Questions tagged [lie-algebras]

For questions about Lie algebras, an algebraic structure whose main use is in studying geometric objects such as Lie groups and differentiable manifolds.

In mathematics, a Lie algebra is an algebraic structure whose main use is in studying geometric objects such as Lie groups and differentiable manifolds. Lie algebras were introduced to study the concept of infinitesimal transformations. The term “Lie algebra” (after Sophus Lie) was introduced by Hermann Weyl in the 1930s. In older texts, the name “infinitesimal group” is used.

Concretely, a Lie algebra $\mathfrak{g}$ over a field $\mathbf{k}$ is a $\mathbf{k}$-vector space equipped with an alternating bilinear multiplication $[{-}\,{-}]\colon \mathfrak{g} \wedge \mathfrak{g} \to \mathfrak{g}$ called the Lie bracket that satisfies the Jacobi identity:

$$\big[x\,[y\,z]\big] + \big[z\,[x\,y]\big] + \big[y\,[z\,x]\big] = 0$$

Examples

  • $\mathbb{R}^3$ endowed with the cross product forms a Lie algebra.

  • For any any associative algebra $A$ with multiplication $\cdot$, you can define a Lie bracket on $A$ as a literal commutator between two elements, $[v\,w]= v\cdot w-w\cdot v\,,$ making $A$ into a Lie algebra.

6730 questions
23
votes
0 answers

Abelian Cartan subalgebras

If a Lie algebra is semisimple or reductive, its Cartan subalgebras are Abelian, and their elements semisimple. Are there non-reductive algebras with Abelian Cartan subalgebras all of whose elements are semisimple? N.B.: I asked this also on MO.
22
votes
1 answer

Prove that every bilinear associative form in a simple Lie algebra is a multiple of a Killing form

Let $L$ be a Lie algebra. I have to prove that if $L$ is a simple Lie algebra every bilinear associative form (e.g. $([x,y],z)= (x,[y,z])$ for all $x,y,z \in L$) is a multiple of Killing form.
ArthurStuart
  • 4,932
14
votes
2 answers

Are there finite-dimensional Lie algebras which are not defined over the integers?

Let $\mathfrak{g}$ be a finite-dimensional complex Lie algebra and let $R \subset \mathbb{C}$ be a subring. Say that $\mathfrak{g}$ is defined over $R$ if there exists a basis $x_1, ... x_n$ for $\mathfrak{g}$ such that the structure constants…
Qiaochu Yuan
  • 419,620
14
votes
2 answers

Prove that $\mathfrak{su}(2)$ is not isomorphic to $\mathfrak{sl}(2,\mathbb R)$

I would like to prove that the two algebras $\mathfrak{su}(2)$ and $\mathfrak{sl}(2,\mathbb R)$ are not isomorphic as Lie algebras. I started considering the following basis for…
13
votes
1 answer

What is the relationship between semisimple lie algebras and semisimple elements?

A Lie algebra $\mathfrak{g}$ is said to be semisimple if its radical is zero. An element $x \in \mathfrak{g}$ is said to be semisimple if $\text{ad} x$ is diagonalizable. A complex semisimple Lie algebra must contain non-zero semisimple elements.…
Tony
  • 6,718
13
votes
1 answer

How to prove these Lie algebra relations for representations of $\mathfrak{sl}_2(\mathbb C)$

This is a bit of a basic computational question concerning Lie algebras, but I'm getting kind of bamboozled so I thought I'd post it. I'm confused about how to perform some computations in Serre's Complex Semisimple Lie algebras. The first case is…
Tony
  • 6,718
12
votes
3 answers

Is it true that the commutators of the gamma matrices form a representation of the Lie algebra of the Lorentz group?

Wikipedia claims (http://en.wikipedia.org/wiki/Gamma_matrices): The elements $\sigma^{\mu \nu} = \gamma^\mu \gamma^\nu - \gamma^\nu \gamma^\mu$ form a representation of the Lie algebra of the Lorentz group. Using the convention $$ \gamma^0 =…
Brian Bi
  • 1,232
  • 1
  • 8
  • 49
12
votes
4 answers

Are Structure Constants of a Lie Algebra always Totally Antisymmetric?

Are the Structure Constants $c^a_{bc}$ of a Lie Algebra always totally antisymmetric so, $$ c_{abc} = c_{bca} = c_{cab} $$ Or is this just the case for semi-simple algebras?
12
votes
1 answer

Sum of nilpotent ideals in a Lie algebra is nilpotent

When trying to show that $I+J$ is nilpotent, whenever $I,J$ are nilpotent ideals of a Lie algebra $L$, I did it brute force: By induction we can show the following: An element of $(I+J)^{2N}$ is a sum of elements of the form…
user42761
11
votes
4 answers

Examples of derivation of Lie algebras

Let $A$ be an algebra over a field $F$. A derivation of $A$ is an $F$-linear map $D : A\to A$ such that $D(ab) = aD(b) + D(a)b$ for all $a, b \in A$. The map $adx : L \to L$ is inner derivation. I'm looking for some examples non-inner derivations of…
negar
  • 143
11
votes
2 answers

$\mathfrak{sl}(3,F)$ is simple

Prove that $\mathfrak{sl}(3,F)$ is simple, unless $\operatorname{char}F=3$. [Use the standard basis $h_1,h_2,e_{ij}(i\neq j)$. If $I\ne 0$ is an ideal, then $I$ is the direct sum of eigenspaces for $\operatorname{ad}h_1$ or $\operatorname{ad}h_2$;…
PJ Miller
  • 8,193
11
votes
1 answer

What is the Killing form of $\mathfrak{gl}_{m}$?

Consider the Killing form of the Lie algebra $\mathfrak{gl}_{m}$. Then $\{e_{ij}\}$ is a basis for this Lie algebra where $e_{ij}$ is a matrix with 1 in the $i$th row, $j$th column and 0 everywhere else. Then every $A = (a_{ij}) \in…
192803
  • 143
10
votes
6 answers

Showing the Lie Algebras $\mathfrak{su}(2)$ and $\mathfrak{sl}(2,\mathbb{R})$ are not isomorphic.

I am working through the exercises in "Lie Groups, Lie Algebras, and Representations" - Hall and can't complete exercise 11 of chapter 3. My aim was to demonstrate that there does not exist a vector space isomorphism $A$ between the two spaces that…
muaddib
  • 8,267
9
votes
1 answer

Center of $\mathfrak{sl}(n,F)$

Prove that $\mathfrak{sl}(n,F)$ (matrices with trace zero) has center $0$, unless $\operatorname{char}F$ divides $n$, in which case the center is $\mathfrak{s}(n,F)$ (scalar multiples of the identity). I don't understand why the center can be $0$.…
PJ Miller
  • 8,193
9
votes
3 answers

Lie algebra isomorphism between $\mathfrak{sl}(2,{\bf C})$ and $\mathfrak{so}(3,\Bbb C)$

I think that this is an exercise. I can not find a solution. We can define Lie bracket multiplication on $\mathbb{C}^3$ : $$ x\wedge y $$ where $x=(x_1,x_2, x_3)$, $y= (y_1,y_2,y_3)$, and $\wedge $ is the wedge product we know. Consider the Lie…
HK Lee
  • 19,964
1
2 3
54 55