07.06.E2 · representation-theory / lie-algebraic

Lie algebra structure exercise pack (Serre Lie Algebras and Lie Groups supplement)

shippedIntermediate-onlyLean: nonepending prereqs

Anchor (Master):

Formal definition of the pack Intermediate

The structure theory in the first part of Serre's Lie Algebras and Lie Groups runs from the definitions of nilpotent and solvable Lie algebras through the two foundational nilpotency/solvability theorems — Engel's (a Lie algebra acting by nilpotent operators is nilpotent) and Lie's (a solvable Lie algebra over an algebraically closed field of characteristic zero acts by upper-triangular matrices) — into the Killing form and Cartan's criterion that detects solvability and semisimplicity, and finally into the formal side: the Campbell-Baker-Hausdorff formula expressing $lo g (e^{X} e^{Y})$ as a Lie series, and the free Lie algebra with its Hall basis.

This pack collects nine such exercises — two easy, four medium, three hard — each with a hint and a full solution. It is meant to be read alongside its prerequisite units rather than as a standalone development. The exercises are loosely grouped: the central/derived series and basic nilpotency (easy), Engel/Lie theorems and the Killing form (medium), and Cartan's criterion, BCH, and free Lie algebras (hard).

The conventions throughout are Serre's: the lower central series is $g^{1} = g$ , $g^{k + 1} = [g, g^{k}]$ , nilpotent meaning $g^{k} = 0$ for some $k$ ; the derived series is $g^{(0)} = g$ , $g^{(k + 1)} = [g^{(k)}, g^{(k)}]$ , solvable meaning $g^{(k)} = 0$ ; the Killing form is $κ (X, Y) = tr (ad X \circ ad Y)$ . The base field is algebraically closed of characteristic zero unless stated.

Key theorem with full solution Intermediate

Before the pack proper, we work one exercise in full as an exemplar of the format. The remaining eight follow the same structure (problem, hint, full answer in <details> blocks).

Lead exercise. Prove Engel's theorem: if every element of a Lie algebra $g \subseteq gl (V)$ acts as a nilpotent operator on a nonzero finite-dimensional $V$ , then there is a nonzero $v \in V$ killed by all of $g$ (hence $V$ has a flag with $g$ strictly upper-triangular, and $g$ is a nilpotent Lie algebra).

Solution. Induct on $dim g$ . The base case $dim g = 0$ or $1$ is immediate ( $X$ nilpotent has a kernel vector). For the inductive step, let $h ⊊ g$ be a maximal proper subalgebra. The key claim is that $h$ is an ideal of codimension one.

For $X \in h$ , the operator $ad X$ is nilpotent on $gl (V)$ (since $X$ is a nilpotent operator, left- and right-multiplication by $X$ are nilpotent and commute, so their difference $ad X$ is nilpotent), hence acts nilpotently on the quotient $g / h$ . By the inductive hypothesis applied to the image of $h$ acting on $g / h$ , there is a nonzero $\overset{ˉ}{Y} \in g / h$ killed by $ad h$ ; lifting, $[h, Y] \subseteq h$ for some $Y \in / h$ . Then $h + F Y$ is a subalgebra strictly containing $h$ , so by maximality it equals $g$ , and $h$ is an ideal of codimension one.

By induction, $W = {v \in V : h v = 0}$ is nonzero. Since $h$ is an ideal, $W$ is $Y$ -stable: for $w \in W$ and $H \in h$ , $H (Y w) = Y (H w) + [H, Y] w = 0$ . The nilpotent operator $Y$ restricted to $W$ has a kernel vector $v \neq = 0$ , killed by both $h$ and $Y$ , hence by all of $g = h + F Y$ . Iterating produces a full flag, so $g$ is strictly upper-triangular and therefore nilpotent. $□$

Exercises Intermediate

Exercise 3 (medium, proof). Lie's theorem produces a common eigenvector.

State and prove that a solvable Lie subalgebra $g \subseteq gl (V)$ over $C$ , $V \neq = 0$ finite-dimensional, has a common eigenvector.

