11 Properties of Roots

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1 Properties of Roots In this section, we fill in the missing details when deriving the properties of the roots of a simple Lie algebra g. We assume that a Cartan algebra h g of simultaneously diagonalizable elements has been chosen, and that g has been decomposed into eigenspaces so that, for any H h, we have g = h g α, α R [H,X α ] = α(h)x α (94) for a finite set of roots R h. Since g is simple, the Killing form B on g is nondegenerate. As argued in the previous section, B(g α,g β ) = 0 = B(g α,h) (95) for α+β 0 (and α 0, since 0 R). Recall further that the Jordan Chevalley decomposition shows that any linear operator can be divided uniquely into the sum of a diagonalizable operator and a nilpotent operator. Thus, nonzero elements of h can not be nilpotent. We begin with a simple lemma about traces. Since the trace is cyclic, so that, for instance, Thus, which implies that tr(xy) = tr(yx), (96) tr(xyz) = tr(zxy). (97) tr([x,y]z) = tr(xyz YXZ) = tr(xyz XZY) = tr(x[y,z]) (98) Preferred Cartan Elements B([X,Y]Z) = B(X[Y,Z]). (99) Since B is nondegenerate, for any X α g α, there must be some Y α g such that B(X α,y α ) 0, (00) and (95) now implies that Y α g α. Thus, if α is a root, so is α, and we have [H,Y α ] = α(h)y α. (0) Furthermore, since α 0, there must be some H h such that, using (99, B([X α,y α ],H) = B(X α,[y α,h]) = B(X α,α(h)y) = α(h)b(x α,y α ) 0 (0) 34

2 so that We now claim that α(h α ) 0. If not, then H α = [X α,y α ] 0. (03) [H α,x α ] = 0 = [H α,y α ]. (04) Consider now any representation of h, and suppose that the image of H α, which we will also call H α, has an eigenspace V = {v : H α v = λv} (05) for some fixed eigenvalue λ. Since X α and Y α commute with H α, they both take eigenvectors to eigenvectors, that is, they take V to itself. So V itself is a representation of the subalgebra generated by {H α,x α,y α }. But this means that, as matrices acting on V, we must have [X α,y α ] = H α = λ (06) since representations preserve commutators (by definition), and H α is a multiple of the identity matrix when acting on V. Taking the trace of both sides immediately forces λ = 0. Thus, the only eigenvalue of H α is 0, which means that H α is nilpotent. But the only nilpotent element of h is 0, and H α 0. This contradiction establishes the claim. We can now rescale X α and Y α if necessary to obtain α(h α ) =. Although this construction does not determine X α and Y α uniquely, H α is uniquely determined. These special elements of h will be referred to as preferred Cartan elements. Root Angles As discussed in the previous section, {H α,x α,y α } form a standard basis for sl(,r), the split real form of su(). Thus, all representations of this Lie subalgebra of g have half-integer eigenvalues, and in particular, β(h α ) Z for all roots β. The restriction of g to real linear combinations of H α, X α, Y α for all α R is therefore a real subalgebra of g, and is in fact the split real form of g. As before, choose T α h to be the element determined by for all H h. We need to verify that T α is a multiple of H α. Using (0) and (03, we have so that in particular α(h) = B(T α,h) (07) B(H α,h) = α(h)b(x α,y α ) (08) B(H α,h α ) = α(h α )B(X α,y α ) = B(X α,y α ) (09) 35

