LIE ALGEBRAS PROF. MAXIM NAZAROV

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1 LIE ALGEBRAS PROF. MAXIM NAZAROV 1. Definition and Examples Any Lie Algebra is a vector space (with extra features). Revision: See Moodle. Definition: Let F be any field (say F = C, which is algebraically closed, which means that any polynomial a o x n + a 1 x n a n, with a o, a 1,..., a n C, has a root). Then a Lie algebra is a vector space g with an operation g g g; (X, Y ) X, Y ] called a Lie bracket, such that: (a) Antisymmetry: Y, X] = X, Y ] (b) Bilinearity: for all a,b F and for all X,Y,Z g (i) ax + by, Z] = ax, Z] + by, Z] (ii) X, ay + bz] = ax, Y ] + bx, Z]. (c) The Jacobi Identity: for all X,Y,Z g, X, Y, Z] ] + Y, Z, X] ] + Z, X, Y ] ] = 0. Remark (Comments on the Axioms). 1. Axiom (a) can be replaced by a (weaker) property X, X] = 0 for all X g. Indeed, (a) X, X] = X, X] for X = Y X, X] = 0 (provided characteristic F 2). Conversely, suppose we know X, X] = 0 for all X g. Then X + Y, X + Y ] = 0 X, Y ] + Y, X] + X, X] + Y, Y ] = 0 X, Y ] + Y, X] = 0 2. In axiom (b) it is enough to assume either (i), or (ii), only. For example, let us show (i) (ii). X, ay + bz] = ay + bz, X] by (a) = ay, X] bz, X] by (i) = ax, Y ] + bx, Z] by (a) Similarly (ii) (i) 3. Axiom (c) holds automatically if dimg = 0, 1, 2. Now taking these case by case we see, 1

2 2 PROF. MAXIM NAZAROV dim g = 0 g = {0}. So we obtain, 30, 0, 0]] = 0 but 0, X] = 0 for all X g and 00, X] = 00, X] = 0. dim g = 1 g = FX with X 0. Now let Y and Z be defined as, Y = ax, Z = bx. So we obtain, X, ax, bx] ] + ax, bx, X] ] + bx, X, ax] ] = 3ab X, X, X] ] dim g = 2. So g is spanned by two vectors, say by X, Y. Then Z = ax + by. X, Y, ax + by ] ] + Y, ax + by, X] ] + ax + by, X, Y ] ] = 0 = a X, Y, X] ] + b X, Y, Y ] ] + a Y, X, X] ] + b Y, Y, X] ] + a X, X, Y ] ] + b Y, X, Y ] ] = 0 Example 1.1 (Abelian Lie Albgebra). Let g be any vector space with X, Y ] 0 for all X, Y g. Then axioms (a), (b) and (c) hold trivially. Example 1.2. Let g = Mat n C = {n n matrices over C}, then dim g = n 2. We define X, Y ] = XY Y X for all X, Y g. Then Y, X] = Y X XY = X, Y ], which is axiom (a). (b) is also easy to check, so I leave it as an exercise. Checking (c), for X, Y, Z g we see X(Y Z ZY ) (Y Z ZY )X + Y (ZX XZ) (ZX XZ)Y + Z(XY Y X) (XY Y X)Z = 0 Example 1.3 (Vector Product). Let g = R 3, where F = R. If we let u, v g then we can define u, v] = u v (the vector cross product). u, v] is a Lie bracket! (a) u v = v u (b) Known (c) please check using the identity v (u w) = (v w)u (v u)w Definition: Take any Lie algebra g. Then a subset h g is a Lie subalgebra if: (a) h is a vector subspace of g, (b) for all X, Y h we have X, Y ] h (i.e. h is closed w.r.t. the Lie bracket).

3 LIE ALGEBRAS 3 Remark. Then h is a Lie algebra (in its own right) when using,] g as the Lie bracket. This is because the Lie Bracket in g is a mapping,, ] g : h h h. As the axioms (a), (b), (c) of a Lie algebra hold for all X, Y, Z g then they will also hold for all X, Y, Z h. Example 1.4 (as in Example 1.2). Let g = Mat n (C) and define X, Y ] = XY Y X. This is the general Lie algebra gl n (C). Example 1.5. The special linear Lie algebra sl n (C) gl n (C) where sl n (C) is the set, {X Mat n (C) trx = 0}. i.e. if we describe X as X = x ij ] n i,j=1 then the condition for tr(x) = 0 can be expressed as x x nn = 0. Claim. The Lie algebra sl n (C) is a Lie subalgebra in gl n (C). Proof. (a) Let X, Y sl n (C) then the tr(x) = 0 = tr(y ). By the linearity of the trace we have tr(x + Y ) = tr(x) + tr(y ) = = 0. Thus X + Y sl n (C). Also if X sl n (C) we have tr(zx) = z tr(x) = z 0 = 0 for all z C, so zx sl n (C). Hence sl n (C) is a vector space of gl n (C). (b) Now considering the Lie Bracket. Let X, Y sl n (C) then which gives us that X, Y ] sl n (C). trx, Y ] = tr(xy ) tr(y X) = 0 Example 1.6. gl n (C) sl n (C) so n (C) = {X Mat n (C) X t = X}. This is the orthogonal Lie algebra. (1a) (1b) tr(x t ) = tr(x) tr( X) = tr(x) Subtracting (1b) from (1a) implies tr(x) = 0 and hence X sl n (C). Claim. so n (C) is a Lie subalgebra of sl n (C). Proof. (a) We have that X, Y so n (C) are antisymmetric (X + Y ) t = X t + Y t = X Y = (X + Y ), and hence X + Y so n (C). For any z C we have,

4 4 PROF. MAXIM NAZAROV (zx) t = zx t = zx which implies that zx so n (C). Hence so n (C) is a vector subspace of sl n (C). (b) Now we examine the Lie Bracket. For X, Y so n (C) X, Y ] t = (XY Y X) t = Y t X t X t Y t = Y X XY = X, Y ] X, Y ] so n (C) and hence so n (C) is a Lie Subalgebra of sl n (C). Example 1.7 (The Symplectic Lie algebra). Let Y Mat 2n (C). Then Y is a matrix of the form ( ) P Q R S where P, Q, R and S are n n matrices. Then we define the Symplectic Group to be the set sp 2n (C) = { X Mat 2n (C) S = P t, R t = R, Q t = Q }. Check this is a Lie subalgebra in gl 2n (C). Definition: Let g 1 and g 2 be any two Lie algebras over F (= C). Then a Lie algebra homomorphism is a map (or transformation) ϕ : g 1 g 2, such that (a) ϕ is a linear map, ϕ(ax + by ) = aϕ(x) + bϕ(y ) with a, b F and X, Y g 1. (b) ϕ(x, Y ] g1 ) = ϕ(x), ϕ(y )] g2. Example 1.8 (Trivial). For all g 1, g 2 Lie algebras ϕ(x) 0 defines a Lie algebra homomorphism. Definition: ϕ : g 1 g 2 is a Lie algebra isomorphism if ϕ is a homomorphism and ϕ is bijective (one-to-one and onto). Remark. If ϕ : g 1 g 2 is an isomorphism then we can identify g 1 and g 2 by identifying X and ϕ(x) for all X g 1. If such an isomorphism exists between g 1 and g 2 then we say the two Lie algebras are isomorphic (essentially the same). We will study Lie algebras up to their isomorphisms. Example 1.9. Let V be a vector-space over C such that dim V = n. Then we have a set gl(v ) = {T : V V }, which contains all the possible mappings from V to itself. We define a Lie bracket for T, S gl(v ) as, T, S] = TS ST

