Modular Lie Algebras - Warwick Insitehomepages.warwick.ac.uk/~masdf/modular/ln_8feb.pdf · 2010. 2. 8. · Classiﬁcation of modular Lie algebras 1.1 Basics We recall some basic

Modular Lie Algebras

Dmitriy Rumynin1

February 8, 2010

1 c©Dmitriy Rumynin 2010

2

How to use these notes

The lecture notes are split into 3 chapters, further split into 30 sections.Each section will be discussed on a separate lecture. There will be a cutoffpoint for the exam: a few of the last lectures will not appear on the exam.

Each section consists of subsections. Regular subsection are examinable.Vista and background subsections are not. Vista subsections contain blue-sky material for further contemplation. Background sections contain mate-rial that should be known to students but is still included to make the notesself-contained. I have no intention to examine background subsections butthey are useful for understanding of the course. If you are missing somebackground, please, ask me and I could add a background subsection toremedy it.

Exercise sections contain exercises that you should attempt: they mayappear on the exam. Proofs of propositions are examinable as soon asthe proof is written or covered in class. Overall, 100% of the exam willbe covered by these lecture notes and in-class proofs. Consult these writtennotes because my planning may not be perfect and some material in a sectioncould be skipped during the lecture. You should still learn it because it couldappear on the exam.

It is worth mentioning that I don’t follow any particular book. Further-more, no book is relevant enough. The lecture notes are not fully ready yet.All the updates will be available on the website. Check whether you havegot the latest version. The final version will be available from the generaloffice.

If you see any errors, misprints, oddities of my English, send me an email.Also write me if you think that some bits require better explanation. If youwant to contribute by writing a proof, an example, a valuable observation,please, do so and send them to me. Just don’t use WORD. Send them astext files with or without LATEX: put all the symbols you want me to latexbetween $ $. All the contributions will be acknowledged and your name willbe covered with eternal (sh)fame.

Introduction

We study modular Lie algebras, that is, Lie algebras over fields of pos-itive characteristic in this course. After careful thinking I have chosen thefollowing three topics to discuss:

• classification of simple Lie algebras,

3

• restricted Burnside problem,

• irreducible representations of classical simple Lie algebras.

The reasons to choose these three is my perception of what constitutes animportant development in the area. Complete proofs are often long andcomplicated and the emphasis is definitely on understanding the conceptsinstead of proving all the results. The aim is to introduce various conceptsfrom Algebra, Algebraic Geometry and Representation Theory and see howthey are useful for solving mathematical problems.

4

Contents

1 Classification of modular Lie algebras 7

1.1 Basics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.2 Things that fail in positive characteristic . . . . . . . . . . . . 121.3 Free algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . 181.4 Universal enveloping algebras . . . . . . . . . . . . . . . . . . 211.5 p-th powers . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241.6 Uniqueness of restricted structures . . . . . . . . . . . . . . . 271.7 Existence of restricted structures . . . . . . . . . . . . . . . . 311.8 Schemes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 331.9 Differential geometry of schemes . . . . . . . . . . . . . . . . 361.10 Generalised Witt algebra . . . . . . . . . . . . . . . . . . . . 381.11 Filtrations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 411.12 Witt algebras are generalised Witt algebra . . . . . . . . . . . 44

5

6 CONTENTS

Chapter 1

Classification of modular Lie

algebras

1.1 BasicsWe recall some basic definitions in this lecture. If you know how to asso-

ciate Lie algebra to a Lie group, then all of this should be quite transparent.

1.1.1 Algebras and multiplications

Recall that an algebra over a field K of characteristic p is a vector spaceA with a K-bilinear multiplication operation µ : A × A → A, which weusually write in the standard shorthand notation a ∗ b or even ab instead ofthe long notation µ(a, b).

To an element a ∈ A one can associate two linear operators La, Ra :A → A where La(b) = ab and Ra(b) = ba. The names of this operatorsare intuitive: La is the left multiplication operator and Ra is the rightmultiplication operator.

These operators are useful for many things. One is that the algebraaxioms can be reformulated in their terms. For instance, we need associativealgebras that are characterised by the associativity identity a(bc) = (ab)cfor all a, b, c ∈ A and the presence of identity element 1 ∈ A such thatL1 = R1 = IdA. The associativity can be reformulated as La ◦ Lb = Labfor all a, b ∈ A in terms of the left multiplication. Equivalently, it saysRc ◦Rb = Rbc for all b, c ∈ A in terms of the right multiplication.

Another important class of algebras we use is commutative algebras.These are associative algebras which satisfy the commutativity identityab = ba for all a, b ∈ A. In terms of multiplication operators, it says thatLa = Ra for all a ∈ A.

7

8 CHAPTER 1. CLASSIFICATION OF MODULAR LIE ALGEBRAS

Our main protagonist is Lie algebras. They satisfy the anticommuta-tivity and Jacobi identity. Both identities have subtle points which requiresome discussion. We define the anticommutativty as a ∗ a = 0 for all a ∈ A.Notice that this implies that 0 = (a+ b) ∗ (a+ b)− a ∗ a− b ∗ b = a ∗ b+ b ∗ aand, consequently, a ∗ b = −b ∗ a for all a, b ∈ A. This property can bereformulated as La = −Ra for all a ∈ A. Notice that if p 6= 2 then this isequivalent to anticommutativity: a ∗ a = −a ∗ a implies that 2a ∗ a = 0 anda ∗ a = 0.

On the other hand, if p = 2 then 2a ∗ a = 0 always holds for trivialreasons. Moreover, a ∗ b = −b ∗ a is the same as a ∗ b = b ∗ a since 1 =−1. This identity is commutativity. The anticommutativity is a ∗ a = 0.One unintended consequence of this terminology is that anticommutativityimplies commutativity.

The Jacobi identity is (a ∗ b) ∗ c + (b ∗ c) ∗ a + (c ∗ a) ∗ b = 0. We aregoing to reformulate it twice in the coming two subsections.

1.1.2 Endomorphisms and Derivations

The following terminology should be familiar to you from Algebra-II,although we are going to apply it to general algebras. A homomorphismof algebras is a linear map f : A → B such that f(xy) = f(x)f(y). If Acontains identity, we also require f(1A) = 1B . As a consequence the zeromap is a homomorphism of associative algebras if and only if B = 0.

An isomorphism is a bijective homomorphism. An endomorphism is ahomomorphism f : A → A. Finally, an automorphism is an isomorphismf : A→ A.

A new notion, which you may have seen in Lie Algebras is a derivation.A derivation is a linear map d : A→ A such that d(ab) = d(a)b+ ad(b), i.e.it satisfies Leibniz identity.

Proposition 1.1.1 For an anticommutative algebra A the Jacobi identityis equivalent to the fact that each La is a derivation for each a ∈ A.

Proof: We rewrite Jacobi identity as (a ∗ b) ∗ c+ (c ∗ a) ∗ b = −(b ∗ c) ∗ a.Using anticommutativity we rewrite further as (a∗b)∗c+b∗(a∗c) = a∗(b∗c)that is exactly the fact that La is a derivation. 2

Notice that it is also equivalent to all right multiplications Ra beingderivations.

1.1.3 Modules and Representations

These notions are usually interchangeable but we make an artificial dis-tinction that we will follow through in the lecture notes. Let A be an algebra

1.1. BASICS 9

(no axioms assumed). We consider vector spaces V equipped with bilinearactions maps A×V → V (denoted by (a, v) 7→ av). Such a vector space canbe a module or a representation if certain axioms are satisfied.

The module axiom is (a ∗ b)v = a(bv) for all a, b ∈ A, v ∈ V . If Acontains 1, we also require 1v = v for all v ∈ V .

The representation axiom is (a ∗ b)v = a(bv) − b(av) for all a, b ∈ A,v ∈ V .

The way we set the distinction up ensures that it only makes sense totalk about modules for associative algebras and representations for Lie alge-bras. The notions of subrepresentation, quotient representation, submoduleand quotient module are standard. Homomorphisms, isomorphisms, directsums and isomorphism theorems work in the same way for modules andrepresentations. Following the convention a representation V is irreducibleif V 6= 0 and 0, V are the only subrepresentations, while a module with thesimilar property is called simple.

Proposition 1.1.2 For an anticommutative algebra A the Jacobi identity isequivalent to the fact that A with the multiplication map is a representationof A.

Proof: A is a representation if and only if La is a derivation for each a ∈ A.One does not even need anticommutativity for this as (a∗b)∗c+b∗(a∗c) =a ∗ (b ∗ c) is rewritten as (a ∗ b)c = a(bc)− b(ac). Now use Proposition 1.1.1.2

1.1.4 Simple algebras

A vector subspace I of an algebra A is an ideal if it is stable under allmultiplications, i.e., La(I) ⊆ I ⊇ Ra(I) for all a ∈ A. The significance ofideals is that they are kernels of homomorphisms. Besides they allow todefine quotient algebras A/I.

An algebra A is simple if A 6= 0 and 0 and A are the only ideals. Simplealgebras do not have non-trivial quotients. In particular, any homomor-phism from a simple algebra is injective.

It is a popular problem in Algebra to classify simple algebras in a certainclass of algebras. It often leads to interesting mathematics. For instance,in Rings and Modules you have seen Artin-Wedderburn theorem that statesthat simple associative artinian algebras are matrix algebras over divisionrings. In Lie Algebras you have seen the Cartan-Killing classification ofsimple finite dimensional Lie algebras over an algebraically closed field ofcharacteristic zero. Our goal is to understand the following classificationtheorem, whose complete proof is beyond the scope of the present course.


Theorem 1.1.3 Let K be an algebraically closed of characteristic p ≥ 5.Then a finite dimensional simple Lie algebra over K is either of classicaltype or of Cartan type or of Melikian type.

1.1.5 Commutation

If A is an algebra we define a new algebra A[−] as the same vector spacewith the commutator product: [a, b] = a∗b−b∗a. We call it the commutatoralgebra of A. Apriori, it is not clear why the commutator algebra is par-ticularly significant, why the commutator product leads to more interestingmathematics than a ∗ b+ b ∗ a or a ∗ b− 2010b ∗ a. The following propositiongives some explanation.

Proposition 1.1.4 Let A be an associative algebra. Then A[−] is a Liealgebra.

Proof: The anticommutativity is obvious: [a, a] = a ∗ a − a ∗ a = 0. TheJacobi identity can be done by a straightforward check but we use a slightlymore conceptual method. Let us check that La in A[−] is a derivation of A:La(b)c+bLa(c) = (ab−ba)c+b(ac−ca) = abc−bca = La(bc). It follows thatit is a derivation of A[−]: La([b, c]) = La(bc) − La(cb) = La(b)c − cLa(b) +bLa(c) − La(c)b = [La(b), c] + [b, La(c)]. Now use Proposition 1.1.1. 2

Notice that A[−] can be a Lie algebra without A being associative (seethe vista subsection below).

If M is an R-module then we denote EndR(M) the set of all R-moduleendomorphisms. It is an algebra under the composition of morphisms. Inparticular, if A is a vector space then EndK(A) is the algebra of linear mapsfrom A to A. It is irrelevant if A is an algebra itself: we still consider alllinear operators.

Proposition 1.1.5 Let A be an algebra. Derivations form a Lie subalgebrain EndK(A)[−].

