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Finite Dimensional Lie Algebras andtheir Representations
Michaelmas 2006Prof. I. Grojnowski∗
Chris Almost†
The exam questions will consist of a subset of the in-class exercises.
Contents
Contents 1
1 Motivation and Some Definitions 21.1 Lie Groups and Lie Algebras . . . . . . . . . . . . . . . . . . . . . . . . 21.2 Representations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
2 Representations of sl2 72.1 Classification . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.2 Consequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112.3 Characters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
3 Semi-Simple Lie Algebras: Structure and Classification 123.1 Solvable and Nilpotent Lie Algebras . . . . . . . . . . . . . . . . . . . 133.2 Structure Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
4 Root Systems 224.1 Abstract Root Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 224.2 Dynkin Diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 264.3 Existence and Uniqueness of an Associated Lie Algebra . . . . . . . . 28
5 Representation Theory 305.1 Theorem of the Highest Weight . . . . . . . . . . . . . . . . . . . . . . 305.2 Proof of the Theorem of the Highest Weight . . . . . . . . . . . . . . 335.3 Weyl Character Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . 37
6 Crystals 416.1 Littelmann Paths . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
Index 49∗[email protected]†[email protected]
1
2 Lie Algebras
1 Motivation and Some Definitions
1.1 Lie Groups and Lie Algebras
This class is about the structure and representations of the so-called simple Liegroups
SLn = A∈Matn | det A= 1, SOn = A∈ SLn | AAT = I, Sp2n
and exactly five others. The only technology we will use is linear algebra. Thesegroups are slightly complicated, and they have a topology on them. To study themwe pass to a “linearization,” the Lie algebra, and study these instead.
1.1.1 Definition. A linear algebraic group is a subgroup of GLn, for some n, de-fined by polynomial equations in the matrix coefficients.
1.1.2 Example. All of the groups mentioned so far are linear algebraic groups, asis the collection of upper triangular matrices.
This definition requires an embedding; it is desirable to come up with onewhich does not. It turns out the linear algebraic groups are exactly the affine alge-braic groups, i.e. affine algebraic varieties such that multiplication and inversionare regular functions.
1.1.3 Definition. The Lie algebra associated with G is g= T1G, the tangent spaceto G at the identity.
1.1.4 Example. Let g = 1 0
0 1
+ ε
a bc d
, where ε2 = 0. Then
det g = (1+ εa)(1+ εd)− bcε2 = 1+ ε(a+ d)
Therefore det g = 1 if and only if tr g = 0.
We are working in the ring E = C[ε]/ε2 = 0 (the so-called dual numbers). IfG ⊆ GLn is a linear algebraic group, we can consider
G(E) = A∈Matn(E) | A satisfies the equations defining G.
There is a natural map π : G(E) → G(C) induced by the ring homomorphismπ : ε 7→ 0.
1.1.5 Definition. G(E) is the tangent bundle to G, denoted T G, so
g= T1G = π−1(I) = A∈MatnC | I + εA∈ G(E).
1.1.6 Exercise. Show that this definition is equivalent to the usual definition ofT G (say, from differential geometry).
1.1.7 Examples.1. Let G = GLn = A∈Matn | A−1 ∈Matn. Then
G(E) = GLn ⊕Matnε = A+ Bε | A−1 ∈Matn,
since (A+ Bε)(A−1 − A−1BA−1ε) = I . Therefore
gln := T1GLn =Matn∼= Rn2
.
Lie Groups and Lie Algebras 3
2. As an exercise, show that det(I + Xε) = 1+ tr(X )ε. As a corollary, sln :=T1SLn = X ∈Matn | tr(X ) = 0.
3. Let G = On = A | AT A = I. Then (I + Xε)T (I + Xε) = I + (X T + X )ε, soon := T1On = X ∈Matn(C) | X T + X = 0.
Remark. If X ∈ on then tr X = 0 (when 2 6= 0), so the Lie algebra of On is containedin the Lie algebra of SLn. Also note that on = son, so the Lie algebra does not keeptrack of whether the group is connected.
What structure on g is there coming from the group structure on G? Noticethat
(I + Aε)(I + Bε) = I + (A+ B)ε,
so traditional multiplication only sees the vector space structure. Instead we usethe commutator map
G× G→ G : (P,Q) 7→ PQP−1Q−1,
and consider it infinitesimally. If P = I + Aε and Q = I + Bδ, then P−1 = I − Aεand Q−1 = I − Bδ, so
PQP−1Q−1 = I + (AB− BA)εδ.
Let [A, B] denote the map (A, B) 7→ AB− BA.
1.1.8 Exercises.1. Show that (PQP−1Q−1)−1 =QPQ−1P−1 implies [X , Y ] =−[Y, X ]
2. Show that the fact that multiplication is associative in G implies
[[X , Y ], Z] + [[Y, Z], X ] + [[Z , X ], Y ] = 0
1.1.9 Definition. Let K be a field of characteristic not 2 or 3. A Lie algebra g overK is a vector space over K equipped with a bilinear map [·, ·] : Λ2g → g, the Liebracket, such that for all X , Y, Z ∈ G,
1. [X , Y ] =−[Y, X ] (skew-symmetry); and
2. [[X , Y ], Z] + [[Y, Z], X ] + [[Z , X ], Y ] = 0 (Jacobi identity).
1.1.10 Example. In most of the following the Lie bracket is inherited from gln.The following are Lie algebras.
1. sln = A∈ gln | tr A= 0;
2. son = A∈ gln | AT + A= 0;
3. sp2n = A∈ gl2n | JAT J−1 + A= 0, where
J = anti-diag(−1, . . . ,−1︸ ︷︷ ︸
n
, 1, . . . , 1︸ ︷︷ ︸
n
).
4. Upper triangular matrices;
4 Lie Algebras
5. Strictly upper triangular matrices;
6. Let V be any vector space and define the Lie bracket to be the zero map.This example is known as the Abelian Lie algebra.
1.1.11 Exercises.1. Check directly that gln = Matn is a Lie algebra with Lie bracket [A, B] =
AB− BA, and in fact this works for any K-algebra;
2. Check that all of the above are truly subalgebras, i.e. vector subspaces andclosed under the Lie bracket.
1.1.12 Exercise. Classify all Lie algebras g over K with dimK g≤ 3.
1.2 Representations
1.2.1 Definition. A Lie algebra homomorphism ϕ : g1 → g2 is a linear map suchthat
ϕ([x , y]1) = [ϕ(x),ϕ(y)]2.
A Lie algebra representation (or simply representation) of a Lie algebra g on a vectorspace V is a Lie algebra homomorphism ϕ : g→ glV .
1.2.2 Example. If g ⊆ glV then the inclusion is a representation of g on V . Forexample, Kn is a representation of son ⊆ gln.
1.2.3 Definition. Let x ∈ g and define the adjoint of x to be
ad x : g→ g : y 7→ [x , y],
a linear map on g.
1.2.4 Lemma. The adjoint map ad : g → End(g) is linear and is in fact ad is arepresentation of g on itself called the adjoint representation.
PROOF: We must check that ad[x , y] = ad x ad y − ad y ad x . For any z,
0= [[x , y], z] + [[y, z], x] + [[z, x], y] Jacobi identity
0= [[x , y], z]− [x , [y, z]] + [y, [x , z]] skew-symmetry
[[x , y], z] = [x , [y, z]]− [y, [x , z]]
so ad[x , y](z) = (ad x ad y − ad y ad x)(z) for every z, implying the result.
1.2.5 Definition. The center of g is
Zg = x ∈ g | [x , y] = 0 for every y ∈ g= ker(ad : g→ End(g)).
Warning: the notation Zg is not used in class (in fact, no notation for center isused in class).
By the definition of center, the adjoint homomorphism embeds g into glg if andonly if g has trivial center.
Representations 5
1.2.6 Example. The adjoint homomorphism maps g= 0 K
0 0
to zero in glg.
1.2.7 Theorem (Ado). Any finite dimensional Lie algebra g has a finite dimen-sional faithful representation, i.e. g ,→ gln for some n.
PROOF: Omitted.
1.2.8 Example. Recall that sl2 =
a bc −a
. Its standard basis is
e =
0 10 0
, f =
0 01 0
, h=
1 00 −1
.
Show that [h, e] = 2e, [h, f ] = −2 f , and [e, f ] = h. Whence, a representation ofsl2 on Cn is a triple of matrices E, H, F such that [H, E] = 2E, [H, F] = −2F , and[E, F] = H. How do we find such a thing?
1.2.9 Definition. If G is an algebraic group then an algebraic representation of Gis a group homomorphism ρ : G→ GLV defined by polynomial equations.
We pull the same trick and substitute E = C[ε]/ε2 for C to get a homomor-phism of groups G(E)→ GLV (E). As ρ(I) = I ,
ρ(I + Aε) = I + ε(some function of A).
Call this function dρ, so that ρ(I + Aε) = I + εdρ(A), giving dρ : g→ glV .
1.2.10 Exercises.1. Show that if ρ : G→ GLV is a group homomorphism then dρ : g→ glV is a
Lie algebra homomorphism.
2. Show that dρ truly is the differential of ρ at the identity when ρ is consid-ered as a smooth map between manifolds.
We have a functor ρ 7→ dρ from the algebraic representations of G to theLie algebra representations of g = T1G. We will discuss this further, but first anexample.
1.2.11 Example. Let G = SL2, and let L(n)1 be the collection of homogeneouspolynomials of degree n in variables x and y , so dimK L(n) = n+1. Indeed, it hasa basis xn, xn−1 y, . . . , x yn−1, yn. Now GL2 acts on L(n) via
ρn : GL2→ GL(L(n)) = GLn+1
by (ρn(
a bc d
) f )(x , y) = f (ax + c y, bx + d y). Then ρ0 is the trivial representa-tion, ρ1 is the standard 2-dimensional representation (i.e. ρ1 is the identity mapon GL2), and ρ2
a bc d
has matrix
a2 ab b2
2ac ad + bc 2bdc2 cd d2
.
1 L(n) = Γ(P1,O (n)) and H1(P1,O (n)) = 0 for all n≥ 0.
6 Lie Algebras
Now of course SL2 < GL2 so the restriction is a representation of SL2. Tocompute dρn(e)(x i y j) (recall e =
0 10 0
) we compute:
ρn(1+ εe)xi y j = ρn
1 ε0 1
x i y j = x i(εx + y) j = x i y j + ε j x i+1 y j−1
Therefore dρn(e)(x i y j) = j x i+1 y j−1. Similarly,
ρn(1+ ε f )x i y j = ρn 1 0ε 1
x i y j = (x + ε y)i y j = x i y j + εi x i−1 y j+1
and
ρn(1+ ε f )x i y j = ρn
1+ε 00 1−ε
x i y j = (1+ ε)i x i(1− ε) j y j = x i y j + ε(i− j)x i y j ,
so
e · x i y j = j x i+1 y j−1 f · x i y j = i x i−1 y j+1 h · x i y j = (i− j)x i y j .
Upon closer inspection, these are operators that we know and love,
dρn(e) = x∂
∂ y, dρn( f ) = y
∂
∂ x, dρn(h) = x
∂
∂ x− y
∂
∂ y. (1)
1.2.12 Exercises.1. Check directly that defining the action of e, f , and h on L(2) by (1) gives a
representation of sl2.
2. Check that L(2) is the adjoint representation of sl2 (notice that they are both3-dimensional).
3. If the characteristic of K is zero, check that L(n) is an irreducible represen-tation of sl2, and hence of the group SL2.
1.2.13 Example. Let’s compare representations of G and representations of g
when G = C∗ (and hence g = C with [x , y] = 0 for all x , y ∈ C). Recall thatthe irreducible algebraic representations of G are one dimensional. z ∈ C∗ acts asmultiplication by zn (n ∈ Z). Moreover, any representation of G splits up into afinite direct sum of one dimensional representations (“finite Fourier series”).
In contrast, a representation ρ : g → End(V ) of g = C on V is any matrixA = ρ(1) ∈ End(V ) (ρ : λ 7→ λA). A subrepresentation W ⊆ V is just an A-stable subspace (so AW ⊆ W ). Any linear transformation has an eigenvector,so there is always a 1-dimensional subrepresentation, which implies that all ir-reducible representations are the 1-dimensional ones. V breaks up into a directsum of irreducibles (i.e. is completely reducible) if and only if A is diagonaliz-able. e.g. if A is the Jordan block of size n, then the invariant subspaces are⟨e1⟩, ⟨e1, e2⟩, . . . , ⟨e1, . . . , en⟩, and does not break up into a direct sum of subspaces.Representations up to isomorphism of gl1 are given by the theory of Jordan normalforms.
Representations of G and g look “really different” in this case. Even theirreducibles are different, since for G they are in 1-1 correspondence with Z(z 7→ zn), while for g they are in 1-1 correspondence with C (λ 7→ “1 7→ λ”).(Here Z ,→ C : z 7→ 2πiz.)
Representations of sl2 7
In contrast, the map from representations of G to representation of g is anequivalence of categories if G is a simply connected simple algebraic group. In par-ticular, every representation of the Lie algebra sl2 is a direct sum of irreducibles,and the irreducibles are exactly L(n).
2 Representations of sl2
2.1 Classification
From now on, all representations and Lie algebras are over C unless otherwisespecified.
2.1.1 Theorem.1. There is a unique irreducible representation of sl2 of dimension (n+ 1) for
every n≥ 0.
2. Every finite dimensional representation of sl2 is isomorphic to a direct sumof irreducible representations.
PROOF (FIRST PART): Let V be a representation of sl2. Define
Vλ := v ∈ V | hv = λv,
the λ-weight space. It is the space of eigenvectors of h with eigenvalue λ. Forexample, L(n)λ = Cx i y j if i− j = λ, and in particular L(n)n = Cxn. Suppose thatv ∈ Vλ, and consider ev.
h(ev) = (he− eh+ eh)v = [h, e]v+ ehv = 2ev+λev = (2+λ)ev
Whence v ∈ Vλ if and only if ev ∈ Vλ+2. Similarly, v ∈ Vλ if and only if f v ∈ Vλ−2.A highest weight vector of weight λ is an element of ker(e) ∩ Vλ, i.e. a v ∈ V
such that ev = 0 and hv = λv.
Claim. If v is a highest weight vector then W = spanv, f v, f 2v, . . . is an sl2-invariant subspace of V .
We need to show that W is stable by e, f , h. It is clear that f W ⊆ W . Ashf k v = (λ− 2k) f k v, it is clear that hW ⊆W . By definition, ev = 0, so
e f v = (e f − f e)v + f ev = hv = λv ∈W.
Similarly,
e f k+1v = (e f − f e+ f e) f k v
= hf k v+ f e f k v
= (λ− 2k) f k v + k(λ− k+ 1) f k v by induction
= (k+ 1)(λ− k) f k v,
so e f k v = k(λ − k + 1) f k−1v ∈ W , by induction, so eW ⊆ W and the claim isproved.
8 Lie Algebras
Claim. If v is a highest weight vector of weight λ then λ ∈ 0, 1,2, . . . if V isfinite dimensional.