5 Proof. Same as in Example 1.2. LIE ALGEBRAS 5 Claim. gl(v ) is isomorphic to gl n (C). i.e. Any linear map can be expressed as a matrix. Proof. Choose any basis {v 1,...,v n } in V. For all T gl(v ) consider its matrix X = X T = x ij ] n i,j=1. Now we can describe the mapping onto the basis as, (2) Tv j = n x ij v i, j = 1,..., n. i=1 Define ϕ : gl(v ) gl n (C) to be the map ϕ(t) = X T. We need to check that ϕ is a Lie algebra isomorphism. First we check the homomorphism property. We can see ϕ is linear, i.e. ϕ(at+bs) = aϕ(t) + bϕ(s). So, ϕ ( T, S] gl(v ) ) = ϕ(ts ST) = ϕ(ts) ϕ(st) = ϕ(t)ϕ(s) ϕ(s)ϕ(t) = ϕ(t), ϕ(s) ] gl n (C) See Linear Algebra for verification of these statements. ϕ is bijective by (2). Remark. ϕ depends on the choice of a basis. Remark. Examining special cases of gl n (C) we see n = 1: gl 1 (C) = {x C} and x, y] = 0. Hence abelian. n = 2: gl 2 (C) = {x C} is non-abelian, i.e. there exists X, Y such that Y X XY. Now examining similar cases sl n (C) we see n = 1: sl 1 (C) = {x C} and dim sl 1 (C) = 0. n = 2: sl 2 (C) = {( a c a b ) : a, b, c C} and dim = 3. Now examining similar cases for so n (C) sl n (C) we see n = 1: so 1 (C) = {0}, dim = 0 n = 2: so 2 (C) = {( 0 x x 0 ), x C}. The Lie bracket gives us, ( ) ( ) xy 0 yx 0 X, Y ] = XY Y X = = 0. 0 xy 0 yx ) } n = 3, so 3 (C) = : x, y, z C. We have that dim ( so 3 (C) ) = 3 and non-abelian! {( 0 x y x 0 z y z 0 Proposition: sl 2 (C) = so 3 (C) Remark. sl 2 (R) = so3 (R), so the field F of our Lie algebra does matter.

6 6 PROF. MAXIM NAZAROV Proof. define ϕ : sl 2 (C) so 3 (C) such that ( ) 0 b c i(b + c) a b ϕ = c b 0 2ai c a i(b + c) 2ai 0 We need to show that ϕ is a Lie algebra isomorphism. First we check to see if ϕ is a bijection (i.e. one-to-one and onto). One-to-one: We know ϕ(x) and we know X. We can clearly see that ϕ(x 1 ) = ϕ(x 2 ) X 1 = X 2 and so ϕ is one-to-one. Onto: Let Z be any element of so 3 (C), i.e. such that ϕ(x) = Z. Specifically, So ϕ is a bijection. z 1 = b c z 2 = i(b + c) z 3 = 2ia a = z 3 2i b = z 2 iz 1 2i c = iz 1 + z 2 2i ( 0 z1 z 2 z 1 0 z 3 z 2 z 3 0 X = ) then we can find X sl 2 (C) z 3 2i iz 1 + z 2 2i z 2 iz 1 2i Finally we check the properties of a homomorphism. The entries of ϕ(x) are linear functions in a, b, c therefore this means ϕ is a linear map, this is the first property of a homomorphism satisfied. Left to check that z 3 2i (3) ϕx, X ] = ϕ(x), ϕ(x )]. Now define X sl 2 (C) to be X = ( a b c a ). Then the commutator gives us, X, X ] = XX X X ( ) ( ) ( ) ( ) a b a b = a b c a c a a b c a c a ( ) ( ) aa = + bc ab ba a ca ac cb + aa a + b c a b b a c a a c cb + a a Now returning to (3) we get the left hand side of the equation to be ab ba a b + b a i(ab b a a b + b c ϕ ( X, X ] ) 0 = ca + ac + c a ac +ca a c c a + a c) 0 2i(aa + bc a a + b c) 0 and the right hand side of the equation to be

7 LIE ALGEBRAS 7 ϕ(x), ϕ(x ) ] = ϕ(x)ϕ(x ) ϕ(x )ϕ(x) 0 b c i(b + c) = c b 0 2ai 0 b c i(b + c ) c b 0 2a i i(b + c) 2ai 0 i(b + c ) 2a i 0 ϕ(x )ϕ(x) = LHS = 0 2a (b + c) + 2a(b + c ) 0 0 This method is extremely ugly! We need a more sophisticated method to construct isomorphisms between two Lie algebras. Definition: We define the direct sum of two Lie algebras g 1, g 2 over F to be g 1 g 2 = {(X 1, X 2 ) : X 1 g 1, X 2 g 2 }, with component wise addition and scalar multiplication. The Lie bracket for the direct sum is (X1, X 2 ), (Y 1, Y 2 ) ] g 1 g 2 ( X 1, Y 1 ] g1, X 2, Y 2 ] g2 ) We claim, without proof, that the axioms of a Lie algebra are satisfied by the direct sum. Remark (on Homomorphisms). We have seen that sl 2 (C) = so 3 (C). Also, one can prove (although difficult) that sl 2 (C) sl 2 (C) = so 4 (C) and sp 4 (C) = so 5 (C) and sl 4 (C) = so 6 (C). These are all isomorphisms for so n (C)! (n > 6) Definition: Take any Lie algebra g (over F). Then the derived Lie algebra (of g) is the Lie subalgebra g g such that g = span { X, Y ] g : X, Y g } which has typical element a 1 X 1, Y 1 ] + a 2 X 2, Y 2 ] +... P, Q g P, Q g 1.1. Derivation of Lie Algebras Definition: a derivation of a Lie algebra g is a linear transformation (operator) δ : g g satisfying the Liebniz Rule, δ(x, Y ]) = δ(x), Y ] + X, δ(y )] for all X, Y g

8 8 PROF. MAXIM NAZAROV Caution: a derivation δ of g should not be confused with the derived Lie algebra g. Notation: the set of all derivations of g is denoted by D(g) = {δ δ is a derivation}. Definition: For all Z g we define the adjoint operator adz : g g to be the mapping ad Z(X) = Z, X]. Claim. For all Z g we have adz D(g). Proof. To show that this is true we have to show that the adjoint operator is linear and satisfies the Liebniz rule. So, let a, b F and X, Y g. We first show it is linear ad Z(aX + by ) = Z, ax + by ] = az, X] + bz, Y ] = a adz(x) + b ad Z(Y ) Now we have to prove that the Liebnize rule holds. (adz)x, Y ] = Z, X, Y ] ] = X, Y, Z] ] Y, Z, X] ] by Jacobi identity = X, Z, Y ] ] + Z, X], Y ] = X, ad Z(Y ) ] + adz(x), Y ] = adz(x), Y ] + X, ad Z(Y ) ]. Hence the Liebniz rule holds for the adjoint operator and we have proved it is in D(g). Definition: Let g be any Lie algebra. Then we define the centre to be the set Z(g) = {Z g Z, X] = 0 for all X} or equivalently we can say Z Z(g) ad Z 0. Example Let g be an abelian Lie algebra, i.e. X, Y ] = 0 for all X, Y g. Then Z(g) = g. Example Let us look at the general linear Lie algebra gl n (C). Then the centre is, Z ( gl n (C) ) = {scalar multiples of the n n identity matrix} λ = 0 λ ; λ C λ

