18 NOTES4.2. reflections. These are orthogonal transformations which fix a hy-perplane. For example, switching x and y coordinates is a reflectionthrough the line x = y and “along” the vector (1, −1). Reflection alongthe root αij= ei− ejswitched the i-th and j-th coordinates.Definition 4.2. Suppose that α =(a1,a2, ··· ,an) is any nonzero vec-tor in Rn. Then the perpendicular hyperplane is the set of all vectorsx which are perpendicular to α. In other words x · α = 0, i.e.,a1x1+ a2x2+ ··· + anxn=0We also discussed the equation for the reflection through this hyper-plane.If α is a vector (nonzero) in Rnand H is the perpendicular hyper-plane, then the reflection “through H” (or “along α”) is the mappingwhich sends y ∈ Rnto a point z which is equi-distant to H on the“other side.”H = α⊥α!!!!!!!!!yz"""""""""θBy looking at the picture we decided that the formula must be:z =rα(y)=y −2y · αα · ααI forgot to point out that, when ||α|| =√2 as it is in our case, thisformula becomes really simple:rα(y)=y −(y · α)αBy looking at the picture for A2we see that the reflection of any rootalong any other root is another root. So, I gave the following definitionof a (finite) root system:Definition 4.3. A finite root system is defined a finite set of vectorsin Rn(The set is called Φ, the vectors are called α, β, γ, etc.) so thatthe reflection of any α ∈ Φ along β ∈ Φ is another element of Φ.In the “crystallographic case” the root system is “simply laced” ifall the roots have the same length. I will explain this later.MATH 47A FALL 2008 INTRO TO MATH RESEARCH 19We decided that the following theorem was “obvious” and we didn’ttry to prove it. The correct statement of the theorem is:Theorem 4.4. In R2, a finite root system is given by taking any setof unit vectors so that the angle between any two vectors in the set isπkmwhere m. In general, the lengths of the vectors need not be the samebut the angles must be kπ/m.I should have pointed out in class that, for any root α, −α must bea root since −α is the reflection of α along α .The group generated by these reflections is the dihedral group Dmof order 2m. This is the symmetry group of the regular m-gon. Forexample for A2it is D3= S3, the symmetry group of an equilateraltriangle.Here is a proof:Proof. Take the smallest angle between any two roots. Suppose thatα, β are two roots which form this angle θ. Suppose that β is coun-terclockwise from α. Then −α is also a root and the reflection of −αalong β is a root which is clockwise from β with an angle of θ, i.e., γforms an angle of 2θ with α . If you then reflect −β along γ you get anew root which is 3θ counterclockwise from α. Proceeding in this way,we will go all the way to a and there are two possibilities.(1) Either we hit −α or(2) we jump over −α.But the second case is not possible. If we jump over −α then the root−α will be in the middle of an an angle θ. So, we would get a smallerangle.So, it has to be Case 1. We have to hit −α in a finite number ofsteps, say m steps. Then θ = π/m. !We divide the set of roots into “positive” and “negative” roots:Φ = Φ+!Φ−The way to do this is to choose a hyperplane which does not containany of the roots. Then the roots on one side of the hyperplane arecalled positive and the ones on the other side are called negative. Thisis random but any two choices can b e shown to be equivalent.Definition 4.5. The set of simple roots α1,α2, ·,αnis defined to be aset of positive roots with two properties:20 NOTES(1) The simple roots form a basis for the subspace spanned by allthe roots. In other words every root is a linear combination ofsimple roots.(2) Every positive root is a nonnegative linear combination of sim-ple roots, i.e.,β ="biαiwhere βi≥ 0.These two properties determine the set of simple roots because thereis only one set with this property. In the case of the 2-dimensionalcase, the simple roots are the two which are closest to the