Indeed, the vectors f k v all lie in different eigenspaces of h, and hence arelinearly independent if they are non-zero. If V is finite dimensional then f k v = 0for some k. Let k be minimal with this property. Then
0= e f k v = k(λ− k+ 1) f k−1v,
and k ≥ 1, so λ= k− 1 ∈ 0,1, 2, . . . since f k−1v 6= 0.
Claim. There exists a highest weight vector if V is finite dimensional.
Let v ∈ V be any eigenvector for h with eigenvalue λ (such things exist becauseC is algebraically closed). As before, v, ev, e2v, . . . are all eigenvectors for h witheigenvalues λ,λ+ 2,λ+ 4, . . . . Hence they are linearly independent, so there issome k+ 1 minimal such that ek+1v = 0. Clearly ek v ∈ Vλ+2k is a highest weightvector, proving the claim.
If V is irreducible and dimC V = n+ 1 then there is a basis v0, . . . , vn suchthat
hvi = (n− 2i)vi , f vi =
¨
vi+1 if i < n0 if i = n
, evi = i(n− i+ 1)vi−1.
Indeed, let v0 be a highest weight vector with weight λ ∈ 0, 1,2, . . . . The stringv, f v, . . . , f λv is a subrepresentation of V . As V is irreducible it must be all of V ,so λ= dimC V − 1. Take vi = f i v. This proves the first part of the theorem.
2.1.2 Exercise. C[x , y] is a representation of sl2 by the formulae in 1.2.11. Showthat for λ,µ ∈ C, xλ yµC[x , y] is also a representation, infinite dimensional, andanalyze its submodule structure.
PROOF (SECOND PART): We will show that strings of different lengths “do not inter-act,” and then that strings of the same length “do not interact.”
Let V be a finite dimensional representation of sl2. Let
Ω := e f + f e+ 12h2 ∈ End(V ),
the Casimir of sl2. Also notice that Ω = ( 12h2 + h) + 2 f e.
Claim. (1) Ω is central, i.e. Ωe = eΩ, Ω f = f Ω, and Ωh= hΩ.
The proof of this is an exercise. For example,
Ωe = e(e f + f e+ 12h2)
= e(e f − f e) + 2e f e+ 12heh+ 1
2(eh− he)h
= eh+ 2e f e+ 12heh− eh
= 2e f e+ 12heh
= (e f + f e+ 12h2)e
Classification 9
Claim. (2) If V is an irreducible representation with highest weight n then Ω actson V by multiplication by 1
2n2 + n.
(It follows from Schur’s Lemma that if V is an irreducible representation of sl2then Ω acts on V as multiplication by a scalar.) Let v be a highest weight vector,so hv = nv and ev = 0, so
Ωv = (( 12h2 + h) + 2 f e)v = ( 1
2n2 + n)v.
Hence Ω( f k v) = f k(Ωv) = ( 12n2 + n) f k v, so Ω acts by multiplication by 1
2n2 + n
since v, f v, . . . , f nv is a basis for V .
Claim. (3) If L(n) and L(m) are both irreducible representations and Ω acts onthem by the same scalar ω then n= m (so L(n)∼= L(m)) and ω= 1
2n2 + n.
Indeed, L(n) and L(m) are irreducible representations with highest weight nand m, respectively, so by the last claim 1
2n2+n=ω= 1
2m2+m. But f (x) = 1
2x2+
x is a strictly increasing function for x > −1, so m = n and the representationsare isomorphic.
Let Vω = v ∈ V | (Ω−ωI)dimC V v = 0 be the generalized eigenspace of Ωassociated with ω. The Jordan decomposition for Ω implies that V ∼=
⊕
ω Vω (asvector spaces).
Claim. (4) Each Vω is a subrepresentation of V , i.e. the above is a direct sumdecomposition as representations of sl2.
Indeed, if x ∈ sl2 and v ∈ Vω then since Ω is central,
(Ω−ω)dimC V x v = x(Ω−ω)dimC V v = 0.
Aside: For a g-module W , a composition series is a sequence of submodules
0=W0 <W1 < · · ·<Wr =W
such that each quotient module Wi/Wi−1 is an irreducible g-module. For example,with g= C and W = C2, acting as 1 7→
0 10 0
, then ⟨e1⟩ ⊆ ⟨e1, e2⟩ is a compositionseries (and this is the only one). If the action is instead 1 7→
0 00 0
then any linein C2 gives rise to a composition series.
Recall that if W is finite dimensional then it has a composition series, and thesubquotients are unique up to permutations.
Claim. (5) If Vω 6= 0 then ω = 12n2 + n for some n, and all the subquotients of a
composition series for Vω are L(n), i.e. Vω is a bunch of copies of L(n), possiblystuck together in a complicated way.
Indeed, suppose W is an irreducible subrepresentation of Vω. Then Ω acts onW as multiplication by ω, since ω is the only possible generalized eigenvalue andΩ acts on irreducible representations as scalar mulitplication by Schur’s Lemma.Thusω= 1
2n2+n for some n with W ∼= L(n), by claim (3). By the same reasoning,
Ω will act on any irreducible subrepresentation of Vω/W as multiplication by ω,so any such irreducible subrepresentation must be L(n), for the same n. Whenceevery subquotient in a composition series is isomorphic to L(n).
10 Lie Algebras
To finish the proof we must show that each Vω is a direct sum of L(n)’s. Thekey point will be to show that h is “diagonalizable.” At this stage we may assumethat V = Vω.
Notice that h acts on V with eigenvalues n, n− 2, . . . , 2− n,−n. (It is a factfrom linear algebra that if h acts on a vector space W and W ′ ≤W is an h-invariantsubspace then the eigenvalues of h on W are the eigenvalues of h on W ′ togetherwith the eigenvalues of h on W/W ′. The remark follows from this since V has acomposition series with subquotients all isomorphic to L(n), and we know fromthe first part that the eigenvalues of h on L(n) are n, n−2, . . . , 2−n,−n.) HenceVm = 0 if m /∈ n, n− 2, . . . , 2− n,−n. It follows that h has only one generalizedeigenvalue on ker(e), namely n.Why does this follow?
Claim. hf k = f k(h− 2k) and e f n+1 = f n+1e+ (n+ 1) f n(h− n).
The first equation is trivial, and the second is proved by induction as follows.Clearly e f = f e+ e f − f e = f e+ h, and
e f n+1 = ( f e+ e f − f e) f n
= f n+1e+ nf n(h− n+ 1) + f n(h− 2n) by induction
= f n+1e+ (n+ 1) f n(h− n).
Claim. h acts as multiplication by n on ker(e), and further, Vn = ker(e).
Indeed, if v ∈ Vn (so hv = nv) then ev ∈ Vn+2 = 0, so Vn ⊆ ker(e). For theconverse, consider these statements:
1. If x ∈ ker(e), then (h− n)dim V x = 0 since n is the only generalized eigen-value of h on ker(e).
2. If x ∈ ker(e), then (h−n+2k)dim V f k x = f k(h−n)dim V x = 0 by the previousclaim.
3. If y ∈ ker(e) and y 6= 0 then f n y 6= 0. (Indeed, let (Wi) be a compositionseries for V , as above. Then there is i such that y ∈ Wi \Wi−1. Let y =y +Wi ∈ Wi/Wi−1
∼= L(n). Then y is a highest weight vector of L(n), sof n y 6= 0, hence f n y 6= 0.)
4. f n+1 x = 0, as f n+1 x is in the generalized eigenspace with eigenvalue−n−2,which is zero.
Hence
0= e f n+1 x = (n+ 1) f n(h− n)x + f n+1ex = (n+ 1) f n(h− n)x
implying f n((h− n)x) = 0, so by statement 3, (h− n)x = 0, or hx = nx .Finally, choose a basis w1, . . . , w` of Vn = ker(e). Then
spanw1, f w1, . . . , f nw1 ⊕ spanw2, f w2, . . . , f nw2 ⊕ . . .
⊕ spanw`, f w`, . . . , f nw`
is a direct sum decomposition of V into subrepresentations, each isomorphic toL(n).
Consequences 11
2.2 Consequences
2.2.1 Exercise. If V and W are representations of a Lie algebra g then the map
g→ End(V )⊗ End(W ) = End(V ⊗W ) : x 7→ x ⊗ 1+ 1⊗ x
is a homomorphism of Lie algebras. (Recall that if a group G acts on V and Wthen G acts on V ⊗W as g 7→ g ⊗ g. An action g→ End(V ⊗W ) is obtained fromthis by differentiating.)
2.2.2 Corollary. If V and W are representations of g then so it V ⊗W .
Remark. If A is an associative algebra and V and W are representations of A thenV ⊗W is a representation of A⊗ A and not of A. So to make it a representationof A you need additional structure, an algebra map A→ A⊗ A (making A a Hopfalgebra, apparently).
2.2.3 Example. Now take g = sl2. Then L(n)⊗ L(m) is also a representation ofsl2, so according to our theorem it should decompose into representations L(k)with some multiplicities, but how?
The naive method is to try and find all the highest weight vectors in L(n)⊗L(m). As an exercise, find all highest weight vectors in L(1)⊗L(n) and L(2)⊗L(n).If vr denotes a highest weight vector in L(r) then vn⊗vm is a highest weight vectorin L(n)⊗ L(m). Indeed,
h(vn ⊗ vm) = (hvn)⊗ vm + vn ⊗ (hvm) = (n+m)vn ⊗ vm
ande(vn ⊗ vm) = (evn)⊗ vm + vn ⊗ (evm) = 0.
This is a start, but L(n)⊗ L(m) = L(n+m) + lots of other stuff, since
dim(L(n)⊗ L(m)) = (n+ 1)(m+ 1)> n+m+ 1= dim L(n+m).
It is possible to write down explicit formulae for all the highest weight vectors inL(n)⊗ L(m); they are messy but interesting (compute them as a non-compulsoryexercise).
A better method is to use the weight space decomposition for h, which we willsee later.
2.3 Characters
2.3.1 Definition. Let V be a finite dimensional representation of sl2. Then
ch V =∑
n∈Zdim Vnzn ∈ Z+[z, z−1]
is the character of V .
2.3.2 Proposition (Properties of Characters).1. ch V |z=1 = dim V
2. ch L(n) = zn + zn−1 + · · ·+ z2−n + z−n = zn+1−z−n−1
z−z−1 =: [n+ 1].
12 Lie Algebras
3. ch V = ch W implies that V ∼=W .
4. ch(V ⊗W ) = ch(V ) ch(W ).
PROOF:1. By our theorem, h acts diagonalizably on V , so V =
⊕
n∈Z Vn with integereigenvalues.
2. We saw this in the proof of our theorem.
3. ch L(n) | n ≥ 0 forms a basis for Z[z, z−1]Z/2Z (the action of Z/2Z is thatof swapping z and z−1). Since any V uniquely decomposes as a direct sum⊕
n≥0 an L(n), we have ch V =∑
n≥0 an[n+ 1], which uniquely determinesthe an.
4. Vn ⊗Wm ⊆ (V ⊗W )n+m, so (V ⊗W )p =∑
n+m=p Vn ⊗Wm.
Back to the example. For example we compute
ch(L(1)⊗ L(3)) = ch L(1) ch L(3)
= (z+ z−1)(z3 + z+ z−1 + z−3)
= (z4 + z2 + 1+ z−2 + z−4) + (z2 + 1+ z−2)
so L(1)⊗ L(3)∼= L(4) + L(2). In general, the Clebsch-Gordan rule says
L(n)⊗ L(m) =k=|n+m|⊕
k=|n−m|k≡n+m mod 2
L(k).
3 Semi-Simple Lie Algebras: Structure and Classification
3.0.3 Definition. A Lie algebra g is simple if
1. g is non-Abelian;
2. dimg> 1; and
3. the only ideals of g are 0 and g.
g is semi-simple if it is a direct sum of simple Lie algebras.
3.0.4 Examples.1. C is not a simple or semi-simple Lie algebra.
2. sl2 is a simple Lie algebra.
Solvable and Nilpotent Lie Algebras 13
3.1 Solvable and Nilpotent Lie Algebras
3.1.1 Definition. Let g be a Lie algebra. The commutator algebra of g is [g,g] =span[x , y] | x , y ∈ g.
3.1.2 Exercises.1. [g,g] is an ideal of g; and
2. g/[g,g] is Abelian.
3.1.3 Definition. The central series of g is defined inductively by g0 = g andgn = [gn−1,g] for n ≥ 1. The derived series is defined by g(0) = g and g(n) =[g(n−1),g(n−1)] for n≥ 1.
It is clear that g(n) ⊆ gn for all n.
3.1.4 Definition. A Lie algebra g is said to be nilpotent if gn = 0 for some n > 0(i.e. if the central series ends at zero, eventually), and solvable if g(n) = 0 for somen> 0 (i.e. if the derived series ends at zero, eventually).
Clearly nilpotent implies solvable.
3.1.5 Example. The Lie algebra n of strictly upper triangular matrices is nilpotent,while the Lie algebra b of upper triangular matrices is solvable but not nilpotent.Abelian Lie algebras are nilpotent.
3.1.6 Exercises.1. Compute the derived and central series of the upper triangular and strictly
upper triangular matrices and confirm the claims made in the example.
2. Compute the center of the upper triangular and strictly upper triangularmatrices.
3. Let L be a vector space. Show that W = L ⊕ L∗ is a symplectic vector space(i.e. a vector space together with a non-degenerate alternating form) withthe alternating form ⟨·, ·⟩ defined by ⟨L, L⟩ = ⟨L∗, L∗⟩ = 0 and ⟨v, f ⟩ = f (v)for v ∈ L and f ∈ L∗ (defined to be alternating).
4. Let W be a symplectic vector space. Define the Heisenberg Lie algebra, HeisW ,to be W⊕Kc, where c is a fresh variable, with bracket defined by [c, W] = 0and [x , y] = ⟨x , y⟩c for x , y ∈ W . Show that HeisW is a nilpotent Liealgebra. Is it Abelian?
5. Show that the simplest Heisenberg algebra (taking L = Cp and L∗ = Cq)has a representation on C[x] by p 7→ ∂
∂ x, q 7→ x , and c 7→ 1.
3.1.7 Proposition.1. Subalgebras and quotient algebras of solvable (resp. nilpotent) Lie algebras
are also solvable (resp. nilpotent).
2. If g is a Lie algebra and h ⊆ g is an ideal, then g is solvable if and only ifboth h and g/h are solvable
3. g is nilpotent if and only if the center of g is non-zero and g/Zg is nilpotent.
14 Lie Algebras
PROOF: Exercise. (For the third part, suppose g ) g1 ) · · · ) gn = 0. Then[gn−1,g] = 0, so gn−1 6= 0 is contained in the center of g.)
3.1.8 Corollary. In particular, g is nilpotent if and only if adg⊆ glg is nilpotent.
PROOF: adg= g/Zg.
3.1.9 Theorem (Lie). Let g⊆ glV (V finite dimensional) be a solvable Lie algebra.Suppose K has characteristic zero and is algebraically closed. Then there is a basisv1, . . . , vn of V such that, with respect to this basis, the matrices of all elementsof g are upper triangular.
PROOF: Omitted.