9 LIE ALGEBRAS 9 Example It is now logical to look at the special linear Lie algebra sl n (C). We have that Z ( sl n (C) ) = {0}. It is easy to see that 0 is in the centre, 0, X] = 0 0 Z(g), but we shall prove later that it is the only element in the centre. Theorem 1. D(g) = {δ δ is a derivation of g} is a Lie algebra with addition / scalar multiplication as those of linear transformations and the Lie bracket being the commutator δ 1, δ 2 ] = δ 1 δ 2 δ 2 δ 1. Proof. We need to check all the axioms of a Lie algebra for D(g). Is D(g) a vector space? For δ 1, δ 2 D(g) then δ 1 + δ 2 is another linear map (δ 1 + δ 2 ) : g g, such that (δ 1 + δ 2 )(X) = δ 1 (X) + δ 2 (X) for all X g. Does the Leibniz rule still hold? (δ 1 + δ 2 ) = δ 1 X, Y ] + δ 2 X, Y ] = δ 1 (X), Y ] g + X, δ 1 (Y ) ] g + δ 2 (X), Y ] g + X, δ 2 (Y ) ] g = (δ 1 + δ 2 )(X), Y ] + X, (δ 1 + δ 2 )(Y ) ]. Yes it does. Similarly we can check that for all δ D(g) and for all a F (= C) then aδ D(g) where, (aδ)(x) = a ( δ(x) ) Is δ 1, δ 2 ] = δ 1 δ 2 δ 2 δ 1 a Lie bracket? Also is δ 1, δ 2 ] D(g)? We show this by checking that the Liebniz rule holds for δ 1 δ 2 δ 2 δ 1. (δ 1 δ 2 δ 2 δ 1 )X, Y ] g = δ 1 ( δ2 X, Y ] ) δ 2 ( δ1 X, Y ] ) ( δ2 = δ 1 (X), Y ] + X, δ 2 (Y ) ]) ( δ1 δ 2 (X), Y ] + X, δ 1 (Y ) ]) (δ1 δ 2 )(X), Y ] + δ2 (X), δ 1 (Y ) ] + δ1 (X), δ 2 (Y ) ] + X, (δ 1 δ 2 )(Y ) ] = (δ 2 δ 1 )(X), Y ] δ1 (X), δ 2 (Y ) ] δ2 (X), δ 1 (Y ) ] X, (δ 2 δ 1 )(Y ) ] = (δ 1 δ 2 δ 2 δ 1 )(X), Y ] + X, (δ 1 δ 2 δ 2 δ 1 )(Y ) ] So this means the commutator δ 1 δ 2 δ 2 δ 1 of two derivatives δ 1, δ 2 is a derivation again and hence δ 1, δ 2 ] D(g). Checking the axioms (a), (b), (c) for the commutator δ 1 δ 2 δ 2 δ 1 is similar to checking the axioms for the matrix commutator, AB BA where A, B Mat(C). We define a function ad such that ad :g D(g) Z ad Z,

10 10 PROF. MAXIM NAZAROV where Z g and adz D(g). Proposition. ad is a Lie algebra homomorphism. Proof. We need to show that the ad operator is linear, i.e. satisfies the condition ad(px+ qy ) = p adx + q ady. We show this by applying the identity to any Z g. ad(px + qy )(Z) = px + qy, Z] = px, Z] + qy, Z] = px, Z] + qy, Z] = p ad(x)(z) + q ad(y )(Z) for all X, Y g. However we also need to show that ad ( X, Y ] g ) = ad X, ady ]D(g). We show this by again applying both sides to any Z g. This gives, ad ( X, Y ] g ) (Z) = X, Y ]g, Z ] g = X, Y, Z] g ] g + Y, Z, X] g ]g = X, ad(y )(Z)] g Y, ad(x)(z)] g = ( (adx)(ad Y ) (ady )(ad X) ) (Z) = ad(x), ad(y )] D(g) (Z). The second line is obtained using the Jacobi identity and this holds for all X, Y g. Notation: We use the following notation to describe the Lie bracket of two Lie algebras. So, for g 1, g 2 we have g 1, g 2 ] g span {X 1, X 2 ] : X 1 g 1, X 2 g 2 }. Note that in this notation we have g = g, g]. Definition: (1) a derivation δ is called inner if δ = ad Z for some Z g. (2) if δ is not inner, then it is called an outer derivation. D(g) = {inner} {outer} Example 1.13 (Very Important Example of a Lie Algebra). Consider a Lie algebra g of dimension 2 with basis {E, F }. Define a Lie bracket on g by E, E] = 0, F, F] = 0 and E, F] = E F, E] = E. It can be extended by bi-linearity to ae + bf, ce + df] = ade bce. This satisfies axiom (b) of a Lie algebra. From general remark we know axiom (c) is automatically satisfied because dimg < 3.

11 Observation: g is not Abelian. Proposition. Every derivation of this g is inner. LIE ALGEBRAS 11 Proof. Take any derivation δ D(g). We need to prove that δ = ad Z for some Z g. (1) We can see that the derived Lie algebra is g = CE. If δ(g ) g then δ(e) = de for some d C. (2) If we apply ad(df) to E then we obtain, ad(df)(e) = df, E] = de. Now define ǫ D(g) to be ǫ = δ + ad(df). This gives ǫ the property that when applied to E it is 0, i.e. ǫ(e) = δ(e) + ad(df)(e) = de de = 0. (3) Using the definition of the Lie brackets we can obtain that, 0 = ǫ(e) = ǫ ( E, F] ) = 0 ǫ(e), F] + E, ǫ(f)]. E, ǫ(f)] = 0. Now looking at ǫ applied to F we have that ǫ(f) g ǫ(f) = xe + yf. Using the previous result we get 0 = E, ǫ(f)] = E, xe + yf] = ye, this implies that y = 0 ǫ(f) = xe. (4) Now consider ad(xe) and apply it to E. This gives us ad(xe)(e) = xe, E] = 0. Looking at the operator applied to F gives ad(xe)(f) = xe, F] = xe. However we have ǫ(e) = 0 by step 2 and ǫ(f) = xe by step 3. We have two linear transformations, ad(xe) and ǫ, that are acting in the same way on the basis vectors E, F. So this means we must have ad(xe) = ǫ. (5) Then ad(xe) (2) = δ + ad(df) δ = ad(xe df). }{{} Z Definition: For any Lie algebra g, a subset h g is an ideal if; g h is a vector subspace, X, Y ] h for all X g, Y h. Remark. If h is an ideal then this implies h is a Lie subalgebra of g (we have from the definition that X, Y ] h for all X, Y h), but this is not true the other way round. Consider Example In that example take h = CE. This is an ideal, g X, E] = ae + bf, E] = be h. Now consider f = CF. This is a subalgebra af, bf] = 0. Now ae + bf, F] = ae for any a 0 therefore ae + bf, F] f. Hence f is not an ideal. Definition: A Lie algebra g is called complete if (1) Z(g) = {0}

12 12 PROF. MAXIM NAZAROV (2) and every derivation of g is inner. Example 1.14 (Example 1.13 revisited). We have already proved part (2) of the definition of complete for this example. So, now let us check part (1). Suppose Z Z(g) and define Z to be Z = ae + bf. Then therefore Z = 0 Z(g) = {0}. 0 = Z, E] = ae + bf, E] = be b = 0 0 = Z, F] = ae + bf, F] = ae a = 0 Theorem 2. Let g be any Lie algebra. Suppose that h g is an ideal, which is complete as a Lie algebra (in its own right). Then there exists another ideal f g such that g = h f. Proof. Let f = {Z g X, Z] = 0 for all X h}, i.e. the centraliser of h in g. We first need to show that f is a Lie subalgebra of g. This has two parts. Is f a vector subspace? Well we have that for Z 1, Z 2 f that Z 1 +Z 2 f by linearity of, ]. Also for all Z f, α C then αz f. So f is a vector subspace of g. Now for all Y g and Z f we need to show that Y, Z] f. i.e. X, Y, Z] ] = 0 for all X h. By the Jacobi identity we have, X, Y, Z] ] = Y, Z, X] ] Z, X, Y ] ]. We have X, Y ] h because it s a h ideal, which implies Z, X, Y ] ] = 0 by definition of f. Also by the definition of f we have Z, X] = 0, which implies Y, Z, X] ] = 0. Hence X, Y, Z] ] = 0 as required. So f is a Lie subalgebra of g. We now need to show that g = h f is a Lie algebra. (1) Start by proving that h f = {0}. Take any Z h f. This means Z f Z, X] = 0 for all X h but Z is also in h. Therefore Z, X] = 0 for all X h implies that Z Z(h). We know h is complete so we must have Z = 0. (2) We now show that g = h + f (which is equal to h f because h f = {0}). Take any A g and consider the operator; ada : g g. For all X h we have that ad A(X) = A, X], which is an element of h because h is an ideal. Therefore we have that ada will map any element of h to h, i.e. ada : h h. The ad operator satisfies the Liebniz rule and hence ada D(h). Apriori we may not know that ada is an inner derivation because A may be outside h. However h is complete as a Lie algebra, so any derivation is inner. So ada h is an inner derivation. Now comparing ad A with another operator from the D(h) we see,

