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11. Roots of Lie Algebras 1 11. Roots of Lie algebras 1 11. Roots of Lie algebras With the scalar product defined in terms of the inverse of the Killing form on the Cartan subalgebra, it can be proved that α α> 0, for any root α. It · has been shown that, for a semi-simple Lie algebra, if α is a root, then α is also a root. It can also be shown that 2α is not a root. − [The obvious result [Eα, Eα] = 0 is consistent with this statement, but is not sufficient to prove it, since it is conceivable that Nαα =0 but [Eα β, Eα β] = 0.] + − 6 The only multiples of a root α which are also roots are 0 and α. It is also possible to prove that all non-zero roots are non-degenerate,± i.e. there is only one root vector for each root. The degeneracy of the zero root, which corresponds to the vectors of the Cartan subalgebra, is equal to the rank of the algebra. Suppose α, β are roots, such that α + β is not a root, i.e. [Eα, Eβ] = 0. ′ ′ Then [E−α, Eβ] Eβ−α is proportional to Eβ−α and so [E−α, Eβ−jα] ′ ≡ ≡ Eβ−(j+1)α is proportional to Eβ−(j+1)α. (It is assumed that some normali- sation has been chosen for the root vectors Eα. The normalisation of the ′ vectors Eα is then fixed by their definition.) Since there is only a finite number of roots, there must be a maximum value of j, denoted M, such ′ ′ ′ that [E−α, Eβ−Mα]=0= Eβ−(M+1)α = 0. Now consider [Eα, Eβ−jα] = ′ ⇒ µjEβ−(j−1)α, which defines the factor of proportionality µj, since the normal- ′ isations of Eα and Eα are fixed. It follows that ′ ′ µj+1Eβ−jα = [Eα, Eβ−(j+1)α] ′ = [Eα, [E−α, Eβ−jα]] ′ ′ = [E−α, [Eβ−jα, Eα]] [Eβ−jα, [Eα, E−α]] − ′ − ′ = µj[E−α, Eβ−(j−1)α]+ gα,−α[Hα, Eβ−jα] ′ ′ = µjE + gα, αα (β jα)E , β−jα − · − β−jα which implies µj = µj +(α β jα α)gα, α. This is a recursion relation +1 · − · − for the coefficient µj, with the initial condition µ0 = 0. ′ [This follows from [Eα, Eβ−α]= µ1Eβ = [Eα, [E−α, Eβ]] = [E α, [Eβ, Eα]] [Eβ, [Eα, E α]] = gα, α[Hα, Eβ] − − − − − = gα, αα βEβ = µ = α βgα, α = µ + α βgα, α.] − · ⇒ 1 · − 0 · − The recursion relation has the solution µj = gα,−α (jα β α αj(j 1)/2), ′ ′ · − · − satisfying µ0 = 0. But µM+1Eβ−Mα = [Eα, Eβ−(M+1)α]=0= µM+1 = 0= α β = Mα α/2= M =2α β/α α. ⇒ ⇒ · · ⇒ · · Introductory Algebra for Physicists Michael W. Kirson 11. Roots of Lie algebras 2 To recapitulate, starting from a root β such that β + α is not a root, a sequence of roots β jα was generated, terminating with the root β Mα, where M =2α β/α− α. This sequence is called an α-string of roots.− · · In general, any root γ belongs to an α-string of roots, ranging from γ mα to γ + pα. Here, m and p are non-negative integers, which may be zero.− The “top” of this string is at β = γ + pα, and it ends at β Mα = γ mα, so M = m + p. Hence m + p = M = 2α (γ + pα)/α α−= 2α γ/α−α +2p. Hence · · · · m p =2α γ/α α. − · · This is a major result, with many implications and applications. It will be referred to from now on as the “magic formula”. According to the “magic formula”, for any two roots α, β, the quantity 2α β/α α is an integer. If β is at the top of an α-string (i.e. β + α is not a root,· so p· = 0), then the quantity is non-negative. If α α = 0, then α β =0 ij · · for all β, which implies Pi αig = 0. But this implies det g = 0, which is impossible for semi-simple Lie algebras. If β belongs to an α-string from β mα to β + pα, then the midpoint of − the string is at β (m p)α/2= β (β α/α α)α and the “mirror image” of β on the string− is β −2(β α/α α−)α,· and is· also a root. − · · β (m p)α/2 − 6− sssαα- - s α- α s sss αα β mα β (m p)α× β + pα − − − β (β + pα) (β mα) - − − (p + m)α/2 - This procedure for producing a root from a given root is called a Weyl re- flection, and the set of all such operations constitutes the Weyl group. There can be at most four roots in a string. [Suppose there are five roots, chosen without loss of generality as β 2α, β α,β,β + α, β +2α. Since β +2α β = 2α, and 2α − − − is not a root, and since β +2α + β = 2(β + α), and 2(β + α) is not a root, the root β +2α belongs to a β-string of length 1, so m =0= p and (β +2α) β = 0. Similarly, (β 2α) β = 0, hence β β = 0, which is impossible.]