Proof: Let c, d be two derivations. It suffices to check that a linear combi-nation αc+βd for some α, β ∈ K and the commutator [c, d] are both deriva-tions. We leave the linear combination to a reader and check the commuta-tor. Pick x, y ∈ A. Then [c, d](xy) = cd(xy) − dc(xy) = c(d(x)y + xd(y)) −d(c(x)y + xc(y)) = c(d(x))y + d(x)c(y) + c(x)d(y) + xc(d(y)) − d(c(x))y −c(x)d(y)−d(x)c(y)−xd(c(y)) = c(d(x))y−d(c(x))y+xc(d(y))−xd(c(y)) =[c, d](x)y + x[c, d](y). 2

We denote this Lie algebra of derivations by Der(A). The geometricnature of this Lie algebra are vector fields under their Lie bracket. To showthis in familiar surroundings, let U be a connected open subset in R

n. Let A

1.1. BASICS 11

be the algebra of all smooth functions U → R. It is a commutative algebraunder pointwise multiplication and addition. It will require some analysis toshow that every derivation of A is a vector field (and we will avoid this) butthe reverse statement is clear. Indeed, every vector field d =

∑i φi(x)∂/∂xi

can be applied to functions so that d(Ψ(x)) =∑

i φi(x)Ψxi(x) where we use

superscripts to denote partial derivatives. The product rule for d followsfrom the familiar product rule for derivatives. It is instructive to compute[c, d] where c =

∑i θi(x)∂/∂xi:

[c, d] =∑

i

(∑

j

(θjφixj− φjθixj

))∂/∂xi

which can be directly checked by applying to a function Ψ.

1.1.6 Vista: affine structures on Lie algebras

An algebra is called Lie-admissible if A[−] is a Lie algebra. It is a wideclass of algebras. We saw that it includes associative algebras but not onlythem. Let us define (a, b, c) = (ab)c − a(bc) for elements a, b, c ∈ A. Thiselement (a, b, c) (or this triple product A × A × A → A) is called an asso-ciator. It measure the failure of associativity, similarly as the commutatormeasure the failure of associativity. An algebra is called left symmetric ofthe associator is left symmetric, i.e., (a, b, c) = (b, a, c) for all a, b, c ∈ A. Itis easy to check that left symmetric algebras are Lie-admissible.

In fact, left symmetric are important in Geometry. An affine structure ona Lie algebra g is a left symmetric multiplication on g whose commutatoris the usual Lie multiplication. Algebraically, affine structure representsg = A[−] for a left symmetric algebra A. Geometrically, it corresponds toleft invariant affine connections on the Lie group. More to follow.

1.1.7 Exercises

Let A be a finite-dimensional algebra over complex numbers C. Provethat if f : A→ A is a derivation then ef = I+f +f2/2!+ . . . =

∑∞k=0 f

k/k!is an automorphism. Prove that if eαf : A→ A is an automorphism for eachα ∈ C then f is a derivation.

State and prove the first isomorphism theorem for arbitrary algebras.Prove the analogue of Proposition 1.1.1 for right derivations.Prove that [θ∂/∂x, φ∂/∂y] = θφx∂/∂y − φθy∂/∂x.


1.2 Things that fail in positive character-

isticHow come so many new Lie algebras slipped into the classification theo-

rem? The reason is that many usual results in characteristic zero suddenlyfail in positive characteristic.

1.2.1 Complete Reducibility

Define gln = gln(K) = Mn(K)[−], sln its subalgebra of matrices with

zero trace. The standard basis of sl2 is e =

(0 10 0

), f =

(0 01 0

), h =

(1 00 −1

). The products in the standard basis are computed commuting

the matrices: e ∗ f = h, h ∗ f = −2f , h ∗ e = −2e. These three productscompletely determine the Lie algebra structure on sl2.

Proposition 1.2.1 If p ≥ 3 then sl2 is a simple Lie algebra.

Proof: Pick a nonzero element x and generate an ideal I = (x). It sufficesto show that I = sl2. Pick a non-zero element and multiply it by e twotimes. This gives you e ∈ I and now multiplying by f , you get I = sl2. 2

Let us now consider the polynomial algebra K[x, y]. We define the actionof sl2 on A by formulas

eφ(x, y) = x∂φ

∂y, hφ(x, y) = x

∂φ

∂x− y

∂φ

∂y, fφ(x, y) = y

∂φ

∂x

and then extended by linearity to an arbitrary element.

Proposition 1.2.2 Under these formulas A is a representation of sl2.

Proof: To conclude that A is a representation we must check that (a∗b)φ =a(bφ)−b(aφ) for all a,b ∈ sl2, φ ∈ A. Because of linearity it is sufficient todo only for the three basic products. We do it for one of them and leave theother two to the reader: h(eφ(x, y))−e(hφ(x, y)) = h(xφy)−e(xφx−yφy) =x(xφy)x − y(xφy)y − x(xφx)y + x(yφy)y = x(φy + xφyx) − xyφyy − x2φxy +x(φy + yφyy) = 2xφy = 2eφ. 2

It is worse pointing out that elements of sl2 act by differentiations andwe get injective homomorphism of Lie algebras sl2 → Der(A). We interpretthese facts geometrically: A are functions on the affine space K

2, Der(A) arevector fields on K

2, and the representation is a realization of sl2 as vectorfields on K

2.

1.2. THINGS THAT FAIL IN POSITIVE CHARACTERISTIC 13

Let An be the subspace of homogeneous polynomials of degree n, i.e., itis a span of all xn−kyk. As the actions of e, f and h preserve the degree,each An is a subrepresentation of A. We claim that Ap is not completelyreducible.

Let us recall that a representation U is completely reducible if for any sub-representation W ⊆ U there exists another subrepresentation V ⊆ U suchthat U = V ⊕W . A module with the same property is called semisimple.According to Weyl’s complete reducibility, a finite dimensional representa-tion of a finite dimensional simple Lie algebra over an algebraically closedof characteristic zero is completely reducible. We can see now that it failsin characteristic p > 2.

Let us check the claim. Let W be the span of xp and yp. This clearlysubmodule with the trivial action axp = 0 for any a ∈ sl2 because ∂xp/∂x =pxp−1 = 0. Suppose there is a complementary submodule V and pick non-zero φ ∈ V . We can write φ =

∑pk=n αkx

kyp−k with αn 6= 0 and 0 6= n 6= p.Then ep−n(φ) = αnx

p and xp ∈ V , contradiction.

1.2.2 Lie’s Theorem

Let us recall some standard material from Lie algebras to make thesenotes self-contained. If A is a linear subspace of a Lie algebra g, we defineits commutant [A,A] as the span of all products x ∗ y, x, y ∈ A. Thename comes from thinking of the Lie multiplication as commutation. Theninductively we define the derived series: A(0) = A, A(n+1) = [A(n), A(n)].

Proposition 1.2.3 If I is an ideal of g then I(n) is an ideal of g for eachn.

Proof: We prove it by induction on n. The basis of induction (n = 0)is our assumption. For the induction step, we have to show that [J, J ] isan ideal whenever J is an ideal. It is clear by the product rule for Lx:x(ab) = (xa)b + a(xb) ∈ [J, J ] if x ∈ g, a,b ∈ J . 2

We say that an ideal I is n-soluble if I(n). An ideal is soluble if it isn-soluble for some n. The Lie algebra g is soluble if g is a soluble ideal.

Proposition 1.2.4 (i) If g is soluble then every subalgebra of g and everyquotient algebra of g is soluble.

(ii) If I is an ideal of g and both I and g/I are soluble then g is soluble.

Proof: The first statement is obvious. For the second statement, observethat the cosets of elements of g(n) are in (g/I)(n). This, if g/I is n-solublethen g(n) ⊆ I. Now if I is m-soluble then g(n+m) = (g(n))(m) ⊆ I(m) = 0. 2

The following proposition immediately follows.


Proposition 1.2.5 A sum of two soluble ideals is soluble.

Proof: By the second isomorphism theorem, I + J/J ∼= I/I ∩ J is solubleas the quotient of I. Hence I + J is soluble by Proposition 1.2.4. 2

This allows to conclude that the radical Rad(g) of a finite-dimensionalLie algebra, defined as the sum of all soluble ideals, is itself soluble.

Lie’s theorem states that over an algebraically closed field of zero char-acteristic, a finite-dimensional irreducible representation of a finite dimen-sional soluble Lie algebra must be one-dimensional. This fails in positivecharacteristic.

Let us consider a two dimensional Lie algebra g with basis x, y. Themultiplication is x ∗ y = y. It is 2-soluble as g(1) = Ky and g(2) = 0. Nowlet us consider a vector space V with a basis ei, i ∈ F where F is a primesubfield with the action of g defined by xei = iei, yei = ei+1 and extendedby bilinearity.

Proposition 1.2.6 V is an irreducible representation of g of dimension p.

Proof: To conclude that V is a representation it suffices to follow throughthe following calculation x(yei)−y(xei) = xei+1−yiei = (i+1)ei+1−iei+1 =ei+1 = yei.

To show irreducibility consider a nonzero subrepresentation U ⊆ V andan element 0 6= z ∈ U . Write z =

∑p−1i=0 αiei. Then xz =

∑i iαiei ∈ U

and xkz =∑

i ikαiei ∈ U for each positive k. These conditions can be now

written in the matrix form as

1 1 1 · · · 1 10 1 2 · · · p− 2 p− 102 12 22 · · · (p− 2)2 (p− 1)2

......

... · · ·...

...0p−1 1p−1 2p−1 · · · (p− 2)p−1 (p− 1)p−1

α0e0α1e1

...αp−1ep−1

∈ Up

Notice that the column has the coefficients in V while the matrix has thecoefficients in the field K. Hence the product apriori has the coefficients inV and the condition tells us that the coefficients are actually in U . Now thematrix is Vandermonde’s matrix and, in particular, is invertible. Multiplying

by its inverse gives

α0e0α1e1

...αp−1ep−1

∈ Up. We will refer to this consideration

later on as Vandermonde’s trick. As all αiei are in U and one of αi is


nonzero, one of ei is in U . Applying y p− 1 times we get that all ei are inU . 2

1.2.3 Cartan’s Criteria

Recall that any finite dimensional algebra A admits Killing form K :A × A → K defined as K(x,y) = Tr(LxLy). Observe that it is a sym-metric bilinear form. In characteristic zero, there are two Cartan’s criteria.Semisimple criterion states that a Lie algebra is semisimple if and only if itsKilling form is non-degenerate, i.e., of maximal possible rank. The solublecriterion states that of K vanishes on g(1) then g is soluble. Both criteriafail in positive characteristic.

Let F be the finite subfield of K of size pn. The Witt algebra W (1, n)is a pn-dimensional vector space with basis ei, i ∈ F and multiplication byei ∗ ej = (i− j)ei+j .

Proposition 1.2.7 (i) W (1, n) is a Lie algebra.(ii) If p ≥ 3 then W (1, n) is simple.(iii) If p ≥ 5 then the Killing form on W (1, 1) is zero.

Proof: The anticommutativity is obvious because i − j = −(j − i). TheJacobi identity is verified directly (ei ∗ ej) ∗ ek + (ej ∗ ek) ∗ ei + (ek ∗ ei) ∗ej = ((i − j)(i + j − k) + (j − k)(j + k − i) + (k − i)(k + i − j))ei+j+k =(i2 − j2 − ik + jk + j2 − k2 − ji+ ki+ k2 − i2 − kj + ij)ei+j+k = 0.