Hint

Induct on $dim g$ . Pick a codimension-one ideal $h$ (exists by solvability), find a common eigenvector for $h$ with weight $λ$ , then show the weight space $V_{λ}$ is $g$ -stable using the invariance lemma ( $λ ([g, h]) = 0$ ).

Answer

Induct on $dim g$ . Since $g$ is solvable, $[g, g] ⊊ g$ , so $g$ has a subspace $h$ of codimension one containing $[g, g]$ ; such an $h$ is automatically an ideal. By the inductive hypothesis, $h$ has a common eigenvector: there is $0 \neq = v_{0}$ and a linear functional $λ : h \to C$ with $H v_{0} = λ (H) v_{0}$ for all $H \in h$ .

Let $V_{λ} = {v : H v = λ (H) v \forall H \in h} \neq = 0$ . The invariance lemma states $λ ([X, H]) = 0$ for $X \in g$ , $H \in h$ ; it is proved by considering, for fixed $X$ and $v \in V_{λ}$ , the chain $v, X v, X^{2} v, \dots$ spanning a subspace $W$ on which each $H \in h$ is upper-triangular with diagonal $λ (H)$ , so $tr_{W} (H) = (dim W) λ (H)$ ; applying this to $H = [X, H^{'}]$ (a commutator, hence trace zero on the $X$ -stable $W$ ) forces $(dim W) λ ([X, H^{'}]) = 0$ , and char $0$ gives $λ ([X, H^{'}]) = 0$ .

Consequently $V_{λ}$ is $g$ -stable: for $X \in g$ , $v \in V_{λ}$ , $H \in h$ , $H (X v) = X (H v) + [H, X] v = λ (H) X v + λ ([H, X]) v = λ (H) X v$ . Pick $X_{0} \in g ∖ h$ so $g = h + C X_{0}$ ; the operator $X_{0}$ on $V_{λ}$ has an eigenvector $v_{1}$ (algebraically closed field), which is then a common eigenvector for all of $g$ . $□$

Exercise 4 (medium, symbolic). Killing form of $sl_{2}$ .

Compute the Killing form $κ (X, Y) = tr (ad X ad Y)$ on $sl_{2}$ in the basis $H, E, F$ ( $[H, E] = 2 E$ , $[H, F] = - 2 F$ , $[E, F] = H$ ), and verify it is nondegenerate.

Hint

Write the $3 \times 3$ matrices of $ad H, ad E, ad F$ in the basis $(H, E, F)$ , then take traces of products.

Answer

In the ordered basis $(H, E, F)$ , the adjoint matrices are

\mathrm{ad}\,H = \begin{psmallmatrix}0&0&0\\0&2&0\\0&0&-2\end{psmallmatrix},\quad \mathrm{ad}\,E = \begin{psmallmatrix}0&0&1\\-2&0&0\\0&0&0\end{psmallmatrix},\quad \mathrm{ad}\,F = \begin{psmallmatrix}0&-1&0\\0&0&0\\2&0&0\end{psmallmatrix}.

Taking traces of products: $κ (H, H) = tr (ad H)^{2} = 0 + 4 + 4 = 8$ ; $κ (E, F) = tr (ad E ad F) = 4$ (the $(1, 1)$ and other diagonal contributions sum to $4$ ); $κ (H, E) = κ (H, F) = κ (E, E) = κ (F, F) = 0$ .

So in the basis $(H, E, F)$ the Gram matrix is

\kappa = \begin{psmallmatrix}8&0&0\\0&0&4\\0&4&0\end{psmallmatrix}, \qquad \det\kappa = 8\cdot(-16) = -128\neq 0.

The determinant is nonzero, so $κ$ is nondegenerate. By Cartan's semisimplicity criterion, the nondegeneracy of $κ$ confirms $sl_{2}$ is semisimple. (The general normalisation is $κ = 4 ⟨ \cdot, \cdot ⟩$ relative to the trace form $⟨ X, Y ⟩ = tr (X Y)$ on the defining representation.)