13 LIE ALGEBRAS 13 (ad A)(X) = (adb)(x) A, X] = B, X] for some B h for all X h A B, X] = 0 for all X h hence A B f by definition of f. Now defining C = A B, we have that A = B+C for any A g, B h, C f. (3) Now only left to prove that g = h f is a Lie algebra. We do this by constructing an isomorphism, ϕ : g h f A (B, C) a trivial isomorphism. We know that ϕ is bijective and linear by definition of vector spaces. Need to show that ϕ satisfies the homomorphism property, ϕ ( A, A ] g ) = ϕ(a), ϕ(a ) ] h f where A (B, C), A (B, C ) and A = B + C, A = B + C. We start with B, B h and C, C f then ϕ ( B + C, B + C ] ) = ϕ ( B, B ] + C, C ] ) = ϕ ( B, B ] ) + ϕ ( C, C ] ) because B, B ] h and C, C ] f we have that ϕ maps them to, = ( B, B ], 0 ) + ( 0, C, C ] ) = ( B, B ], C, C ] ) = (B, C), (B, C ) ] = ϕ(b + C), ϕ(b + C ) ] 2. Classification of Lie algebras of dimensions 1,2,3 For the next couple of lectures we will be trying to classify Lie algebras of small dimension. This will help us a lot in the theory of Lie algebras. We will be focusing on Lie algebras of dimension 1, 2 or 3 and will take them case by case. I) Take a Lie algebra g such that dim g = 1. Let X, Y g = CE where E g and E 0 then we must have X = xe, Y = ye. So the Lie bracket of these two elements is, X, Y ] = xe, ye] = xye, E] = 0. This is an abelian Lie algebra and hence for one dimension we must have, ] 0.

14 14 PROF. MAXIM NAZAROV II) Take a Lie algebra g such that dim g = 2. We consider the possible cases of g g but in fact there are only two. a) g = {0}, which from above gives us X, Y ] = 0 for all X, Y g and hence g is an abelian Lie algebra. b) g {0}. Then g must be the span of two basis vectors (as dimg = 2). So, choose any basis in g, say {E, F }. Then, by E, E] = F, F] = 0 g = span { X, Y ] X, Y g } = span { E, E], E, F], F, E], F, F] } = span { E, F] } = CE, F] but g 0 by assumption and hence E, F] 0. Also we note that dimg = 1. Up to this point our choice of E, F was arbitrary, however let us now choose a specific E and F. We choose E and F such that g = CE. This means that E, F] = ce for some c C \ {0}. Thus we have, E, F ] = E c but we can just set F to be F. Doing this gives us E, F] = E. Now g has a c basis, {E, F }, which corresponds to the Lie bracket relationships, E, F] = 0 F, F] = 0 E, F] = E F, E] = E. We notice that this is our Very Important Example, i.e. Example So, this set of Lie algebras are the only non-abelian Lie algebras of 1 or 2 dimensions. III) Take a Lie algebra g such that dim g = 3. We will again consider g and its various cases. (a) dim g = 0 then g = {0} and hence g is an abelian Lie algebra. (b) dim g = 1 and let us assume that g Z(g) (the centre of g). Now choose an element E g such that g = CE with E 0. We now complete E to a basis, {E, F, G} in G. So we have E g Z(g) and hence E, X] = 0 for all X g. Then, again using that E Z(g) we obtain the Lie bracket relationships; (4) E, F] = E, G] = 0. We know g is a span of the basis so we obtain,

15 LIE ALGEBRAS 15 g = span {E, F], E, G], F, G]} = span {F, G]} = CE. So we must have that F, G] = de for some d 0. Just as before we obtain a new E by setting E to de. We now obtain our final Lie bracket relationship to be (5) F, G] = E (N.B. this also gives us G, F] = E to satisfy antisymmetry). Hence there is only one Lie algebra g such that dim g = 3, dim g = 1 and g Z(g). Exercise: it is still left to check that (4) and (5) define a Lie algebra. (c) dim g = 1 and let us assume this time that g Z(g). Now choose an element E g such that g = CE with E 0. We assume E Z(g) as it implies that there exists an F g such that E, F] 0 otherwise E, F] = 0 for all F g. Also we need F xe otherwise E, F] = E, xe] = xe, E] = x.0 = 0. So we use E, F as a two basis vector. By the definition of g we know E, F] g = CE and hence E, F] = be for some, non-zero, b C. Just as before we have that the Lie bracket of E and F will be b E, F ] = E. b So setting F to F b we get the Lie bracket relationship E, F] = E. Now consider the Lie subalgebra h g spanned by E, F. From above we have that E, F] = E, E, E] = F, F] = 0 but this is just our Very Important Example, i.e. Example Moreover, h is an ideal of g for all X g and for all Y h. By definition we have that X, Y ] g = CE h. However, by Example 1.14, we know that h is complete as a Lie algebra. Thus by Theorem 2 we have that there exists an f g such that g = h f. Looking at the dimensions of our Lie algebras we see that dim g = dim h + dim f 3 = 2 +? Hence f must have dimension 1 and by above results is abelian. Hence we must have for some G g that f = CG. Now {E, F, G} is a basis of g, with Lie bracket relationships,

16 16 PROF. MAXIM NAZAROV E, F] = E E, G] = 0 F, G] = 0. We note again that there is only one Lie algebra g with dim g = 3, dim g = 1 and g Z(g). (d) dim g = 2. We start by proving a useful remark about g. Remark. g must be abelian. Proof. Suppose, for a contradiction, that g is not Abelian. Then g would be the Lie algebra from example It is known that g g is always an ideal of g. So, for all X g, Y g we have that X, Y ] g and by example 1.14 g is complete. Then by Thereom 2 we can express g as, g = g f = (CE CF) CG, letting f = CG. However, g is also g = span { X, Y ] } X, Y g = span { E, F], E, G], F, G] } = span{e} by the Lie bracket relationships in part (c), thus dim g contradiction. Hence g must be abelian. = 1 but this is a Now choose any basis in g, say {E, F } then E, F] = 0 by above remark. We complete {E, F } to a basis {E, F, G} in g. So, we have to consider two new Lie brackets G, E], G, F] g, which we define to be G, F] = ae + bf and G, E] = ae + bf G, E] = ae + bf where a, b, c, d C. Firstly g = span { E, F], G, E], G, F] } = span{ae + bf, ce + df } and the dim g = 2 thus we must have that the elements of the span ae + bf, ce + df are linearly independent. We could show this by saying, ( ) a b T = det 0. c d So it is seen that g is completely described by a 2 2 non-singular matrix T. FACT: any such T does define a Lie algebra. That is, we define a Lie bracket on g by E, F] = 0, G, E] = ae + bf and G, F] = ce + df. We know the first two axioms of a Lie algebra hold for this

17 LIE ALGEBRAS 17 Lie bracket, all is left is to check the Jacobi identity for any X, Y, Z g. This is left, however, as an exercise. We first pose a question. Can different matrices T define the same Lie algebra (i.e. isomorphic)? The answer is yes! Let us change the basis {E, F } in g to another basis { E, F }, notice we still have E, F ] = 0. Then ( a b d c ) is the matrix corresponding to the operator adg on g. By definition of the adjoint operator we have that (adg)(e) = G, E] { and (ad } G)(F) = G, F]. If C is the coordinate change matrix from {E, F } to E, F then the new T is T such that T = T = C 1 T C. So by changing the basis {E, F } in g we can replace T by C 1 T C for all C. Also I can choose another G, say G, such that G = tg + X for all t C (t 0) and X g. This new G gives the following bracket relationships with E and F G, E ] = tg + X, E] = tg, E] = t(ae + bf) G, F ] = tg + X, F] = tg, F] = t(ce + df). So, this corresponds to T being multiplied by any t 0. Conclusion: two different matrices, T and T, define the same g iff T can be obtained from T by a sequence of operations. i.e. T tt, t 0 and T (C )T(C ) 1 (e) There is only one remaining case to consider, which is dimg = 3. So (g g g = g ). Theorem 3. If dim g = dim g = 3 then g = sl 2 C. Proof. The plan is that we will choose a basis {H, E, F } in g such that H, E] = 2E, H, F] = 2F and E, F] = H. Then we can define a linear map ϕ : g sl 2 C by ( ) 1 0 H 0 1 ( ) 0 1 E 0 0 ( ) 0 0 F 1 0 They re all in sl 2 C a bijection. Why ϕ is a Lie algebra homomorphism? Need to show

