Lecture 16 - Dynkin Diagrams from Lie Algebras, and Vice Versa (Schuller's Geometric Anatomy of Theoretical Physics)

September 9, 2017 | Author: Simon Rea | Category: Lie Algebra, Lie Groups, Vector Space, Algebra, Mathematical Structures

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Lecture notes for lecture 16 of Frederic Schuller's Lectures on the Geometric Anatomy of Theoretical Physics. Conte...

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16 16.1

Dynkin diagrams from Lie algebras, and vice versa The simplicity of sl(2, C)

We have seen that the non-zero structure constants of T( 1 0 ) SL(2, C) are 01 C 212 = 2,

C 313 = −2,

C 123 = 1,

plus those related by anti-symmetry. Proposition 16.1. Two Lie algebras A and B are isomorphic if, and only if, there exists a basis of A and a basis of B in which the structure constants of A and B are the same. Since we have already proved that Te G ∼ =Lie alg L(G) for any Lie group G, we can deduce the existence of a basis {X1 , X2 , X2 } of sl(2, C) with respect to which the structure constants are those listed above. In other words, we have [X1 , X2 ] = 2X2 , [X1 , X3 ] = −2X3 , [X2 , X3 ] = X1 . In this basis, the Killing form of sl(2, C) has components κij = C min C njm , with all indices ranging from 1 to 3. Explicitly, we have κ11 = C m1n C n1m n 2 n 3 n  = C 1 1n C 11 + C 1n C 12 + C 1n C 13



= C 212 C 212 + C 313 C 313 = 8. Since κ is symmetric, we only need to determine κij for i ≤ j. By writing the components in a 3 × 3 array, we find   8 0 0   [κij ] = 0 −8 0 , 0 0 8 which is just shorthand for κ(X1 , X1 ) = 8,

κ(X2 , X2 ) = −8,

κ(X3 , X3 ) = 8,

and κ(Xi , Xj ) = 0 whenever i 6= j. Proposition 16.2. The Lie algebra sl(2, C) is semi-simple. Proof. Since the diagonal entries of κ are all non-zero, the Killing form is non-degenerate. By Cartan’s criterion, this implies that sl(2, C) is semi-simple.

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Remark 16.3. There is one more thing that can be read off from the components of κ, namely, that it is an indefinite form, i.e. the sign of κ(X, X) can be positive or negative depending on which X ∈ sl(2, C) we pick. A result from Lie theory states that the Killing form on the Lie algebra of a compact Lie group is always negative semi-definite, i.e. κ(X, X) is always negative or zero, for all X in the Lie algebra. Hence, we can conclude that SL(2, C) is not a compact Lie group. In fact, sl(2, C) is more than just semi-simple. Proposition 16.4. The Lie algebra sl(2, C) is simple. Recall that a Lie algebra is said to be simple if it contains no non-trivial ideals, and that an ideal I of a Lie algebra L is a Lie subalgebra of L such that ∀ x ∈ I : ∀ y ∈ L : [x, y] ∈ I. Proof. Consider the ideal of sl(2, C) I := {αX1 + βX2 + γX3 | α, β, γ restricted so that I is an ideal}. Since the bracket is bilinear, it suffices to check the result of bracketing an arbitrary element of I with each of the basis vectors of sl(2, C). We find [αX1 + βX2 + γX3 , X1 ] = −2βX1 + 2γX3 , [αX1 + βX2 + γX3 , X2 ] = 2αX2 − γX1 , [αX1 + βX2 + γX3 , X3 ] = −2αX3 + βX1 . We need to choose α, β, γ so that the results always land back in I. Of course, we can choose α, β, γ ∈ C and α = β = γ = 0, which correspond respectively to the trivial ideals sl(2, C) and 0. If none of α, β, γ is zero, then you can check that the right hand sides above are linearly independent, so that I contains three linearly independent vectors. Since the only n-dimensional subspace of an n-dimensional vector space is the vector space itself, we have I = L. Thus, we are left with the following cases: i) if α = 0, then I ⊆ spanC ({X2 , X3 }) and hence we must have β = γ = 0 as well; ii) if β = 0, then I ⊆ spanC ({X1 , X3 }), hence we must have α = 0, so that in fact I ⊆ spanC ({X3 }), and hence γ = 0 as well; iii) if γ = 0, then I ⊆ spanC ({X1 , X2 }), hence we must have α = 0, so that in fact I ⊆ spanC ({X2 }), and hence β = 0 as well. In all cases, we have I = 0. Therefore, there are no non-trivial ideals of sl(2, C).