Equivalently, Lie’s theorem says that there is a common eigenvector for g, i.e.there is λ : g→ K linear and v ∈ V such that x v = λ(x)v for all x ∈ g. This is thesame as saying that V has a 1-dimensional subrepresentation.
3.1.10 Exercises.1. Prove the equivalence of the two formulations of Lie’s Theorem (essential
exercise).
2. Show that it is necessary that K is algebraically closed and that the charac-teristic is zero. (Hint: let K = Fp and g be the 3-dimensional Heis( ∂
∂ x, x , c).
Show that K[x]/x p is an irreducible representation of g.)
3.1.11 Corollary. Suppose K has characteristic zero. If g is a solvable Lie algebrathen [g,g] is a nilpotent Lie algebra.
PROOF: Apply Lie’s Theorem to the adjoint representation adg ⊆ glg. Then adg
is solvable, and Lie’s theorem implies that it sits inside the collection of uppertriangular matrices for some choice of basis. Hence [adg, adg] consists of strictlyupper triangular matrices, so it is nilpotent. But ad[g,g] = [adg, adg], so ad[g,g]is nilpotent, so [g,g] is nilpotent by 3.1.8.
3.1.12 Exercise. Construct a counterexample in characteristic p.
3.1.13 Theorem (Engel). Let g ⊆ glV (V finite dimensional), and suppose thatx : V → V is a nilpotent transformation for all x ∈ g. Then there is a commoneigenvector v for g.
PROOF: Omitted.
Here “nilpotent” means that all the generalized eigenvalues are 0. WhenceEngel’s theorem says that V has a 1-dimensional subrepresentation, or that x v = 0for all x ∈ g.
3.1.14 Corollary. g is nilpotent if and only if adg consists of nilpotent endomor-phisms of g.
PROOF: Exercise.
Cryptic Remark: Solvable and nilpotent Lie algebras are garbage.
Solvable and Nilpotent Lie Algebras 15
3.1.15 Definition. A symmetric bilinear form (a.k.a. inner product) (·, ·) : g×g→K is invariant if ([x , y], z) = (x , [y, z]) for all x , y, z ∈ g.
3.1.16 Exercise. If h ⊆ g is an ideal and (·, ·) is an invariant inner product on g
thenh⊥ := x ∈ g | (x ,h) = 0
is an ideal. In particular, g⊥ is an ideal.
3.1.17 Example (Trace forms). The most important class of invariant inner prod-ucts are the so-called trace forms, defined as follows. If (V,ρ) is a representationof g, define (·, ·)V : g× g→ K by
(x , y)V = tr(ρ(x)ρ(y) : V → V ).
Clearly (·, ·)V is symmetric and bilinear. Show that it is invariant (you will need touse the fact that ρ is a representation).
The Killing form is (x , y)ad = tr(ad x ad y : g→ g).
3.1.18 Theorem (Cartan’s Criterion). Let V be a finite dimensional representa-tion of g (i.e. g ,→ glV ) and K have characteristic zero. Then g is solvable if andonly if (x , y)V = 0 for all x ∈ g and y ∈ [g,g].
PROOF: The forward direction is immediate from Lie’s Theorem. The other direc-tion is where the content lies, and we omit the proof.
3.1.19 Corollary. g is solvable if and only if (g, [g,g])ad = 0.
PROOF: The forward direction follows from Lie’s Theorem. For the reverse direc-tion, Cartan’s Criterion implies that g/Zg = adg is solvable. But then g is solvableas well.
If g is a solvable non-Abelian Lie algebra then all trace forms are degenerate(as g⊥ ⊇ [g,g]). Warning/Exercise: Not every invariant inner product is a traceform.
3.1.20 Exercise. Let h= ⟨c, p, q, d⟩ be defined by [c, h] = 0, [p, q] = c, [d, p] = p,and [d, q] = −q (we think of p = ∂
∂ x, q = z, and d = −x ∂
∂ x). Then h is a Lie
algebra. Show that it is solvable, and construct a non-degenerate invariant innerproduct on it.
3.1.21 Definition. The radical of g is the maximal solvable ideal of g, denotedR(g).
3.1.22 Exercises.1. Show that the sum of solvable ideals is solvable, and hence R(g) is the sum
of all solvable ideals of g (so there is none of this nasty Zorn’s Lemma crap).
2. R(g/R(g)) = 0.
Recall that g is semi-simple (s.s.) if it is a direct sum of simple (non-Abelian)Lie algebras.
16 Lie Algebras
3.1.23 Theorem. Suppose the characteristic of K is zero. The following are equiv-alent.
1. g is semi-simple;
2. R(g) = 0;
3. The Killing form (·, ·)ad is non-degenerate (ironically, this is the “Killing cri-terion”).
Moreover, if g is s.s. then every derivation D : g → g (i.e. map D satisfyingD[a, b] = [Da, b] + [a, Db]) is inner (i.e. D = ad x for some x ∈ g), but notconversely.
PROOF: Begin by observing that R(g) = 0 if and only if g has no non-zero commu-tative ideals.
(i) implies (ii): If g is s.s. then it has no non-zero commutative ideals, bydefinition.
(iii) implies (ii): It suffices to show that if a is a commutative ideal then it iscontained in the kernel of the Killing form, i.e. g⊥. Write g = a+ h for h somevector space complement. If a ∈ a then ad a has matrix
0 ∗0 0
, as a is Abelian andan ideal. For x ∈ g, ad x has matrix
∗ ∗0 ∗
, so (a, x)ad = tr(ad(a)ad(x)) = 0.
(ii) implies (iii): Let r= g⊥, an ideal. We have a map rad−→ glg and (x , y)ad = 0
for all x , y ∈ r, by definition. By Cartan’s Criterion r/Z(r) is solvable, and hence r
is solvable and so r⊆ R(g) = 0.(iii), (ii) imply (i): Suppose (·, ·)ad is non-degenerate. Let a ⊆ g be a non-zero
minimal ideal. Then (·, ·)ad|a is either zero or non-degenerate. Indeed, the kernelof (·, ·)ad|a is an ideal of g and a is minimal. But if (·, ·)ad|a = 0 then the CartanCriterion implies that a is solvable. But R(g) = 0, so there should be no solvableideals. Therefore (·, ·)ad|a is non-degenerate, and we have g = a ⊕ a⊥, a directsum decomposition as ideals with (·, ·)ad|a⊥ non-degenerate. a is not solvable, so itmust be simple (non-Abelian). By induction on dimK g, we may write g as a directsum of simple Lie algebras, so it is s.s.
Finally, if D is a derivation then we get a linear function ` : g → K : x 7→trg(D · ad x). Since g is s.s., (·, ·)ad is non-degenerate, so `(x) = (y, x)ad for aunique y ∈ g. This implies that trg((D − ad y) · ad x) = 0 for all x ∈ g. LetE = D− ad y , a derivation. We must show E = 0. Notice that
ad(Ex)(z) = [Ex , z] = E[x , z]− [x , Ez] = [E, ad x](z)
whence
(Ex , z)ad = trg(ad(Ex) · ad z)= trg([E, ad x] · ad z)= trg(E[ad x , ad z])= trg((D− ad y)ad[x , z]) = 0
for all z ∈ g, so Ex = 0 for all x ∈ g.
3.1.24 Exercises.1. Prove that R(g)⊇ ker(·, ·)ad ⊇ [R(g), R(g)].
Structure Theory 17
2. Show that g s.s. implies that g can be written uniquely as a direct sum of itsminimal ideals.
3. Show that a nilpotent Lie algebra always has a non-inner derivation.
4. Show that the Lie algebra defined by h = ⟨a, b⟩, where [a, b] = b, has onlyinner derivations.
Remark. The exact sequence 0 → R(g) → g → g/R(g) → 0 shows that every Liealgebra has a maximal s.s. quotient, and is an extension of this by a maximalsolvable ideal. In fact we have the following theorem
3.1.25 Theorem (Levi’s Theorem). If K has characteristic zero then the exactsequence splits, i.e. there is a subalgebra h ,→ g such that g ∼= ho R(g) (indeed,we may take h= g/R(g)).
PROOF: Omitted.
3.1.26 Exercise. Show R(slp(Fp)) = Fp I , but there does not exist a splittingg/R(g) ,→ g.
3.1.27 Exercises.1. Let g be a simple Lie algebra and (·, ·)1 and (·, ·)2 be non-degenerate invari-
ant forms. Then there is λ ∈ K∗ such that (·, ·)1 = λ(·, ·)2. (Apply Schur’sLemma.)
2. (essential exercise) Let g = sln(C) and put (A, B) = tr(AB), so (A, B) =λ(A, B)ad. Find λ (you may assume g is simple).
3.2 Structure Theory
3.2.1 Definition. A torus t ⊆ g is an Abelian subalgebra such that ad t : g→ g issemi-simple (i.e. diagonalizable) for all t ∈ t. A maximal torus is a torus whichis maximal with respect to inclusion. A maximal torus is also known as a Cartansubalgebra.
3.2.2 Exercises.1. Let G be an algebraic group, T a subgroup isomorphic to C∗×· · ·×C∗. Then
t= T1T is a torus.
2. (essential exercise) Let g= sln, or
sp2n = A∈ gl2n | JA+ AT J = 0
orson = A∈ gln | MA+ AT M = 0,
where M = anti-diag(1, . . . , 1). Prove that t = diagonal matrices in g is amaximal torus.
3.2.3 Lemma. Let t1, . . . , t r : V → V , where V is finite dimensional, be pair-wise commuting, semi-simple, linear maps. Let ~λ = (λ1, . . . ,λr) ∈ Cr , and setV~λ = v ∈ V | t i v = λi v, i = 1, . . . , r be the simultaneous eigenspace. ThenV =
⊕
~λ∈Cr V~λ.
18 Lie Algebras
PROOF: By induction on r. The case r = 1 is clear since t1 is assumed to be semi-simple, and hence diagonalizable. If r > 1 then V =
⊕
Cr−1 V(λ1,...,λr−1) by induc-tion. As t r commutes with t1, . . . , t r−1, t r(V(λ1,...,λr−1)) ⊆ V(λ1,...,λr−1), so decomposeV(λ1,...,λr−1) as a direct sum of eigenspaces.
We now rephrase the above lemma in the language we have developed sofar. Let t :=
⊕ri=1Ct i be an Abelian Lie algebra isomorphic to Cr . Then V is
a representation of t, by definition. Since a basis of t acts on V semi-simply,V =
⊕
~λ∈Cr V~λ is the decomposition of V as a representation of t, and is a directsum of irreducibles. Hence all t ∈ t act semi-simply (as on each V~λ they act asmultiplication by a scalar).
Moreover, it is convenient to think of λ as being in the dual space t∗. For sucha λ ∈ t∗, the λ-weight space of V is Vλ = v ∈ V | t v = λ(t)v. (In the abovelemma we would take λ(
∑
i x i t i) =∑
x iλi .)
3.2.4 Proposition. Let g be a semi-simple Lie algebra and t a maximal torus. Thent 6= 0, and the collection of elements of g which commute with t is exactly t itself.
PROOF: Omitted.
We may write g= g0 +⊕
λ∈t∗,λ 6=0 gλ, where for λ ∈ t∗,
gλ := x ∈ g | [t, x] = λ(t)x for all t ∈ t.
This is known as the root-space decomposition of g. The proposition implies thatwhen g is s.s., g0 = t.
3.2.5 Definition. If λ 6= 0 and gλ 6= 0 then we say that λ is a root of g. LetR= Rg := λ ∈ t∗ | λ 6= 0,gλ 6= 0, the set of roots of g.
3.2.6 Example. Let g = sln(C), so that t is the collection of diagonal matrices ing. Then for t ∈ t, [t, Ei j] = (t i − t j)Ei j . Let εi : t→ C : t 7→ t ii . Then [t, Ei j] =(εi − ε j)(t)Ei j . Note that ε1 + · · ·+ εn = 0 since sln is the collection of matriceswith trace zero. In this case we have that gεi−ε j
= CEi j is one-dimensional, andR= εi − ε j | i 6= j. We will see that this holds in general.
3.2.7 Proposition. sln(C) is simple (and hence semi-simple).
The Killing criterion can be used to show that sln(C) is semi-simple (do so asan exercise).
PROOF: Suppose that r ⊆ sln is a non-zero ideal. Let r ∈ r, so by the root-spacedecomposition, r = h +
∑
α∈R eα for some h ∈ t and eα ∈ gα. Choose such anon-zero r with a minimal number of non-zero terms.
If h 6= 0 then choose a generic h0 (i.e. choose h0 so that its diagonal elementsof are all different, or equivalently so that α(h0) 6= 0 for all α ∈ R). As r is anideal,
∑
α∈Rα(h0)eα = [h0, r] ∈ r and if this is a non-zero element then it hasfewer non-zero terms than r, contradicting our choice of r. Therefore [h0, r] = 0,and so r = h by 3.2.4. As h 6= 0, there are i and j such that c := (εi − ε j)(h) 6= 0,so [h, Ei j] = cEi j ∈ r, implying Ei j ∈ r. From this we will now show that r containsa basis for sln and hence r = sln. As [Ei j , E jk] = Eik if i 6= k and [Esi , Ei j] = Eik if
Structure Theory 19
j 6= s, we have Eab ∈ r for all a 6= b. But now [Ei,i+1, Ei+1,i] = Eii − Ei+1,i+1 ∈ r aswell. All these matrices together form a basis for the trace zero matrices.
On the other hand, suppose h= 0, so r =∑
α∈R eα. If r = eα = cEi j with c 6= 0then we are done as before. Suppose r = cEα+dEβ+
∑
γ∈R\α,β eγ, where c, d 6= 0.Choose h0 ∈ t such that α(h0) 6= β(h0), so [h0, r] = cα(h0)Eα + dβ(h0)Eβ +∑
γ(h0)eγ. Therefore a linear combination of [h0, r] and r is in r and has fewernon-zero terms, contradicting that r was chosen with the minimal number. There-fore sln is simple.
3.2.8 Exercises.1. Show that A∈ son if and only if A is anti-skew-symmetric (i.e. skew-symmetric
across the anti-diagonal).
2. Show that A ∈ sp2n if and only if A = A1 A2
A3 A4
with A2 and A3 anti-skew-symmetric and A4 =−AaT
1 (anti-transpose).
3. Prove that the maximal tori in son and sp2n are exactly the diagonal matricesin those algebras.
4. Show that the root spaces for son areC(Ei j−En− j+1,n−i+1), where i+ j < n+1and i 6= j. Also show that
R=
¨
±εi ± ε j | 1≤ i, j ≤ `, i 6= j if n= 2`
±εi ± ε j ,±εi | 1≤ i, j ≤ `, i 6= j if n= 2`+ 1
5. Show that the root spaces for sp2n are C(Ei j−En− j+1,n−i+1), where 1≤ i, j ≤n and i 6= j and C(Ei j + En− j+1,n−i+1), where n < i ≤ 2n and j ≤ n or i ≤ nand n< j ≤ 2n. Also show that
R= ±εi ± ε j ,±2εi | 1≤ i, j ≤ n, i 6= j.
6. Show that sp2n is simple, and son is simple if n > 4 or n = 3, and show thatso4 = so3 + so3 and that so3 = sl2.
Notice that the root spaces are all one-dimensional.