18 18 PROF. MAXIM NAZAROV ( ) ϕ(x, Y ] g ) = ϕ(x), ϕ(y )] sl2 C = ϕ(x)ϕ(y ) ϕ(y )ϕ(x) Because of linearity of ϕ it is enough to check ( ) only for X, Y {E, H, F } 0 = ϕ(e, E]) = ϕ(e)ϕ(e) ϕ(e)ϕ(e) = 0 If I know that ( ) is OK for X = E, Y = F then it is OK for X = F, Y = E as well, ϕ(f, E]) = ϕ(f)ϕ(e) ϕ(e)ϕ(f) by antisymmetry of, ] g ϕ(e, F]) = ϕ(e)ϕ(f) + ϕ(f)ϕ(e) Enough to check ( ) for (X, Y ) = (E, F), (E, H), (F, H). Then ( ) can be checked directly. Remark. (i) For all X g, X 0 we have rank(adx) = dim(im(adx)) = 2. Complete X to a basis {X, Y, Z}. Then g = span {X, Y ], X, Z], Y, Z]}. All the Lie brackets are linearly independent in g. So, Im(adX) = span {ad X(X), adx(y ), ad X(Z)} = span {X, X], X, Y ], X, Z]} = span {X, Y ], X, Z]} (= V ) So dim(im ad(x)) = 2 = rank(adx). (ii) dim(im ad(x))+dim(ker(ad X)) = 2+1 = dimg = 3. X ker(ad(x)) = CX = span {X}, ad(x)(x) = X, X] = 0. Take any X g, X 0. Consider ad(x) : g g. (A) Suppose ad(x) has at least one non-zero eigenvalue a C, a 0. Let E be the corresponding eigenvector. Then X, E] = ae 2X a, E] = 2E Let 2X = H by definition, then H, E] = 2E. a (B) Now suppose that X does not have a non-zero eigenvalue all eigenvalues of ad(x) are 0. FACT: From the theory of matrices we know. Suppose T : V V (ad X : g g), dim V = 3 such hat all eigenvalues of T are 0 and

19 LIE ALGEBRAS 19 rankt = 2, rank(adx) = 2. Then there exists a basis {v 1, v 2, v 3 } such that the matrix of T relative to this basis is Consequence of Jordan normal form of any n n matrix. Now T = ad(x), v 1 ker T, T(v 1 ) = 0, T(v 2 ) = v 1. So ad(x)(x) = X, X] = 0 v 1 = X. Let g Y = v 2. Then X, Y ] = ad(x)(y ) = X = v 1. So X, Y ] = X. Put H = 2Y and E = X. Then H, E] = 2Y, X] = ( 2)( X) = 2E. We have H, E] = 2E for some H, E g. Then ad(h) : g g. We know H g = g H is a linear combination of brackets A, B] where A, B g. Also ad(h) is a linear combination of ad(a, B]) = ad(a), ad(b)] = ad(a) ad(b) ad(b) ad(a). trace(ad(h)) is a linear combination of trace(ad(a, B])) which is trace(ad(a) ad(b)) trace(ad(b) ad(a)) = 0. Eigenvectors of ad H such that H, H] = 0 = 0 H and ad(h)(e) = H, E] = 2E. So H has eigenvalue 0 and E has eigenvalue 2. This implies there exists F with eigenvalue -2 (as sum of eigenvalues must equal 0). H, F] = 2F. Need to get the third relation. Now, H, E, F]] + E, F, H]] + F, H, E]] = 0 the Jacobi identity. So rearranging gives, H, E, F]] = E, H, F]] + H, F], F] = E, 2F] + 2E.F] = 0 using the relations H, E] = 2E and H, F] = 2F. So E, F] an eigenvector for adh with zero eigenvalue (adh)(e, F]) = 0 E, F] E, F] = ch for some c C. Can c = 0? No, otherwise E, F] = 0 and F ker(ad(e)) but ker(ad(e)) = C E (remark at the beginning of the proof). E, F] = ch for some c 0 E, F] = c H. Set Ẽ = E then Ẽ, F] = H. c

20 20 PROF. MAXIM NAZAROV the relation H, E] = 2E still holds for Ẽ = E. i.e. H, Ẽ] = 2Ẽ c and now we have Ẽ, F] = H, the final relation required. (Note H, F] = 2F also holds for Ẽ). End proof. 3. Solvable Lie algebras Remark. Differential equations Lie algebras and solvable D.E. solvable Lie algebras Quotient (of factor) vector space V - any vector space and V U a vector subspace. As a set V/U = { x } + {{ U } : x V } {x+u:u U} cosets of the Abelian group V with respect to U. We define the operation as; (x 1 + U) + (x 2 + U) def = (x 1 + x 2 ) + U Suppose x 1 such that x 1 + U = x 1 + U (1) and x 2 such that x 2 + U = x 2 + U (2). Then, (x 1 + U) + (x 2 + U) def = (x 1 + x 2 + U) x 1 x 1 U by (1) and x 2 x 2 U by (2). Adding these two statements together gives us x 1 x 1 + x 2 x 2 U x 1 + x 2 + U = x 1 = x 2 + U. Similarly α C and x +U V/U such that if x + U = x + U then α(x + U) def = αx + U α(x + U) def = αx + U x x U α(x x ) U αx + U = αx + U. Exercise: Check that these operations of vector addition and scalar multiplication satisfy all the axioms of a vector space for V/U. In fact, they will follow from the respective axioms for V itself. 0 in V/U is the class (coset) 0 + U = u + U for all u U where 0 is in V. Definition: Let g be any Lie algebra over F (= C). Let h g be an ideal of g. The quotient Lie algebra g/h is g/h as a vector space as defined above. Take X + h, Y + h g/h where X, Y g. Then, X + h, Y + h] g/h def = X, Y ] g + h Suppose X + h = X + h (3) and Y + h = Y + h (4) for some X, Y g. Then,

21 LIE ALGEBRAS 21 X + h, y + h] g/h = X, Y ] g + h Now (3) X X = A h for X = X +A and (4) Y Y = B h with Y = Y +B. Now, X + h, Y + h] g/h = X, Y ] g + h = X + A, Y + B] g + h = X, Y ] + X, B] }{{} h = X, Y ] + h + A, Y ] }{{} h + A, B] + h }{{} h as required. Need to check that the definition of a Lie bracket on g/h satisfies the three axioms of a Lie algebra. Claim: they follow from the respective axioms for g itself. Check this for revision. Example 3.1. Take any Lie algebra g such that h = g g/g X + g, Y + g. Now, g is an ideal of g. Now the zero coset. Hence g/g is Abelian. X + g, Y + g ] def = X, Y ] g }{{} g + g = g = 0 + g Let g be any Lie algebra over F(= C). Then, Definition: the derived series of g is g 0 = g g (1) = g g (2) = (g )... Definition: g is solvable, if there exists p such that g (p) = {0} so that the derived series terminates. Remark. g (p+1) = g (p+2) = = {0} Example 3.2. g = {X = x ij ] n i,j=1 Mat n C x ij = 0 if i j}. Typical X may be, a basis of g is {E ij i < j} where E ij, E kl ] = δ ij E il δ li E kj for all i, j, k, l = 1,..., n. Lemma 1. g (p) span{e ij : j i 2 p } ( ) Proof. We prove this by induction (1) p = 0, g (p) = g where g = span{e ij i < j}, Now, i < j j i > 0 j i 1 = 2 0.