3 which is furthermore nonzero by assumption. Thus, B(H α,h) = B(T α,h)b(h α,h α ) = B ( B(H α,h α )T α,h ) (0) for all H h, and we conclude that T α = H α B(H α,h α ). as claimed previously. It now follows immediately that α(h β ) = B(T α,h β ) = B(H α,h β ) B(H α,h α ) Z leading to the angles and ratios discussed in the previous section. Multiples of Roots, and Multiplicity It remains to show that the only multiples of a root α that are roots are ±α, and that each root only occurs once, that is, that g α =. We have shown that {H α,x α,y α } is a standard basis for sl(,r), so that X α, Y α act as raising and lowering operators, respectively, for H α. Suppose that Z g α, so that [H α,z] = α(h α )Z = Z. () We know that all representations of sl(, R) consist of integer or half-integer ladders. So how does sl(,r) act on Z? Moving down the ladder, Z 0 = [Y α,z] () is an element of h, since Y α decreases all the eigenvalues of Z by α, that is [H,Z 0 ] = [ H,[Y α,z] ] = [ [H,Y α ],Z ] + = [ Y α,[h,z] ] = ( α+α)[y α,z] = 0 (3) for any H h. Since B is positive-definite on h, we can expand Z 0 as Z 0 = zh α +H (4) where But so that B(H α,h ) = 0. (5) B(H α,h ) = 0 = B(T α,h ) = 0 = α(h ) = 0 (6) α(z 0 ) = zα(h α )+0 = z. (7) 36

4 Putting this all together, we have [X α,z 0 ] = [Z 0,X α ] = α(z 0 )X α = zx α. (8) But moving back up the ladder has to yield a multiple of Z, and we conclude that Z is itself a multiple of X α, thus confirming that g α =. The argument against any other multiples of α other than ±α being roots is similar. If cα is a root, then choosing Z g cα leads to [H α,z] = (cα)(h α )Z = cz (9) which forces c to be half-integer. Suppose c Z. Then we can move up or down the ladder from Z to some element Z g α. By the argument above, Z must be a multiple of X α. But then the ladder collapses, and c = ±. In particular, α can not be a root. But now α also can not be a root, which rules out half-integer values of c. 37

5 Dynkin Diagrams We have seen that any simple Lie algebra can be decomposed into a Cartan subalgebra h and its eigenspaces g α. Furthermore, the commutators of elements in g ±α determine a collection of preferred elements H α h, and the angles between the H α are tightly constrained. We show in this section that these constraints in turn impose additional constraints on the possible types of simple Lie algebras. The roots α h share the same angles as the corresponding H α. We first reduce the number of roots to an independent set, {α i }. Any other root can, of course be expressed as a linear combination of our chosen basis. The symmetry of the roots allows us to choose the basis so that the nonzero coefficients are always either strictly positive or strictly negative; the two types of roots are then called positive and negative roots, respectively, and the basis elements are called simple roots. For each root α, one of ±α is positive. The positive roots have the property that they are all sums, not differences, of the simple roots; in fact, the coefficients turn out to be positive integers. It is not hard to see that the positivity property means that the angle between any two simple roots must be obtuse, rather than acute. Recall that all pairs of roots satisfy α β α α Z so that the angle θ between the roots must satisfy 4cos θ {0,,,3,4}. Since the simple roots α i R n are independent, they satisfy and the angles θ ij between roots must satisfy 0 > α i α j α i α j Z (0) 4cos θ ij = 0,,,3. () A Coxeter graph is a graph with one dot representing each simple root, and with each pair of dots connected by 4cos θ ij lines. Equation (0) also constrains the relative magnitudes of (non-orthogonal)roots; a Dynkin diagram is a Coxeter graph with the addition of arrows on the double and triple connections pointing from the longer root to the shorter. We now derive several restrictions on the possible Dynkin diagrams. To do so, the only assumptions we need are that the α i R n satisfy the conditions above. Most of these conditions apply both to Coxeter graphs and Dynkin diagrams. There are at most k connections between k simple roots. For the purpose of this assertion, a connection refers to a pair of roots with at least one line connecting them; multiple lines between a given pair of roots still count as a single connection. 38

