1 Attachment(s)

Reflections - Kane's Book

I am reading Kane - Reflection Groups and Invariant Theory and need help with property A-4 of the properties of reflections stated on page 7

(see attachment - Kane _ Reflection Groups and Invariant Theory - pages 6-7)

On page 6 Kane mentions he is working in http://latex.codecogs.com/png.latex? \ell dimensional Euclidean space ie http://latex.codecogs.com/png.latex? E = R^{\ell} where http://latex.codecogs.com/png.latex? R^{\ell} has the usual inner product (x,y).

In defining reflections with respect to vectors Kane writes:

" Given http://latex.codecogs.com/png.latex? 0 \ne \alpha \in E let http://latex.codecogs.com/png.latex?...lpha}\subset E be the hyperplane

http://latex.codecogs.com/png.latex?...alpha ) = 0 \}

We then define the reflection http://latex.codecogs.com/png.latex?...ngrightarrow E by the rules

http://latex.codecogs.com/png.latex?...a} \cdot x = x if http://latex.codecogs.com/png.latex? x \in H_{\alpha}

http://latex.codecogs.com/png.latex?...pha = - \alpha "

Then Kane states that the following property follows:

(A-4) If $\displaystyle \phi $ is an orthogonal automorphism of E then

$\displaystyle \phi \cdot H_{\alpha} = H_{\phi \cdot \alpha} $

$\displaystyle \phi s_{\alpha} {\phi}^{-1} = s_{\phi \cdot \alpha $

**Can someone explicitly and formally demonstrate A-4?**

I am not really sure of the meaning of $\displaystyle H_{\phi \cdot \alpha} $ but I assume

$\displaystyle H_{\phi \cdot \alpha} = \{ x \ | \ (x, {\phi \cdot \alpha ) = 0 \}$

And I assume that

$\displaystyle \phi \cdot H_{\alpha} = \{ \phi \cdot x \ | \ (x, \alpha \} = 0 \}$

Can anyone help me formulate an explicit and detailed proof of A-4

Peter

Peter

Re: Reflections - Kane's Book

first of all, realize that $\displaystyle \phi(H_\alpha) = H_{\phi(\alpha)}$ is an equality of two sets.

so suppose $\displaystyle y \in \phi(H_\alpha)$. then $\displaystyle y = \phi(x)$ for some $\displaystyle x \in H_\alpha$.

so $\displaystyle (x,\alpha) = 0$ and because $\displaystyle \phi$ is orthogonal, $\displaystyle (\phi(x),\phi(\alpha)) = 0$, that is $\displaystyle (y,\phi(\alpha)) = 0$

so $\displaystyle y \in H_{\phi(\alpha)}$, so $\displaystyle \phi(H_\alpha) \subseteq H_{\phi(\alpha)}$.

now suppose we have $\displaystyle z \in H_{\phi(\alpha)}$, so that $\displaystyle (z,\phi(\alpha)) = 0$.

since $\displaystyle \phi$ is an orthogonal automorphism, it is bijective, and its inverse $\displaystyle \phi^{-1}$ is also orthogonal.

so $\displaystyle (\phi^{-1}(z),\phi^{-1}(\phi(\alpha)) = (z,\phi(\alpha)) = 0$. but $\displaystyle (\phi^{-1}(z),\phi^{-1}(\phi(\alpha)) = (\phi^{-1}(z),\alpha)$.

this means that $\displaystyle \phi^{-1}(z) \in H_\alpha$ so $\displaystyle z = \phi(\phi^{-1}(z)) \in \phi(H_\alpha)$,

which shows that $\displaystyle H_{\phi(\alpha)} \subseteq \phi(H_\alpha)$, thus the two sets are equal.

the second equality, is an equality of functions, so we must show that the two functions have the same value at every point of $\displaystyle E$.

so let $\displaystyle v \in E$. since $\displaystyle \phi$ is surjective, we can write $\displaystyle v = \phi(w)$ for some $\displaystyle w = x + k\alpha$,

where $\displaystyle x \in H_\alpha ,k \in \mathbb{R}$. so then we have:

$\displaystyle \phi s_\alpha \phi^{-1}(v) = \phi s_\alpha \phi^{-1}(\phi(w)) = \phi s_\alpha(w) = \phi s_\alpha(x + k\alpha) = \phi(s_\alpha(x) + ks_\alpha(\alpha))$

$\displaystyle = \phi(x - k\alpha) = \phi(x) - k\phi(\alpha) = s_{\phi(\alpha)}(\phi(x) + k\phi(\alpha)) = s_{\phi(\alpha)}(\phi(x+k\alpha))$

$\displaystyle = s_{\phi(\alpha)}(\phi(w)) = s_{\phi(\alpha)}(v)$ for all $\displaystyle v \in E$.

this shows the two functions are indeed equal on all of $\displaystyle E$.