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16.2

The roots and Dynkin diagram of sl(2, C)

By observing the bracket relations of the basis elements of sl(2, C), we can see that H := spanC ({X1 }) is a Cartan subalgebra of sl(2, C). Indeed, for any h ∈ H, there exists a ξ ∈ C such that h = ξX1 , and hence we have ad(h)X2 = ξ[X1 , X2 ] = 2ξX2 , ad(h)X3 = ξ[X1 , X3 ] = −2ξX3 . Recall that in the section on Lie algebras, we re-interpreted these eigenvalue equations in terms of functionals λ2 , λ3 ∈ H ∗ λ2 :

H− →C

λ3 :

ξX1 7→ 2ξ,

H− →C ξX1 7→ −2ξ

whereby ad(h)X2 = λ2 (h)X2 , ad(h)X3 = λ3 (h)X3 . Then, λ2 and λ3 are called the roots of sl(2, C), so that the root set is Φ = {λ2 , λ3 }. Of course, we are mainly interested in a subset Π ⊂ Φ of fundamental roots, which satisfies i) Π is a linearly independent subset of H ∗ ; ii) for any λ ∈ Φ, we have λ ∈ span,N (Π). We can choose Π := {λ2 }, even though Π := {λ3 } would work just as well. Since |Π| = 1, the Weyl group is generated by the single Weyl transformation sλ2 : HR∗ → HR∗ µ 7→ µ − 2

κ∗ (λ2 , µ) λ2 . κ∗ (λ2 , λ2 )

Recall that we can recover the entire root set Φ by acting on the fundamental roots with Weyl transformations. Indeed, we have sλ2 (λ2 ) = λ2 − 2

κ∗ (λ2 , λ2 ) λ2 = λ2 − 2λ2 = −λ2 = λ3 , κ∗ (λ2 , λ2 )

as expected. Since there is only one fundamental root, the Cartan matrix is actually just a 1 × 1 matrix. Its only entry is a diagonal entry, and since sl(2, C) is simple, we have C = (2). The Dynkin diagram of sl(2, C) is simply

Hence, with reference to the Cartan classification, we have A1 = sl(2, C).

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16.3

Reconstruction of A2 from its Dynkin diagram

We have seen an example of how to construct the Dynkin diagram of a Lie algebra, albeit the simplest of this kind. Let us now consider the opposite direction. We will start from the Dynkin diagram

We immediately see that we have two fundamental roots, i.e. Π = {π1 , π2 }, since there are two circles in the diagram. The bond number is n12 = 1, so the two fundamental roots have the same length. Moreover, by definition 1 = n12 = C12 C21 and since the off-diagonal entries of the Cartan matrix are non-positive integers, the only possibility is C12 = C21 = −1, so that we have   2 −1 C= . −1 2 To determine the angle ϕ between π1 and π2 , recall that 1 = n12 = 4 cos2 ϕ, and hence | cos ϕ| = 21 . There are two solutions, namely ϕ = 60◦ and ϕ = 120◦ . y cos x

1 1 2

360◦

120◦ 0 − 12

60◦

180◦

x

−1

By definition, we have cos ϕ = and therefore 0 > C12 = 2

κ∗ (π1 , π2 ) , |π1 | |π2 |

κ∗ (π1 , π2 ) |π1 | |π2 | cos ϕ |π2 | =2 ∗ =2 cos ϕ. ∗ κ (π1 , π1 ) κ (π1 , π1 ) |π1 |

It follows that cos ϕ < 0, and hence ϕ = 120◦ . We can thus plot the two fundamental roots in a plane as follows.