· − · · Introductory Algebra for Physicists Michael W. Kirson 11. Roots of Lie algebras 3 The only possible values of m p in a 4-string are 3 and 1; in a 3- string they are 2 or 0; in a 2-string− they are 1; and in± a 1-string± the only possible value is± 0. So, from the “magic formula”,± 2α β/α α can only take · · on the values 0, 1, 2, 3. ± ± ± It is possible to adopt a simple geometrical picture in which α α is the square of the length of the root α and α β is proportional to the cosine· of the · 2 2 angle between the roots α and β. In fact, cos φαβ =(α β) /(α α)(β β)= rs/4, where r, s are integers, the appropriate values of m· p for· the relevant· α- and β-strings. − 2 [Note that cos φαβ 1, because of the Schwarz inequality (α β)2 (α α)(β ≤β). This inequality also correlates the values of m· p≤for α· - and· β-strings.] − 2 It may be concluded that cos φαβ takes only the values 0, 1/4, 1/2, 3/4, 1, so the only possible angles between two roots are 90◦, 60◦ or 120◦, 45◦ or 135◦, ◦ ◦ 2 30 or 150 , with cos φαβ =1= β = α. Since r/s = β β/α α, it is easy to deduce that the ratio⇒ of lengths± of two roots is 1· if the· angle between them is 60◦ or 120◦, √2 if the angle is 45◦ or 135◦, √3 if the angle is 30◦ or 150◦ and indeterminate if the angle is 90◦. Each Lie algebra can then be represented by a vector diagram of roots in an ℓ-dimensional space, where ℓ is the rank of the algebra. If a basis set of roots α(i) is chosen in the root space, then any root can {(i) } be written as β = Pi qiα , where the expansion coefficients qi are rational numbers. (j) (j) (j) (i) (j) (j) (j) [2β α /α α = P qi2α α /α α is a set of linear · · i · · equations for the qi , with integer coefficients, so the solutions are rational.] { } This is consistent with the earlier comment that real linear combinations of roots constitute a sufficient root space. The scalar products α α and α β · · are also rational numbers. ij ik lj ik lj [α α = Pij αig αj = Pijkl αig gklg αj = Pijkl Pβ αig βkβlg αj = · 2 2 2 Pβ(α β)(β α) = Pβ(α β) = Pβ(m p)β(α α) /4 = · · 2 · − · ⇒ α α = 4/ Pβ(m p)β, which is rational and manifestly pos- itive,· justifying the− use of the term “scalar product”. Then α β =(m p)βα α/2, also rational.] · − · Introductory Algebra for Physicists Michael W. Kirson 11. Roots of Lie algebras 4 An ordering can be introduced into the real root space by choosing an arbitrary basis and defining a root as positive if the first non-zero coefficient in its expansion in terms of the basis is positive. Then α > β if α β > 0. − Since α> 0 α< 0, half of the non-zero roots will be positive. ⇐⇒ − A positive root is said to be simple if it cannot be written as the sum of two positive roots. If two roots α, β are simple, then their difference α β − is not a root. [If α β is a root, then α β > 0 or β α> 0. But α =(α β)+β and −β =(β α)+ α, so− either α or −β is not simple.] − − For the α-string containing β, m p =2α β/α α. But β α is not a root, so m = 0. Therefore α β 0 for− simple roots.· · − · ≤ The set of simple roots is linearly independent. (i) (i) [If the simple roots α were linearly dependent, then Pi ciα = { } (i) 0 for some coefficients ci . Since all α > 0, some of the ci { } (i) {(j)} are negative. Rewrite the dependence as Pi aiα = Pj bjα , (i) (i) (i) where all ai, bj > 0. Then (Pi aiα ) (Pi aiα )=(Pi aiα ) (j) (i) (j) · · (Pj bjα ) = Pi=6 j aibjα α . The left hand side is positive, while the right hand side is· a sum of negative terms.] It can be shown that the number of simple roots is equal to ℓ, the rank of the algebra, so the simple roots form a basis for the real root space ∗.
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