3.2.9 Theorem (Structure Theorem for S.S. Lie Algebras). Let g be a s.s. Lie al-gebra, t a maximal torus, and let g= t+
⊕
λ∈R gλ be the root-space decomposition.Then
1. CR= t∗, i.e. the roots of g span t∗;
2. dimgα = 1 for all α ∈ R;
3. For α,β ∈ R, if α+β ∈ R then [gα,gβ] = gα+β , and if α+β /∈ R and β 6=−αthen [gα,gβ] = 0;
4. [gα,g−α] ⊆ t is one-dimensional, and gα + [gα,g−α] + g−α is isomorphic tosl2.
PROOF:1. If R does not span t∗ then there is 0 6= h ∈ t such that α(h) = 0 for all α ∈ R.
But then if x ∈ gα then [h, x] = α(h)x = 0, so [h,gα] = 0 for all α ∈ R. But[h, t] = 0 as t is Abelian, so [h,g] = 0, contradicting that g is semi-simpleand has no center (since it has no Abelian ideals at all).
20 Lie Algebras
2,3,4. Observe that [gλ,gµ]⊆ gλ+µ as for t ∈ t, x ∈ gλ, and y ∈ gµ,
[t, [x , y]] = [[t, x], y] + [x , [t, y]] = λ(t)[x , y] +µ(t)[x , y]
by the Jacobi identity.
Claim. (gλ,gµ)ad = 0 if λ+µ 6= 0. Moreover, if g is s.s. then (·, ·)ad restrictedto gλ + g−λ is non-degenerate.
The moreover part follows from the first part of the claim because, as (·, ·)adis non-degenerate if g is s.s., it must be non-degenerate on each gλ+g−λ. Toprove the first part, if x ∈ gλ and y ∈ gµ then (ad x ad y)Ngα ⊆ gα+N(λ+µ).But g is finite dimensional, so if λ+ µ 6= 0 then gα+N(λ+µ) = 0 for N suffi-ciently large. But this implies that (ad x ad y) is nilpotent, and so has zerotrace.
Note in particular that (·, ·)ad|t is non-degenerate. (Warning: (·, ·)ad|t is notequal to the Killing form of t, which is obviously zero). Therefore we havean isomorphism
ν : t→ t∗ : t 7→ “t ′ 7→ (t, t ′)ad”.
This induces an inner product on t∗, denoted (·, ·), by (ν(t),ν(t ′)) := (t, t ′)ad.
Claim. α ∈ R implies −α ∈ R.
Indeed, (·, ·)ad is non-degenerate on gα + g−α. But (gα,gα)ad = 0 as 2α 6= 0,so if gα 6= 0 then it must be that g−α 6= 0 also. Moreover, (·, ·)ad defines aperfect pairing between gα and g−α, namely gα
∼= g∗−α.
Claim. If x ∈ gα and y ∈ g−α then [x , y] = (x , y)adν−1(α).
It is clear that [x , y] ∈ t= g0, so to check the claim it is enough to check that(t, [x , y])ad = (x , y)adα(t) for all t ∈ t, as (ν−1(α), t)ad = α(t) by definition.But both sides are equal to ([t, x], y)ad since x ∈ gα.
Let eα ∈ gα be non-zero, and pick e−α ∈ g−α such that (eα, e−α)ad 6= 0. Then[eα, e−α] = (eα, e−α)adν
−1(α), and [ν−1(α), eα] = α(ν−1(α))eα = (α,α)adeα.Therefore eα, e−α,ν−1(α) span a Lie subalgebra m of dimension three.
3.2.10 Key Lemma. (α,α)ad 6= 0.
Suppose otherwise. If (α,α)ad = 0 then [m,m] = Cν−1(α), which impliesthat m is solvable. Lie’s theorem implies that ad[m,m] acts by nilpotentelements, i.e. that adν−1(α) acts nilpotently. But adν−1(α) ∈ t and allelements of t act semisimply, so these together imply that α = 0. This con-tradiction proves the claim.
Let hα =2ν−1(α)(α,α)ad
, and rescale e−α so that (eα, e−α)ad =2
(α,α)ad. Then eα, hα, e−α
span a copy of sl2.
Now we use the representation theory of sl2 to deduce the structure of R, g,etc.
Claim. dimg−α = 1 for all α ∈ R.
Structure Theory 21
Choose eα, hα, e−α as above, so that we have a map g−α→ Cν−1(α) = Chα :x 7→ [eα, x]. If dimg−α > 1 then this map has a kernel, i.e. there is a vectorv such that ad(eα)v = 0 and ad(hα)v = −α(hα)v = −2v. But then v is ahighest weight vector with negative weight, implying that the sl2-submodulegenerated by v is infinite dimensional, contradicting that dimg is finite.
3.2.11 Proposition. With the notation from the proof above,
1. 2 (α,β)(α,α)
∈ Z for all α,β ∈ R;
2. if α ∈ R and kα ∈ R then k =±1;
3.⊕
k∈Z gβ+kα is an irreducible sl2-submodule for all α,β ∈ R, so the set β +kα ∈ R | k ∈ Z ∪ 0 is of the form β − pα, . . . ,β ,β +α, . . . ,β + qα, wherep− q = 2 (α,β)
(α,α), and is called the “α string through β”;
4. if α,β ,α+ β ∈ R then [gα,gβ] = gα+β .
PROOF:3. Let q = maxk | β + kα ∈ R. If v ∈ gβ+kα is non-zero then ad eαv ∈
gβ+(q+1)α = 0 by choice of q and ad hαv = (β + qα)(h)v = (β +α) 2ν−1(α)(α,α)
sov is a highest weight vector for sls = ⟨eα, e−α, hα⟩, so its weight is in N, i.e.N contains 2(α,β)
(α,α)+ 2q, which implies that 2(α,β)
(α,α)∈ Z.
1. It follows from the representation theory of sl2 that β+qα,β+(q−1)α, . . . ,β−(q+ 2(α,β)
(α,α))α are all roots. We must show that there are no more roots. Let
p =maxk | β−kα ∈ R (start at the bottom and squeeze). We now get thatβ − pα,β − (p− 1)α, . . . ,β +(p− 2(α,β)
(α,α))α are roots. Then q+ 2(α,β)
(α,α)≤ p by
the definition of p, and p− 2(α,β)(α,α)
≤ q by the definition of q.
2. If kα ∈ R then 2(kα,α)(kα,kα)
= 2k∈ Z and 2(α,kβ)
(α,α)= 2k ∈ Z. Hence k = ± 1
2,±1,±2,
so it is enough to show that α ∈ R implies that −2α /∈ R. Suppose −2α ∈ Rand take v ∈ g−2α non-zero. Then ([eα, v], eα)ad = (v, [eα, eα])ad = 0 byinvariance. But ad eαv = λe−α for some λ, and λ = 0 since the Killing formis non-degenerate on g−α + gα. Thus ad eαv = 0, by hαv = −4v, once againa contradiction (negative highest weight).
4. Suppose that α,β ,α+ β are roots. Since root spaces are one-dimensional,it is enough to show that [gα,gβ] 6= 0. Now
⊕
k∈Z gβ+kα is an irreduciblesl2-module, so ad eα : gβ+kα → gβ+(k+1)α is an isomorphism if −p ≤ k < q,but α+ β ∈ R, so q ≥ 1 and we are done.
3.2.12 Lemma. sα(β) := β − 2(α,β)(α,α)
α is a root when α,β ∈ R.
PROOF: Let r = 2(α,β)(α,α)
. If r ≥ 0 then p = q+ r ≥ r and if r ≤ 0 then q = p− r ≥−r,so β − rα is in the string in either case.
Recall that R spans t∗.
22 Lie Algebras
3.2.13 Proposition.1. If α,β ∈ R then (α,β) ∈Q.
2. If we choose a basis β1, . . . ,β` of t∗ with each βi ∈ R and if β ∈ R thenβ =
∑
i ciβi where ci ∈Q for all i.
3. (·, ·) is positive definite on the Q-span of R.
PROOF:1. As 2(α,β)
(α,α)∈ Z, it is enough to show that (α,α) ∈ Q for all α ∈ R. Let t, t ′ ∈ t.
Then(t, t ′)ad = trg(ad t ad t ′) =
∑
α∈R
α(t)α(t ′)
as each gα is one-dimensional. So if λ,µ ∈ t∗ then
(λ,µ) = (ν−1(λ),ν−1(µ))ad
=∑
α∈R
α(ν−1(λ))α(ν−1(µ)) =∑
α∈R
(α,λ)(α,µ).
In particular, (β ,β) =∑
α∈R
α,β2. Dividing by (β ,β)2/4, we get
4
(β ,β)=∑
α∈R
(2(α,β)(β ,β)
)2 ∈ Z
which implies (β ,β) ∈Q.
2. Exercise: if (·, ·) is a non-degenerate symmetric bilinear form, β1, . . . ,β` arelinearly independent vectors then det(βi ,β j)i j 6= 0. Set B = (βi ,β j)i j , thematrix of inner products. Then B is a matrix with rational entries, B−1
exists and is rational, so if β =∑
i ciβi then (β ,β j) =∑
i ci(βi ,β j) and(c j) j = B−1(β ,β j) j is a rational vector.
3. If λ ∈ spanQ(R) then (λ,λ) =∑
α∈R(λ,α)2 ≥ 0 since it is a sum of squaresof rational numbers.
4 Root Systems
4.1 Abstract Root Systems
Let V be a vector space over R (though the theory could be developed over Qas well), and let (·, ·) be a positive definite symmetric bilinear form. If α ∈ V isnon-zero, define α∨ = 2α
(α,α), so (α,α∨) = 2. Define sα : V → V : v 7→ v − (v,α∨)α,
a linear map.
4.1.1 Lemma. sα is the reflection in the hyper-plane perpendicular to α. In par-ticular, all its eigenvalues are 1, except for one which is −1, i.e. s2
α = id, (sα −1)(sα + 1) = 0. Moreover,
sα ∈ O(V ) = ϕ : V → V | (ϕv,ϕw) = (v, w) for all v, w ∈ V.
Abstract Root Systems 23
PROOF: V = Rα⊕α⊥, and sαv = v if v ∈ α⊥, sαα= α− (α,α∨)α=−α.
4.1.2 Definition. A root system R in V is a finite set R⊆ V such that
1. RR= V and 0 /∈ R;
2. (α,β∨) ∈ Z for all α,β ∈ R;
3. sαR= R for all α ∈ R;
(4.) R is reduced if kα ∈ R implies k =±1 for all α ∈ R.
Most of the time we will be concerned with reduced root systems.
4.1.3 Example. Let g be a s.s. Lie algebra over C and t a maximal torus. Theng = t+
⊕
α∈R gα and (R,RR ⊆ t∗) is a reduced root system (this example is thewhole point of looking at root systems).
4.1.4 Definition. Let W ⊆ GLV be the group generated by the reflections sα ∈ Rfor α ∈ R, the Weyl group of the root system R.
4.1.5 Lemma. The Weyl group W is finite.
PROOF: By property 3, W permutes R, so W → SR. By property 1 this map isinjective.
4.1.6 Definition. The rank of a root system (R, V ) is the dimension of V . If (R, V ),(R′, V ′) are root systems then the direct sum root system is (R q R′, V ⊕ V ′). Anisomorphism of root systems is a bijective linear map (not necessarily an isometry)ϕ : V → V ′ such that ϕ(R) = R′. A root system which is not isomorphic to a directsum of root systems is an irreducible root system.
4.1.7 Examples.1. When the rank is 1, V = R and (x , y) = x y . Therefore A1 = α,−α (α 6= 0)
is the only root system (up to isomorphism). The Weyl group is Z/2Z.
2. When the rank is 2, V = R2 and the inner product is the usual inner product.Root systems are A1 × A1, A2, B2 = C2 and G2 (insert diagrams).
4.1.8 Lemma. If (R, V ) is a root system then (R∨, V ) is a root system.
PROOF: Exercise.
4.1.9 Definition. A root system is simply laced if all the roots have the same length(i.e. (α,α) is the same for all α ∈ R).
4.1.10 Example. A1, A2, and A1 × A1 are simply laced, B2 and G2 are not.
4.1.11 Exercise. If (R, V ) is simply laced then (R, V )∼= (R′, V ′), where R′ is a rootsystem such that (α,α) = 2 for all α ∈ R′, (i.e. α = α∨ and |α| =
p2). (Hint: if R
is irreducible then just rescale).
4.1.12 Definition. A lattice L is a finitely generated free abelian group Zn, equippedwith a bilinear form (·, ·) : L × L → Z. (We will usually assume that it is positivedefinite and non-degenerate.)
24 Lie Algebras
It follows that (L⊗ZR, (·, ·)) is an inner product space, and it is positive definiteif (·, ·) is positive definite.
4.1.13 Definition. A root of a lattice L is an ` ∈ L with (`,`) = 2. Write
RL = ` ∈ L | (`,`) = 2= ` ∈ L | `= `∨.
If α ∈ RL then sα(L)⊆ L. (Indeed, sαβ = β − (β ,α∨)α ∈ L.)
4.1.14 Lemma. RL is a root system in RRL and it is simply laced.
PROOF: RL is finite since it is the intersection of a compact set ` ∈ RL | (`,`) = 2with a discrete set L. The rest of the axioms are obvious (exercise).
4.1.15 Definition. L is generated by roots if ZRL = L.
If L is generated by roots then L is an even lattice, i.e. (`,`) ∈ 2Z for all ` ∈ L.
4.1.16 Example. Let L = Zα with (α,α) = λ. Then RL = ±α if λ = 2, and L isgenerated by roots. If k2λ 6= 2 for all k ∈ Z then RL =∅.
4.1.17 Examples.An: Consider the square lattice Zn+1 =
⊕n+1i=1 Zei , where (ei , e j) = δi j . Let An be
the sublattice
` ∈ Zn+1 | (`, e1 + · · ·+ en+1) = 0= n+1∑
i=1
aiei | ai ∈ Z,n+1∑
i=1
ai = 0
.
Then RAn= ei− e j | i 6= j and |RAn
|= n(n+1). If α= ei− e j then sα swapsthe ith and jth coordinate, so WAn
= ⟨si j | i 6= j⟩=Sn+1.
Dn: Consider the square lattice Zn =⊕n
i=1Zei , where (ei , e j) = δi j . Take RDn=
±ei ± e j | i 6= j and let
Dn = ZRDn= n∑
i=1
aiei | ai ∈ Z,n∑
i=1
ai ∈ 2Z
.
Then |RDn| = 2n(n − 1). It is seen that sei−e j
swaps the ith and jth coor-
dinate and sei+e jchanges the sign of the ith and jth coordinate so WDn
=(Z/2Z)n−1 o Sn, ((Z/2Z)n−1 is the subgroup of an even number of signchanges). RDn
is irreducible if n≥ 3, and RD3= RA3
and RD2= RA1×A1
.
E8: Let Γn = (k1, . . . , kn) | either ki ∈ Z for all i or ki ∈12+ Z for all i and
∑ni=1 ki ∈ 2Z. Consider α= ( 1
2, . . . , 1
2), so that (α,α) = n
4. Then α ∈ Γn and
Γn is an even lattice only if 8 | n. As an exercise, prove that if n> 1 then theroots of Γ8n form a root system of type D8n (this requires also proving thatΓ8n is a lattice). For n= 1, show that
RΓ8= ±ei ± e j ,
12(±e1 ± · · · ± e8) | i 6= j, the number of minuses is even.