22 22 PROF. MAXIM NAZAROV (2) Induction step: assume ( ) is true. Then, g (p+1) = (g (p) ) = span{x, Y ] X, Y g (p) } span{e ij, E kl ] j i 2 p, l k 2 p } = span{δ jk E il δ li E kj j i, l k 2 p } j i, l k 2 p j i + l k 2, where 2 p = 2 p+1 {E ab b a 2 p+1 } Remark. In fact, we have equality instead of in ( ) (need to prove in ( )). This is an exercise. The Lemma implies that g is solvable. Indeed, suppose p is so large that n < 2 p. Then there is no pair (i, j) with i, j = 1,...,n such that j i 2 p (j i < n anyway). Then, g (p) span() = {0}. Example 3.3. g = gl n C, n > 1 (g = gl 1 (C)-abelian, g = {0} solvable). Now n > 1. Then (gl n C) = sl n C ((sl n C) ) = sl n C g (0) = gl n C, g (1) = sl n C; g (1) = g (2) = = sl n C {0}. The derived series does not terminate and therefore g is not solvable Theorems on solvable Lie algebras A) Let g be solvable and g h a subalgebra. Then h is solvable as well. B) Let ϕ : g h be a Lie algebra homomorphism onto h. let g be solvable. Then h is solvable as well. C) g h an ideal of g. Suppose that g/h is a solvable Lie algebra and that h is solvable. Then g is solvable as well. Proof. A) Now we have,

23 LIE ALGEBRAS 23 g (0) = g h = h (0) g h g h. g (p) h (p) p large g (p) = {0} h (p) = {0}. h = span{x, Y ] X, Y h} = span{ϕ( X), ϕ(ỹ )] X, Ỹ g}. = span{ϕ X, Ỹ ] X, Ỹ g} = ϕ(g ) So h = = ϕ(g ) and so on h (p) = ϕ(g (p) ) for all p = 0, 1, 2,... there exists p such that g = {0} h (p) = ϕ({0}) = {0} Proof. C) g/h solvable. There exists p such that (g/h) (p) = {0}. Consider the canonical homomorphism π :g g/h X X + h (Linear map by definition of the vector space g/h). Homomorphism property of π, π(x, Y ]) = def π X, Y ] + h = def g/h X + h, Y + h] = π(x), π(y )] g/h π a homomorphism arguments of B. Show that π(g (p) ) = π(g) (p) (ϕ = π) (only the homomorphism property of ϕ = π is required for this). Now π(g (p) ) = (g/h) (p) = {0} because π is onto by definition for some p. Now, π(g (p) ) = {0} g (p) ker π ker π = {X g X + h = h} = {X h} = h. So g (p) h for p large enough h is solvable there exists q such that h (q) = {0}. Then, which implies g is solvable. g (p+q) h (q) = {0} g (p+q) = {0}

24 24 PROF. MAXIM NAZAROV Lie Theorem (on solvable Lie algebras) Let V be any finite dimensional vector space, over C. Consider the general linear Lie algebra of V, gl(v ) = {X : V V linear}; X, Y ] = XY Y X. We know that gl(v ) = gl n (C) where n = dim(v ), so gl(v ) not solvable for n > 1. Let gl(v ) g be a Lie subalgebra which is sovable. Theorem 4. then V contains a vector v, which is an eigenvector for all X g; that is X Xv = λ(x)v for all X g. Comment: take any set S gl(v ). Suppose we look for a common eigenvector for all X S. Take a Lie subalgebra gl(v ) g spanned by S (i.e. g closed with respect to X, Y ] = XY Y X). Then if g is solvable then there exists v. Corollary 1 (to Lie Theorem). there is a suitable basis of V such that the matrices of all X g relative to this basis are upper triangular. For example, λ 0. λ..... X = , v = Xv = λv Remark. The corollary implies the Theorem by the rules of matrix presentation of X g. Proof. By induction on dim V = n. If dim V = 0. Then V = {0} and gl(v ) = 0. So the statement to prove is trivial. Now suppose that the Corollary is true for all vector spaces U (instead of V ) such that the dimension of U < n = dim V. (On the other hand, by the Lie theorem itself there exists v 1 0 V such that Xv 1 = λ(x)v 1 for all X g. dim Cv 1 = 1). Consider the quotient space U = V/Cv 1, now we have dim U = dim(v ) 1 = n 1. So induction assumption applies to U. Consider the map where ϕ : g gl(u) X X X(v + Cv 1 U ) = def Xv + Cv 1. This map ϕ is linear. Moreover, this mapping ϕ is a Lie algebra homomorphism. That is, need to check that ϕ(x, Y ] g ) = ϕ(x), ϕ(y )] gl(u), where we have ϕ(x, Y ] g ) = ϕ(xy Y X) = XY Y X and by definition ϕ(x), ϕ(y )] gl(u) = XY Y X. Apply the LHS and RHS to any v + Cv 1.

25 LIE ALGEBRAS 25 similarly Y X(v + Cv 1 ) = Ȳ Xv + Cv 1. XY (v + Cv 1 ) = XY v + Cv 1 X Ȳ (v + Cv 1) = X(Y v + Cv 1 ) = XY v + Cv 1 So homomorphism property is OK for ϕ. We need to check that ϕ is defined correctly. Suppose v + Cv 1 = v + Cv 1 ( ). Then, X(v + Cv 1 ) = Xv + Cv 1 Xv + Cv 1 = X(v + Cv 1 ) Need to show that Xv + Cv 1 = Xv + Cv 1 where X(v v ) Cv 1 but ( ) implies v v Cv 1 X(Cv 1 ) Cv 1 X(v v ) Cv 1 because v 1 is an eigenvector for X.) Now we can use Thereom B on solvability. The image of ϕ in gl(u) is a solvable Lie subalgebra in gl(u). By the induction assumption, there is a good basis u 2,...,u n of U such that all elements of ϕ(g) gl(u) get represented by upper triangular matrices. ϕ(g) ϕ(x) = X. Then, ( ) ϕ(x)u j = n a ij u j i=2 i j j = 2,...,n u j = v j + Cv 1 for j = 2,...,n. Now the left hand side of ( ) is, X(u j ) = X(v j + Cv 1 ) = Xv j + Cv 1 = which implies Xv j Cv 1 Cv j. Then, ( ) Xv j = j b ij v i i=1 n i=2 i j for all j = 2,...,n a ij ( v i + Cv 1 Cv 1 Cv j ) Also Xv 1 = λv 2, so ( ) also holds for j 1. So the matrix of X is upper triangular relative to the basis of V, {v 1, v 2,...,v n } where v 2,...,v n are representatives of u 2,...,u n. Lemma 2. Let ϕ : g h a Lie algebra homomorphism and f h an ideal. Then ϕ 1 (f) g is an ideal of g. Proof. ϕ 1 (f) = {X g ϕ(x) f}. Take any X ϕ 1 (f), Y g. We want to prove that X, Y ] ϕ 1 (f) ϕ ( X, Y ] ) f ] ϕ(x), ϕ(y ) f f h but this is true because f is an ideal.

26 26 PROF. MAXIM NAZAROV Remark. ϕ 1 (f) is a subspace of g linearity of ϕ. Lemma 3 (Dynkin s Lemma). Let V be a vector space, dim V < and gl(v ) g any subalgebra. Let h g be any ideal and let λ : h C be a linear function. Consider W = {v V Y v = λ(y )v Y h}, note 0 W W and also W is a subspace, i.e. W V. Then W is g-invariant, that is for all v W, X g then Xv W. Proof. later but non-examinable. Proof. (of Lie Theorem) By induction on dim g. g gl(v ) dim g dim gl(v ) = (dim V ) 2 <. Case dim g = 0, the induction base. Then g = {0}, 0.v = 0v. So any non-zero v V is an eigenvector. Assume now that dim g > 0 and that the Lie Theorem holds for Lie algebras of dim < dim g. Three steps now, (1) g is solvable g g. Otherwise the derived series of g would be g (0) = g, g (1) = g = g, g, g, {0}. So would not terminate. So g g, g g g/g {0} and dim g/g 1. Tke any subspace f g/g such that dim f = dim(g/g ) 1 0. We know that g/g is an Abelian Lie algebra, which implies that f is an ideal of g/g for all X g/g, Y f we have X, Y ] = 0 f. Now apply Lemma 1 to g = g, h = g/g and f = f. ϕ = π : g g/g ; X X + g a canonical homomorphism. Then π 1 (f) g is an ideal by Lemma 1. So we have π 1 (f) is solvable by Theorem A on solvability and g solvable. We have, and π 1 (f) g gl(v ) dim π 1 (f) = dim(g/g ) 1 + dim(ker π) = (dim g dim g ) 1 + dim g = dim g 1 < dim g (2) Induction assumption applies to π 1 (f) gl(v ). Then there exists a ṽ V with ṽ 0 such that Xṽ = λ(x)ṽ for all X π 1 (f). Here λ : π 1 (f) C is a linear function. So, consider W = {v V Xv = λ(x)v for all X π 1 (f)} because ṽ W. Apply Dynkin Lemma to λ and h = π 1 (f). Then this W is g-invariant by Dynkin Lemma. (3) dim π 1 (f) = dim g 1. Write g = π 1 (f) C.X 0 (X 0 is a complementary vector to π 1 (f), i.e. X 0 π 1 (f), by definition of direct sum) for some X 0 g (this is a direct sum of vector spaces, not necessarily of Lie algebras). So, X 0 v W for all v W. So X 0 W : W W a linear transformation. Because V is a vector space over C so is W and dim W > 0, X 0 W is a linear transformation of W. Hence there is an eigenvector v 0 W for X 0 W. So X 0 v 0 = µ.v 0, v 0 0. Claim this v 0 is an

