MATH 47A FALL 2008 INTRO TO MATH RESEARCH 154. Roots and reflections4.1. roots of type An−1.Definition 4.1. The roots of type An−1are the vectors in Rnof theformei− ejFirst I had to explain to students what Rnmeans. The standardnotation is:R = the set of all real numbers.Z = the set of all integers.N = the set of all nonnegative integers. Unfortunately, some peoplethink that 0 is not in this set. So, I will remind you each time I usethis symbol.Rn:= {(x1, x2, ··· , xn) |xi∈ R}This represents n-dimensional space. Every point is specified by its ncoordinates.(1) How many roots are there?(2) What is the length of each root?(3) Which roots are perpendicular?(4) What are the angles between the roots?(5) What is the formula for the angle between two vectors?4.1.1. number of roots. There are n(n − 1) roots αijsince there are nchoices for i and after that there are n −1 choices for j (since i, j mustbe distinct).I forgot to say that the set of all roots is denoted Φ. I defined thepositive roots to be the roots αijwhere i > j. The simple roots are theones where i, j differ by 1. These are denotedαi:= αi+1,i, −αi= αi,i+14.1.2. length. OK, this is a dumb question. The roots all have length#X# :=!"x2i=√2.4.1.3. Which roots are ⊥? Two vectors are perpendicular if and onlyif their dot product:X ·Y =n"i=1xiyiis zero. But, the dot product of two vectors αij, αk!is zero if and onlyif the indices i, j, k, " are distinct.16 NOTES4.1.4. formula for angle. You take the second formula for the dot prod-uctX ·Y = ||X|| · ||Y ||cos θand solve for θ:θ = cos−1X ·Y||X|| · ||Y ||The dot product is 0, ±1, ±2 and the denominator is always 2. So, theangles that you get arecos−1(0) = 90◦= π/2cos−1(1/2) = 60◦= π/3cos−1(−1/2) = 120◦= 2π/3cos−1(1) = 0cos−1(−1) = 180◦= π(1) θ = π/2 (= 90◦) or α · β = 0. This means the two roots α, βare perpendicular and this happens when i, j, k, " are distinct.(2) θ = π. Then α, β are parallel and point in opposite directions.Since all roots have the same length the too roots are negativesof each other: α and −α.(3) θ = 0. This means α = β. (They have the same length andpoint in the same direction.)(4) θ = π/3. Then α, β form two sides of an equilateral triangle.(5) θ = 2π/3. Two consecutive (positive) simple roots αi, αi+1havethis angle.4.1.5. Andiagram. The graph Anconsists of n vertices connected byn −1 edges in a straight line. For example, A4stands for the followinggraph:•• • •Each vertex represents a simple root. You draw a line between tworoots if they are not perpendicular. When there is an edge connectingtwo roots, unless otherwise stated, the angle is 2π/3.The root system A2is drawn:• •There are two simple roots forming an angle of 120◦. There are alsothe negatives of these roots. On Wednesday we looked at the list of allroots in A2:(1, −1, 0), (−1, 1, 0), (0, 1, −1), (0, −1, 1), (1, 0, −1), (−1, 0, 1)MATH 47A FALL 2008 INTRO TO MATH RESEARCH 17None of these roots are ⊥. So the picture must be:These are vectors in R3. But I only drew the plane given by:x + y + z = 0Recall that the equation of a plane in R3is:ax + by + z = dwhere (a, b, c) is a vector which is perpendicular to the plane and d tellsyou how far away the plane is from the origin. In our case all rootshave coordinates adding up to zero. So they lie in the n−1 dimensionalhyperplane given by the equation"xi= 0The perpendicular vector is (1, 1, 1, ··· , 1).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 p oint 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