6 We let α = α i α i 0. () where the inequality follows from the independence of the simple roots. The magnitude of α is given by 0 < α α = i,j α i α j α i α j = k + i<j α i α j α i α j = k + i<j cosθ ij. (3) But either α i α j = 0, in which case there are no connections, or 4cos θ ij {,,3}, (4) in which case Thus, cosθ ij. (5) 0 < k # of connections (6) which is what we were trying to show. An immediate corollary is that there are no closed loops in a Coxeter graph. There are at most 3 lines at each point. Suppose that a simple root α is connected to k other simple roots α i. Since there are no cycles, none of the α i can be connected to each other. Thus, i j = α i α j = 0. (7) Since the simple roots are independent, we can extend {α i } to an orthogonal basis {α 0,α i } of the span α,α i. We can expand α in this basis, yielding α = 0 α α i α i α i α i. (8) Since α is independent of the α i, we must have α α 0 0 (9) so that α = 0 (α α i ) α i α i > (α α i ) α i. (30) 39

7 Thus, # of lines = 4 cos θ i = 4 so that the number of lines is strictly less than 4, as claimed. Simple chains of roots can be replaced by a single root. (α α i ) α α i < 4 (3) Suppose that there is a chain of single lines connecting α i to α i+ for i k, in which case We claim that the entire chain can be replaced by α i α i+ = α i α i, (3) α α =... = α k α k = Q. (33) α = α i (34) with the result still being a valid Coxeter graph. We first compute α α = i,j α i α j = i α i α i + i<j α i α j = k α i α i + α i α i+ = kq (k )Q = Q (35) so that α has the same magnitude as each of the α i. Furthermore, if β is any other root, β can be connected to at most one of the α i, in which case β α = β α i ; (36) if not, β α = 0. In either case, all of the conditions on the original roots continue to hold if the k roots α i are replaced by the single root α, as claimed. Allowed Diagrams We can use these three properties to rule out several Coxeter graphs. First of all, we consider only connected graphs, as only such graphs correspond to simple Lie algebras. The simplest graphs correspond to n roots, connected in a single chain by single lines. This diagram is 40

8 of type A; since there are n roots, h is n-dimensional. The corresponding Lie algebras are therefore called a n. All of the roots have the same magnitude. Consider instead a single chain, but with at least one double line. If there are more than one, then collapsing the roots in between must yield an allowable graph. But such a graph has four lines at a single point, which is not allowed. Thus, there is at most one double line, and hence exactly one. In this case, the roots have one of two magnitudes, depending on which side of the double connection they are on. Now suppose that there is a branch point in the graph. Again, there can not be more than one, since otherwise the chain between them could be collapsed, again resulting in a graph with four lines. Similarly, there can not be both a branch point and a double line. We will discuss these cases in further detail below; the corresponding Lie algebras are of types B F. Finally, if two roots are connected by three lines, then no other lines are possible. Thus, there is only one Coxeter graph with three lines. This diagram is of type G; since there are two roots, h is -dimensional. The corresponding Lie algebra is therefore called g. Diagrams with a double link. Consider two simple chains, p roots α m on one side of a double link, and q roots β k on the other, with the double link connecting α p and β q. Since all but one of the links are single lines, we know that α m α m = P, β n β n = Q (37) and we can assume without loss of generality that there are p short roots and q long roots, so that Q = P. We also know that with a similar relation for the roots β k. Setting α = α m α m+ α m α m =, (38) p m α q m α m, β = k β k β k, (39) and using similar techniques as in the previous calculations, we can compute α α = p p m m(m+) p ( = p + m m(m+) ) p = p m = p p(p ) = p(p+) (40) 4

