# Thread: Reflections - Kane's Book

1. ## 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 $\ell$ dimensional Euclidean space ie $E = R^{\ell}$ where $R^{\ell}$ has the usual inner product (x,y).

In defining reflections with respect to vectors Kane writes:

" Given $0 \ne \alpha \in E$ let $H_{\alpha}\subset E$ be the hyperplane

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

We then define the reflection $s_{\alpha} : E \longrightarrow E$ by the rules

$s_{\alpha} \cdot x = x$ if $x \in H_{\alpha}$

$s_{\alpha} \cdot \alpha = - \alpha$ "

Then Kane states that the following property follows:

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

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

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

Can someone explicitly and formally demonstrate A-4?

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

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

And I assume that

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

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

Peter

Peter

2. ## Re: Reflections - Kane's Book

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

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

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

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

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

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

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

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

which shows that $H_{\phi(\alpha)} \subseteq \phi(H_\alpha)$, 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 $E$.

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

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

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

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

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

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