To check simplicity, let us look at the ideal I generated by a non-zeroelement x =

∑i∈F

αiei =∑

i∈S αiei where S is the subset of all i such thatαi 6= 0. It suffices to show that I = g.

By the ideal property x ∗ e0 =∑

i∈S iαiei ∈ I. Applying Re0 severalmore times, we get Rke0

(x) =∑

i∈S ikαiei ∈ I. Using Vandermonde trick,

ei ∈ I for all i ∈ S.

As soon as j 6= 2i for some i ∈ S, it follows that ej = (2i−j)−1ei∗ej−i ∈I. If p > 2 , we get I = g in two iterations: ei ∈ I implies ej ∈ I for allj 6= 2i, hence at least for one more j 6= i. Then 2i 6= 2j and we concludethat e2i ∈ I.

Let us compute K(ei, ej). Since (ei ∗(ej ∗ek)) = (j−k)(i−j−k)ei+j+k,it is immediate that K(ei, ej) = 0 unless i = −j. Finally, K(ei, e−i) =∑

k∈F(−i − k)(2i − k) =

∑p−1k∈F

(−2i2 − ik + k2). This can computed usingthe usual formula

∑nt=0 t

2 = n(n + 1)(2n + 1)/6 if the characteristic is atleast 5: K(ei, e−i) = −2i2p− ip(p − 1)/2 + p(p+ 1)(2p + 1)/6 = 0. 2

Thus, W (1, 1) is a counterexample to both Cartan criteria. In fact, soare W (1, n), which is the first example of Cartan type Lie algebra.


1.2.4 Vista: Semisimple Lie algebras

One usually defines a semisimple Lie algebra as a finite dimensional Liealgebra such that Rad(g) = 0. In characteristic zero, such an algebra is adirect sum of simple Lie algebras. It fails in characteristic p too.

Let A = K[z]/(zp) be the commutative algebra of truncated polynomi-als. In characteristic p this algebra admits the standard derivation ∂(zk) =kzk−1. One needs characteristic p for this to be well-defined: ∂ is always well-defined on K[z] turning K[z] into a representation of the one-dimensional Liealgebra. The ideal (zp) needs to be a subrepresentation. Since ∂(zp) = pzp−1

it happens only in characteristic p. Notice that the product rule for ∂ on Ais inherited from the product rule for ∂ on K[z].

Now extend ∂ to the derivation of the Lie algebra sl2(A): ∂

(f(z) g(z)h(z) −f(z)

)=

(f ′(z) g′(z)h′(z) −f ′(z)

). Finally, we define the Lie algebra g as a semidirect prod-

uct of a one dimensional Lie algebra ans sl2(A).Abstractly speaking, we say that a Lie algebra k acts on a Lie algebra h

if h is a representation of k and for each x ∈ k the action operator a 7→ xa isa derivation of k. The semidirect product is the Lie algebra g = k ⊕ h withthe product

(x, a) ∗ (y, b) = (x ∗ y, xb− ya+ a ∗ b)

.

Proposition 1.2.8 The semidirect product is a Lie algebra, h is an ideal,and the quotient algebra g/his isomorphic to k.

Proof: 2

Hence g = Kd⊕sl2(A) is 3p+1-dimensional Lie algebra. Let us describeits properties.

Proposition 1.2.9 g admits a unique minimal ideal isomorphic to sl2(K).The commutant g(1) is equal to sl2(A).

Proof: 2

The first statement implies that g is semisimple. The second statementimplies that g is not a direct sum of simple Lie algebras.

1.2.5 Exercises

Prove that if the Killing of g is nondegenerate then the radical of g iszero.

Compute the Killing form for W (1, 1) in characteristic 3 (note it is non-degenerate there).


Show that in characteristic 2 all ei with i 6= 0 span a nontrivial ideal ofW (1, n).

Show that W (1, 1) is solvable in characteristic 2 and that its Killing formhas rank 1.


1.3 Free algebras1.3.1 Free algebras and varieties

Let X be a set. We define a free “thingy” (X) to be a free set witha binary operation without any axioms. Informally, it consists of all non-associative monomials of elements of X. To define it formally we can userecursion:

(basis) if a ∈ X then (a) ∈ (X),(step) if a, b ∈ X then (ab) ∈ (X).

We will routinely skip all unnecessary brackets. For instance, if X = {a, b},then the recursion basis will give us two elements of X: a and b. Using therecursion step for the first time gives 4 new elements: aa, ab, ba, bb. Using itfor the second time promises to give 36 new elements, but since 4 productsare already there, it gives only 32. We will not attempt to list all of themhere but we will list a few: a(aa), (aa)a, (aa)b, a(ab), etc.

The fact that this defines a set is standard but not-trivial: it is calledRecursion Theorem. Notice the difference with induction. Induction is anaxiom and it is used only to verify a statement. Recursion is a theorem andit is used to construct a set (or a function or a sequence). Have you heardof recursive sequences?

Now the binary operation on (X) is defined by the recursive step: a∗b =(ab) as in the recursion step. Let K(X) be the vector space formally spannedby (X). It consists of formal finite linear combinations

∑t∈(X) αtt, finite

means that αt = 0 except for finitely many elements t. The vector spaceK(X) is an algebra: the multiplication on (X) is extended by bilinearity.We call it the free algebra of X because of the following universal property.

Proposition 1.3.1 (Universal property of the free algebra) For each alge-bra A and a function f : X → A there exists a unique algebra homomorphismf : K(X) → A such that f(z) = f(z) for each z ∈ X.

Proof: We start by extending f to a function f : (X) → A preserving themultiplication:

(basis) if a ∈ X define f(a) = f(a),(step) if a, b ∈ X define f(ab) = f(a)f(b).

This is well-defined by recursion and it preserves the multiplication becauseof the recursive step. Any other such extension must also satisfy the recur-sion basis and steps. Hence, the extension is unique.

f is defined on the basis of K(X) and we extend it to f : K(X) → A bylinearity. Uniqueness is obvious. 2

1.3. FREE ALGEBRAS 19

We can formulate Proposition 1.3.1. Think of X as a subset of K(X).Then we have the natural restriction functionR : Hom(K(X), A) → Fun(X,A)where Fun(X,A) is just the set of all functions from X to A. Proposi-tion 1.3.1 essentially says that R is a bijection.

The elements of K(X) are nonassociative polynomials. They can beevaluated on any algebra. Formally let X = {x1, . . . xn}, w(x1, . . . xn) ∈K(X), A an algebra, a1, . . . an ∈ A. Consider a function f : X → A definedby f(xi) = ai. It is extended to a homomorphism f : K(X) → A byProposition 1.3.1. Finally, we define w(a1, . . . an) = f(w).

We say that w(x1, . . . xn) is an identity of A if w(a1, . . . an) = 0 for alla1, . . . an ∈ A. Let us write IdX(A) for the set of all identities of A in K(X).Now we say that an ideal I of K(X) is verbal if w(s1, . . . sn) ∈ I for all w ∈ I,s1, . . . sn ∈ K(X)

Proposition 1.3.2 The set IdX(A) is a verbal ideal for any algebra A.

Proof: Let us first check that it is a vector subspace. If v,w ∈ IdX(A),α, β ∈ K and f : X → A is an arbitrary function then f(αv + βw) =αf(v) + βf(w) = 0. Hence αv + βw is also an identity of A.

The ideal property is similar. If w ∈ IdX(A), v ∈ K(X) then f(vw) =f(v)f(w) = f(v)0 = 0. Hence vw is an identity. Similarly, wv is an identity.

The verbality is intuitively clear since of w(a1, . . . an) = 0 on A thenw(s1(a1, . . . an), . . . sn(a1, . . . an)) will also be zero. Formally the substitu-tion xi 7→ ai corresponds to a homomorphism f : K(X) → A while the sub-stitution xi 7→ si corresponds to a homomorphism g : K(X) → K(X) andw(s1(a1, . . . an), . . . sn(a1, . . . an)) = f(g(w)). Define h = f ◦ g : X → K(X).Since h(z) = f(g(z)) for each z ∈ X, we conclude that h = f ◦ g andf(g(w)) = h(w) = 0. 2

1.3.2 Varieties of algebras

A variety of algebras is a class of algebras where a certain set of identitiesis satisfied. We can choose enough variables so that the set of identities isa subset S of K(X). Let (S)v be the verbal ideal generated by S. One canformally define (S)v as the intersection of all verbal ideals containing S. Itfollows from Proposition 1.3.2 that (S)v consists of the identities in variablesof X satisfied in all the algebras of the variety. We have seen the followingvarieties already.

Ass is given by the identity x(yz) − (xy)z. It consists of all associativealgebras, not necessarily with identity.


Com consists of all commutative algebras in Ass. Its verbal ideal is gener-ated by x(yz) − (xy)z and xy − yx.

Lie is the variety of Lie algebras and the identities are xx x(yz) + y(zx) +z(xy).

n− Sol consists of all n-soluble Lie algebras. Its additional identity wnrequires 2n variables. We define it recursively: w1(x1, x2) = x1x2.Thus, 1-soluble Lie algebras are just algebras with zero multiplication.Then wn+1 = wn(x1, . . . x2n) ∗ wn(x2n+1, . . . x2n+1).

Let V be a variety. For each set X, the variety will have a verbal idealIV in K(X) of all identities in variables X that hold in V. We define the freealgebra in the variety V as KV < X >= K(X)/IV .

Proposition 1.3.3 (Universal property of the free algebra in a variety) Forevery algebra A in a variety V the natural (restriction) function Hom(KV <X >,A) → Fun(X,A) is a bijection.

Proof: Take a function f : X → A and consider its unique extensionhomomorphism f : K(X) → A. It vanishes on IV because elements of IVare identities of A while f(w) are results of substitutions. Hence, f : KV <X >→ A where f(w + IV) = f(w) is well-defined. This proves that therestriction is surjective.

The injectivity of the restriction is equivalent to the uniqueness of thisextension, which follows from the fact that X generates KV < X >. 2

1.3.3 Vista: basis in free Lie algebra and powers of repre-

sentations

Although we have defined the free Lie algebra, it is a pure existencestatement. Fortunately, there are ways to put hands on the free Lie algebra,for instance, by constructing a basis. Contrast this with the free alternativealgebra: this is given by the verbal ideal generated by two identities x(xy)−(xx)y and x(yy)− (xy)y. No basis of the free alternative alternative algebraKAlt(X) is known explicitly if X has at least 4 elements.

The most well-known basis of the free Lie algebra is Hall basis but herewe construct Shirshov basis. More to follow.

1.3.4 Exercises

Let K(X)n be the space of homogeneous polynomials of degree n. Given|X| = m, compute the dimension of K(X)n.

1.4. UNIVERSAL ENVELOPING ALGEBRAS 21

1.4 Universal enveloping algebras1.4.1 Free associative algebra and tensor algebra

The free algebra KAss(X) has associative multiplication but no identityelement. This is why it is more conventional to introduce new algebra K <X >= K1 ⊕ KAss(X) with the obvious multiplication: (α1,a) ∗ (β1,b) =(αβ1, αb+βa+ab). This algebra K < X > will be called the free associativealgebra. Its basis is the free monoid (a set with associative binary andidentity) product < X >, generated by X.

Proposition 1.4.1 (The universal property of K < X >) For every as-sociative algebra A and every function f : X → A there exists a uniquehomomorphism of algebras f : K < X >→ A such that f(x) = f(x) for eachx ∈ X.