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π2

π1

We can determine all the other roots in Φ by repeated action of the Weyl group. For instance, we easily find that sπ1 (π1 ) = −π1 and sπ2 (π2 ) = −π2 . We also have sπ1 (π2 ) = π2 − 2

κ∗ (π1 , π2 ) π1 = π2 − 2(− 12 )π1 = π1 + π2 . κ∗ (π1 , π1 )

Finally, we have sπ1 +π2 (π1 + π2 ) = −(π1 + π2 ). Any further action by Weyl transformations simply permutes these roots. Hence, we have Φ = {π1 , −π1 , π2 , −π2 , π1 + π2 , −(π1 + π2 )} and these are all the roots. π2

π1 + π2

−π1

π1

−π2

−(π1 + π2 )

Since H ∗ = spanC (Π), we have dim H ∗ = 2, thus the dimension of the Cartan subalgebra is also 2. Since |Φ| = 6, we know that any Cartan-Weyl basis of the Lie algebra A2 must have 2 + 6 = 8 elements. Hence, the dimension of A2 is 8. To complete our reconstruction of A2 , we would now like to understand how its bracket behaves. This amounts to finding its structure constants. Note that since dim A2 = 8, the structure constants C kij consist of 83 = 512 complex numbers (not all unrelated, of course). Denote by {h1 , h2 , e3 , . . . , e8 } a Cartan-Weyl basis of A2 , so that H = spanC ({h1 , h2 }) and the eα are eigenvectors of every h ∈ H. Since A2 is simple, H is abelian and hence [h1 , h2 ] = 0

C k12 = C k21 = 0,

∀ 1 ≤ k ≤ 8.

To each eα , for 3 ≤ α ≤ 8, there is an associated λα ∈ Φ such that ∀ h ∈ H : ad(h)eα = λα (h)eα .

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In particular, for the basis elements h1 , h2 , [h1 , eα ] = ad(h1 )eα = λα (h1 )eα , [h2 , eα ] = ad(h2 )eα = λα (h2 )eα , so that we have C 11α = C 21α = 0, C α1α = λα (h1 ),

∀ 3 ≤ α ≤ 8,

C 12α = C 22α = 0, C α2α = λα (h2 ),

∀ 3 ≤ α ≤ 8.

Finally, we need to determine [eα , eβ ]. By using the Jacobi identity, we have [hi , [eα , eβ ]] = −[eα , [eβ , hi ]] − [eβ , [hi , eα ]] = −[eα , −λβ (hi )eβ ] − [eβ , λα (hi )eα ] = λβ (hi )[eα , eβ ] + λα (hi )[eα , eβ ] = (λα (hi ) + λβ (hi ))[eα , eβ ], that is, ad(hi )[eα , eβ ] = (λα (hi ) + λβ (hi ))[eα , eβ ]. If λα + λβ ∈ Φ, we have [eα , eβ ] = ξeγ for some 3 ≤ γ ≤ 8 and ξ ∈ C. Let us label the roots in our previous plot as λ3

λ4

λ5

λ6

λ7

λ8

π1

π2

π1 + π2

−π1

−π2

−(π1 + π2 )

Then, for example ad(h)[e3 , e4 ] = (π1 + π2 )(h)[e3 , e4 ], and hence [e3 , e4 ] is an eigenvector of ad(h) with eigenvalues (π1 + π2 )(h). But so is e5 ! Hence, we must have [e3 , e4 ] = ξe5 for some ξ ∈ C. Similarly, [e5 , e7 ] = ξe3 , and so on. If λα + λβ ∈ / Φ, then in order for the equation above to hold, we must have either [eα , eβ ] = 0 (so both sides are zero), or λα (h) + λβ (h) = 0 for all h, i.e. λα + λβ = 0 as a functional. In the latter case, we must have [eα , eβ ] ∈ H. This follows from a stronger version of the maximality property of the Cartan subalgebra H of a simple Lie algebra L, namely that  ∀ h ∈ H : [h, x] = 0 ⇒ x ∈ H. Summarising, we have [eα , eβ ] =

   ξeγ

if λα + λβ ∈ Φ

∈H   0

if λα + λβ = 0 otherwise

and these relations con be used to determine the remaining structure constants of A2 .

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