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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Borel and radical

I’m a little bit confused, could anyone explain what is the difference between the Borel subalgebra of $\mathfrak{g}$ and a radical of $\mathfrak{g}$?
Matthew Willow
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Maximal solvable ideals in infinite dimensional Lie algebras

For finite dimensional Lie algebras we now that there exists a unique maximal solvable ideal. Now for infinite dimensional Lie algebras this breaks down because of the existence part. I have a rough idea on why that is true but i want to have a…
Adronic
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Question regarding the relationship of solvability of Lie algebras and short exact sequences.

This is about the fact that for a short exact sequence of Lie algebras $$0\rightarrow K\rightarrow L\rightarrow Q\rightarrow 0$$ we have that L is sovable iff K and Q are solvable. In essence i would like to know why the following argument doesent…
Adronic
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Proof that nilspaces are orthogonal with respect to Killing form

I'm reading Serre's book about semisimple complex Lie algebras and having problem to understand one part of a proof. The theorem states, that the restriction of the Killing form $B$ of a semisimple Lie algebra $\mathfrak{g}$ to a Cartan subalgebra…
user1268702
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On the Killing splitting of some algebras

It is well-known that the algebra $ so(4,1) $ admits the Killing splitting as vector spaces as follows: $ so(4,1)=so(3,1) \oplus \mathbb{R}^{3,1} $. In the same context, is there a Killing splitting of the algebra $ gl(5,\mathbb{R}) $ of the group $…
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Semisimple elements in compact Lie groups

Suppose that $\mathfrak{g}$ is a Lie algebra. We call $\mathfrak{g}$ a compact Lie algebra if there exists an inner product $\langle\cdot,\cdot\rangle$ on $\mathfrak{g}$ such that $$\langle[Z,X],Y\rangle+\langle X,[Z,Y]\rangle=0,\forall…
Hdd
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The notation for a non-abelian two dimensional Lie algebra

Most of the introductions into Lie algebras start with the notion of a two-dimensional non-abelian Lie algebra $\mathfrak{g} = \langle x,y\rangle$ such that $[x,y]=x$. Is there a common notation for this one?
Matthew Willow
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About Reductive Lie Algebra

Question: Let $L$ be an reductive lie algebra,$H$ is an ideal of $L$.Let $S$ be the centralizer of $H$ in $L$, If $S$ is abelian, proof that $\left[ L,L\right] \subseteq H$. My attempt:Since $S$ be the centralizer of $H$ in $L$ and abelian, so it…
zeyu hao
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Universal enveloping algebra of an ideal of a Lie algebra

Let $I$ be an ideal of a Lie algebra $L$. By the PBW theorem, we can consider $U(I)$ as a subring of $U(L)$. Is it true that $U(I)$ is an ideal of the ring $U(L)$? If not, what kind of object it is?
Luka
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Matrix lie algebra with nilpotent element

Let $\mathfrak{g}$ be a matrix Lie algebra of dimension $n$ containing elements $x,y,z$ such that: $z=[x,y], [x,z]=[y,z]=0$. Show that $z$ only has $0$ as an eigenvalue. Immediately from $z=[x,y]=xy-yx$ we get that $tr(z)=0$ but I don't know how…
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What is the definition of Hamiltonian derivation in Poisson algebra?

Let consider a Poisson algebra P with Poisson bracket $\{,\}$. Let consider a derivation map $d_{a}: P \to P$ defied by $d_{a}=\{a,x\}$ for $a \in P$ which is both a Lie algebra and associative algebra derivations. What is called this type pf…
Nil
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Adjoint representation is completely reducible

Let $L$ be a Lie algebra. Let $Ad$ be the adjoin representation of $L$. If $Ad$ is completely reducible, then $$Ad(L)\cong L/Z(L) \cong L_1 \oplus \dots \oplus L_t$$ Now from this I think it's clear than $L$ it's completely reducible. So if $I$ is…
Mario
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Question about a proof theorem 5.2 of Humphreys

I am reading through the proof of every semisimple Lie Algebra being a direct sum of simple ideals. Firstly, we take an arbitrary ideal $I$ and define $I^{\bot} = $ {$x \in L | \kappa(x,y) = 0 $ for all $ y \in I$}. Now I get why $I \cap I^{\bot} =…
abbaaaa
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A confusion about a linear basis of Lie subalgebra of Lie algebra

I have already tried to understand this good explanation: Basis of free Lie algebras But still I have a confusion in the following case: Let $L(a,b)$ be a free Lie algebra of rank 2. How is it possible to say that the elements $a, [b,a], [b,[b,a]],…
Nil
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Logic behind one step of the proof of set of roots of a semisimple Lie algebra $\mathfrak{g}$ must span the dual vector space of its Cartan subalgebra

Suppose $\mathfrak{g}$ is a semisimple Lie algebra, and denote $\mathfrak{h}$ one of its Cartan subalgebras. I would like to prove the set of roots of $\mathfrak{g}$ necessarily spans $\mathfrak{h}^\star$. A linear form $\alpha\in\mathfrak{h}^\star$…
Rescy_
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