This is the root system of type E8. Notice that |RE8| = 7·8
24 + 128 = 240.
Compute WE8(answer: |WE8
|= 21435527).
Abstract Root Systems 25
E6, E7: As an exercise, prove that if R is a root system and α ∈ R then α⊥∩R is also aroot system. In Γ8 let α= ( 1
2, . . . , 1
2) and β = e7+ e8. Then α⊥∩RΓ8
is a rootsystem of type E7 and (α,β)⊥ ∩ RΓ8
is a root system of type E6. Show that|RE7|= 126 and |RE6
|= 72 and describe the cooresponding root lattices.
As an essential exercise, check this whole example and draw the low dimen-sional systems (e.g. draw An ⊆ Zn+1 and RAn
for n= 1, 2).
4.1.18 Theorem (ADE classification). The complete list of simply laced irreducibleroot systems is An (n≥ 1), Dn (n≥ 4), E6, E7, and E8. No two of these root systemsare isomorphic.
4.1.19 Theorem. The remaining root systems are B2 = C2, Bn (n≥ 3), Cn (n≥ 3),F4, and G2, where F4 is described below, G2 has been drawn, and
RBn= ±ei ,±ei ± e j | i 6= j ⊆ Zn
andRCn= ±2ei ,±ei ± e j | i 6= j ⊆ Zn.
(Notice that RBn= R∨Cn
and WBn=WCn
= (Z/2Z)oSn.)
4.1.20 Example. Set Q = (k1, . . . , k4) | either all ki ∈ Z or all ki ∈12+Z. Then
F4 = α ∈Q | (α,α) = 2 or (α,α) = 1.
We have RF4= ±ei ,±ei ± e j ,
12(±e1 ± e2 ± e3 ± e4) | i 6= j.
We want to choose certain “good” bases of V . We make an auxiliary choice ofa linear function f : V → R such that f (α) 6= 0 for all α ∈ R. A root α ∈ R is saidto be “positive” if f (α)> 0 and “negative” if f (α)< 0. Write R+ for the collectionof positive roots, so that R= R+ q (−R+).
4.1.21 Definition. α ∈ R+ is a simple root if it is not the sum of two other positiveroots. Write Π = α1, . . . ,αn for the collection of simple roots.
4.1.22 Proposition.1. If α,β ∈ Π then α− β /∈ R;
2. If α,β ∈ Π and α 6= β then (α,β∨)< 0.
3. Every α ∈ R+ can be written α=∑
αi∈Πkiαi with ki ∈ N.
4. Simple roots are linearly independent, so the decomposition above is unique.
5. If α ∈ R+ \Π then there is αi ∈ Π such that α−αi ∈ R+.
6. Π is indecomposable (i.e. it cannot be written as a diagonal union of orthog-onal sets) if and only if R is reducible.
PROOF: Exercise: check all these properties for the above examples, or find a prooffrom the definition of root system (this is harder).
Write Π = α1, . . . ,α`, where ` is the rank of R, which is dim V . Set ai j =(αi ,α
∨j ) and A= (ai j)`×`, the Cartan matrix.
26 Lie Algebras
4.1.23 Proposition.1. ai j ∈ Z, aii = 2, and ai j ≤ 0 if i 6= j;
2. ai j = 0 if and only if a ji = 0;
3. det A> 0
4. All principal subminors have positive determinant.
PROOF: (written down)
A matrix which satisfies the conditions of 4.1.23 is an abstract Cartan matrix.
4.2 Dynkin Diagrams
Special graphs called Dynkin diagrams are used to summarize the information inthe Cartan matrix. The vertices are the set of simple roots, and αi is joined to α jby ai ja ji lines. But ai ja ji is always 1 if the root system is simply laced, while Bnand Cn have a double edge, and G2 has a triple edge. If ai ja ji is 2 or 3 then put anarrow in the direction of the shorter root.
An : · · · n≥ 1
Bn : · · · > n≥ 2
Cn : · · · < n≥ 3
Dn : · · ·
n≥ 4
G2 : >
F4 : >
E8 :
E7 :
E6 :
Dynkin Diagrams 27
4.2.1 Exercises.1. Show that these really are the Dynkin diagrams of the named root systems.
2. Show that the Dynkin diagram of AT is that of A, except with any arrowsreversed. Also check that AT is the Cartan matrix of (A∨, V ).
3. Compute the determinants of the Cartan matrices.
4. For type An, compute the inverse of the Cartan matrix. (Hint: A1 has ma-trix [ 2 ], A2 has matrix
2 −1−1 2
, so the answer is something of the form1
det A[ pos. ints ]. What do the positive integers represent?)
5. Redo the above exercise for all other types.
Remark. SLn = A ∈ GLn | det A = 1, and ZSLn= λIn | λn = 1 ∼= Z/nZ. Here
n = det A. In general, the determinant of the Cartan matrix is |ZG |, where G is asimply connected compact or algebraic group of the given type.
Notice that R is indecomposable if and only if the Dynkin diagram is connected.
4.2.2 Theorem. An abstract Cartan matrix which is indecomposable is, after re-ordering simple roots, has Dynkin diagram which is one of the given types.
PROOF: We will classify connected Dynkin diagrams of the Cartan matrix.First, it is immediate from 1–4 of the definition of an abstract Cartan matrix
(4.1.23) that any subdiagram of a Dynkin diagram is a disjoint union of Dynkindiagrams. All of the rank 2 Cartan matrices are A =
2 −a−b 2
, where 4 − ab =det A> 0, so (a, b) ∈ (1,2), (2, 1), (1,3), (3, 1), (1,1).
Second, Dynkin diagrams contain no cycles. Write Π = α1, . . . ,αn, and put
α=n∑
i=1
αip
(αi ,αi).
Then
0< (α,α) = n+n∑
i=1
2(αi ,α j)p
(αi ,αi)(α j ,α j)= n−
∑
i< j
p
ai ja ji
as (αi ,α j) ≤ 0 by definition of ai j . Therefore∑
i< j
p
ai ja ji < n. If there were acycle on α1, . . . ,αn then there would be at least n edges, and if there is an edgefrom αi to α j then
p
αi ja ji ≥ 1. A cycle would contradict the identity.
4.2.3 Exercise. If (R, V ) is an indecomposable root system and R+ is a choice ofroots, then there exists a unique θ ∈ R+ such that θ + αi /∈ R for all αi ∈ Π. (e.g.for An we have θ = e1 − en+1 when αi = ei − ei+1, for Bn, θ = e2 + e1, for Cn,θ = 2e1, and for Dn, θ = e1 + e2.) θ is the highest root.
Third, the extended Cartan matrix A is (ai j)i, j=0,...,`, where α0 = −θ (from theexercise). Then A satisfies properties 1 and 2, but det A = 0 since α0, . . . ,α` arelinearly dependent. Property 4 still holds if one takes proper subminors, as anysubset of the extended simple roots is linearly independent. A Dynkin diagramcannot contain one of the extended Dynkin diagrams as a subdiagram (draw theseout).
28 Lie Algebras
Fourth, if the diagram is simply laced (i.e. has simple edges) then it is one ofA, D or E. Indeed, if it is not A` then it must have a branch point. But it cannothave D4 as a subdiagram, so it can only have triple branches. Further, it has onlyone branch point since otherwise it has D` as a subdiagram. Since it does containany E, it must be one of A, D, or E.
Fifth (this part is an exercise), if a Dynkin diagram has G2 as a subdiagram thenit is G2. (Hint: G2 is prohibited and the rest do not have positive determinant.)If there is a double bond then there is only on double bond (because of Cn), andthere are no branch points (because of Bn). If the double bond is in the middlethen it is F4 (by F4), and if it is on an end then it is B or C series. (Warning: we areoften using the fact that the direction of the arrows don’t matter from computingthe determinant.)
4.3 Existence and Uniqueness of an Associated Lie Algebra
So far we have accomplished the following:
g a s.s. Lie algebra
choosing t a maximal torus
R root system
choosing f : V → R a linear map
R+ positive roots, Π simple roots, Cartan matrix A
and we have classified all possible Cartan matrices A.But the choices made are inconsequential.
4.3.1 Theorem. All maximal tori are conjugate. More precisely, for K algebraicallyclosed and characteristic zero, if t1 and t2 are maximal tori of g then there is
g ∈ (Autg)0 = g ∈ GL(g) | g : g→ g is a Lie algebra homomorphism0
(the “0” means “connected component of the identity”) such that g(t1) = t2.
4.3.2 Theorem. All R+ are conjugate. More precisely, let (R, V ) be a root systemand f1, f2 : V → R be two linear maps which define positive roots R+(1) and R+(2).Then there is a unique w ∈WR such that wR+(1) = R+(2) and we have wΠ(1) = Π(2).
4.3.3 Corollary. g determines the Cartan matrix, i.e. the Cartan matrix is inde-pendent of the choices made.
There is a one-to-one correspondence between Cartan matrices and s.s. Liealgebras.
4.3.4 Theorem. Let gi (i = 1, 2) be s.s. Lie algebras over C, and ti maximal tori,Ri , R+i , Πi , Ai be choices of associated objects. If, after reordering the indices,A1 = A2 then there is a Lie algebra isomorphism ϕ : g1 → g2 such that ϕ(t1) = t2,ϕ(R1) = R2, etc.
Existence and Uniqueness of an Associated Lie Algebra 29
4.3.5 Theorem. Let A be an abstract Cartan matrix. Then there is a s.s. Lie algebrag over C with Cartan matrix A.
Remark.1. We already know this for An, Bn, Cn, and Dn (namely sln+1, so2n+1, sp2n, and
so2n, respectively).
2. It remains to check that E6, E7, E8, F4, G2 exist. We can do this case by case,e.g. G2 = Aut(O). There are abstract constructions in the literature.
Existence and uniqueness will also follow from the following.
4.3.6 Definition. A generalized Cartan matrix is a matrix A such that
1. ai j ∈ Z, aii = 2, and ai j ≤ 0 for i 6= j; and
2. ai j = 0 if and only if a ji = 0.
Let An×n be such a matrix. Define g to be a Lie algebra with generatorsEi , Fi , Hi | i = 1, . . . , n and relations
1. [Hi , H j] = 0 for all i, j;
2. [Hi , E j] = ai j E j for i 6= j;
3. [Hi , F j] =−ai j F j for i 6= j;
4. [Ei , F j] = 0 for i 6= j; and
5. [Ei , Fi] = Hi for all i.
Let g be the quotient of g by the additional relations
1. (ad Ei)1−ai j E j = 0 for i 6= j; and
2. (ad Fi)1−ai j F j = 0 for i 6= j.
For example, if ai j = 0 then [Ei , E j] = 0, and if ai j = −1 then [Ei , [Ei , E j]] = 0.These are known as the Serre relations, though they are due to Harish-Chandraand Chevalley.
4.3.7 Theorem.1. Let A be indecomposable. Then g has a unique maximal ideal and g is its
quotient. Therefore g is a simple Lie algebra (which need not be finite di-mensional).
2. Hence if g is a finite dimensional s.s. Lie algebra with Cartan matrix A, thenthe map g→ g that takes Ei 7→ eαi
, Fi 7→ e−αi, and Hi 7→
2ν−1(αi)(αi ,αi)ad
is surjective,and hence factors as g→ g→ g.
3. g is finite dimensional if and only if A is a Cartan matrix (i.e. all principalsubminors has positive determinant). (This last part is equivalent to theExistence and Uniqueness Theorem.)
4.3.8 Definition. g is the Kac-Moody algebra associated to the abstract Cartanmatrix A.
30 Lie Algebras
5 Representation Theory
5.1 Theorem of the Highest Weight
For this section let g be a s.s. Lie algebra, g= t+⊕
α∈R gα be the root space decom-position, R+ a choice of positive roots, and Π = α1, . . . ,α` be the correspondingsimple roots. Let V be a (finite dimensional) representation of g. Set, for λ ∈ t∗,Vλ = v ∈ V | t v = λ(t)v for all t ∈ t, the λ-weight space of V .
Define n+ =⊕
α∈R+ gα and n− =⊕
α∈R+ g−α, so that g = n+ + t+ n−. Noticethat n+, t, and n− are all Lie subalgebras of g (and n+ and n− are nilpotent?).Recall also for each α ∈ R, gα + [gα,g−α] + g−α
∼= sl2, and we write (sl2)α :=⟨eα, hα, e−α⟩ ≤ g.
5.1.1 Proposition. For g s.s. and V f.d.,
1. V =⊕
λ∈t∗ Vλ, i.e. each t ∈ t acts semi-simply.
2. If Vλ 6= 0 then λ(hα) ∈ Z for all α ∈ R.
PROOF: For each α ∈ R, (sl2)α acts on V , and V is finite dimensional, so hα actssemi-simply and has eigenvalues in Z by the representation theory of sl2. Fromthis 2 is immediate. As for 1, the hα span t as the roots span t∗, in fact hα1
, . . . , hα`is a basis of t.
5.1.2 Definition. The character of V is ch(V ) =∑
dim Vλeλ, where eλ ∈ Z[t∗] isa formal symbol intended to remind you of the exponential map.
5.1.3 Examples.1. Recall the representations L(n) of sl2 have characters
ch(L(n)) = zn + zn−2 + · · ·+ z2−n + z−n =zn+1 − z−(n+1)
z− z−1 ,
where z = eα and Rsl2= ±α.
2. Take g= sl3 and consider its adjoint representation. Then
ch(g) = dim t+∑
α∈R
eα = zw+ z+w+ 2+w−1 + z−1 + z−1w−1,
where w = eα and z = eβ , recalling Rsl3= ±α,±β ,±(α+ β).
5.1.4 Definition. The lattice of roots is Q := ZR, and the lattice of weights is
P := γ ∈QR | (γ,α∨) ∈ Z for all α ∈ R.
Recall that (γ,α∨) = γ(hα) by definition of the normalization of hα, so thesecond part of 5.1.1 says that if Vλ 6= 0 then λ ∈ P. In particular, ch(V ) ∈ Z[P].Notice that if α,β ∈ R then (α,β∨) = 2(α,β)
(β ,β)∈ Z, so Q ⊆ P (i.e. each root is a
weight).
5.1.5 Examples.1. In the case of sl2, Q = Zα and P = Z(α
2) since (α,α) = 2. Therefore |P/Q|=
2.
Theorem of the Highest Weight 31
2. For sl3, Q = Zα+Zβ and P = Z( 2α+β3) +Z(α+2β
3) so |P/Q|= 3.
5.1.6 Exercises.1. Prove that |P/Q|= det A, the determinant of the Cartan matrix.
2. Show that P is preserved under the action of the Weyl group on t∗.
5.1.7 Proposition. If V f.d. then dim Vλ = dim Vwλ for all w ∈W .
PROOF: Since the Weyl group is generated by sα | α ∈ R (by definition), it sufficesto prove that dim Vλ = dim Vsαλ. We give sketches of three possible ways to provethe result.
