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
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Lie algebra with additional property

Let $[\cdot,\cdot]$ be the Lie bracket of some Lie algebra. Assume the following holds $$y \neq \lambda x \iff [x,y] \neq 0.$$ What is the name of Lie algebra with this property? This property holds, i.e., for the cross product of 3D vectors.…
Fizikus
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Degrees of freedom in Lie algebra structure constants

[Disclosure: As of 1/17/23, a version of this has been posted on mathoverflow] This may be a very naive question, but I'm having some trouble counting degrees of freedom for Lie algebraic structure constants. Given an $n$-dimensional Lie algebra,…
Kensmosis
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Algebra but not a Lie Algebra.

I'm looking for an example to show that there can be Algebras over a field $F$ of characteristic 2, which follows $[x,y] = -[y,x]$ and the Jacobi identity, but is not a Lie Algebra.
Mike
  • 99
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Proof that the Killing form on a simple Lie algebra is non-degenerate

This is obvious since there is a more general result for semi-simple Lie algebras, but I would like to prove it directly. So suppose that the Killing form $\kappa$ is degenerate and let $X\in L$ be such that $\kappa(X,Y)=0$ for all $Y \in L$. I…
prt13463
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Showing the Weyl element is sent to its negative by a certain element of the Weyl group

Let $\mathfrak g$ be a complex semi-simple Lie algebra with a choice of positive root system. Let $\rho$ be the Weyl element, i.e. the sum of the fundamental weights (or half the sum of the positive roots). There is a unique element $w_0$ of the…
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Proof or Counter example: Every non zero Lie algebra has an ad-nilpotent element?

I am trying to understand a definition in chapter $1$ of Humphreys Lie Algebra. I think I need the following result to understand the definition properly: Let $\mathcal L$ be a non zero Lie Algebra over some (algebraically closed) field $F$. Recall…
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Are the Lie algebras $\mathfrak{sl}_2(\Bbb{R})$ and $\mathfrak{su}(2)$ simple?

I believe I have proven that the real Lie algebras $\mathfrak{sl}_2(\Bbb{R})$ and $\mathfrak{su}(2)$ are simple. However I have tried searching for results that mention this but they only talk of these Lie algebras as being semi-simple. Is what I…
user38268
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restricted lie algebras definition

Jacobson (Lie algebras, p.187) defines what is meant by a restricted Lie algebra: Def4: A restricted Lie algebra, $L$, of characteristic $p\not = 0$ is a Lie algebra of characteristic $p$ in which there is defined a mapping $a\rightarrow a^{[p]}$…
userabc
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Direct sum as vectors but not Lie algebra direct sum

Let $L$ be a Lie algebra and let $I$ be an ideal in $L$ and $K$ be a subalgebra in $L$ such that $L=I\oplus K $. Why this sum is direct as vector subspaces but not Lie algebra direct sum? Can't we define a Lie bracket on $I\oplus K $ in someway to…
Ronald
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What role does abstract Jordan decomposition play in the study of semisimple Lie algebras?

I have seen the notion of the abstract Jordan decomposition be given for semisimple Lie algebras over an algebraically closed field of char = 0 in a number of places, a few theorems proved about it (The abstract and usual Jordan decompositions of…
john
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The center of a nilpotent Lie algebra intersects each ideal

If $\cal h$ is a nonzero ideal in a nilpotent Lie algebra $\cal g$. How to prove that $\mathcal h\cap Z(\mathcal g)\not =0$, where $Z(\mathcal g)$ is the center of $\mathcal g$?
Ronald
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Why, for nilpotent Lie Algebras, is the inclusion to the derivations $x \mapsto {ad}_x$ not surjective?

Let $\mathfrak g$ be a finite dimensional nilpotent Lie Algebra over a field $F$ (is characteristic zero necessary?). Why is the map $ad: \mathfrak g \to Der (\mathfrak g), x \mapsto {ad}_x$ not surjective? One result, which is simple to show, is…
s.harp
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Computing information about a Lie algebra from cartan matrix

Let's consider the Cartan matrix : $$ \begin{pmatrix} 2 & -2 \\ -1 & 2 \end{pmatrix} $$ I am asked to find the number of roots and then to compute character of the adjoint representation. I know that from this answer : How to determine the number…
J. Doe
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Lie Algebra: Vector subspace is an ideal of its normalizer

Let $S$ be a vector subspace of the Lie algebra $\mathfrak{g}$. Is $S$ an ideal of the normalizer $N_\mathfrak{g}(S)$, I would say yes since: $\forall n\in N_\mathfrak{g}(S), [n,S]\subseteq S\implies [N_\mathfrak{g}(S),S]\subseteq S$ which is the…
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Identification of $H$ with $H^{*}$ relative the Killing-form

Let $H$ be a maximal toral subalgebra of a semisimple Lie Algebra $L$. The identification of $H^{*}$ and H relative the Killing-form says, that to $\phi\in H^{*}$ corresponds the unique element $t_{\phi}\in H$ satisfying $\phi(h)=\kappa(t_{\phi},…
Idun E.
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