Proof: The homomorphism of associative algebras takes 1 to 1 and theextension to f : KAss(X) → A exists and is unique by Proposition 1.3.3).Thus, f(α1,a) = α1A + f(a) is uniquely determined. 2

It is sometimes convenient to reinterpret the free associative algebra astensor algebra. Let V be a vector space. We define the tensor algebraTV as the free associative algebra K < B > where B is a basis of V .Notice that TV depends on B up to a canonical isomorphism. If C isanother basis, the change of basis matrix ci =

∑i αi,jbj gives a function

C → K < B >, ci 7→∑

i αi,jbj, which can be extended to the canonicalisomorphism K < C >→ K < B >.

We are paying a price here for not having defined tensor products ofvector spaces. It is acceptable for a student to use tensor products instead.

Notice that the natural embedding B → K < B > of sets gives a naturallinear embedding V → TV .

Proposition 1.4.2 (The universal property of TV ) For every associativealgebra A the natural (restriction) function Hom(TV,A) → Lin(V,A) is abijection.

Proof: Let B be a basis of V . The basis property can be interpreted asthe fact the restriction function Lin(V,A) → Fun(B,V ) is bijective. Noweverything follows from Proposition 1.4.1. 2

1.4.2 Universal enveloping algebra

If the vector space V = g is a Lie algebra there is a special subsetHom(g, A) ⊆ Lin(g, A) consisting of Lie algebra homomorphisms. This mo-tivates the definition of a universal enveloping algebra. The universal en-veloping algebra of a Lie algebra g is Ug = Tg/I where I is generated by all


ab−ba−a∗b for a, b ∈ g. Notice that ab is the product in Tg while a∗b isthe product in g. Notice also that the natural linear map ω = ωg : g → Ug

defined by ω(a) = a + I is a Lie algebra homomorphism from g to Ug[−]:ω(a)∗ω(b) = (a+I)(b+I)−(b+I)(a+I) = ab−ba+I = a∗b+I = ω(a∗b).

Proposition 1.4.3 (The universal property of Ug) For every associativealgebra A and every Lie algebra homomorphism f : g → A[−] there ex-ists a unique homomorphism of associative algebras ϕ : Ug → A such thatϕ(ω(x)) = f(x) for each x ∈ g.

Proof: It follows from Proposition 1.4.2 and the fact that f(I) = 0 isequivalent to f being Lie algebra homomorphism for a linear map f . 2

Corollary 1.4.4 For a vector space V and a Lie algebra g there is a bi-jection between the set of Ug-module structures on V and the set of Ug-representation structures on V .

Proof: Ug-module structures on V are the same as associative algebra ho-momorphisms Hom(Ug,EndKV ) that are in bijection with Lie algebra ho-momorphisms Hom(Ug,EndKV

[−]) that, in their turn, are g-representationstructures on V . 2

Bijections in Corollary 1.4.4 leave the notion of a homomorphism ofrepresentations (modules) unchanged. In fact, it can be formulated as anequivalence of categories, which we will discuss later on. This seems to makethe notion of a representation of a Lie algebra obsolete. Why don’t we juststudy modules over associative algebras? A point in defense of Lie algebrasis that g is finite-dimensional and easy to get hold of while Ug is usuallyinfinite-dimensional.

The following proposition follows immediately from the Poincare-Birkhoff-Witt Theorem that will not be proved in this lectures.

Proposition 1.4.5 ωg is injective

1.4.3 Free Lie algebra

Let L(X) be the Lie subalgebra of KAss(X)[−] generated by

Theorem 1.4.6 (i) The natural map φ : KLie(X) → KAss(X) gives anisomorphism of Lie algebras between KLie(X) and L(X).

(ii) The natural map ψ : UKLie(X) → K < X > is an isomorphism ofassociative algebras.

1.4. UNIVERSAL ENVELOPING ALGEBRAS 23

(iii) The following diagram is commutative.

UKLie(X)ψ

−−−−→ K < X >xωKLie(X)

xi

KLie(X)φ

−−−−→ L(X)

Proof: Let us first construct both maps. Using the function f : X →KAss(X) and interpreting KAss(X) as a Lie algebra KAss(X)[−] gives a Liealgebra homomorphism φ : KLie(X) → KAss(X)[−] by the universal propertyof the free Lie algebra. Since KLie(X) is generated by X as a Lie algebra,the image of φ is exactly L(X). Hence we have a surjective Lie algebrahomomorphism φ : KLie(X) → L(X).

Now L(X) is a Lie subalgebra in K < X >[−], thus φ gives a Lie algebrahomomorphism KLie(X) → K < X >[−] that can be lifted to a homo-morphism ψ : UKLie(X) → K < X > of associative algebras. Since bothalgebras are generated by X, the homomorphism ψ is surjective.

The commutativity of the diagram follows from the construction becauseψ(ω(z)) = φ(z) for each z ∈ X. Consequently, it should be true for anyelement z ∈ KLie(X) as both ψ◦ω and φ are homomorphisms of Lie algebras.

The inverse of ψ is constructed from the universal property of K < X >:let θ be the extension of function X → UKLie(X) to a homomorphismK < X >→ UKLie(X). The composition θψ is an identity on X, henceidentity on UKLie(X). Similarly, the composition ψθ is an identity on X,hence identity on K < X >.

Finally, ω is injective by Proposition 1.4.5. Hence, ψω = φ is injective.2

Now we get a good grasp on the free Lie algebra: it is just L(X) and itsuniversal enveloping algebra is K < X >.

1.4.4 Vista: Baker-Campbell-Hausdorff formula

1.4.5 Exercises

State and prove the universal property of the free commutative algebra.Prove the universal enveloping algebra of an abelian Lie algebra is iso-

morphic to the free commutative algebra of any basis.


1.5 p-th powersWe discuss various p-th powers and introduce restricted structures on

Lie algebras. The field K has characteristic p > 0 throughout the lecture.

1.5.1 In commutative algebra

Proposition 1.5.1 (Freshman’s dream binomial formula) Let A be an as-sociative algebra and x, y ∈ A such that xy = yx. Then (x+ y)p = xp + yp.

Proof: Since x and y commute, they satisfy the binomial formula (x+y)p =∑p

n=0

(pn

)xnyp−n. It remains to observe that for any n between 1 and

p − 1 the binomial coefficient

(pn

)= p!/n!(p − n)! vanish since only the

denominator is divisible by p. 2

Corollary 1.5.2 If T is a matrix with coefficients in K then Tr(T p) =Tr(T )p.

Proof: If λi are eigenvalues of T then λpi are eigenvalues of T p. Hence,Tr(T )p = (

∑i λi)

p =∑

i λpi = Tr(T p). 2

Corollary 1.5.3 Let A be an associative algebra and x, y ∈ A such thatxy = yx. Then (x− y)p−1 =

∑p−1n=0 x

nyp−n.

Proof: Let us check this identity in the free com-algebra KCom(X) whereX = {a, b}. There (a−b)p = ap−bp = (a−b)

∑p−1n=0 a

nbp−n. Since KCom(X)

is a domain, we can cancel (a− b) proving (a− b)p−1 =∑p−1

n=0 anbp−n. Since

x, y commute the function f : X → A, f(a) = x, f(b) = y can be extended toa homomorphism f . Hence, (x−y)p−1 = f((a−b)p−1) = f(

∑p−1n=0 a

nbp−n) =∑p−1n=0 x

nyp−n. 2

Now look at a Lie algebra g = sln. It is a Lie subalgebra of two differentassociative algebras, U(g) and Mn(K). Let us denote the p-th power inthe former algebra by xp and in the latter algebra by x[p]. Observe that inMn(K), Tr(x[p]) = Tr(x)p as follows from (1.5.1 ).

1.5.2 In associative algebra

Let us prove that

Theorem 1.5.4 If y, z ∈ X then (z + y)p − zp − yp ∈ K < X > belongs toL(X) and contains homogeneous components of degree at least 2.

Proof: We consider polynomials K < X > [T ] in one variable T overK < X >. One can differentiate them using the derivation ∂ : K < X >

1.5. P -TH POWERS 25

[T ] → K < X > [T ] where ∂(fT n) = nfT n−1, f ∈ K < X >. One can alsoevaluate them at α ∈ K using homomorphism evα : K < X > [T ] → K <X >, evα(fT n) = αnf .

Before going further we observe that the associativity that operatorsLs and Rs commute for any s ∈ K < X > [T ]. Hence, (Ls − Rs)

p−1 =∑p−1n=0 L

nsR

p−ns by Corollary 1.5.3.

Now we write (zT + y)p = zpT p + yp +∑p−1

n=1 Fn(z, y)Tn for some

Fn(z, y) ∈ K < X >. Differentiating this equality we get∑p−1

n=1 nFn(z, y)Tn−1 =∑p−1

n=0(zT+y)nz(zT+y)p−n−1 =∑p−1

n=0 LnzT+y(R

p−nzT+y(z)) = (LzT+y−RzT+y)

p−1(z) =

[zT + y, . . . [zT + y, z] . . .]. Comparing coefficients at T n−1 we express eachFn(z, y) as a sum of commutators, hence Fn(z, y) ∈ L(X) with homogeneouscomponents of degree 2 and higher.

Finally, using ev1, we get (z+ y)p− zp− yp =∑p−1

n=1 Fn(z, y) ∈ L(X). 2

This polynomial Λp(z, y) = (z+y)p−zp−yp plays a crucial role in whatfollows. Since it is a Lie polynomial it can be evaluated on Lie algebras.

1.5.3 In Lie algebra

A restricted Lie algebra is a pair (g, γ) where g is a Lie algebra andγ : g → g is a function (often denoted γ(x) = x[p] such that

(i) γ(αx) = αpγ(x),(ii) Lγ(x) = Lpx,(iii) γ(x + y) − γ(x) − γ(y) = Λp(x,y)

for all α ∈ K and x,y ∈ g.The following proposition gives us a tool to produce examples of re-

stricted Lie algebras.

Proposition 1.5.5 Let g be a Lie subalgebra of A[−] where is an associativealgebra. If xp ∈ g for all x ∈ g then g admits a structure of a restricted Liealgebra.

Proof: The restricted structure is given by the associative p-th power:γ(x) = xp. Axiom (i) is obvious. Axiom (iii) follows from the fact that itis the p-th power in an associative algebra: Λp(z, y) = (z + y)p − zp − yp ∈K < y, z >, hence, in any associative algebra.

To prove axiom (ii), we distinguish Lie multiplication operators by adding∗. Then L∗

z = Lz − Rz, i.e., the Lie multiplication is the difference of twoassociative multiplications for any z ∈ A. By associativity Lz and Rz com-mute, hence we can use Proposition 1.5.1: (L∗

z)p = (Lz −Rz)

p = Lpz −Rpz =Lzp −Rzp . 2

Corollary 1.5.6 If A is an algebra then Der(A) is a restricted Lie algebra.


Proof: Thanks to Proposition 1.5.5, it suffices to check that ∂p is a deriva-tion whenever ∂ is a derivation. Using induction on n one establishes that

∂n(xy) =∑n

k=0

(nk

)∂k(x)∂(y)n−k. If n = p all the middle terms vanish,

hence ∂p is a derivation. 2

Corollary 1.5.7 Classical Lie algebras sln, sp2n and son are restricted.