1. Suppose that G is an algebraic or compact group with Lie algebra g and letT be the torus corresponding to t. Then W = N(T )/T . (e.g. in the case ofg = sln, G = SLn and T is the collection of diagonal matrices. Then N(T )is the collection of “monomial matrices” and it is seen that Wsln
= Sn =N(T )/T .) Whence for any w ∈ T there is w ∈ N(T ) such that w = wT , sofor any t ∈ T and v ∈ Vλ,
twv = ww−1 twv = wλ(w−1 tw)v = (wλ)(t)wv,
so w(Vλ) = Vwλ.
2. We can mimic the above construction in g. Observe that in sl2 we maydecompose
0 −11 0
=
1 01 1
1 −10 1
1 01 1
= exp( f )exp(−e)exp( f ).
In general, for each root α ∈ R define sα = exp(e−α)exp(−eα)exp(e−α).
Exercise:
a) Each eα acts nilpotently on V (as V is f.d.) so Sα is a finite sum on Vand is hence well-defined.
b) s2α =: εα : Vλ→ Vλ, where ε2
α = 1 and εα is multiplication by a constanton Vλ. (Which constant?)
c) sα(Vλ) = Vsαλ.
3. But we don’t really need to do any of this. Consider V as a t×sl2-module. Bythe representation theory of sl2, it breaks up as a direct sum of irreducibles,each of which is a string µ,µ − α,µ − 2α, . . . ,µ − mα, where m = λ(hα).These strings are sα-invariant by the sl2-theory.
It follows that when V is f.d., ch(V ) ∈ Z[P]W .
5.1.8 Definition. For µ,λ ∈ P, write µ≤ λ if λ−µ=∑`
i=1 kiαi with ki ∈ N.
5.1.9 Exercise. Prove that “≤” is a partial order on P and show that µ ∈ P | µ≤λ is an obtuse cone in P.
32 Lie Algebras
5.1.10 Definition. µ ∈ P is a highest weight if Vµ 6= 0 and Vλ 6= 0 implies λ ≤ µ.v ∈ Vµ is a singular vector if v 6= 0 and eαv = 0 for all α ∈ R+. (Notice that theweight of eαv is µ + α, so if µ is a highest weight then v is singular.) µ is anextremal weight if wµ is an highest weight vector for some w ∈W .
It is clear that highest weights exist when V is f.d.
5.1.11 Theorem (Theorem of the Highest Weight).Let g be a s.s. Lie algebra over an algebraically closed field K of characteristic zero.
1. a) If V is a f.d. representation of g then V is a direct sum of irreduciblerepresentations.
b) Let P+ = λ ∈ P | (λ,α∨) ≥ 0 for all α ∈ R+ = λ ∈ P | (λ,αi) ≥0 for all αi ∈ Π, the cone of dominant weights. The f.d. irreduciblerepresentations of g are in bijection with P+ via λ 7→ L(λ).
2. More precisely, let V be a f.d. representation of g and let v ∈ Vλ be a singularvector. Then
a) dim Vλ = 1, i.e. Vλ = Kv.
b) If Vµ 6= 0 then µ≤ λ, i.e. λ is a highest weight.
c) λ(hαi) ∈ N for all αi ∈ Π, i.e. λ ∈ P+.
Moreover, if V ′ is another f.d. irreducible representation of g and v′ ∈ V ′ is asingular vector of weight λ then there is a unique isomorphism of g-modulesV → V ′ sending v to v′. We say that V has highest weight λ and denote thisg-module by L(λ).
3. Given λ ∈ P+ there is a f.d. representation of g with highest weight λ.
5.1.12 Corollary.1. For a highest weight λ,
ch(L(λ)) = eλ +∑
µ≤λ
aµλeµ,
and furtherch(L(λ)) = mλ +
∑
µ≤λ,µ∈P+aµλmµ,
where mµ =∑
γ∈Wµ eγ. The mµ | µ ∈ P+ are a basis of Z[P]W , and in factch(L(λ)) | irreducible representations L(λ) are a basis as well.
2. If V and W are f.d. representations of g and ch(V ) = ch(W ) then V ∼=W .
We will (slowly) prove (most of) this theorem and corollary.
5.1.13 Definition. Let ω1, . . . ,ω` be a basis of P which is dual to the basis ofsimple co-roots, i.e. (ωi ,α
∨j ) = δi j for i, j = 1, . . . ,`. These are the fundamental
weights and we could have defined P+ =⊕
i Nωi .
Yet another way to think of P+, the parameters of irreducible representations,is as
P+ = λ ∈ t∗ | ⟨λ, hi⟩ ∈ N.
Proof of the Theorem of the Highest Weight 33
5.1.14 Examples.1. Consider the adjoint representation of g on itself. If g = t+
⊕
α∈R gα thenthe character of this representation is ` +
∑
α∈R eα. What is the highestweight? Recall that a highest weight is a weight λ such that λ+ αi /∈ R forall αi ∈ Π. This implies that λ = θ , the highest root, is the highest weight,and g= L(θ).
2. Representations of sln. There is the standard representation over Cn, whichhas character eε1+· · ·+eεn . The highest weight is ε1 since ε2+(ε1−ε2) = ε1,ε3 + (ε2 − ε3) = ε2, etc.
3. If V and W are representations then so is V ⊗W . Whence V ⊗ V is a repre-sentation, but σ : V ⊗ V → V ⊗ V commutes with the g-action, so S2V and∧2V are g-modules. These need not be irreducible in general, but ∧sCn andSmCn are irreducible sln-modules.
∧sCn: for s ≤ n− 1 has basis ei1 ∧ · · · ∧ eis | i1 < · · · < is, with associatedweights εi1 + · · ·+ εis . The action of an element x ∈ g on v1 ∧ · · · ∧ vs is∑s
i=1 v1 ∧ · · · ∧ x(vi)∧ · · · ∧ vs. Notice Eie j = δ j,i+1e j−1, so Ei(ei1 ∧ · · · ∧eis) = 0 for all i if and only if ei1 ∧ · · · ∧ eis = e1 ∧ · · · ∧ es. Whence thehighest weight isωs = ε1+ · · ·+εs, and ∧sCn = L(ωs) for all 1≤ s < n.(Exercise: compute ch L(ωs).)
SmCn: Exercise: Show that SmCn has basis ei1 ∧ · · · ∧ eim | i1 ≤ · · · ≤ im,and the unique highest weight vector is em
1 (so SmCn = L(mω1) isirreducible) and compute its character.
5.1.15 Exercises.1. Show that V ⊗ V = S2V +∧2V .
2. Show that V ⊗V ∗→ Hom(V, V ) : v⊗w∗ 7→ (x 7→ w∗(x)v) is an isomorphismof g-modules, and that Homg(V, V ) is a submodule, which is 1-dimensionalif V is irreducible.
3. If g = sln then show that (Cn)⊗ (Cn)∗∼−→ sln ⊕C (i.e. the adjoint represen-
tation plus the trivial representation).
4. Let algebras so2n, sp2n act on C2n
a) Compute the highest weight of C2n;
b) Show that (C2n)∼= (C2n)∗ as g-modules;
c) Show that (C2n)⊗ (C2n) has 3 irreducible components and find them.
5.2 Proof of the Theorem of the Highest Weight
Let g be any Lie algebra with a non-degenerate invariant bilinear symmetric form(·, ·) (e.g. the Killing form if g is s.s.). Let x1, . . . , xN be a basis of g, with dualbasis x1, . . . , xN so that (x i , x j) = δi j . Define Ω =
∑Ni=1 x i x
i , the Casimir of g,an operator on any representation of g. (Exercise: show that Ω is independent ofthe choice of basis (though it does depend on the form).)
Claim. Ω is central, i.e. Ωx = xΩ for all x ∈ g.
34 Lie Algebras
Indeed,
[Ω, x] =h∑
x i xi , xi
=∑
x i[xi , x] +
∑
[x i , x]x i ,
as [·, x] is a derivation. Now write [x i , x] =∑
ai j xj and [x i , x] =
∑
bi j x j , so
ai j = ([xi , x], x j) =−([x , x i], x j) =−(x , [x i , x j])
= (x , [x j , x i]) = ([x , x j], x i) =−([x j , x], x i) =−b ji .
Thus [Ω, x] = 0.Now let g be s.s., g= t+
⊕
α∈R gα, and (·, ·) be the Killing form. Choose a basisu1, . . . , u` of t and eα ∈ gα, and a dual basis u1, . . . , u` of t∗ and e−α ∈ g−α suchthat (eα, e−α) = 1. We may normalize the eα so that (eα, e−α) = 1, and uniquenessof the dual basis implies that e−α = eα, so [eα, e−α] = ν−1(α). Then
Ω =∑
i=1
uiui +∑
α∈R+eαe−α +
∑
α∈R+e−αeα
=∑
i=1
uiui + 2
∑
α∈R+eαe−α +
∑
α∈R+ν−1(α).
Write ρ = 12
∑
α∈R+ α, so that
Ω =∑
i=1
uiui + 2ν−1(ρ) + 2
∑
α∈R+e−αeα.
Claim. Let V be a g-module with singular vector v ∈ V of weight λ. Then Ωv =(|λ+ρ|2 − |ρ|2)v.
Indeed, eαv = 0 for all α ∈ R+ and t v = λ(t)v for all t ∈ t, so
Ωv =
∑
i=1
λ(ui)λ(ui) +λ
2ν−1(ρ)
v
=
(λ,λ) + 2(λ,ρ)
v =
(λ+ρ,λ+ρ)− (ρ,ρ)
v.
Claim. If V is also irreducible then Ω acts as (λ+ρ,λ+ρ)− (ρ,ρ) on all of V .
This claim follows from Schur’s Lemma.
5.2.1 Definition. Let g be any Lie algebra over a field K . The universal envelopingalgebra U g is the free associative algebra generated by g modulo the relationsx y − y x − [x , y] | x , y ∈ g.
5.2.2 Exercises.1. An enveloping algebra for g is an associative algebra A and a map ι : g→ A
such that ιxι y − ι yιx = ι[x , y]. Show that U g is the initial object in thecategory of enveloping algebras. For example, if V is a representation of g
then A= End(V ) is an enveloping algebra. Naturally, Ω ∈ U g. Let (U g)n bethe collection of sums of products of at most n things from g, e.g. (U g)0 = K ,(U g)1 = K + g, etc.
Proof of the Theorem of the Highest Weight 35
2. Show that UnUm = Un+m, i.e. that U is a filtered algebra. Also show thatif x ∈ Un and y ∈ Um then x y − y x ∈ Un+m−1.
3. Prove that these monomials span U g (i.e. that Sg→ grU g is surjective).
4. Show that the previous exercise gives a well-defined map Sg→ grU g.
5.2.3 Theorem (PBW). grU g =⊕
nUn/Un−1∼= Sg. The natural map is an iso-
morphism.
Equivalently (and in a language I can understand), the PBW theorem statesthat if x1, . . . , xN is a basis of g then xa1
1 · · · xaNN | a1, . . . , aN ∈ N is a basis of
U g.Let V be a representation of g and v ∈ Vλ a singular vector of weight λ (so
n+v = 0 and t v = λ(t)v for all t ∈ t). It is clear that the g-submodule generatedby v is U gv. If U gv = V then we say that V is a highest weight module withhighest weight λ.
Claim. If v is a singular vector then U gv =U n−v.
Indeed, since g = n− + t+ n+ we have U g = U n− ⊗U t⊗U n+, so the easypart of the PBW theorem implies that we can write a basis for U g as a product ofbases for n−, t, and n+ (in that order). Whence applying v kills the last bit, leavingonly the n− part.
Claim. If V is irreducible and f.d. then V is a highest weight module.
Indeed, V f.d. implies a highest weight exists, call it λ. Then each v ∈ Vλ is asingular vector, so U gv is a submodule, whence it is V since V is irreducible.
5.2.4 Proposition. Let V be a (not necessarily f.d.) highest weight module for g
and v ∈ V be a highest weight vector with highest weight Λ. Then
1. t acts diagonalizably, so V =⊕
λ∈D(Λ) Vλ, where D(Λ) = µ | µ ≤ Λ = µ ∈t∗ | µ= Λ−
∑
kiαi , ki ∈ N;
2. VΛ = Kv; all weight spaces are f.d.;
3. V is irreducible if and only if all singular vectors are in VΛ;
4. Ω = |Λ+ p|2 − |p|2 on V ;
5. if vλ ∈ V is a singular vector then |λ+ p|2 = |Λ+ p|2;
6. if Λ ∈QR, then there are only finitely many λ such that Vλ contains a singu-lar vector;
7. V contains a unique maximal proper submodule I , where I is graded withrespect to t, i.e. I =
⊕
(I ∩ Vλ) and I is the sum of all proper submodules.
PROOF:1. Write V =U n−v for some v, so it is spanned by
e−β1e−β2· · · e−βr
v | eβ ∈ gβ ,β ∈ R+ basis vectors.
But for x = e−β1e−β2· · · e−βr
v, t(x) = (Λ− β1 − · · · − βr)(t)x , whence part1.
36 Lie Algebras
2. There are only finitely many β ∈ R+ which appear in the sum of a givenweight λ ∈ NR+.
3. The third part follows because if vλ ∈ Vλ is a singular vector then U gvλ =U n−vλ =:U is a submodule and a highest weight module. By 1, all weightsof U are in D(λ) = λ−
∑
βi | βi ∈ R+. But λ 6= Λ implies that D(λ) (D(Λ), so in particular Λ /∈ D(λ). Whence U is a proper submodule and V isnot irreducible if there is a singular vector with weight λ 6= Λ.
Conversely, if U ( V is a proper submodule, then tU ⊆ U , so U is a t-graded,so its weights are in D(Λ) and U =
⊕
λ∈D(Λ) Uλ. Let λ be maximal such thatUλ 6= 0. If w ∈ Uλ then eβw ∈ Uλ+β = 0 by maximality for β ∈ R+. Whencew is a singular vector. λ 6= Λ since U 6= V .
7. Any proper submodule U is t-graded, and U =⊕
λ∈D(Λ),λ 6=Λ Uλ, and thesum of all such submodules is still of this form, so I is a maximal propersubmodule.
4. We know that for any singular vector of weight λ, Ωvλ = (|λ+ρ|2−|ρ|2)vλ,so apply to v ∈ VΛ (recall that Ω is central).
6. If λ is singular then 5 (which is obvious) implies that |λ+ ρ|2 = |Λ+ ρ|2.This defines a sphere in RR, a compact set, but D(Λ) is discrete. The singularvectors are the intersection of these, so there are finitely many.
5.2.5 Lemma. If λ is a dominant weight (i.e. λ ∈ P+), λ+ µ ∈ P+, and λ− µ =∑
i kiαi with ki ≥ 0 for all i then |λ+ρ|2 = |µ+ρ|2 implies λ= µ.
PROOF: We have
0= (λ+ρ,λ+ρ)− (µ+ρ,µ+ρ)= (λ−µ,λ+µ+ 2ρ)
=∑
i=1
ki(αi ,λµ+ 2ρ)
so since λ+ µ+ 2ρ ∈ P+ and (αi , 2ρ) = (αi ,αi) > 0, we get that 0 is a positivelinear combination of the ki , so each ki = 0 since they are all non-negative.
