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
2
votes
1 answer

Weyl group and Dynkin diagram

Can somebody help me with following questions: 1)Prove that two simple roots in a Dynkin diagram that are connected by a single edge are in the same orbit under the Weyl group. and 2)For an irreducible root system, prove that all roots of a…
Robert
  • 21
2
votes
1 answer

Counter-example of Jacobi identity for antisymmetric bilinear operation

For a bilinear, antisymmetric, alternating operator to be a Lie bracket, it must satisfy the Jacobi identity. I assume this is because a bilinear, antisymmetric, alternating operator does not always satisfy the Jacobi identity. If I consider this…
Nugi
  • 560
2
votes
0 answers

Are Hom-Lie algebras Schreier variety?

Lie algebras are Schreier variety, that is, a variety of algebraic systems (i.e., a class of algebras defined by identical relations) in which the subalgebras of free algebras are again free. The variety of abelian groups, commutative and…
Nil
  • 1,306
2
votes
1 answer

Lie group of orthogonal matrix group - what would this mean?

In Lie group terms, this means that the Lie algebra of an orthogonal matrix group consists of skew-symmetric matrices. Going the other direction, the matrix exponential of any skew-symmetric matrix is an orthogonal matrix (in fact, special…
Zeus
  • 815
  • 9
  • 16
2
votes
1 answer

Show $\dim\{v\in V\mid bv=8v\}=0$ in $A_1$ module $V$

If $\{a,b,c\}$ is a basis for the Lie algebra $A_1$ with $[b,a]=2a, [b,c]=-2c$ and $[a,c]=b$. Suppose $V$ is a 13-dimensional $A_1$ module and that $\dim\{v\in V\mid bv=4v\}=\dim\{v\in V\mid bv=5v\}=1$. I want to show that $\dim\{v\in V\mid…
njlieta
  • 411
2
votes
0 answers

Proof of Kac-Peterson formula

The Kac-Peterson formula arises from affine Lie algebras. My goal is to prove the formula $$(R(S))_{\Lambda,\Lambda'}=(-i)^{(d-r)/2}|L_w/L^\vee|^{-1/2}(k^\vee+g^\vee)^{-r/2}\sum_{\bar w\in\bar W} (\text{sign }\bar w)\exp\left(\dfrac{-4\pi…
wilsonw
  • 1,016
2
votes
1 answer

Tensor product of two irreducible $\mathfrak sl_2$ modules

I know that given an irreducible $\mathfrak sl_2$ module $V$, it is $V\cong V(m), m\in \mathbb Z_+$, whose basis $(v_0,\cdots,v_m$) is such that: $$H\cdot v_j = (m-2j)v_j, Y\cdot v_j = (j+1)v_{j+1}, X\cdot v_j = (m-j+1)v_{j-1}, \mbox{for } j\geq 0…
user2345678
  • 2,885
2
votes
0 answers

Lie Bracket - Lie derivative: computational experiments

Can anyone, with a simple and illustrative example, explain the meaning of such concepts as the Lie derivative and the Lie bracket and demonstrate a simple computational experiment for a function of 3-4 variables? Theoretically, everything is more…
dtn
  • 731
  • 1
  • 6
  • 19
2
votes
2 answers

Definition of trace of an adjoint representation.

What does the trace of an adjoint representation mean? I was asked to prove the following If $z \in L'$, then $\operatorname{tr}(\operatorname{ad}z)=0$. I know what a trace of a matrix is, but trace of an adjoint map is not making any sense to…
2
votes
2 answers

Regarding basis vectors of a Lie algebra.

From the book "Introduction to Lie Algebras" by Erdmann & Wildon: If $L$ is a Lie algbra over a field $F$ with basis $(x_1,\cdots, x_n)$, then $[-,-]$ is complete determined by the products $[x_i,x_j]$. We define scalars $a_{ij}^k\in F$ such that…
2
votes
0 answers

Is a stabilizer subgroup a symmetric subalgebra?

Consider the Lie algebra $\mathfrak{su}(n)$ and the set of operators that do not change the direction of the vector $\psi$, $$ K:=\{ s\in \mathfrak{su}(n)\ :\ s \psi \propto \psi \}. $$ Let $P$ be the orthogonal complement of $K$,…
Georg
  • 201
2
votes
0 answers

Applications of the PBW theorem

What are some nice corollaries or applications of the PBW theorem? There's this immediate corollary that a Lie algebra sits inside the universal enveloping algebra so in particular, the Lie algebra structure comes from an associative algebra. Why…
nobody
  • 593
2
votes
1 answer

Show $\left_{\mathbb{C}}$ is a Cartan subalgebra of $\mathfrak{sl}\left( 2,\mathbb{C}\right)$.

Using the standard basis elements $$e=\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, f=\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix},h=\begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}$$ I want to show that the span of $h$,…
2
votes
0 answers

Integer domain enveloping algebra

I must prove that if $L$ is a Lie algebra and denoting $U(L)$ the enveloping algebra, then $U(L)$ hasn't zero divisions (e.g. if $ab=0 \,\,\, a,b \in U(L)$ then $a=0$ or $b=0$). Some ideas?
2
votes
2 answers

Any Lie monomial may be written as a linear combination of "simpler" Lie monomials

Some definitions: A Lie monomial in the elemens of a set $X$ is a finite product of elements of $X$ bracketed by Lie brackets in any manner, e.g. $[[[x_3,[x_1,x_2]],x_3],[x_2,[x_1,x_1]]] $. A simpler Lie monomial is a Lie monomial bracketed…
user2345678
  • 2,885