# Thread: Relection Groups - Kane - Reflection Groups and Invariant Theory

1. ## Relection Groups - Kane - Reflection Groups and Invariant Theory

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. ## Re: Relection Groups - Kane - Reflection Groups and Invariant Theory

first of all, we have a direct sum decomposition:

E = Hα⊕span(α).

so we can write any v in E uniquely as v = x + kα, for some x in Hα, and k in R.

therefore, sα(v) = sα(x+kα) = sα(x) + ksα(α) = = x - kα.

so now let's look a bit more carefully at what "k" is.

note that v.α = x.α + k(α.α). but by the very definition of Hα, x.α = 0. so

v.α = k(α.α), and therefore k = v.α/(α.α).

putting this all together, we get:

sα(v) = x - kα = x + kα - 2kα = v - [2(v.α)/(α.α)]α, which is what (1) states.

so then we compute:

(sα(x)).(sα(y)) = (x - [2(x.α)/(α.α)]α).(y - [2(y.α)/(α.α)]α)

= x.y - 2(x.α)(α.y)/(α.α) - 2(y.α)(x.α)/(α.α) + 4(x.α)(y.α)(α.α)/[(α.α)2]

= x.y - 4(x.α)(y.α)/(α.α) + 4(x.α)(y.α)/(α.α) = x.y, which is what (2) states.

Thanks

Peter

4. ## Re: Relection Groups - Kane - Reflection Groups and Invariant Theory

Thanks for the previous post, but I am having difficulties understanding some fundamentals - possibly I do not have a good intuitive understanding of the geometry of $H_a$ and $span (a)$.

I am having difficulties seeing how $E = H_a \oplus span(a)$

It seems to me that span(a) is just a line - see my diagrammatic understanding in the attachement "Reflections and Reflection Groups"

Hope I have made myself clear?

Do I need to get a simple book on Differential geonetry and do some reading?

Peter

5. ## Re: Relection Groups - Kane - Reflection Groups and Invariant Theory

yes, it IS just a line. and a hyperplane is "all the rest of the dimensions". it's easiest to see what's going on in 3-dimensional space.

let's pick a vector that's non-zero. i'm going to use a special one, just to keep the arithmetic easy. the vector i will choose is α = (1,0,0) in R3.

so what is Hα in this case? well suppose that:

(x,y,z).(1,0,0) = 0. clearly this happens if and only if x = 0. so Hα = {(0,y,z): y,z in R}.

that is: the hyperplane orthogonal to the x-axis is the yz-plane. this should be clear geometrically. and note that we can write any vector in R3 like so:

(x,y,z) = (0,y,z) + x(1,0,0).

in general, a direct sum decomposition works sort of like a basis: V = U⊕W means two things:

1. V = U+W (this is like a basis spanning, we can write any v in V as v = u+w).
2. U∩W = {0} (this is like linear independence, there's no redundancy, so the u and w in v = u+w are unique).

in general, in an inner product space V, for any subspace U, there is a canonical way to find a W so that V = U⊕W (such a W is called a complement for U). and that is to let W = U┴, the orthogonal complement of U, which is defined as {v in V: v.u = 0 for all u in U}.

what one typically does, is find a basis for U, and then orthogonalize it using a process called Gram-Schmidt orthogonalization. then we extend that to a basis for V, and orthogonalize the extended basis (using Gram-Schmidt again). since every basis element is orthogonal to every other, the linear combinations of the basis elements NOT in U, are orthogonal to every linear combination of basis elements IN U. in my example above U = span{(1,0,0)} and U┴ = span{(0,1,0),(0,0,1)} (i cheated and started with a basis that was already orthogonal).

what is happening in your example, is we are choosing a 1-dimensional subspace (a line through the origin) as our "U", in which case U┴ is the hyperplane Hα.

Thank you

Peter

7. ## Re: Relection Groups - Kane - Reflection Groups and Invariant Theory

Just checking a minor point:

"first of all, we have a direct sum decomposition:

E = $H_a \oplus span(a)$.

so we can write any v in E uniquely as v = x + kα, for some x in $H_a$ , and k in R.

therefore, $s_a (v) = s_a (x+ ka) = s_a (x) + k s_a (a) = x - ka$ "

How do we know that:

$s_a(x+ka) = s_a(x) + ks_a(a)$

Kane (see attachement] does not seem to eplicitly define $s_a$ as a linear transformation - as he does with $\ s_H : \ E \rightarrow E$ but maybe he means us to understand this implicitly

Can you clarify?

Peter

8. ## Re: Relection Groups - Kane - Reflection Groups and Invariant Theory

yes, he should mention that reflections ARE linear transformations. in fact, they are more than just mere linear transformations, they are actually linear isometries.

from the definitions he gives, and the observation he makes just below the definitions, it should be clear that:

$s_\alpha = s_{H_\alpha}$, and the map on the right is explicitly said to be linear.

9. ## Re: Relection Groups - Kane - Reflection Groups and Invariant Theory

Thanks ... that makes things clearer and gives me confidence to go on ... great!

Thanks again

Peter