5.2.6 Definition. Let Λ ∈ t∗. A Verma module M(Λ) is an initial object in thecategory of highest weight modules with highest weight Λ and highest weightvector vΛ, i.e. if V is any highest weight module with highest weight Λ and highestweight vector v then there is a g-module map M(Λ)→ V : vΛ 7→ v.
Such a map is obviously surjective and unique.
5.2.7 Proposition. Given Λ ∈ t∗ there exists a unique Verma module M(Λ) andthere is a unique highest weight module L(Λ) which is irreducible.
PROOF: Uniqueness is clear from the definition of Verma module via universalproperty. Existence requires the PBW theorem. (M(Λ) = U g ⊗Ub kΛ, whereb = n+ + t and kΛ is the 1-dimensional b-module such that t x = λ(t)x for t ∈ t
Weyl Character Formula 37
and n+x = 0 (it is IndgbkΛ). SinceU g=U n−⊗U t⊗U n+ =U n−⊗U b as vector
spaces (by PBW), so basically M(Λ) =U n− (as K-vector spaces).)If V is any highest weight module, V = M(Λ)/J for some submodule J . If
it is an irreducible highest weight module, J must be maximal. But previousproposition implies that the maximal proper submodule of M(Λ) is unique, soL(Λ) is unique.
5.2.8 Exercise. Show that Homg(U g⊗Ub W, V ) = Homb(W, ResgbV ).
5.2.9 Corollary. In particular, any irreducible finite dimensional module is L(Λ),where Λ(hi) ∈ N, i.e. Λ ∈ P+.
We still need to show that if Λ ∈ P+ then L(Λ) is f.d. We also need to showthat every f.d. module is a direct sum of irreducibles. The proof has been omittedfor lack of time, but all the ingredients are in place. (As with sl2, the Casimir Ωshows that the different irreducible representations do not interact. The fact thata representation whose composition series consists of a single irreducible is in facta direct sum follows almost immediately from the fact that t acts diagonalizably.)
5.2.10 Exercises.1. For g = sl2, there is an M(λ) for each λ ∈ C (where Hv = λv). Show that
M(λ) is irreducible if and only if λ 6∈ N. Moreover, if λ ∈ N then M(λ)contains a unique proper submodule, the span of Fλ+1v, Fλ+2v,. . .
2. Show that if g is an arbitrary s.s. Lie algebra and Λ(hi) ∈ N then FΛ(hi)+1i vΛ is
a singular vector and Fi = e−αi(working in the sl2 attached to some simple
root αi).
3. Compute ch M(Λ). Note for sl2 the characters are zλ
1−z−2 .
5.3 Weyl Character Formula
Recall that W := ⟨sα | α ∈ R⟩.
5.3.1 Lemma. If α is a simple root then sα(R+ \ α) = R+ \ α. Note that sαα=−α /∈ R+.
PROOF: Write si = sαifor αi ∈ Π. If α ∈ R+ then α=
∑
i kiαi where ki ≥ 0, and
siα=∑
j 6=i
k jα j +
− ki +∑
j 6=i
(α j ,α∨i )k j
αi .
If α 6= αi then some k j > 0, so the coefficient of α j in siα is greater than zero.But R = R+ q (−R+), so one positive coefficient implies that all coefficients arepositive, so siα ∈ R+.
If g ∈ O(V ) (the orthogonal group) then gsαg−1 = sgα, so in particular wsαw−1 =swα for all w ∈W . It is a theorem that all Weyl groups are of the form
W = ⟨s1, . . . , s` | s2i = 1, (sis j)
mi j = 1⟩
where the mi j are given by the Cartan matrix according to the following table.
38 Lie Algebras
ai ja ji 0 1 2 3mi j 2 3 4 6
5.3.2 Exercise. Check for each root system that the above holds (note that it isenough to check for the rank two root systems (why?)). Compare this to theexistence and uniqueness of g which was generated by ei , hi , fi, αi ∈ Π.
5.3.3 Lemma. ρ(hi) = 1 for i = 1, . . . ,`, i.e. ρ =∑`
i=1ωi .
PROOF: By 5.3.1,
siρ = si
1
2αi +
1
2
∑
α∈R+\αi
α
=−1
2αi +
1
2
∑
α∈R+\αi
α= ρ−αi
but also siρ = ρ− ⟨ρ,α∨i ⟩αi , so ⟨ρ,α∨i ⟩= 1 for all i, implying ρ =∑
iωi .
For sln,
2ρ = (ε1 − ε2) + · · ·+ (εn−1 − εn)+ (ε1 − ε3) + · · ·+ (εn−2 − εn)+ · · ·+ (ε1 − εn)
= (n− 1)ε1 + (n− 3)ε2 + · · ·+ (3− n)εn−1 + (1− n)εn.
5.3.4 Lemma. For λ ∈ t∗ arbitrary, if M(λ) is the Verma module with highestweight λ then
ch M(λ) = eλ∏
α∈R+(1− e−α)−1 = eλ
∏
α∈R+(1+ e−α + e−2α + · · · )
PROOF: Write R+ = β1, . . . ,βs. Then ek1
−β1· · · eks
−βsvλ is a basis of M(λ) of weight
λ − k1β1 − · · · − ksβs, so dim M(λ)λ−β is the number of ways which β can bewritten as a sum
∑
i kiβi of roots βi with non-negative coefficients ki , which isthe coefficient of e−β in
∏
α∈R+(1− e−α)−1. Indeed, ch(V ⊗W ) = ch V ch W , andchC[x] = 1
1−x.
Let ∆=∏
α∈R+(1− e−α), so ch M(λ) = eλ/∆.
5.3.5 Lemma. w(eρ∆) = det(w)eρ∆ for all w ∈ W . (Notice that det(w) ∈ ±1since det sα =−1 for all α ∈ R.)
PROOF: It is enough to show that si(eρ∆) = −eρ∆ as W is generated by the si .But
eρ∆= eρ(1− e−αi )∏
α∈R+\αi
(1− e−α)
si(eρ∆) = eρ−αi (1− eαi )
∏
α∈R+\αi
(1− e−α)
= eρ(e−αi − 1)∏
α∈R+\αi
(1− e−α)
=−eρ∆.
Weyl Character Formula 39
Recall that P = λ ∈ t∗ | λ(hi) ∈ Z for all αi ∈ Π and P+ = λ ∈ t∗ | λ(hi) ∈N for all αi ∈ Π.
5.3.6 Theorem (Weyl Character Formula).Let λ ∈ P+, so L(λ) is finite dimensional. Then
ch L(λ) =1
∆
∑
w∈W
det(w)ew(λ+ρ)−ρ
=∑
w∈W
w
eλ
∆
i.e. ch L(λ) =∑
w∈W det(w) ch M(w(λ+ρ)−ρ).
PROOF: Omitted, but again, all the ingredients are in place. The second equalityis immediate from 5.3.5.
For sl2, set z = e−α/2, so C[P] = C[z, z−1], ρ = α
2, ∆ = (1− z−2)−1, W = ⟨s |
s2 = 1⟩, and sz = z−1. Then
ch L(mα
2) =
zm − z−(m+1)−1
1− z−2
=zm+1 − z−(m+1)
z− z−1
= zm + zm−2 + · · ·+ z2−m + z−m.
5.3.7 Corollary. ch L(0) = 1, as L(0) = C, so
∆=∏
α∈R+(1− e−α) =
∑
w∈W
det(w)ew(ρ)−ρ.
This is the Weyl denominator formula.
5.3.8 Exercise. For g = sln, show that the Weyl denominator formula is just theidentity (rewriting allowed)
det
1 . . . 1x1 . . . xn...
...xn−1
1 . . . xn−1n
=∏
i< j
(x j − x i),
i.e. the Vandermond determinant formula, and that Z[P] = Z[x±11 , . . . , x±1
n ] andW =Sn.
Given µ ∈ P, define a homomorphism Fµ : C[P] → C((q)) by eλ 7→ q−(λ,µ).Then F0(eλ) = 1 and F0(ch L(λ)) = dim L(λ). Let’s apply Fµ to the Weyl denomi-nator formula. We get
q−(ρ,µ)∏
α∈R+(1− q(α,µ)) =
∑
w∈W
det(w)q−(w(ρ),µ) =∑
w∈W
det(w)q−(ρ,w(µ))
so
Fµ(ch L(λ)) =
∑
w∈W det(w)q−(w(λ+ρ),µ)
q−(ρ,µ)∏
α>0(1− q(α,µ))
if (α,µ) 6= 0 for all α ∈ R+ (e.g. µ= ρ satisfies this condition).
40 Lie Algebras
5.3.9 Proposition (q -dimension formula).
dimq L(λ) :=∑
β
dim L(λ)βq−(β ,ρ) =q−(ρ,λ+ρ)
∏
α>0(1− q(α,λ+ρ))
q−(ρ,ρ)∏
α>0(1− q(α,ρ))
PROOF: Immediate from the above discussion.
5.3.10 Corollary (Weyl dimension formula).
dim L(λ) =∏
α>0
(α,λ+ρ)(α,ρ)
PROOF: Use l’Hopital, letting q→ 1.
5.3.11 Example. For sl3 (a.k.a. A2), R+ = α,β ,α+β, ρ = α+β =ω1+ω2, soif λ= m1ω1 +m2ω2 (mi ∈ N) then
α β α+ β(·,λ+ρ) m1 + 1 m2 + 1 m1 +m2 + 2(·,ρ) 1 1 2
whence dim L(λ) = 12(m1 + 1)(m2 + 1)(m1 + m2 + 2). As an exercise, compute
dimq L(λ), and do this for B2 and G2.
5.3.12 Proposition. The q-dimension Fρ(ch L(λ)) = dimq L(λ) is a (symmetric)unimodal polynomial in q.
PROOF: If V is an representation of sl2, then ch V is a symmetric unimodal poly-nomial. If we put H = ν−1(ρ) then Fρ(ch L(λ)) =
∑
dim L(λ)nqn, where L(λ)n =x ∈ L(λ) | H x = nx, so we need to extend H to an sl2. Set E = eα1
+ · · ·+ eα` .Observe that H =
∑
i cihi for some ci ∈ Q, as the hi are a basis of t. DefineF =
∑
i cie−αi(where ⟨eα, hα = α∨, e−α⟩ are the root sl2’s). Exercise: ⟨E, H, F⟩ is
an sl2.
5.3.13 Exercise. Deduce the following polynomials are unimodal.
1. n
k
= [n]![k]![n−k]!
, where [n] = qn−q−n
q−q−1 and [n]! = [n][n− 1] · · · [1]. (Hint:
g= sln and V = SkCn or g= sln+k and V = ∧kCn+k.)
2. (1+ q)(1+ q2) · · · (1+ qn). (Hint: spin representation of Bn.)
If λ,µ ∈ P+, then L(λ)⊗ L(µ) =∑
ν∈P+ mνµλL(ν) for some mν
µλ ∈ N. Let’s de-termine these coefficients using the Weyl character formula. Define a conjugationon Z[P] by eλ = e−λ and C T : Z[P]→ Z by C T (eλ) = 1 if λ= 0 and 0 otherwise.Define (·, ·) : Z[P]×Z[P]→Q by ( f , g) = 1
|W |C T ( f g∆∆), and let χλ = ch L(λ).
5.3.14 Lemma. (χλ,χµ) = δλµ if λ,µ ∈ P+.
PROOF: We have
(χλ,χµ) =1
|W |C T
∑
x ,w∈W
det(x)det(w)ew(λ+ρ)−ρex(µ+ρ)−ρ
,
but λ,µ ∈ P+ so w(λ+ ρ) = µ+ ρ if and only if µ = λ and w = 1 (exercise).Whence C T (ew(λ+ρ)−x(µ+ρ)) = δw,xδµλ.
Crystals 41
5.3.15 Corollary. mνλµ = (χµχλ,χν) = dimHomg(L(ν), L(µ)⊗ L(λ)).
6 Crystals
The following chapter consists mostly of results of Kashiwara.Let g be a s.s. Lie algebra, Π = α1, . . . ,α` a choice of simple roots, and P the
corresponding weight lattice.
6.0.16 Definition. A crystal for g is a set B (not containing 0, by convention) withfunctions wt : B→ P and ei : B→ Bq 0 and fi : B→ Bq 0 such that
1. if ei b 6= 0 then wt(ei b) = wt(b)+αi and if fi b 6= 0 then wt( fi b) = wt(b)−αi;
2. for all b, b′ ∈ B, ei b = b′ if and only if b = fi b′; and
3. ϕi(b)− εi(b) = ⟨wt(b),α∨i ⟩ for all i and b ∈ B, where
εi(b) =maxn≥ 0 | eni b 6= 0 and φi(b) =maxn≥ 0 | f n
i b 6= 0.
The coloured, directed graph with vertices B and edge bb′ coloured by i if ei b′ = b
is the crystal graph, i.e. the i-arrows of the crystal graph point along the action offi .
6.0.17 Example. With g= sl2, for any n
n // n− 2 // n− 4 // . . . // 2− n // −n
is a crystal, as is
n // n− 2 // n− 4 // . . . ,
where all the edge colours are 1. For sl2, if b is of weight n−2k (so b = (n−2k)ω1,where ω1 =
12α), in the finite crystal ε1(b) = k and ϕ1(b) = n− k, so ϕ1(b)−
ε1(b) = n− 2k = ⟨wt(b),α∨1 ⟩ as required. Also observe that ϕ1(b) + ε1(b) = n,the length of the string.
6.0.18 Definition. For crystals B1 and B2, define B1 ⊗ B2 as B1 ⊗ B2 = B1 × B2 assets, wt(b1 ⊗ b2) = wt(b1) +wt(b2), and
ei(b1 ⊗ b2) =
¨
ei b1 ⊗ b2 if ϕi(b1)≥ εi(b2)b1 ⊗ ei b2 if ϕi(b1)< εi(b2)
,
and
fi(b1 ⊗ b2) =
¨
fi b1 ⊗ b2 if ϕi(b1)> εi(b2)b1 ⊗ fi b2 if ϕi(b1)≤ εi(b2)
.
42 Lie Algebras
Try it and see: in the case of sl2 B1 ⊗ B2 breaks up into strings just as sl2-representations do.
• // • // • // • // •
•
• // • // • // • // •
•
• // • // • // •
•
•
• // • // •
•
•
• • // • • • •
6.0.19 Exercises.1. B1 ⊗ B2 is a crystal with these associated functions.
2. Via the obvious map, (B1 ⊗ B2)⊗ B3∼= B1 ⊗ (B2 ⊗ B3) as crystals.
6.0.20 Definition. B∨, the dual crystal of B, is obtained from B by reversing ar-rows of the crystal graph. Formally, wt(b∨) = −wt(b), ei(b∨) = ( fi b)∨ andfi(b∨) = (ei b)∨, and εi(b∨) = ϕi(b) and ϕi(b∨) = εi(b).
6.0.21 Exercise. (B1 ⊗ B2)∨ ∼= B∨2 ⊗ B∨1 via the obvious map.
6.0.22 Theorem. Let L(λ) be a f.d. irreducible highest weight representation withhighest weight λ. Then there is a crystal B(λ) in one-to-one correspondence witha basis of L(λ).