Proof: It suffices to check closeness under p-th power thanks to Proposi-tion 1.5.5. For sln, it follows from Corollary 1.5.2. Symplectic or orthogonalmatrices satisfy the condition XJ = −JX where J is the matrix of theform. Clearly, XpJ = −Xp−1JX = . . . = (−1)pJXp = −JXp 2

A homomorphism between Lie algebras is called restricted if it commuteswith γ. An ideal or a subalgebra is restricted if they are closed under γ. Arepresentation V is restricted if x[p]v = x(vx(. . . xv)) where v ∈ V and x ∈ g

appears p times. A direct sum of restricted Lie algebras is the direct sum ofLie algebras with the direct sum of restricted structures of the componentsas the restricted structure.

1.5.4 Vista: restricted Lie algebroids

1.5.5 Exercises

Prove by induction on n that ∂n(xy) =∑n

k=0

(nk

)∂k(x)∂(y)n−k.

Prove that the kernel of a restricted homomorphism is a restricted ideal.Formulate and prove the isomorphism theorem for restricted Lie alge-

bras.

1.6. UNIQUENESS OF RESTRICTED STRUCTURES 27

1.6 Uniqueness of restricted structuresWe develop more efficient methods for working and constructing re-

stricted structures.

1.6.1 Uniqueness of restricted structures

Recall the centre of a Lie algebra g consists of all x ∈ g such that x∗y = 0for all y ∈ g. The centre of a Lie algebra is an ideal.

A function f : V → W between vector spaces over the field K of positivecharacteristic p is semilinear if f is an abelian group homomorphism andf(αa) = αpa. The following proposition addresses uniqueness.

Proposition 1.6.1 If γ and δ are two distinct restricted structures on g

then δ−γ is a semilinear map g → Z(g). On the opposite, if γ is a restrictedstructure and φ : g → Z(g) is a semilinear map then γ + φ is a restrictedstructure too.

Proof: Let φ = δ − γ where γ is a restricted structure. It is clear thatφ(αx) = αpφ(x) if and only if δ(αx) = αpδ(x).

Then φ(x)∗y = (δ(x)−γ(x))∗y = Lpx(y)−Lpx(y) = 0, hence φ(x) ∈ Z(g).In the opposite direction, if φ : g → Z(g) is any function then γ+φ satisfiesaxiom (ii) of the restricted structure by the same argument.

Finally, φ(x+y)−φ(x)−φ(y) = Λp(x,y)−Λp(x,y) = 0. In the oppositedirection, semilinearity of φ implies axioms (iii) for δ. 2

We have proved that the restricted structures form an affine space overthe vector space of all semilinear maps from g to Z(g). In particular, ifZ(g) = 0 then the restricted structure is unique if it exists. To the existencewe turn our attention now.

1.6.2 Restricted abelian Lie algebras

Let us consider an extreme case Z(g) = g, i.e. the Lie algebra is abelian.Since Λp has no degree 1 terms, the zero map 0(x) = 0 is a restrictedstructure. By Proposition 1.6.1, any semilinear map γ : g → g is a restrictedstructure and vice versa. We consider the following abelian restricted Liealgebras: a0 is a one-dimensional Lie algebra with a basis x and x[p] = x;

an is an n-dimensional Lie algebra with a basis x1, . . . xn and x[p]k = xk+1,

x[p]n = 0. The following theorem is left without a proof in the course.

Theorem 1.6.2 Let g be a finite-dimensional abelian restricted Lie algebraover an algebraically closed field K. Then g is a direct sum of an. Themultiplicity of each an is uniquely determined by g.


While uniqueness in Theorem 1.6.2 is relatively straightforward, the ex-istence is similar to existence of Jordan normal forms (proofs will be eitherelementary and pointless or conceptual and abstract). One way you canthink of this is in terms of matrices: on a basis a semilinear map γ is de-fined by a matrix C = (γi,j) so that on the level of coordinate columnsγ(αi) = (γi,j)(α

pj ). The change of basis Q will change C into F (Q)CQ−1

where F (βi,j) = (βpi,j) is the Frobenius homomorphism F : GLn → GLn,which is an isomorphism of groups in this case. Hence, we are just classifyingorbits of this action, very much like in Jordan forms.

One interesting consequences of the theorem is that all nondegeneratematrices form a single orbit: its canonical representative is identity matrixI. The orbit map L : GLn → GLn, L(Q) = F (Q)IQ−1 = F (Q)Q−1 issurjective. This map is called Lang map and its surjectivity is called Lang’stheorem.

The conceptual proof is in terms of the modules over the algebra oftwisted polynomials: A = K[z] but z and scalars do not commute: αz = zαp.The module structure on V comes from linear action of K and the action ofz by the semilinear map. The algebra A is non-commutative left Euclideandomain and the theorem follows from the classification of indecomposablefinite-dimensional modules over A.

1.6.3 PBW-Theorem and reduced enveloping algebra

After careful consideration, I have decided to include the PBW-theoremwithout a proof. My original plan was to spend two lectures to give a com-plete proof but then I decided against as a partial proof (modulo Grobner-Shirshov’s basis) has featured in Representation Theory last year.

Theorem 1.6.3 (PBW: Poincare, Birkhoff, Witt) Let X be a basis of a Liealgebra g over any field K. If we equip X with a linear order ≤ (we say thatx < y if x ≤ y and x 6= y) then elements x

k11 x

k22 . . .xkn

n , with ki ∈ {1, 2 . . .}and x1 < x2 < . . . < xn ∈ X form a basis of Ug.

Now we are ready to establish the following technical proposition. Recallthat the centre of an associative algebra A is a subalgebra Z(A) = {a ∈A | ∀b ∈ A ab = ba}.

Proposition 1.6.4 Let the field be of characteristic p. Let xi (as i runsover linearly ordered sets) form a basis of g. Suppose for each i we have anelement zi ∈ Z(Ug) such that zi − x

pi ∈ g + K1 ⊆ Ug. Then the elements

zt11 zt22 . . . ztnn x

k11 x

k22 . . . xkn

n , with ki ∈ {0, 1, 2 . . . , p − 1}, ti ∈ {0, 1, 2 . . . , },and x1 < x2 < . . . < xn ∈ X form a basis of Ug.

1.6. UNIQUENESS OF RESTRICTED STRUCTURES 29

Proof: We refer to the PBW-basis elements as PBW-terms and to the newbasis elements as new terms. To show that the new terms span Ug it sufficesto express each PBW-term x

s11 x

s22 . . .xsn

n , si ≥ 0 as a linear combination ofthe new terms. We do it by induction on the degree S =

∑i si. If all si

are less than p then the PBW-term is a new term. This gives a basis ofinduction. For the induction step we can assume that si ≥ p for some i. Letyi = x

pi − zi. Then

xs11 xs22 . . .xsnn = xs11 . . . xsi−p

i yixsi+1

i+1 . . .xsnn + zix

s11 . . .xsi−p

i xsi+1

i+1 . . .xsnn

The second summand is a new term and the first summand has degreeS + 1 − p < S, i.e. it is a linear combination of PBW-terms of smallerdegrees. Using induction to the terms of the first summand completes theproof.

To show the linear independence we consider Um, the linear span of allPBW-terms of degree less or equal thanm. If a new term zt11 z

t22 . . . ztnn x

k11 x

k22 . . .xkn

n ,has degree m+ 1 =

∑i(pti + ki) then

zt11 . . . ztnn xk11 . . .xkn

n = xk11 (xp1 − y1)

t1 . . .xknn (xpn − yn)

tn

and, consequently,

zt11 zt22 . . . ztnn xk11 xk22 . . .xkn

n + Um = xpt1+k11 . . .xptn+kn

n + Um .

Note that this gives a bijection f between the new terms and PBW-terms.Consider a non-trivial linear relation on new terms with highest degreem+1.It gives a non-trivial relation on cosets of Um. This is a relation on cosets ofthe corresponding PBW-terms but this is impossible: PBW-terms of degreem+ 1 are linearly independent modulo Um by PBW-theorem. 2

We can interpret this proposition in several ways. Clearly, it implies thatA = K[z1 . . . zn] is a central polynomial subalgebra of Ug. Moreover, Ug isa free module over A with basis xk11 . . . xkn

n , with ki ∈ {0, 1, 2 . . . , p− 1}. Inparticular, if the dimension of g is n then the rank of Ug as an A-moduleis pn. Another interesting consequence is that elements zi generate an idealI which is spanned by all zt11 . . . ztnn ,x

k11 . . .xkn

n with at least one positiveti. Hence the quotient algebra U ′ = Ug/I has a basis x

k11 . . .xkn

n ,+I withki ∈ {0, 1, 2 . . . , p− 1}.

The quotient algebra U ′ is called the reduced enveloping algebra and weare going to put some gloss over it slightly later.


1.6.4 Vista: modules over skew polynomial algebra

1.6.5 Exercises

Prove uniqueness part in Theorem 1.6.2. (Hint: kernels of semilinearmaps are subspaces)

Prove Theorem 1.6.2 for one dimensional Lie algebras.Using PBW-theorem, prove that the homomorphism ω : g → Ug is

injective.Using PBW-theorem, prove that the universal enveloping algebra Ug is

an integral domain.

1.7. EXISTENCE OF RESTRICTED STRUCTURES 31

1.7 Existence of restricted structures1.7.1 Existence of restricted structures

We are ready to prove the following important theorem.

Proposition 1.7.1 Let g be a Lie algebra with a basis xi. Suppose for eachi there exists an element yi ∈ g such that Lpxi

= Lyi. Then there exists a

unique restricted structure such that x[p]i = yi.

Proof: If γ and δ are two such structure then φ = γ − δ is a semilinearmap such that φ(xi) = yi−yi = 0 for all i. Thus, φ = 0 and the uniquenessis established.

To establish existence, observe that zi = xpi − yi ∈ Z(g). The reason is

that the left Lie multiplication operator is Lzi= Lx

pi−Lyi

== Lpxi−Lyi

= 0.

This allows to construct the reduced enveloping algebra U ′ and g is its Liesubalgebra. It remains to observe that the p-th powers in U ′ behave in theprescribed way: (xi + I)p = x

pi + I = yi + I on the basis. Off the basis, we

need to observe that xp ∈ g if x ∈ g to apply Proposition 1.5.5. It follows byinduction on the length of a linear combination using the standard formula(αx + βy)p = αpxp + βpyp + Λp(αx, βy). 2

Let us describe the restricted structure onW (1, n): Lpei(ej) = −Rpei

(ej) =

−Rp−1ei

((j − i)ei+j) = −Rp−2ei

((j − i)je2i+j) = . . . = −∏p−1t=0 (j + ti)ej =

−f(j/i)ipej. The polynomial f(z) =∏p−1t=0 (z + t) has p distinct roots,

all elements of the prime subfield. Elements of the prime subfield satisfyzp − z = 0. Comparing the top degree coefficients, f(z) = zp − z andLpei(ej) = −(jp− jip−1)ej. If n = 1 then i and j lie in the prime subfield. If

i 6= 0 then ip−1 = 1 and jp − jip−1 = jp − j = 0 and e[p]i = 0. If i = 0 then

jp − jip−1 = jp = j and e[p]0 = e0.

Now if n ≥ 2 then consider i = 0 and j 6= 0. As jp − jip−1 = jp it

forces e[p]0 = jp−1e0. It depends on j and is not well-defined. Thus, W (1, n)

admits no restricted structure in this case.

1.7.2 Glossing reduced enveloping algebra up

Let (g, γ) be a restricted Lie algebra. It follows that x 7→ xp − x[p] is asemilinear map g → Z(Ug). Its image generates a polynomial subalgebraof the centre of Ug which we call the p-centre from now one and denoteZp(Ug).