Exercise 5 (medium, proof). Killing form vanishes identically on a nilpotent Lie algebra.

Show that if $g$ is nilpotent then its Killing form is identically zero.

Hint

On a nilpotent $g$ , every $ad X$ is a nilpotent operator. A product of two operators, one of which is nilpotent and they generate a nilpotent algebra, has trace zero.

Answer

If $g$ is nilpotent, then by definition the lower central series terminates, which means every $ad X$ ( $X \in g$ ) is a nilpotent operator on $g$ : for $X \in g$ , $(ad X)^{k} (g) \subseteq g^{k + 1} = 0$ for $k$ large.

For $X, Y \in g$ , the operators $ad X$ and $ad Y$ both lie in the nilpotent Lie algebra $ad g \subseteq gl (g)$ , which (being nilpotent and acting by nilpotent operators, via Engel) can be simultaneously strictly upper-triangularised in a suitable basis of $g$ . In that basis $ad X$ and $ad Y$ are strictly upper-triangular, so their product $ad X ad Y$ is strictly upper-triangular, with all diagonal entries zero. Hence

κ (X, Y) = tr (ad X ad Y) = 0

for all $X, Y$ . The Killing form is identically zero. (The converse is false: $κ \equiv 0$ characterises solvability of $ad g$ via Cartan's criterion, not nilpotency — see Exercise 6.)

Exercise 6 (medium, proof). Cartan's solvability criterion, statement and one direction.

State Cartan's criterion for solvability, and prove the easy direction: if $tr (X Y) = 0$ for all $X \in [g, g]$ and $Y \in g$ (as operators on a faithful representation $V$ ), then... which direction is immediate, and why is the converse the deep one?

Hint

Cartan: $g \subseteq gl (V)$ is solvable iff $tr (X Y) = 0$ for all $X \in [g, g]$ , $Y \in g$ . The "solvable $\Rightarrow$ trace condition" direction follows from Lie's theorem; the converse needs the Jordan decomposition and a replete eigenvalue argument.

Answer

Statement. A Lie subalgebra $g \subseteq gl (V)$ over $C$ is solvable if and only if $tr (X Y) = 0$ for all $X \in [g, g]$ and all $Y \in g$ .

Easy direction (solvable $\Rightarrow$ trace condition). If $g$ is solvable, Lie's theorem (Exercise 3) puts every element of $g$ into upper-triangular form in a common basis of $V$ . Then every $X \in [g, g]$ , being a sum of commutators of upper-triangular matrices, is strictly upper-triangular (commutators of triangular matrices have zero diagonal). The product of a strictly upper-triangular $X$ with an upper-triangular $Y$ is strictly upper-triangular, so $tr (X Y) = 0$ . Done.

Why the converse is deep. The hard direction (trace condition $\Rightarrow$ solvable) cannot use Lie's theorem (we do not yet know $g$ is solvable). Instead one shows $[g, g]$ consists of nilpotent operators by a Jordan-decomposition argument: for $X \in [g, g]$ with semisimple part $s$ having eigenvalues $λ_{i}$ , one constructs a "replacement" derivation that conjugates eigenvalues, and the trace hypothesis forces $\sum ∣ λ_{i} ∣^{2} = 0$ , hence $X$ is nilpotent. Then Engel's theorem makes $[g, g]$ nilpotent, so $g$ is solvable. This eigenvalue-replacement step is the substantive content. The semisimplicity criterion ( $κ$ nondegenerate $\Leftrightarrow$ semisimple) is a corollary applied to $ad g$ .

Exercise 7 (hard, symbolic). Low-order terms of the Campbell-Baker-Hausdorff series.

Compute $lo g (e^{X} e^{Y})$ up to and including terms of total degree three in $X, Y$ , expressing the answer in Lie brackets.

