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
0
votes
0 answers

Proof for a corollary from PBW theorem

I need to know how we can prove the following corollary : If $x_1, \ldots, x_n$ is a vector space basis for Lie algebra $L$ then a vector space basis for $U(L)$, $U(L)$ is universal enveloping algebra, is given by all monomials of the form …
0
votes
0 answers

What is the image of an automorphism of Lie algebra?

Let $L$ be a simple Lie algebra over ${\rm GF}(2)$. If $α$ is an automorphism of $L$ then for any element of $L$ we must have $α[a,b]=[α(a),α(b)]$. Now I want to have a clear understanding of image of $α$. Since $L$ is a simple Lie algebra, how can…
0
votes
1 answer

Is adjoint map invertible?

I've already studied the group of automorphisms of a simple lie algebra on a finite field, but according to the definition of an adjoint representation of a Lie algebra, can we claim an adjoint map is invertible? Can we say that an adjoint map is a…
0
votes
1 answer

Question to Roger Carter's "Lie Algebras of Finite and Affine Type"

In the proof of Proposition 7.31 in Roger Carter's Lie Algebras of Finite and Affine Type, Carter notes that the sets $H_\mu$ and $H_\alpha$ are distinct. Can someone find a good argument why that is the case? I have included the page in question…
Gaussler
  • 2,766
0
votes
1 answer

Why the square of ideal in Lie algebra is also ideal?

Let $L$ be a Lie algebra over field $F$, $I$ - ideal in this algebra. It's stated that $I^2$ (and so any item of central series) is also ideal in $L$. 1) For any $a, b \in I^2: [a,b] \in I^2$. True, since $a, b \in I$. 2) For any $k \in F, a \in…
Glinka
  • 3,182
0
votes
1 answer

Existence of a basis of common eigenvectors

Suppose that L is a complex semisimple Lie algebra containing an abelian subalgebra H consisting of semisimple elements. I am wondering how to see that L has a basis of common eigenvectors for the elements of ad(H).
Yuan
  • 553
0
votes
1 answer

Killing form for a non-abelian Lie Algebra of dimension $2$

Aratati ca forma Killing pentru algebra Lie ne-abeliana de dimensiune 2 nu este zero. How can I prove that the Killing forme of a non-abelian Lie algebra is not equal with $0$? Thanks
Iuli
  • 6,790
0
votes
1 answer

Isomorphism between Lie algebras

This is an exercise (cf. exe 2.11 in the Erdmann and Wildon's book) Define $$ gl_S(n, F) = \{ x\in gl(n,F)|\ x^t S = -Sx \} $$ Here $t$ is transpose. Then let $T=P^tSP$ and show that $$ gl_T(n,F)=gl_S(n,F) $$ wher $P$ is invertible. Thank you.
HK Lee
  • 19,964
0
votes
0 answers

irreducible representations of lie algebras

We have the following criterion for the irreducibility of a Lie algebra representation (we work with $L$-modules here). Let $L$ be a Lie algebra, $V$ a finite dimensional vector space, and let $L \times V \to V$ be an $L$-module. Then $V$ is…
nigel
  • 3,214
0
votes
0 answers

Is the matrix of every set of base vectors of $\Bbb{C}^n$ symmetric?

The book "Theory of Lie Groups" by Chevalley says A linear endomorphism $\alpha$ of $C^n$ is determined when the elements $\alpha e_i=\sum\limits_{j=1}^n a_{ji}e_j$ are given. There corresponds to this endomorphism a matrix $(a_{ij})$ of degree…
user67803
0
votes
1 answer

How to prove a property of Lie derivatives

I know that there are five properties for Lie derivative. But one of them I don't know how to prove. It is $ L_x[\omega(Y)]=(L_x \omega)(Y)+\omega(L_x Y)$ Note : Here $\omega$ is a covariant vector field while Y is a vector field. Is there anyone…
Jack
  • 91
0
votes
1 answer

The associated Lie algebra of the unital associative algebra $\mathcal{A}$ in the Universal Enveloping Algebras

Definition: Let $\mathfrak{g}$ be a Lie algebra. A universal enveloping algebra of $\mathfrak{g}$ is a pair $(\mathfrak{U},i)$ consisting of a unital associative algebra $\mathfrak{U}$ and a Lie morphism $i: \mathfrak{g} \rightarrow \mathfrak{U}_L$…
Elianna
  • 59
  • 4
0
votes
0 answers

Show that $C_{B}$ is a quadratic central element of $\mathfrak{U(g)}$

Let $\mathfrak{g}$ be a finite dimensional Lie algebra and $B:\mathfrak{g} \times \mathfrak{g} \rightarrow \mathbb{K}$ be a non-degenerate invariant symmetric bilinear form. Then there exists a unique $C_{B} \in \mathfrak{U(g)}$, called the Casimir…
Elianna
  • 59
  • 4
0
votes
1 answer

Why $B^{\sharp}e_{1}^{*},...,B^{\sharp}e_{n}^{*}$ is a basis of $\mathfrak{g}$?

Let $\mathfrak{g}$ be a finite dimensional Lie algebra and $B:\mathfrak{g} \times \mathfrak{g} \rightarrow \mathbb{K}$ be a non-degenerate invariant symmetric bilinear form. Then there exists a unique $C_{B} \in \mathfrak{U(g)}$, called the Casimir…
Elianna
  • 59
  • 4
0
votes
0 answers

Regular elements of $G_2$

An element of Lie algebra $L$ is called $\textbf{regular}$ if dimension of its centralizer is equal to rank of Lie algebra $L$. I work over field of characteristics zero. In case of $G_2$, regular elements are those that have $\dim(C(x)) = 2$, where…
Dibidus
  • 31