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.

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How to show that $\mathfrak{sl}_n(\mathbb{R})$ and $\mathfrak{sl}_n(\mathbb{C})$ are simple?

We defined a Lie algebra to be simple, if it has no proper Lie ideals and is not $k$ (the ground field). We have the proposition that $\mathfrak{sl}_n(\mathbb{R})$ and $\mathfrak{sl}_n(\mathbb{C})$ are simple $\mathbb{R}$-Lie algebras, and…
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Is Cartan subalgebra of a Lie algebra unique?

Let $\mathfrak{g}$ be a Lie algebra. Is Cartan subalgebra of $\mathfrak{g}$ unique? I see in some places it is written "Let $\mathfrak{h}$ be a Cartan subalgebra of $\mathfrak{g}$".
LJR
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Classify Lie algebra with 1 dimension derived algebra

Problem 3.10 from Erdmann, Wildon ask: Find, up to isomorphism, all Lie algebras with 1-dimensional derived algebra. (also, this book assume finite dimension Lie algebra only; I'm not sure whether it's over a general field, or just $\mathbb{C}$, but…
Gina
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Tensoring with induced representation

On J. Humphreys' book "Representations of Semisimple Lie Algebras in the BGG Category O", Theorem 3.6, a Tensor Identity is quoted: $$ (U(\mathfrak{g}) \otimes _{U(\mathfrak{b})} L) \otimes M\simeq U(\mathfrak{g}) \otimes _{U(\mathfrak{b})} (L…
waikit
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Semisimple Lie algebras are perfect.

Can anyone explain why a semi-simple finite dimensional Lie algebra $\mathfrak{g}$ has to be perfect ? The natural way to prove something like that would be to look to the algebra generated by the Lie brackets, which when $\mathfrak{g}$ is not…
Selim Ghazouani
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Lie algebra of dimension $3$ and $L=[L,L]$ must be simple

Suppose $\dim L=3$ and $L=[L,L]$. Prove that $L$ must be simple. [Observe first that any homomorphic image of $L$ also equals its derived algebra.] Recover the simplicity of $\mathfrak{sl}(2,F)$, $\operatorname{char}F\ne 2$. First, the part in…
PJ Miller
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Are (semi)-simple Lie algebras not solvable?

Let $L\ne \{0\}$ be a non-abelian simple Lie algebra, then the only ideals are $L$ and $\{0\}$. We know that $L'$ is the smallest ideal of $L$ such that $L/L'$ is abelian. If $L'=\{0\}$, then $L/L'=L$, but $L$ is non-abelian. So $L'=L$ and hence…
Phil-ZXX
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Automorphism group of a Lie algebra

I read Nolan Wallach's Geometric Invariant Theory book. In a chapter, the author talks about the group $G$ in Aut$(\mathfrak{g})$ corresponding to ad$(\mathfrak{g})$ (here $\mathfrak{g}$ is a complex Lie algebra). I do not understand what this group…
Levent
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When is the Killing form null?

When is the Killing form $\kappa$ of a Lie algebra $\mathfrak g$ null, i.e. $\kappa(\cdot,\cdot)=0$? Surely this is true for any Lie algebra with trivial bracket, but is this the only case? I can't seem to find any nontrivial examples.
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Why Lie algebras of type $B_2$ and $C_2$ are isomorphic?

both of Lie algebras of type $B_2$ and $C_2$ have dimension 10 and we can find two basis of them on page 3 in the book: Introduction to Lie algebras and representation theory . How could we show that these two Lie algebras are isomorphic by…
user
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Is there any example of a Lie Algebra, who has nontrivial radical but contains no abelian ideal?

Is there any example of a Lie Algebra, who has nontrivial radical but contains no abelian ideal? Here, the radical of a Lie algebra means its maximal solvable ideal. This question occurs in the proof of the theorem which states that "A Lie algebra…
ShinyaSakai
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Does the proof of the Poincare-Birkhoff-Witt theorem need the Jacobi identity?

The title says it. Suppose I have a vector space $V$ equipped with a bilinear bracket such that $[x,y]=-[y,x]$, and define the universal enveloping algebra $U$ as usual: namely the tensor algebra on $V$ modulo the 2-sided ideal generated by…
h__
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Classification results for solvable lie algebras.

According to the Levi decomposition every (real or complex) Lie algebra $g$ can be written as the semidirect product of a solvable and a semisimple Lie algebra. The semisimple Lie algebras can be classified. But how does one deal with solvable Lie…
Niklas
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An exercise in Serre's Lie algebra book

Let $k$ be a commutative ring. Prove that a Lie $k$-algebra $\mathfrak{g} = 0$ iff $U\mathfrak{g} = k$. Use the adjoint representaion. Here is my attempt at it: The only non-trivial statement is that if $U\mathfrak{g} = k$, then $\mathfrak{g} =…
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No Lie algebra over $\Bbb R$ or $\Bbb C$ can have a unit element.

I want to prove: No Lie algebra over $\Bbb R$ or $\Bbb C$ can have a unit element. Now I am not sure how to take this in regard to the Lie bracket. I.e. I have now idea where to start. $[x,e]=[e,x]=x$ or something? That doesn't make sense to me,…
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