# Euclidean Reflection Groups _ Kane's text

1. Mar 28, 2012

### Math Amateur

I am reading Kane - Reflection Groups and Invariant Theory and need help with two 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 two properties follow:

(1) $s_{\alpha} \cdot x = x - [2 ( x, \alpha) / (\alpha, \alpha)] \alpha$ for all $x \in E$

(2) $s_{\alpha}$ is orthogonal, ie $( s_{\alpha} \cdot x , s_{\alpha} \cdot y ) = (x,y)$ for all $x, y \in E$

I would appreciate help to show (1) and (2) above.

Peter

2. Mar 28, 2012

### morphism

Note that $H_\alpha$ is codim 1 in E with perp spanned by $\alpha$. So any linear operator defined on E is completely determined by its action on $H_\alpha$ and $\alpha$. In particular, there is a unique linear operator satisfying the defining conditions for $s_\alpha$. Since the mapping defined by the RHS of (1) is linear and satisfies these conditions as well, it must be $x \mapsto s_\alpha x$.

Of course what's going on geometrically is that $s_\alpha$ is simply reflection in the hyperplane $H_\alpha$. Try this out in $R^2$, with $H_\alpha$ a line through the origin that is perpendicular to $\alpha$: what is the formula for reflecting about $H_\alpha$?

As for (2), it suffices to check that it holds in the 3 cases: (a) $x,y \in H_\alpha$; (b) $x\in H_\alpha$ and $y=\alpha$; and (c) $x=y=\alpha$. Or you could use (1).