these "new coordinates" the matrix for T is simply C. so C is in some sense "the canonical reflection", since any other reflection T is just C followed by a rotation R.

in geometry, the points of the plane are all "uniform", the only "distinguished" point is the origin. the "coordinates" we introduce, in order to be able to use linear algebra, are somewhat arbitrarily chosen: we "pick" a direction and call it the x-axis, and then call the perpendicular direction the y-axis. this choice of axes is somewhat arbitrary, the geometry holds true no matter how we "label" our points.

now orthogonality, or "right-angled-ness" is a geometric property. it shouldn't depend on the coordinate system we are using. but to define it in the context of vector spaces (so as to define linear transformations that PRESERVE it), we need a way of defining orthogonality on vector spaces. this is normally done by defining an inner product (the USUAL inner product is the "dot product" in euclidean n-space). then T preserves orthogonality if Tu.Tv = u.v. note that we get something "for free" here, if we define the length of the vector u as:

|u| = √(u.u),

then if T preserves orthogonality: |Tu| = √(Tu.Tu) = √(u.u) = |u|, so T preserves lengths as well.

and we can go even further: we can define the DISTANCE between the vectors u and v as: |u - v|. then we have:

|Tu - Tv| = |T(u - v)| = √(T(u - v).T(u - v)) = √((u - v).(u - v)) = |u - v|, so T preserves distances as well.

in the 2x2 case (the euclidean plane) if T =

[a b]
[c d]

then T(x,y) = (ax + by,cx + dy). if T preserves the dot product, then a2+c2 = 1, and b2+d2 = 1, and ac+bd = 0

now consider the transpose of T:

[a c]
[b d].

note that TTT =

[a2+b2 ac+bd]
[ac+bd c2+d2] = I

hence det(TTT) = det(T)det(TT) = det(T)2 = 1, so evidently det(T) is a square root of 1, so either 1 or -1. on the other hand, if det(T) = 1 or -1, running the argument in reverse shows T preserves inner products.

3. ## Re: Finite Reflection Groups in Two Dimensions - R2

Thank you for this extensive guidance ... I find it so helpful ...

Working through all you have said ...

Peter

4. ## Re: Finite Reflection Groups in Two Dimensions - R2

Followed you where you establish the following:

" $T x_1 = x_1$ and ...

now, for any real number λ, we have: T(λx1) = λ(Tx1) (since T is linear)

= λx1, so any vector λx1 is also an eigenvector for T (with eigenvalue 1).

this means that the line L = {λx1: λ in R} is an invariant subspace of T: T(L) ⊂ L (in fact, since T has non-zero determinant, T(L) = L)."

BUT

how do you conclude "in fact, since T has non-zero determinant, T(L) = L"

I cannot find this result in my linear algebra texts ... can you demonstrate formally why this is true ...

By the way ... do you know of a good reference that gives a straightforward account of the basic ideas/theory of rotations and reflections along the lines of your help?

Peter

5. ## Re: Finite Reflection Groups in Two Dimensions - R2

T has non-zero determinant = T is invertible = T is bijective.

remember, orthogonal linear mappings are a subgroup of the general linear group.

6. ## Re: Finite Reflection Groups in Two Dimensions - R2

Thanks so much for that help

So T is bijective and hence surjective and so T(L) = L ... OK, yes ... thanks

Peter