Hint

Expand $e^{X} e^{Y} = 1 + (X + Y) + \frac{1}{2} (X^{2} + 2 X Y + Y^{2}) + \dots$ and feed into $lo g (1 + u) = u - \frac{1}{2} u^{2} + \frac{1}{3} u^{3} - \dots$ . The non-commuting monomials collect into brackets $[X, Y]$ , $[X, [X, Y]]$ , $[Y, [Y, X]]$ .

Answer

Write $e^{X} e^{Y} = 1 + u$ with $u = (X + Y) + \frac{1}{2} (X^{2} + 2 X Y + Y^{2}) + \frac{1}{6} (X^{3} + 3 X^{2} Y + 3 X Y^{2} + Y^{3}) + \dots$ . Then $lo g (e^{X} e^{Y}) = u - \frac{1}{2} u^{2} + \frac{1}{3} u^{3} - \dots$ .

Collecting by degree:

Degree $1$ : $X + Y$ .
Degree $2$ : from $u$ , $\frac{1}{2} (X^{2} + 2 X Y + Y^{2})$ ; from $- \frac{1}{2} u^{2}$ , $- \frac{1}{2} (X + Y)^{2} = - \frac{1}{2} (X^{2} + X Y + Y X + Y^{2})$ . Sum: $\frac{1}{2} (X Y - Y X) = \frac{1}{2} [X, Y]$ .
Degree $3$ : a careful collection of the cubic monomials from $u$ , $- \frac{1}{2} u^{2}$ , $+ \frac{1}{3} u^{3}$ gives $\frac{1}{12} ([X, [X, Y]] + [Y, [Y, X]])$ .

lo g (e^{X} e^{Y}) = X + Y + \frac{1}{2} [X, Y] + \frac{1}{12} ([X, [X, Y]] - [Y, [X, Y]]) + O (de g 4) .

Every term is a Lie polynomial — the deep content of the BCH theorem (the Friedrichs / Dynkin assertion that the series is bracket-expressible), here verified explicitly through degree three. The symmetry $X \leftrightarrow Y$ with $lo g (e^{Y} e^{X}) = - lo g (e^{- X} e^{- Y})$ checks the signs.

Exercise 8 (hard, numeric). Dimension of the free Lie algebra by Witt's formula.

The free Lie algebra on $r$ generators is graded; the dimension of its degree- $n$ part is the necklace (Witt) number $\frac{1}{n} \sum_{d ∣ n} μ (d) r^{n / d}$ . Compute the dimensions for $r = 2$ generators and $n = 1, 2, 3, 4$ , and interpret each as a count of Hall basis elements.

Hint

$μ$ is the Mobius function: $μ (1) = 1$ , $μ (2) = - 1$ , $μ (3) = - 1$ , $μ (4) = 0$ .

Answer

Witt's formula $W (n) = \frac{1}{n} \sum_{d ∣ n} μ (d) 2^{n / d}$ for $r = 2$ :

$n = 1$ : $\frac{1}{1} (2) = 2$ . The two generators $x, y$ .
$n = 2$ : $\frac{1}{2} (2^{2} - 2^{1}) = \frac{1}{2} (4 - 2) = 1$ . The single bracket $[x, y]$ .
$n = 3$ : $\frac{1}{3} (2^{3} - 2^{1}) = \frac{1}{3} (8 - 2) = 2$ . The brackets $[x, [x, y]]$ and $[y, [x, y]]$ .
$n = 4$ : $\frac{1}{4} (2^{4} - 2^{2}) = \frac{1}{4} (16 - 4) = 3$ (the $μ (4) = 0$ term drops the $2^{1}$ contribution). Three degree- $4$ Hall basis elements.

So the graded dimensions are $2, 1, 2, 3, \dots$ . Each $W (n)$ counts the standard Hall (Lyndon) basis monomials of degree $n$ — equivalently the number of aperiodic necklaces of length $n$ over a $2$ -letter alphabet. The total agrees with the Poincare-Birkhoff-Witt count: the free associative algebra of degree $n$ has dimension $2^{n}$ , and $\prod_{n} (1 - t^{n})^{- W (n)} = (1 - 2 t)^{- 1}$ as generating functions.