Pick a basis xi of g. Now for each χ ∈ g∗ we define zi = xpi−x

[p]i −χ(xi)

p1.Finally, we define reduced enveloping algebra Uχg the algebra U ′ for thisparticular choice of basis. Clearly, it is just the quotient of Ug by the idealgenerated by xp − γ(x) − χ(x)p1 for all x ∈ g.


Proposition 1.7.2 (Universal property of reduced enveloping algebra) Forevery associative algebra A and every Lie algebra homomorphism f : g →A[−] such that f(x)p − f(γ(x)) = χ(x)p1 for all x ∈ g there exists a uniquehomomorphism of associative algebra f : Uχg → A such that f(x) = f(x)for all x ∈ g.

We say that a representation V of g admits a p-character χ ∈ g∗ ifxp(v) − x[p](v) = χ(x)pv for all v ∈ V .

Corollary 1.7.3 For a vector space V there is a natural bijection betweenstructures of a representation of g with a p-character χ and structures of aUχg-module.

One particular important p-character is zero. Representations with zerop-character are called restricted while the reduced enveloping algebra U0g iscalled the restricted enveloping algebra.

Example. Clearly, Ua0∼= Ua1

∼= K[z], making their representa-tions identical. However, Uχa0 and Uχa1 are drastically different. Uχa0

∼=K[z]/(zp−z−χ(z)p1) and the polynomial zp−z−χ(z)p1 has p distinct rootsbecause (zp− z−χ(z)p1)′ = −1. Hence Uχa0

∼= Kp is a semisimple algebra.

Consequently, representations of Uχa0 with any p-character are completelyreducible.

On the other hand, Uχa1∼= K[z]/(zp − χ(z)p1) and the polynomial zp−

χ(z)p1 = (z − χ(z)1)p has one multiple root. Hence Uχa1∼= K[z]/(zp) is a

local algebra. Consequently, there is a single irreducible representation ofUχa1 with any given p-character.

1.7.3 exercises

Prove Proposition 1.7.2 and Corollary 1.7.2.Choose a basis of sln and describe the restricted structure on the basis.

1.8. SCHEMES 33

1.8 Schemes1.8.1 Schemes, points and functions

A scheme X is a commutative algebra A. To be precise we are talkingabout proaffine schemes. To be even more precise, “the category of schemesis the opposite category to the category of commutative algebras”. Thismeans that if a scheme X is an algebra A and Y is an algebra B thenmorphisms from X to Y are algebra homomorphisms from B to A. Toemphasize this distinction between schemes and algebras we say that X isthe spectrum of A and write X = Spec A. We are going to study geometryof schemes.

We will regard algebra A as functions on the scheme X = Spec A.Functions can be evaluated at points and the scheme has various notions ofa point. We will just use only one: a point is an algebra homomorphismA → R to another commutative algebra R. If R is fixed we call them R-points and denote them X (R). Notice that a morphism of schemes φ : X →Y gives rise to a function on points φ(R) : X (R) → Y(R) via compositionφ(R)(a) = a ◦ φ.

We have to learn how to compute the value of a function F ∈ A at a pointa ∈ X (R). Well, it is just F (a) = a(F ) as a : A → R is a homomorphismand F is an element of A.

As an example, let us consider A = K[X,Y ]. We can regard the corre-sponding scheme as “an affine plane”. What are its R-points? A homomor-phism a is given by any two elements a(X), a(Y ) ∈ R. Hence, X (R) = R2.

Let us consider “a circle” X = Spec K[X,Y ]/(X2 + Y 2 − 1). Indeed,what are the R-points? To define a homomorphism a(X), a(Y ) ∈ R mustsatisfy a(X)2 + a(Y )2 = 1. Hence, X (R) = {(x, y) ∈ R2 | x2 + y2 = 1}.

1.8.2 Product of schemes

A product of schemes X = Spec A and Y = Spec B is the spectrumof the tensor product of algebras A and B, i.e., X × Y = Spec A ⊗ B.The tensor product of algebras is the standard tensor product of vectorsspaces with the product defined by a⊗ b · a′ ⊗ b′ = aa′ ⊗ bb′. Observe thatX × Y(R) = X (R) × Y(R) for each commutative algebra R.

1.8.3 Algebraic varieties and local schemes

The circle and the affine plane are very special schemes. Their algebrasare finitely generated domains (with an exception of the circle in character-istic 2, which is no longer a domain). If A is a finitely generated domain,the scheme is called an affine algebraic variety.

We are more interested in local schemes, i.e., spectra of local rings. Recall


that a commutative ring A is local if it accepts an ideal I such that everyelement a ∈ A\ I is invertible. One particular scheme of interest is the localaffine line A

1,n. Let us emphasize that the characteristic of K is p. Thealgebra A has a basis X(k), 0 ≤ k ≤ pn − 1 and the product

X(k)X(m) =

(m+ kk

)X(m+k).

We think that X(k) = 0, if k ≥ pn.

As a scheme, A1,n = (A1,1)n. All we need is to construct an algebra

isomorphism between K[Y1, . . . Yn]/(Ypi ) and A. It is given by φ(Yi) = X(pi).

Hence points A1,n(R) are just collections (r1, . . . rn) ∈ Rn such that rpi = 0

for all i. However, we want to distinguish A1,n and (A1,1)n by equipping

them with different PD-structure.

1.8.4 PD-schemes

Let R be a commutative ring, I it ideal. A PD-structure on I is asequence of functions γn : I → A such that

[i] γ0(x) = 1 for all x ∈ I,[ii] γ1(x) = x for all x ∈ I,[iii] γn(x) ∈ I for all x ∈ I,[iv] γn(ax) = anγn(x) for all x ∈ I, a ∈ A,[v] γn(x+ y) =

∑ni=0 γi(x)γn−i(y) for all x, y ∈ I,

[vi] γn(x)γm(x) =

(n+mn

)γn+m(x) for all x, y ∈ I,

[vii] γn(γm(x)) = An,mγnm(x) where An,m = (nm)!/(m!)nn! ∈ K forall x ∈ I,

The number An,m needs to be integer for the scalar An,m ∈ K to make

sense. It follows from the formula An,m =∏n−1j=1

(jm+m− 1

jm

).

A PD-structure on a scheme X = Spec A is an ideal I of A and a PD-structure on I. Let us describe some basic properties of it. We think thatx0 = 1 for any x ∈ A including zero.

Proposition 1.8.1 [i] γn(0A) = 0A if n > 0.[ii] n!γn(x) = xn for all x ∈ I and n ≥ 0.

Proof: Statement [i] follows from axiom [iv]. Statement [ii] follows fromaxioms [i], [ii] and [vi] by induction. See lecture notes for details. 2

Statement [ii] has profound consequences in all characteristics. In charac-teristic zero it implies that each ideal has a canonical DP -structure γn(x) =

1.8. SCHEMES 35

xn/n!. Observe how the axioms and the intuition agrees with this structure.In characteristic p it implies that xp = p!γp(x) = 0 for each x ∈ I. Thus, itis a necessary condition for an ideal I to admit a p-structure.

1.8.5 Local affine n-space

Let φ : {1, . . . n} → Z≥1 be a function. We introduce the local affinespace A

n,φ including its PD-structure. It is the spectrum of an algebraAφ whose vector space basis consists of commutative monomials X(a) =

X(a1)1 . . . X

(an)n which we will write in multi-index notation. The multi-index

entries vary in the region determined by φ: 0 ≤ ai < pφ(i) and we will thinkthat X(a) = 0 if one of entries falls outside this region.

The multiplication is defined similarly to the space A1,n above:

X(k)i X

(m)i =

(m+ kk

)X

(m+k)i

while the different variables are multiplied by concatenation (don’t forgetcommutativity). The element X(0,0,...0) is the identity of Aφ. If n = 1,we just write An for Aφ. In the general case, Aφ is isomorphic to thetensor product Aφ(1) ⊗ . . . ⊗ Aφ(n). Hence, on the level of schemes A

n,φ ∼=∏ni=1 A

1,φ(i).Finally, we describe the PD-structure. The ideal I is the unique max-

imal ideal. It is generated by all non-identity monomials X(a). Finally,

γn(X(1)i ) = X

(n)i determines the PD-structure. Indeed,

γn(X(m)i ) = γn(γm(X

(1)i )) = An,mX

(nm)i

from axiom [vii]. Axiom [iv] allows us to extend this to multiples of mono-mials γn(λX

(a)) = λn(n!)sX(na) where n(ai) = (nai) and s+1 is the numberof nonzero elements among ai. Finally, axiom [v] extends γn to any elementof I.

One should ask why this is well-defined. Rephrasing this question, wecan wonder why the repeated use of the axioms on a element x ∈ I willalways produce the same result for γn(x). This will not be addressed in thelecture notes (maybe in the next vista section).

1.8.6 Vista: divided powers symmetric algebra of a vector

space

1.8.7 Exercises

Prove that An,m =∏n−1j=1

(jm+m− 1

jm

).


1.9 Differential geometry of schemes1.9.1 Tangent bundle and vector fields on a scheme

For a scheme X = Spec A we discuss its tangent bundle TX . It is ascheme but it is easier to describe its points than its functions.

As far as points are concerned TX (R) = X (R[z]/(z2)). A homomor-phism φ : A→ R[z]/(z2) can be written as φ(a) = x(a)+τ(a)z using a pair oflinear maps x, τ : A → R. The homomorphism condition φ(ab) = φ(a)φ(b)gets rewritten as x(ab) + τ(ab)z = (x(a) + τ(a)z)(x(b) + τ(b)z) or as a pairof conditions

x(ab) = x(a)x(b), τ(ab) = x(a)τ(b) + x(b)τ(a).

The former condition means x ∈ X (R), i.e., it is a point where the tangentvector is attached. The latter condition is a relative (to x) derivation con-dition. All such elements belong to a vector space Derx(A,R) which we canthink of as the tangent space TxX .

To see that the tangent bundle is a scheme we have to understand dif-ferentials. At this point we judge fudge them up. Associated to an algebraA we consider a new commutative algebra A generated by elements a andda for each a ∈ A. We are going to put three kinds of relations on A. Thefirst kind just says that A → A defined by a 7→ a is a homomorphism ofalgebras. The second kind says that A → A defined by a 7→ da is a linearmap. Finally, we add relations d(ab) = adb+ bda for all a, b ∈ A.

Proposition 1.9.1 If X = Spec A then we can define TX as Spec A.

Proof: We need to come up with a “natural” bijection between Hom(A,R[z]/(z2))and Hom(A,R). Naturality means that the bijection agrees with homomor-phisms R→ in a sense that the following diagram commutes:

Hom(A,R[z]/(z2)) −−−−→ Hom(A,R)y

y

Hom(A,S[z]/(z2)) −−−−→ Hom(A, S)

We describe the bijections and leave necessary details (well-definedness,inverse bijections, naturality) to the student. An element x+τz ∈ Hom(A,R[z]/(z2))defines φ ∈ Hom(A, R) by φ(a) = x(a) and φ(da) = τ(a). In the op-posite direction φ ∈ Hom(A, R) defines q ∈ Hom(A,R[z]/(z2)) by q(a) =φ(a) + φ(da)z. 2

1.9. DIFFERENTIAL GEOMETRY OF SCHEMES 37

Finally derivations ∂ ∈ Der(A) can be thought of as global vector fieldsbecause they define give sections of tangent sheaf. We leave a proof of thefollowing proposition as an exercise.