27 LIE ALGEBRAS 27 eigenvector for all X g. Take any X g = π 1 (f) CX 0. So X = Y + zx 0, for z C and Y π 1 (f). So, Xv 0 = (Y + zx 0 )v 0 = Y v 0 + zx 0 v 0 = λ(y )v 0 + µzv 0 = (λ(y ) + µz)v 0 Remark. (1) By Lie theorem (by its Corollary) any solvable subalgebra of gl(v ) can be represented by upper triangular matrices (choosing a good basis in V ). (2) Also, take any basis {v 1,...,v n } in V and identify gl(c) = gl n C. Consider u = {A gl n (V ) A ij = 0 for i > j} (Exercise: check that u is solvable). Now, take any g u a subalgebra. By Theorem A, g is solvable. (3) This means that g gl(v ) is solvable iff g can be represented by upper triangular matrices. 4. Semi-simple Lie Algebras Let g be any finite dimensional Lie algebra over C. Lemma 4. Let h 1, h 2 g be any ideals of g. Then, h 1 + h 2 h 2 = h 1 h 1 h 2. Proof. Define a map ϕ : h 1+h 2 h 2 h 1 h 1 h 2 as follows; take any X 1 + (X 2 + h 2 ) h 1 + h 2 h 1 h 2 h 2 N.B. X 2 + h 2 = X 1 + h 2 X 1 + (h 1 h 2 ). Then ϕ is linear and onto. To show it is a bijection we only need to show it is 1-1. So, if for some X 1, X 1 h 2. Then, ϕ(x 1 + h 2 ) = ϕ(x 1 + h 2 ) X 1 + (h 1 h 2 ) = X 1 + (h 1 h 2 ) or X 1 X 1 h 1 h 2. X 1 X 1 h 2 in particular. Then X 1 + h 2 = X 1 + h 2 and hence ϕ is 1-1. Now we need to show that ϕ is a homomorphism. So, take any X 1, Y 1 h 1 so X 1 + h 2, Y 1 + h 2 h 1+h 2 h 2. Their Lie bracket is then, X 1 + h 2, Y 1 + h 2 ] = X 1, Y 1 ] + h 2

28 28 PROF. MAXIM NAZAROV X 1, Y 1 ] X 1, Y 1 ] + (h 1 + h 2 ) h 1 h 1 h 2 ϕ(x1 + h 2 ), ϕ(y 1 + h 2 ) ] = X 1 + h 1 h 2, Y 1 + h 1 h 2 ] = X 1, Y 1 ] g + (h 1 h 2 ) Hence X 1 + h 2, Y 1 + h 2 ] = ϕ(x 1 + h 2 ), ϕ(y 1 + h 2 ) ] and so ϕ is a homomorphism. More Theorems on solvable Lie Algebras D) Let h 1, h 2 g be any solvable ideals of g. Then h 1 + h 2 = {X 1 + X 2 g X 1 h 1, X 2 h 2 } is also a solvable ideal of g. Proof. Let Y g. Then, examining the Lie bracket we see X 1 + X 2 h 1 +h 2, Y ] = X 1 h 1, Y ] + X 2 h 2, Y ] h 1 + h 2. Now proving the solvability of h 1 + h 2 ; consider π :h 1 h 1 h 1 h 2 X 1 X 1 + (h 1 h 2 ) h Theorem B (of solvability) implies that 1 h 1 h 2 is solvable because h 1 is solvable, h 1 h 2 h 2. By Theorem A we have that h 1 h 2 is solvable. Note, h 1 is solvable by assumption. Next step is to use the Lemma; Now, h 1 h 1 + h 2 = h 1 h 2 h 2 solvable solvable. h 1 + h 2 is solvable h 2 h 1 + h 2 is solvable. Thm C h 2 is solvable Definition: The Radical of any Lie algebra g, is the maximum solvable Lie algebra of g, R(g). That is, if h g is any solvable ideal, then R(g) h Proposition. R(g) exists for any finite dimensional Lie algebra g.

29 LIE ALGEBRAS 29 Proof. Let {h α : α A} = H be all solvable ideals of g. Pick up h = h 1 H. Then, take another, say h 2 H. Consider h 1 +h 2 H. Take another, h 3 H, then, (h 1 +h 2 )+h 3 H and so on... At each step, there are 2 possibilities either; or (h h k 1 ) + h k (h h k 1 ) (h h k 1 ) + h k = h 1... h k 1 obsolete step. Therefore no need to make steps of the second kind. So, assume, h 1 h 1 + h 2 h 1 + h 2 + h 3 g dim d 1 < d 2 < d 3 < < d = dim g. So process will terminate, can t add anything else, then h h k = R(g) Definition: a Lie algebra g is semi-simple if R(g) = {0}. That is, the only solvable ideal of g is {0}. Theorem 5. g R(g) g R(g) is always semi-simple. Proof. Suppose is not semi-simple, so there exists h g Canonical homomorphism, R(g) a solvable ideal; h {0}. then π :g g R(g) π 1 (h) h ideal ideal solvable π : π 1 (h) h h = π 1 (g) ker π = π 1 (h) R(g)

30 30 PROF. MAXIM NAZAROV solvable. R(g) is solvable. By Theorem C we have that π 1 (h) is solvable, π 1 (h) R(g) maximal. π ( π 1 (h) ) = {0} π ( R(g) ) = {0} not solvable. Theorem 6 (on semi-simple Lie algebras). Suppose g 1 and g 2 are both semi-simple Lie algebras. Then g 1 g 2 is also semi-simple. Proof. Consider g = g 1 g 2 h - solvable ideal. Need to show that h = {0}. So that R(g) = {0}. Consider π 1 : g 1 g 2 g 1 ; (X 1, X 2 ) X 1. claim: π 1 (h) is an ideal of g 1 then h-ideal for all (X 1, X 2 ) h and for all (Y 1, Y 2 ) g 1 g 2 = g. h (X 1, X 2 ), (Y 1, Y 2 ) ] g = ( X 1, Y 1 ] g1, X 2, Y 2 ] g2 ) π 1 (h) X 1, Y 1 ] g1 for all X 1 π 1 (h) and for all Y 1 g 1 π 1 (h) is a vector subspace of g 1 because π 1 is linear. So π 1 (h) is an ideal. Next, π 1 is a homomorphism (Exercise). Then, by Theorem B on solvability, π 1 (h) g 1 is solvable. But g 1 is semi-simple π 1 (h) = {0} for all (X 1, X 2 ) h, X 1 = 0 (and we prove that X 2 = 0). Similarly, using π 2 : g 1 g 2 g 2 ; (X 1, X 2 ) X 2 h = {0} Definition: g is simple, if dim g > 1, and g has no ideals except for g itself and {0}. Example 4.1. g = sl 2 C (Exercise 15, proof given already). dimg = 3 > 1. Proposition. g 1,...,g k are simple g = g 1 g k is semi-simple. Proof. by induction on k. (1) k = 1. Suppose g = g 1 is simple. Need to show that g is semi-simple. We know that the only ideals of g are g and {0}. Take any solvable ideal, h, of g. If h = {0} we are done. So, we are left with the case when h = g. Can g = h be solvable? Consider g g g... Then g is an ideal of g. If g = {0} then g is Abelian, dim g > 1 because g is simple. Take f g a subspace such that dim f = 1. Then f g is an ideal. So f ideal of g different from g, {0} but this is a contradiction. Hence g {0} g = g = g = g =..., does not terminate. g is not solvable. This completes the proof for k = 1

