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Finite Reflection Groups in Two Dimensions - Grove & Benson Section 2.1

I am seeking to understand Finite Reflection Groups and am reading Grove and Benson (G&B) : Finite Reflection Groups

Grove and Benson, in Section 21 Orthogonal Transformations in Two Dimensions define T as a linear transformation belonging to the groups of all orthogonal transformations .

G&B point out that the vector is an eigenvector having eigenvalue 1 for T and that similarly is an eigenvector with eigenvalue -1 and **[see attachment]**

G&B then state that if , then and T sends x to its mirror image with respect to the line L (see Figure 2.2(b) in attachement)

The transformation T is called the refection through L or the reflection along .

G&B then say "observe that for all "

Can someone help me show (explicitly and formally) that for all ?

And further (and possibly more important) can someone help me get a geometric sense of what the formula above means? **ie why do G&B highlight this particular relationship?**

Would very much appreciate such help

Peter

Re: Finite Reflection Groups in Two Dimensions - Grove & Benson Section 2.1

Tx = T(λ_{1}x_{1} + λ_{2}x_{2}) = λ_{1}T(x_{1}) + λ_{2}T(x_{2})

= λ_{1}x_{1} - λ_{2}x_{2} = λ_{1}x_{1} + λ_{2}x_{2} - 2λ_{2}x_{2}

= x - 2λ_{2}x_{2} (this is just simple algebra, to this point).

now (x,x_{2}) = (λ_{1}x_{1} + λ_{2}x_{2},x_{2}) = λ_{1}(x_{1},x_{2}) + λ_{2}(x_{2},x_{2})

and since x_{1},x_{2} are orthogonal, (x_{1},x_{2}) = 0, therefore:

(x,x_{2}) = λ_{2}(x_{2},x_{2}) = λ_{2}|x_{2}|^{2}.

but x_{2} lies on the unit circle, therefore it has length (and thus length squared) of 1.

so (x,x_{2}) = λ_{2}, and we have:

Tx = x - 2λ_{2}x_{2} = x - 2(x,x_{2})x_{2}.

so why?

the quantity (x,x_{2})x_{2} is a little misleading (because we are dealing with UNIT vectors x_{1},x_{2}).

normally, this is written as: [(x,x_{2})/(x_{2},x_{2})]x_{2} which is known as:

the projection of the vector x in the direction of the vector x_{2}. geometrically this is: "the part of x that lies on the line generated by x_{2}".

normally, we project a vector onto the unit vectors e_{1} = (1,0) and e_{2} = (0,1).

that is: if x = (x_{1},x_{2}) = x_{1}e_{1} + x_{2}e_{2}, then:

the projection of x in the direction of e_{1} is [(x.e_{1})/(e_{1}.e_{1})]e_{1}

= [(x_{1}*1 + x_{2}*0)/(1*1 + 0*0)]e_{1} = x_{1}e_{1} = (x_{1},0).

Re: Finite Reflection Groups in Two Dimensions - Grove & Benson Section 2.1

Thanks ... that post was most helpful, particularly the bit about the misleading nature of the formula as stated ... that clarified a few issues for me.

Geometrical interpretation is most helpful

Peter