Proposition 1.9.2 If x ∈ X (R), ∂ ∈ Der(A) then x ◦ ∂ ∈ Derx(A,R).

1.9.2 PD-derivations

We already saw that derivations form a restricted Lie algebra. Now wesay that ∂ is a PD-derivation if ∂(γn(x)) = γn−1(x)∂(x) for all x ∈ I. Thisdefinition is motivated by the following calculation using the chain rule orrepeated product rule: ∂(fn/n!) = ∂(f)fn−1/(n − 1)!. It also show that incharacteristic zero every derivation is a PD-derivation.

It is no longer true in characteristic p even if the PD-structure is trivial.Say A = K[z]/(zp). The PD-structure is γk(f) = fk/k! for k < p andγk(f) = 0 for k ≥ p. Then ∂ = ∂/∂z is not a PD-derivation as it marginallyfails the definition: ∂(γp(x)) = 0 6= γp−1(x)∂(x). Let PD − Der(A, I, γ) bethe set of all PD-derivations.

Proposition 1.9.3 PD − Der(A, I, γ) is a Lie algebra.

Proof: For the Lie algebra (∂ ∗ δ)(γn(x)) = ∂(δ(γn(x))) − δ(∂(γn(x))) =∂(γn−1(x)δ(x)) − δ(γn−1(x)∂(x)) = γn−1(x)∂(δ(x)) + γn−2(x)∂(x)δ(x) −γn−1(x)δ(∂(x)) − γn−2(x)∂(x)δ(x) = γn−1(x)(∂ ∗ δ)(x). 2

Notice that PD − Der(A) is not restricted, in general. One can show that∂p(γn(x)) = γn−1(x)∂

p(x) + γn−p(x)∂(x)p The best one can conclude fromthis that is that PD-derivations preserving I, i.e ∂(I) ⊆ I form a restrictedLie algebra.

1.9.3 Exercises

Prove that if A is finitely generated then A is finitely generated.Prove Proposition 1.9.2.Compute PD − Der(K[z]/(zp), (z), γ).


1.10 Generalised Witt algebra1.10.1 Definition

We define a derivation ∂i of Aφ by ∂i(X(k)i ) = X

(k−1)i . On a general

monomial ∂i(X(a)) = (X(a−ǫi) where ǫi is the multi-index with the only

nonzero entry 1 in position i. A special derivation is a derivation of Aφ ofthe form

∑ni=1 fi∂i for some fi ∈ Aφ. We define the generalised Witt algebra

as W (n, φ) as the Lie algebra of special derivations of Aφ.We start by stating some basic properties.

Proposition 1.10.1 (i) W (n, φ) is a free Aφ-module with a basis ∂i, i =1, . . . n.

(ii) W (n, φ) is a Lie subalgebra of Der(Aφ).(iii) W (n, φ) is a restricted Lie algebra if and only if φ(i) = 1 for all i.

Proof: (i) By definition∑n

i=1 fi∂i(Xi) = fi. Hence∑n

i=1 fi∂i = 0 if andonly if all fi are zero.

(ii) It follows from the standard commutations formula f∂i ∗ g∂j =f∂i(g)∂j − g∂j(f)∂i.

(iii) If all φ(i) = 0 then W (n, φ) = Der(Aφ) is restricted. In the oppositedirection, suppose φ(i) ≥ 2 for some i. Observe that L∂i

(F∂j) = ∂i(F )∂j .

Suppose W (n, φ) is restricted. Then Lp∂i= Lx for x = ∂

[p]i =

∑j fj∂j .

Hence, Lp∂i(X

(p)i ∂i) = ∂i while, on the other hand, Lx(X

(p)i ∂i) = (

∑j fj∂j) ∗

X(p)i ∂i = fiX

(p−1)i ∂i −

∑j X

(p)i ∂i(fj)∂j . The resulting equation

∂i = fiX(p−1)i ∂i −

∑

j

X(p)i ∂i(fj)∂j

contains a contradiction as the degrees of the monomials in the right handside are at least p− 1 ≥ 1 while it is zero in the left hand side. 2

It follows that the dimension of W (n, φ) is n · dim(A) = npP

φ(t). Tounderstand the Lie multiplication we write the standard formula

X(n) ∂

∂X∗X(m) ∂

∂X= (X(n)X(m−1) −X(n−1)X(m))

∂

∂X=

(

(n+m− 1

n

)−

(n+m− 1n− 1

))∂

∂X

Let us clear the connection between special and PD-derivations. The spe-cial derivation ∂i is not a PD-derivation as ∂i(γpφ(i)(Xi)) = 0 6= γpφ(i)−1(Xi)∂i(Xi).

Proposition 1.10.2 Every PD-derivation is special.

1.10. GENERALISED WITT ALGEBRA 39

Proof: Every element δ ∈ PD − Der(A1,n) is determined by n elementsδ(Xi). Indeed, δ(X(a)) =

∑iX

(a−ǫi)δ(Xi). Hence δ =∑

i δ(Xi)∂i is special.2

1.10.2 Gradings

A graded algebra is an algebra A together with a direct sum decompo-sition A = ⊕n∈ZAn such that An ∗ Am ⊆ An+m. In a graded algebra wedefiance subspaces A≤n = ⊕k≤nAk and A≥n = ⊕k≥nAk. Let us observesome obvious facts. A0 is a subalgebra. If n ≥ 0 then A≥n is a subalgebraand an ideal of A≥0.

A graded Lie algebra is a Lie algebra and a graded algebra. Observe thatin a graded Lie gi is a representation of g0.

The algebra Aφ is a graded commutative algebra. We define the degreeof X(a) as |a| =

∑i |ai|. Now we extend this grading to W (n, φ) by defining

the degree of ∂i to be −1. Let |φ| =∑

t(|φ(t)| − 1).

Proposition 1.10.3 (i) W (n, φ) = ⊕|φ|−1k=−1W (n, φ) is a graded Lie alge-

bra.(ii) W (n, φ)0 ∼= gln.(iii) W (n, φ)−1 is the standard representation of gln. In particular, it is

irreducible.(iv) W (n, φ)k = W (n, φ)−1 ∗W (n, φ)k+1 for every k 6= |φ| − 1.(v) W (n, φ)|φ|−1 = W (n, φ)0 ∗W (n, φ)|φ|−1 unless p = 2 and n = 1

Proof: (i) follows from the standard Lie bracket formula..

(ii) The isomorphism is given by Xi∂j 7→ Ei,j.

(iii) gln has two standard representations coming from the actions onrows or columns. For the column representation A · v = Av where A is amatrix, v is a column. For the row representation A · v = −AT v where A isa matrix, v is a column. It follows from the standard representation formulathat W (n, φ)−1 is the standard row-representation of gln.

(iv) Since ∂i ∗X(a)∂j = X(a−ǫi)∂j , the statement is obvious as you can

increase one of the indices.

(v) In the top degree |φ|−1 we consider the multi-index a = (φ(1)−, . . . φ(n)−1). The elements X(a)∂i span the top degree space and Xi∂i ∗ X

(a)∂j =XiX

(a−ǫi)∂j − δi,jX(a)∂i = (pφ(i) − 1)X(a)∂j − δi,jX

(a)∂i. It is now obviousif p > 2 and we leave the case of p = 2 as an exercise. 2

1.10.3 Simplicity

Theorem 1.10.4 If p > 2 or n > 1 then W (n, φ) is a simple Lie algebra.


Proof: Pick a non-zero ideal I � W (n, φ) and a non-zero element x ∈ I.Multiplying by various ∂i, elements of degree -1, we get a nonzero elementy ∈ I ∩ g−1.

Since W (n, φ)−1 is an irreducible W (n, φ)0-representation, we, multiply-ing by elements of degree 0, conclude that g−1 ⊆ I.

Using part [iv] of proposition 1.10.3, we get gk ⊆ I for each k < |φ| − 1.Finally, g|φ|−1| = g0 ∗ g|φ|−1| ⊆ I by part [v] of proposition 1.10.3. 2

1.10.4 Exercises

Finish the proof of part (v) of Proposition 1.10.3 in characteristic 2.Demonstrate a non-trivial ideal of W (1, φ) in characteristic 2.Let p > 2. Describe W (n, φ)1 as a representation of W (n, φ)0. Is it

irreducible?

1.11. FILTRATIONS 41

1.11 Filtrations1.11.1 Consequences of two descriptions

We intend to establish an isomorphism W (1, φ) ∗ W (1, φ(1)) betweenthe generalised Witt algebra and the Witt algebra. This isomorphism isnon-trivial. Here we discuss four consequences of this isomorphism.

First, W (1, φ) is defined over the prime field F(p) whileW (1, k) is definedover the field F(pk) of cardinality pk. The field of definition is the field wherethe multiplication coefficients µki,j belong. The multiplication coefficients

appear when you write the product on a basis: ei ∗ ej =∑

k µki,jek.

Second, W (1, φ) is graded while there is no obvious grading on W (1, k).To be more precise, the natural grading on W (1, k) (eα ∈ W (1, k)α) is agrading by the additive group of the field F(pk), not by Z.

Third, in characteristic zero there is an analogue of both Lie algebrasand the proof of isomorphism is nearly obvious. One sets it up by φ(ei) =−Xi+1∂. While this map works for some elements in characteristic p it isnot useful, in general. The product ei ∗ ej is never zero unless i = j whileX(a)∂ ∗X(b)∂ = 0 if a+ b > pφ(1).

Finally, these two realisations will lead to two completely different Car-tan decompositions. Recall that a Cartan subalgebra is a soluble subalgebrawhich coincides with its normaliser. The normaliser of a subalgebra l ≤ g

consists of all x ∈ g such that x ∗ l ⊆ l. A Cartan subalgebra l leads toCartan decomposition, which is essentially considering g as a representationof l.

In a simple Lie algebra in characteristic zero, a Cartan subalgebra isabelian and unique up to an automorphism. It is no longer the case in char-acteristic p. For the Witt algebra realisation W (1, k) we see one-dimensionalCartan subalgebra Ke0. From the generalised Witt algebra realisationW (1, φ)we see another Cartan subalgebra ⊕p|nW (1, φ)n. Notice that these coincidefor W (1, 1).

1.11.2 Filtrations

Gradings on an algebra are very helpful because of the way they structuremultiplication. However, they may be difficult to find. For instance, W (1, k)has no obvious grading as far as we can see. The second best thing after agrading is a filtration.

We fix ε ∈ {1,−1}. A filtration on an algebra A is a collection of vectorsubspaces FnA, n ∈ X ⊆ Z where X is either Z or Z≤a or Z≥a. It hasto satisfy two properties: FnA ∗ FmA ⊆ Fn+mA and FnA ⊆ Fn+εA for alln,m ∈ X such that n +m,n + ε ∈ X. The filtration is ascending if ε = 1.


It is descending if ε = −1.Every graded algebra has two associated filtrations: ascending FnA =

⊕k≤nAk and descending FnA = ⊕k≥nAk. Notice that sometimes these fil-trations are defined only on a subset of Z. Consider the ascending filtrationof a grading that stops at some positive number N , i.e., A = ⊕N

−∞Ak andAN 6= 0. Then FNA = A. However, we cannot define FN+1A = A becauseFN+1A ∗ FmA will be in FN+mA but not FN+m+1A, in general. Thus, thisfiltration is defined over Z≤N . Similarly, the descending filtration can bedefined only on Z≥a in similar circumstances.