1. It follows that ch L(λ) =∑
b∈B ewt(b).
2. Moreover, for each simple root αi , there is (sl2)i ,→ g, and the decompositionof L(λ) as an sl2-module is exactly given by the i-strings. In particular, thecrystal graph of B(λ) is connected (when colours are ignored), and it isgenerated by the fi ’s applied to the unique element of B(λ)λ.
3. The crystal B(λ)⊗B(µ) is precisely the crystal of L(λ)⊗ L(µ), i.e. it decom-poses into connected components, each component a B(ν) for some ν ∈ P+,and we get as many B(ν)’s as the multiplicity with which L(ν) occurs inL(λ)⊗ L(µ).
PROOF: Omitted.
Remark. B(λ)∨ is the crystal of the g-module L(λ)∗ = L(µ), a f.d. irreduciblemodule with lowest weight −λ. B 7→ B∨ corresponds to the automorphism of Liealgebras g→ g mapping ei 7→ fi , fi 7→ ei , and hi 7→ −hi . (Exercise: compute µ interms of λ.)
Remark. If h⊆ g is a subalgebra generated by (sl2)i , for i ∈ I ⊆ Π, then the crystalfor L(λ)|h is just B(λ)|forget stuff i for i /∈ I .
Crystals 43
Notice that L(λ)⊗ L(µ)∼−→ L(µ)⊗ L(λ), so B(λ)⊗B(µ)
∼−→ B(µ)⊗B(λ) by thetheorem, but B⊗ B′ 6∼= B′ ⊗ B for arbitrary crystals.
6.0.23 Example. For g = sln and λ = ω1 = ε1, L(λ) = Cn is the standard repre-sentation of sln, and it has crystal
•
1
λ = ε1
•
2
λ−α1 = ε2
...
n−1
• λ−α1 − · · · −αn−1 = εn
Taking the specific example of n = 3, computing C3 ⊗C3 is done by multiplyingand decomposing.
• 1 // • 2 // •
•
1
• 1 // • 2 //
1
•
1
•
2
•
2
• 2 // •
2
• • 1 // • •
From the diagram we see that C3 ⊗C3 ∼= S2C3 ⊕∧2C3.
6.0.24 Definition. A crystal B is an integrable crystal if it is the crystal of a f.d.representation for g. b ∈ B is a highest weight vector if ei b = 0 for all i.
B(λ) has a unique highest weight vector. If B is an integrable crystal then itsdecomposition into connected components is in one-to-one correspondence withhighest weight vectors in B. We now apply this to B1 ⊗ B2.
6.0.25 Lemma. ei(b1 ⊗ b2) = 0 if and only if ei b1 = 0 and εi(b2)≤ ϕi(b1).
PROOF: If εi(b2) > ϕ(b1) then εi(b2) > 0, so ei b2 6= 0. The other direction isclear.
6.0.26 Corollary. b ⊗ b′ ∈ B(λ)⊗ B(µ) is a highest weight vector if and only ifb ∈ B(λ)λ is a highest weight vector for B(λ) and εi(b′)≤ ⟨λ,α∨i ⟩ for all i.
PROOF: εi(b′) ≤ ϕi(b) = ⟨λ,α∨i ⟩ (which is the length of the i-string through thehighest weight vector b, and by condition 3 since ei(b) = 0).
44 Lie Algebras
Whence we have a Clebsch-Gordan type rule
L(λ)⊗ L(µ) =⊕
b′∈B(µ)εi(b′)≤⟨λ,α∨i ⟩
L(λ+wt(b′)).
Recall that ∧iCn has highest weight ωi = ε1 + · · · + εi , so if λ ∈ P+ withλ= k1ω1 + · · ·+ kn−1ωn−1, then L(λ) is a summand of
L(ω1)⊗k1 ⊗ · · · ⊗ L(ωn−1)
⊗kn−1 ,
as v⊗k1ωi⊗ · · ·⊗ v⊗kn−1
ωn−1is a singular vector in this space since vωi
is a highest weightin each L(ωi), and the weight of this vector is λ = k1ω1 + · · ·+ kn−1ωn−1. But inthe specific case of sln, ∧iCn is a summand of (Cn)⊗i , i.e. B(ωi) can be determinedfrom B(ω1)⊗i . Therefore the combinatorial rule for product of crystals, togetherwith the standard n-dimensional crystal, already determine the crystal of all sln-modules.
Semi-standard Young tableaux
Write
B(ω1) = 11 // 2
2 // . . . n−1 // n ,
and considerbi := 1 ⊗ 2 ⊗ · · · ⊗ i ∈ B(ω1)
⊗i ,
which corresponds to e1 ∧ e2 ∧ · · · ∧ ei ∈ ∧iCn ⊆ (Cn)⊗i .
6.0.27 Exercises.1. Show that bi is a highest weight vector in B(ω1)⊗i , of highest weight ω1 =ε1 + · · ·+ εi . Hence the connected component of B(ω1)⊗i containing bi isB(ωi).
2. Show that this component consists precisely of
a1 ⊗ · · · ⊗ ai | 1≤ a1 < a2 < · · ·< ai ≤ n ⊆ B(ω1)⊗i .
Write this vector as
a1...ai
, so that the highest weight vector is
1...i
.
Now let λ=∑
i kiωi , and embed
B(λ) ,→ B(ω1)⊗k1 ⊗ · · · ⊗ B(ωn−1)
⊗kn−1
by the highest weight vector bλ 7→ b⊗k11 ⊗ · · · ⊗ b⊗kn−1
n−1 . Represent an element ofthe righthand side as (picture of a Young tableaux with no relations for rows andstrictly increasing going down each column)
6.0.28 Definition. A semi-standard tableaux is a tableaux which strictly increasesdown columns and decreases (not necessarily strictly) along rows.
Crystals 45
6.0.29 Theorem.1. The collection of semi-standard tableaux of shape λ is the connected com-
ponent of B(λ).
2. ei and fi are given by. . . (determine this as an exercise).
PROOF: Exercise.
6.0.30 Example. The crystals of the standard representations of the classical Liealgebras are as follows.
An: We have seen that the standard representation of sln has crystal
11 // 2
2 // . . . n−1 // n .
Bn: For so2n+1,
11 // 2
2 // . . . n−1 // n n // 0n // n
n−1 // . . . 2 // 21 // 1 .
Cn: For sp2n,
11 // 2
2 // . . . n−1 // n n // nn−1 // . . . 2 // 2
1 // 1 .
Dn: For so2n,
nn
AAA
AAAA
A
11 // 2
2 // . . . n−2 // n−1
n−1>>
n@
@@@@
@@@ n−1
n−2 // nn−1 // . . . 2 // 2
1 // 1
n
n−1
??~~~~~~~~
.
6.0.31 Exercises.1. Check that these are the crystals of the standard representations.
2. What subcategory of the category of representations of g do they generate,i.e. what do you get by taking the tensor product any number of times anddecomposing (as we did above for An)? (Hint: consider the center of thesimply connected group G with Lie algebra g and how it acts on the standardrepresentation, i.e. consider the sublattice of P which all summands of ten-sor powers of the standard representation lie in. The centers are as follows:Z/2×Z/2 for D2n, Z/4 for D2n+1, and Z/2 for Bn and Cn. The center doesnot act faithfully on the standard representation for Bn, but does for Cn.)
46 Lie Algebras
6.0.32 Example (Spin Representation). Recall the a Lie algebra of type Bn hasDynkin diagram of the form
• α1 = ε1 − ε2
• α1 = ε2 − ε3
...
• αn−1 = εn−1 − εn
• αn = εn
.
The representation L(ωn) is the spin representation. As an exercise, show thatdim L(ωn) = 2n (hint: use the Weyl dimension formula). Set B = (i1, . . . , in) |i j ∈ ±1 and define wt(i1, . . . , in) =
12
∑
j i jε j . Define
e j(i1, . . . , in) =
¨
(i1, . . . , i j−1, 1,−1, i j+2, . . . , in) if (i j , i j+1) = (−1,1)0 otherwise
and
en(i1, . . . , in) =
¨
(i1, . . . , in−1, 1) if in =−1
0 otherwise.
In the case of Dn, the representations V+ := L(ωn) and V− := L(ωn−1) are thehalf-spin representations. Set
B± = (i1, . . . , in) | i j ∈ ±1,n∏
j=1
i j =±1,
and wt and e1, . . . , en−1 as above. Define
en(i1, . . . , in) =
¨
(i1, . . . , in−2, 1, 1) if (in−1, in) = (−1,−1)0 otherwise
.
6.1 Littelmann Paths
6.1.1 Definition. Let PR denote P ⊗Z R, the R-vector space spanned by P. ALittelmann path (or simply path) is a piecewise-linear continuous map π : [0,1]→PR.
Paths are considered up to reparameterization, i.e. define π ∼ π ϕ, whereϕ : [0, 1]→ [0,1] is any piecewise-linear homeomorphism. Let
P = π | π(0) = 0,π(1) ∈ P/∼ .
Littelmann Paths 47
Define a crystal structure on P by wt(π) = π(1) and ei as follows. Let
h=min0,Z∩ ⟨π(t),α∨i ⟩ | 0≤ t ≤ 1= smallest integer in ⟨π([0, 1]),α∨i ⟩ ∪ 0.
and define ei(π) = 0 if h = 0. Otherwise, if h < 0, let t1 be the smallest timet ∈ [0,1] such that ⟨π(t),α∨i ⟩ = h (i.e. the first time that π crosses h) and let t0be the largest t ∈ [0, t1] such that ⟨π(t),α∨i ⟩= h+ 1. Define the path ei(π) by
ei(π)(t) =
π(t) 0≤ t < t0
π(t0) + sαi(π(t)−π(t0)) t0 ≤ t < t1
π(t) +αi t ≥ t1
In words, ei(π) is that path that is π up to t0 (i.e. the last time h+ 1 is hit beforethe first time hitting h), the reflection of π via sαi
from t0 to t1, and then π againfrom t1 to 1, but translated for continuity. (The picture from lecture makes thisclear.)
6.1.2 Exercise. Show that εi(π) =−h (hint: ⟨αi ,α∨i ⟩= 2).
6.1.3 Example. In sl2,
e1
−α2 0oo
= 0 // α2
and applying ei again gives the zero path. Similarly,
e1
−α 0oo
= −α2 0 ,
i.e. the path that goes from 0 to −α2
and back. Applying e1 again gives 0→ α, andapplying it again give the zero path.
6.1.4 Exercise. What do you get when you hit −α−β ← 0 with e1 and e2’s in thecase of sl3?
Let π∨ denote the reverse of the path π, i.e. define π∨ by π∨(t) = π(1− t)−π(1). To complete the definition of the crystal structure let fi(π) = ei(π∨)∨.
6.1.5 Exercise. Prove that P is a crystal with wt, ei , and fi defined in this way.
Let P + = π ∈ P | π[0, 1] ⊆ P+R , where P+R = x ∈ PR | (∀ i)⟨x ,α∨i ⟩ ≥ 0.Observe that if π ∈ P + then eiπ = 0 for all i. Let Bπ be the subcrystal of Pgenerated by π, soBπ = fi1 · · · fir
π | r ≥ 0.
6.1.6 Theorem (Littelmann). If π,π′ ∈ P + thenBπ ∼=Bπ′ if and only if π(1) =π′(1). Furthermore,Bπ is isomorphic to the crystal B(π(1)) (i.e. the crystal asso-ciated with the highest wieght module L(π(1))).
It follows that we need only consider straight line segements 0→ λ and theirimages under the fi . Littelmann described the paths explicitly in this case.
48 Lie Algebras
6.1.7 Definition. For π1,π2 ∈ P , let π1 ∗π2 be their concatenation, i.e.
(π1 ∗π2)(t) =
(
π1(2t) 0≤ t ≤ 12
π2(2t − 1) +π1(1)12≤ t ≤ 1
6.1.8 Exercise. Prove that ∗ : P ⊗ P : π1 ⊗ π2 7→ π1 ∗ π2 is a well-definedmorphism of crystals.
6.1.9 Corollary. Bπ1⊗Bπ2
∼= Bπ1∗π2is an explicit model for the tensor product
of crystals.
6.1.10 Exercises.1. Describe the tensor product of representations of sln in terms of Young
tableaux. This explicit bit of combinatorics is called the Littlewood-Richardsonrule.
2. Compute Lλ ⊗ sl3.
Let B be a crystal with εi(b),γi(b) <∞ for all i. We may define an action ofthe Weyl group on B by
si(b) =
(
f⟨wt(b),α∨i ⟩i b ⟨⟨wt(b),α∨i ⟩ ≥ 0
e−⟨wt(b),α∨i ⟩i b ⟨⟨wt(b),α∨i ⟩ ≤ 0
Then s2i = 1 and clearly wt(si b) = si(wt(b)).
pic ture
6.1.11 Proposition (Kashiwara). If B is an integral crystal then the si ’s satisfythe braid relations, so the above truly gives an action of W on B when extendedin the obvious way (since the si generate W ).
PROOF: It is enough to prove it for rank two Lie algebras, hence for the crystalsof all sl3, sl2 × sl2, so5, and G2 modules. Do this as an exercise using Littelmannpaths.
Warning: for w ∈W , wei 6= e jw in general, even when wαi = α j .
Index
HeisW , 13
Abelian Lie algebra, 4abstract Cartan matrix, 26adjoint, 4adjoint representation, 4algebraic representation, 5
Cartan matrix, 25Cartan subalgebra, 17Casimir, 8, 33center, 4central series, 13character, 11, 30Clebsch-Gordan rule, 12commutator algebra, 13composition series, 9cone of dominant weights, 32crystal, 41crystal graph, 41
derived series, 13direct sum root system, 23dual crystal, 42dual numbers, 2Dynkin diagrams, 26
enveloping algebra, 34even lattice, 24extended Cartan matrix, 27extremal weight, 32
fundamental weights, 32
generalized Cartan matrix, 29generated by roots, 24
half-spin representations, 46Heisenberg Lie algebra, 13highest root, 27highest weight, 32highest weight module, 35highest weight vector, 7, 43Hopf algebra, 11
integrable crystal, 43invariant, 15irreducible root system, 23isomorphism of root systems, 23
Kac-Moody algebra, 29Killing form, 15
lattice, 23lattice of roots, 30lattice of weights, 30Lie algebra, 2, 3Lie algebra homomorphism, 4Lie algebra representation, 4Lie bracket, 3linear algebraic group, 2Littelmann path, 46Littlewood-Richardson rule, 48
maximal torus, 17
nilpotent, 13
radical, 15rank of a root system, 23reduced, 23representation, 4root, 18root of a lattice, 24root system, 23root-space decomposition, 18
semi-simple, 12semi-standard tableaux, 44Serre relations, 29simple, 12simple root, 25simply laced, 23simultaneous eigenspace, 17singular vector, 32solvable, 13spin representation, 46string, 8
tangent bundle, 2torus, 17trace forms, 15
universal enveloping algebra, 34
Verma module, 36
weight space, 7Weyl denominator formula, 39Weyl group, 23
49
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