9 and similarly We also know that so that But or, in other words, (α p β q ) = β β = q(q +). (4) 4cos θ = 4 (α p β q ) P Q = (4) ( pαp P qβ ) q = p q (α p β q ) = Q P Q p q. (43) (α β) < (α α)(β β) (44) p q < p(p+)q(q +) 4 (45) = pq < pq +p+q + (46) = pq < p+q + (47) = (p )(q ) <. (48) There are three ways to satisfy (48). If p =, then there is only one short root; these Lie algebras are of type B, and denoted b p+. If q =, then there is only one long root; these Lie algebras are of type C, and denoted c q+. Finally, if p = q =, we get a single, exceptional case, of type F, and denoted f 4. Diagrams with a branch point. Consider now three simple chains meeting at a branch point ψ, with p roots α m in one chain, q roots β k in the second, and r roots γ l in the third. Since all of the links are single lines, we know that As above, set α = α m α m = β k β k = γ l γ l = ψ ψ = Q. (49) p m α q m α m, β = k β r k β k, γ = l γ l γ l. (50) The magnitudes of α, β, and γ can be computed as in the preceding case, taking into account that the chains (excluding ψ) now contain p, q, and r roots, respectively, yielding α α = p(p ), β β = q(q ), γ γ = r(r ). (5) 4

10 The roots α m, β k, γ l are mutually orthogonal; as before, we can complete this set of roots to an orthogonal basis by adding ψ 0, with ψ ψ 0 0. Expanding ψ in terms of this basis, we get and therefore ψ = p ψ ψ > q ψ α m α m + α m α m p (ψ α m ) α Q m + ψ β k r ψ γ l β k + γ l + ψ ψ 0 ψ 0 (5) β k β k γ l γ l ψ 0 ψ 0 q (ψ β k ) r (ψ γ l ) β Q k + γ Q l. (53) In other words, if θ α is the angle between the vectors ψ and α, etc., then But Thus, cos θ α +cos θ β +cos θ γ <. (54) (α ψ) = (p ) (α p ψ) Q = (p ) (ψ ψ)cos θ p = 4 (p ) ψ. (55) (α ψ) α ψ = (p ) /4 p(p )/ = p p = ( p ) (56) and similarly for β and γ, which, using (54), yields 3 ( p + q + ) < (57) r or in other words p + q + r >. (58) There are several ways to satisfy (58). If q = r =, we get the diagrams of type D, and the Lie algebras denoted d p+. The only other distinct possibilities are r =, q = 3, and p {3,4,5}. These three exceptional cases belong to class E, and are denoted e 6, e 7, and e 8. Special Cases The four infinite families correspond to known symmetry groups. In terms of their Lie algebras, we have a n = su(n+) = su(n+,c), (59) b n = so(n+) = su(n+,r), (60) c n = sp(n) = su(n,h), (6) d n = so(n) = su(n,r). (6) 43

11 There are similar correspondences for four of the five exceptional cases, namely but the last case is somewhat different, namely which at the group level says that f 4 = su(3,o), (63) e 6 = sl(3,o) = su(3,c O), (64) e 7 = sp(6,o) = su(3,h O), (65) e 8 = su(3,o O), (66) g = Der(O), (67) G = Aut(O), (68) that is, G is the group of automorphisms of the octonions. We will discuss all of these correspondences further below. The allowable Dynkin diagrams can easily be found online. Writing out the first few cases, there are several that overlap. For instance, Somewhat more unexpectedly, we have as well as su() = a = b = so(3), (69) so(5) = b = c = su(,h). (70) su(4) = a 3 = d3 = so(6), (7) so(4) = d = a a = so(3) so(3) (7) showing explicitly that d, whose Dynkin diagram consists of two disconnected points, is not simple. To avoid these repetitions, one normally assumes that n > for b n, that n > for c n, and that n > 3 for d n. Finally, the Lie algebras e n correspond to the the cases described above by the parameter values (p,q,r) = (n 3,3,) with n = 6,7,8. We could in principle consider the cases n = 4,5 as (also) being of type E. We would then have the duplications e 5 = d5 = so(0), (73) e 4 = a4 = su(5), (74) so this normally not done. However, it is worth pointing out that the nested sequence of Lie algebras so(0) su(5) e 6 e 7 e 8 (75) has a long history in physics, with each of the corresponding Lie groups being considered as candidates for the symmetry group of a so-called Grand Unified Theory, combining three of the four fundamental forces of nature, namely electromagnetism and the strong and weak nuclear forces. 44

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