In the opposite direction, from a filtered algebra A we can construct anassociated graded algebra grA = ⊕nAn/An−ε with multiplication defined by

(a+An−ε)(b+Am−ε) = ab+An+m−ε.

Ascending filtrations are more common. Given generators xi of an alge-bra A, we define F−1A = 0 and F0A as either 0 or K1, the latter in case ofassociative or commutative algebras. For a positive n we define FnA as thelinear span of all products (any order of brackets) of k generators, for allk ≤ n. This filtration is particularly useful for the universal enveloping alge-bra Ug. Its associated graded algebra is the symmetric algebra Sg which canbe defined as the universal enveloping algebra of g with zero multiplication.This fact follows from the PBW-theorem and we leave it as exercise.

Let us explain a method of obtaining a descending filtration on a Liealgebra. Let g be a Lie algebra, l its subalgebra. We define vector subspacesFng recursively:

F−1g = g, F0g = l, Fn+1g = {x ∈ Fng | x ∗ g ⊆ Fng}.

For example, if l is an ideal then all positive Fng are equal to l. Notice thatthis filtration is define over Z≥−1

Proposition 1.11.1 Subspaces Fng form a descending filtration and ∩Fngis an ideal of g.

Proof: The descending property is clear. The multiplicative property of thefiltration is observe by induction. As an induction basis, F−1g∗Fng ⊆ Fn−1g

be the definition of Fng and F0g∗F0g ⊆ F0g because F0g = l is a subalgebra.Now we assume that we have proved Fig ∗ Fjg ⊆ Fi+jg for i + j ≤ k andconsider the case of i + j = k. Using Jacobi’s identity, (Fig ∗ Fjg) ∗ g =(Fig ∗ g) ∗Fjg +Fig ∗ (Fjg ∗ g) = Fi−1g ∗ Fjg +Fig ∗ Fj−1g ⊆ Fi+j−1g. Thisimplies that Fig ∗ Fjg ⊆ Fi+jg.

Finally, g ∗ ∩jFjg ⊆ ∩jFj−1g ⊆ ∩jFjg. 2

1.11. FILTRATIONS 43

1.11.3 Filtration on Witt algebra

To conclude we exhibit a useful subalgebra of W (1, k) to construct afiltration on.

Proposition 1.11.2 l = {∑

i αiei ∈ W (1, k) |∑

i αi = 0} is a Lie asubalgebra of W (1, k).

Proof: Let∑

i αi = 0 =∑

i βi. Then (∑

i αiei) ∗ (∑

i βiei) =∑

i,j αiβj(i−j)ei+j belongs to l because

∑i,j αiβj(i−j) =

∑i αii

∑j βj−

∑j βjj

∑i αi =

0. 2

1.11.4 Exercises

Prove that the normaliser of a subalgebra (in a Lie algebra) is a subal-gebra.

Prove that the normaliser of a subalgebra a (in a Lie algebra) is theunique maximal subalgebra that contains a as an ideal.

Prove that Ke0 and ⊕p|nW (1, φ)n are both Cartan subalgebras.What is the dimension of ⊕p|nW (1, φ)n? When is it non-abelian?Let ei be a basis of a Lie algebra g. Prove that gr(Ug) is isomorphic to

the (commutative) polynomial algebra K[ei] in variables ei.Let g = ⊕kgk be a graded Lie algebra. We consider the standard filtra-

tion FnU given by g as generators of Ug. Choosing a basis of homogeneouselements of g and using PBW, we can see that Ug is graded. We denote Units n-th graded piece. Let F ′

nU be the span of all FiU ∩Uj for all 2i+ j ≤ n.Prove that this is an ascending filtration1. Prove that if ei is a homoge-neous basis of g then gr(Ug) is isomorphic to the (commutative) polynomialalgebra K[ei] in variables ei.

1called Kazhdan filtration


1.12 Witt algebras are generalised Witt

algebra1.12.1 Graded algebras of certain filtered Lie algebras

Theorem 1.12.1 Let g be a simple Lie algebra of dimension pk with a subal-gebra of codimension 1. Using filtration of Proposition 1.11.1, the associatedgrade algebra gr(g) is isomorphic to W (1, φ) where φ(1) = k.

Proof: We start with two general observations. First, ∩nFng = 0 since g

is simple.

Second, the codimension of Fn+1g in Fng is 1 unless both spaces arezero. If the codimension is zero then Fng = Fn+1g = Fn+2g = . . . andFng = ∩kFkg is an ideal. Hence, Fng = Fn+1g = 0. To prove that thecodimension is always less than 2, suppose the contrary. Pick the smallestn when it is at least 2. For each k = −1, 0, . . . n− 1 choose ak ∈ Fkg so thatak = ak+Fk+1g forms a basis of Fkg/Fk+1g. When we reach n we choose atleast two elements b1,b2, . . . ∈ Fng so that bj = bj+Fn+1g forms a basis ofFng/Fn+1g. In the graded algebra a−1 ∗bj = λjan−1, which is equivalent toa−1 ∗bj ∈ λjan−1 +Fng in g itself. Notice that λj 6= 0 because λj = 0 wouldimply bj ∈ Fn+1g and bj = 0 contradicting the basis property. Finally,a−1 ∗ (λibj − λjbi) = 0 implying as just above that λibj − λjbi ∈ Fn+1g,contradiction.

Since the codimension is always 1 unless things degenerate into zero,we can choose choose a−1 ∈ g so that a−1 = a−1 + F0g forms a basis ofF−1g/F0g. Also for the top degree N = |φ| − 1 we choose nonzero aN ∈Fng. With this choices, we define all the elements in the middle recursively:an = a−1 ∗ an+1. Notice that an 6= 0 because if an ∈ Fn+1g then Fn+1g isan ideal that contradicts simplicity of g.

When we reach the bottom degree the element is already predefinedthere. Nevertheless, we can conclude that λa−1 + x0 = a−1 ∗ a0 for somex0 ∈ F0g. Without loss of generality λ = 1 because we can replace aN byλ−1aN . In the associated graded algebra, ai ∗ aj = αi,jai+j and we cancompute the coefficients αi,j recursively:

α−1,j = 1 = −αj,−1, αi,j = αi−1,j + αi,j−1.

The second formula follows from αi,jai+j−1 = a−1 ∗ αi,jai+j = (a−1 ∗ ai) ∗aj + ai ∗ (a−1 ∗ aj) = ai−1 ∗ aj + ai ∗ aj−1. By induction, it follows that

αi,j =

(i+ j + 1i+ 1

)−

(i+ j + 1

i

)

1.12. WITT ALGEBRAS ARE GENERALISED WITT ALGEBRA 45

and ak 7→ X(k+1)∂ is an isomorphism since X(n)∂ ∗X(m)∂ = (X(n)X(m−1)−

X(n−1)X(m))∂ = (

(n+m− 1

n

)−

(n+m− 1n− 1

))X(n+m−1)∂. 2

We have used the dimension assumption only at the end. In fact, it is notrequired. One can use the properties of binomial coefficients to fiddle theargument at the end to show a stronger statement that any finite dimensionalsimple Lie algebra with subalgebra of codimension 1 and the associatedfiltration has gr(g) isomorphic to W (1, φ) for some φ.

A more interesting question is whether we can change a−1 to force b0 =0. This would lead to an even stronger statement that g ∼= W (1, φ). Whilethe answer to this question is positive we are in position to tackle it here. Itinvolves subtleties of the deformation theory2 of W (1, φ). In characteristiczero, a simple Lie algebra is rigid, i.e., all its deformations are isomorphicto itself. W (1, φ) is no longer rigid but its non-trivial deformations are nolonger simple.

1.12.2 Filtration properties of the Witt algebra

We are ready to describe the filtration on g = W (1, k) described inProposition 1.11.1,

Proposition 1.12.2 If n ≥ 0 then Fng = {∑

i αiei ∈ W (1, k) |∀k ∈{0, . . . , n}

∑i αii

k = 0}.

Proof: We go by induction on n. The basis of the induction is the definitionof l.

Suppose we have proved it for n−1. Consider an element x =∑

i αiei ∈Fn−1g. By induction assumption,

∑i αii

k = 0 for k < n. Now x ∗ et =∑i αi(i − t)ei+t. Clearly, x ∈ Fng if and only if x ∗ et ∈ Fn−1g for all t if

and only if∑

i αi(i − t)(i + t)k = 0 for all k < n and t. Now observe that∑i αi(i− t)(i+ t)n−1 =

∑i αii

n while∑

i αi(i− t)(i+ t)k = 0 for k ≤ n− 2by induction assumption. 2

1.12.3 Main theorem

Let F(pk) be the field of pk elements.

Lemma 1.12.3 If 0 < m < pk − 1 then∑

α∈F(pk) αm = 0. If m = pk − 1

then∑

α∈F(pk) αm = −1.

Proof: Let d be the greatest common divisor of m and pk − 1. Writem = dm′. Since the multiplicative group of F(pk) is the cyclic group3 of order

2never mind, it is not examinable3In general, any finite subgroup A of the multiplicative group of any field is finite as

you have learnt in Algebra-II. This follows from the unique factorisation property of the


pk−1, a 7→ am′

is a permutation of the field and∑

α∈F(pk) αm =

∑α∈F(pk) α

d.

Now let pk − 1 = dq. The map a 7→ ad is an endomorphism of themultiplicative group whose image consists of all the elements with ordersdividing q. Calling this image A we get

∑α∈F(pk) α

d = d∑

α∈A α.

Observe that A is a cyclic group of order q. If m = pk − 1 then q = 1,d = pk − 1 and we get the second statement. If 0 < m < pk − 1 then the qelements of A are the roots of zq − 1. Hence, zq − 1 =

∏α∈A(z − α) and,

consequently,∑

α∈A α = 0 as the coefficients at zq−1 is zero. 2

Theorem 1.12.4 Let n = 1 and φ(1) = k. Then W (1, φ) ∼= W (1, k).

Proof: (partial, k = 1 only) Lemma 1.12.3 and Proposition 1.12.2 give usa non-zero element in the top of the filtration: apk−2 =

∑i ei. It remains

to choose a−1 and I can do this painlessly only if k = 1. Let a−1 = −e1.Observe that by induction

apk−2−t = a−1 ∗ apk−1−t =∑

i

(−1)t(1 − i)(0 − i)(−1 − i) . . . (2 − t− i)ei+t

for non-negative grades. If k = 1 each consecutive element has one less termas zeros will appear in the product which will give a−1 ∗ a0 = (−1)p−1(p −1)!e1 = a−1 because (p− 1)! = −1 by Wilson’s theorem4 2

1.12.4 Exercises

Prove that∏α∈F(pk) α = −1.

polynomials as every z − α, α ∈ A divides zm− 1 where m is the exponent of A. Thus,

the exponent A must be equal to the order of A that characterises cyclic groups.4Observe that the product of all elements in an abelian group is equal to the product

of elements of order 2 as the rest of the elements can be paired up with their inverses.Now the multiplicative group of F(p) is cyclic and has only one element −1 of order 2.

Documents

Modular Lie Algebras - Warwick Insitehomepages.warwick.ac.uk/~masdf/modular/ln_8feb.pdf · 2010. 2. 8. · Classiﬁcation of modular Lie algebras 1.1 Basics We recall some basic