31 LIE ALGEBRAS 31 (2) Induction step. g 1...g k are simple and let the induction assumption be g 1 g k 1 is semi-simple. g = (g 1 g k 1 ) }{{} g }{{} k semi-simple semi-simple semi-simple by Theorem Theorem 7 (without proof). Let g be any semi-simple Lie algebra. finite dimensional over C (or R). Then there exist ideals g 1,...,g k of g such that each of g 1,...,g k is a simple Lie algebra and g = g 1 g k. Moreover, these g 1,..., g k are unique up to permuting them. 5. Classification of Simple Lie algebras Theorem 8 (without proof). Any simple finite dimensional Lie algebra over C is isomorphic to one in the following table. Case A l, l 1 B l, l 1 C l, l 1 D l, l 3 G 2 F 4 E 6 E 7 E 8 Description sl l+1 C so 2l+1 C sp 2l C so 2l Dimension (l + 1) 2 (2l)(2l+1) 2l(2l 1) 1 l(2l + 1) Comments (1) all algebras in the table are not isomorphic, except for B 2 = C2 ; D 3 {( ) = A3 0 x (2) D 1 = so 2 = x C}, Abelian with dim = 1. x 0 D 2 = so 4 C = A 1 A 1 = sl 2 C sl 2 C (already mentioned.) (3) About proof: any simple Lie algebra can be described by a Dynkin diagram, with l vertices. Then classify diagrams. Lemma 5. Let gl(v ) g h be an ideal of a Lie algebra g and let λ : h C be any linear function. Consider the subspace of V W = W λ = {v V Y v = λ(y )v for all Y h}. Then W is g-invariant, i.e. Xv W for all X g, v W. Example 5.1. it may be that W λ = {0}. 0 = X 0 {0}. Proof. (non exminable). (1) Take any X g, Y h and v W = W λ ; v 0. Then Y v = λ(y )v. Need to show that }{{} Xv W that is Y Xv = λ(y )Xv ( ). Y Xv = (XY + Y X XY )v = XY v + Y, X] v }{{} h

32 32 PROF. MAXIM NAZAROV and h is an ideal = λ(y )Xv + λ ( Y, X] ) v (2) Let us prove that λ ( Y, X] ) = 0, then ( ) OK. Fix X g. Consider the vectors v 0 = v, v 1 = Xv, v 2 = X 2 v,..., v i = X i v and consider U i = span(v 0, v 1,...,v i ) U 0 U 1 U 2 U i U i+1 V. Can it happen that U i U i+1, dim U i < dim U i+1 for all i. No, because dim U 0 dim U 1 dim V <. So, let k = min{i U i = U i+1 } - exists. U 0 U 1 U 2 U k = U k+1 = U k+2 = U k+3 = V For instance, let us check that U k+1 = U k+2. So, U k+2 = span(v 0, v 1,...,v k+1, v k+2 ) U k+1 = span(v 0, v 1,...,v k+1 ) We have that U k+1 U k+2 V. Need to show that v k+2 = X k+2 v U k+1. We know that U k+1 = U k =span(v 0, v 1,..., v k ) v k+1 = X k+1 v U k U k+1 X k+2 v = X(X k+2 v) X span(v 0, v 1,...,v k ) =span(xv 0, Xv 1,...,Xv k ) =span(v 1, v 2,...v k+1 ). So indeed, U k+1 = U k+2 = U k+3 =..., a similar proof. U 0 = span(v 0 ) U 1 = span(v 0, v 1 ) {v 0, v 1 } linearly independent. U 2 = span(v 0, v 1, v 2 ) {v 0, v 1, v 2 } linearly independent.. U k = span(v 0, v 1, v 2,...v k ) {v 0, v 1, v 2,...v k } linearly independent. So, {v 0, v 1,..., v k } is a basis for U k, so that dim U k = k + 1. (Note: U 0 U 1 U 2 U k ) (3) We will show by induction on i = 0, 1,..., k that (a) U i is Y -invariant, i.e. Y (U i ) U i.

33 LIE ALGEBRAS 33 (b) the matrix of Y Ui relative to the basis {v 0, v 1,...,v i } is upper triangular with λ(y ) on the diagonal. In particular Y v i = λ(y )v i + U i 1 λ(y ) λ(y ) So, now proving this using induction. (I) i = 0 then U 0 = Cv W. Y v = λ(y )v so U 0 is Y -invariant. Y v 0 λ(y )v 0 (II) The induction step. Assume that parts (a) and (b) are true for indices 0, 1,..., i 1 instead of i. Need to prove parts (a) and (b) for i itself. Y v j λ(y )v j +U j 1 ( ) for j = 0, 1,...,i 1 - by our assumption. Y V i = Y X i v = Y XX i 1 v = Y Xv i 1 = XY v i 1 + Y, X] h g }{{} h X ( λ(y )v i 1 + U i 2 ) + λ ( Y, X] ) vi 1 + U i 2 =... But inductive assumption is made for all Y h, in particular for Y, X] instead of Y. v i 1 λ(y )v i + U i 1 + λ ( Y, X] ) v i 1 + U i 2 }{{} U i 1 so Y v i λ(y )v i + U i 1, i.e. ( ) is true for j = 0,...,i. ( ) is true for v 0,...,v i - basis in U i (a) Y (v j ) span(v j, v j 1,...,v 0 ) true. (b) also true. (4) trace(y Ui ) = (i + 1)λ(Y ) true for all Y h also ( ) trace ( ) ( ) Y, X] }{{} Ui = (i + 1)λ Y, X] h put i = k. X preserves U k - by construction and Y preserves U k - induction. Y, X] Uk = (Y X XY ) Uk = Y Uk X Uk X Uk Y Uk trace ( ) Y, X] Uk = 0 = (k + 1) λ ( Y, X] ) >0 λ ( Y, X] ) = 0 by ( )

34 34 PROF. MAXIM NAZAROV 6. Representations of Lie Algebras Definition: For any Lie algebra g a representation of g, over C, is any Lie algebra homomorphism ϕ : g gl(v ) where V is any vector space over C. Remark. if dim V = n < then gl(v ) = gl n C. Remark. it is possible that ker ϕ = {X g ϕ(x) = 0} {0}. For instance, one can define a representation by setting ϕ(x) = 0 for all X g. (called trivial representation). Definition: ϕ is faithful if ker ϕ = {0}. Theorem 9 (Ado). If dim g < then there exists a faithful representation of g (without proof). Definition: suppose ϕ 1 : g gl(v 1 ) and ϕ 2 : g gl(v 2 ) are representations of g. Then ϕ 1 ϕ 2 is a representation of g defined as follows: V = V 1 V 2 ϕ : g gl(v }{{} 1 V 2 ) X ϕ(x)(u 1, u 2 ) = ( ) ϕ 1 (X)u 1, ϕ 2 (X)u 2 Comment: suppose ϕ : g gl(v ) a representation and U V such that ϕ(x)u U or ϕ(x)u U (for u U). Then X g ϕ(x) U gl(u); X ϕ(x) U is a homomorphism g gl(u) another representation. Definition: ϕ is an irreducible representation if the only subspaces U V such that ϕ(x)u U for all X g are the trivial ones, U = V, {0}. For example, question 17 gives a full list of all irreducible representations of g = sl 2 C. 7. Revision 7.1. Main definitions in order of appearance Lie algebra g (3 axioms) Abelian Lie algebra (X, Y ] 0) Homomorphism of Lie algebras (ϕ : g 1 g 2 ) Isomorphism (homomorphism & bijection) Direct sum of Lie algebras (g 1 g 2 ) Derivations of Lie algebras (δ : g g linear, Leibniz rule) - inner (ad Z : g g), outer derivations. A central element Z of g, the centre Z(g) = {Z g Z central}. Lie subalgebra h g then X, Y h] for all X, Y h

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