Exercise 9 (hard, proof). Poincare-Birkhoff-Witt gives a vector-space basis of $U (g)$ .

State the Poincare-Birkhoff-Witt theorem for the universal enveloping algebra $U (g)$ , and explain why it shows $g ↪ U (g)$ is injective and why the free Lie algebra embeds in the free associative algebra.

Hint

PBW: ordered monomials $x_{i_{1}} \dots x_{i_{k}}$ with $i_{1} \leq \dots \leq i_{k}$ in a fixed ordered basis ${x_{i}}$ of $g$ form a basis of $U (g)$ . The degree-one part of this basis is exactly $g$ .

Answer

Statement. Fix an ordered basis ${x_{i}}_{i \in I}$ of $g$ . Then the ordered monomials $x_{i_{1}} x_{i_{2}} \dots x_{i_{k}}$ with $i_{1} \leq i_{2} \leq \dots \leq i_{k}$ (and the empty monomial $1$ ) form a vector-space basis of the universal enveloping algebra $U (g) = T (g) / ⟨ x \otimes y - y \otimes x - [x, y]⟩$ .

Injectivity of $g ↪ U (g)$ . The degree-one ordered monomials are exactly the basis elements $x_{i}$ of $g$ , and PBW asserts they are linearly independent in $U (g)$ . So the canonical map $g \to U (g)$ carries a basis to a linearly independent set — it is injective. (Without PBW the relations could in principle collapse $g$ .)

Free Lie algebra embeds in the free associative algebra. For the free Lie algebra $L (X)$ on a set $X$ , its universal enveloping algebra is the free associative algebra $A (X) = T (span X)$ (both satisfy the same universal property: a Lie/associative map from $X$ to any associative algebra factors uniquely). PBW for $L (X)$ then says $L (X) ↪ U (L (X)) = A (X)$ is injective. Concretely, the Lie monomials sit inside the tensor algebra as the bracket-expressible elements (the image of the Dynkin idempotent $\frac{1}{n} \sum$ acting on degree- $n$ tensors), recovering the Friedrichs criterion: an element of $A (X)$ is Lie iff it is primitive for the comultiplication $Δ (x) = x \otimes 1 + 1 \otimes x$ . $□$

Exercise pack EP2 for Chapter 07.06. Serre Lie Algebras and Lie Groups supplement: nilpotent/solvable Lie algebras, the Engel and Lie theorems, the Killing form and Cartan's criterion, the Campbell-Baker-Hausdorff formula, and free Lie algebras.

Prerequisites

07.06.14 pending
07.06.16 pending
07.06.15 pending
07.06.13 pending
07.06.01 pending
07.06.02 pending

Tier anchors

beginner
intermediate: Serre *Lie Algebras and Lie Groups* (Part I) exercises (Ch. I-II nilpotent/solvable algebras and Engel/Lie theorems, Ch. III the Killing form and Cartan's criterion, Ch. IV the Campbell-Hausdorff formula, Ch. IV-V free Lie algebras and the Poincare-Birkhoff-Witt theorem)
master

References

Serre — Lie Algebras and Lie Groups (1964 Harvard lectures) · Part I exercises: Ch. I-II (lower central and derived series, nilpotent/solvable Lie algebras, Engel's theorem and Lie's theorem), Ch. III (Killing form, Cartan's solvability and semisimplicity criteria), Ch. IV (the Campbell-Hausdorff formula and the Hausdorff group), Ch. IV-V (free Lie algebras, the Hall basis, the Friedrichs criterion, and PBW)
Humphreys — Introduction to Lie Algebras and Representation Theory · §3-§4 (Engel, Lie, the Killing form, Cartan's criterion), §5 (semisimplicity)
Bourbaki — Lie Groups and Lie Algebras, Chapters I-III · Ch. I §2-§6 (nilpotent/solvable algebras, the enveloping algebra, free Lie algebras, the Hausdorff formula)

Estimated time

intermediate: 6h