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## Hot answers tagged lie-algebras

3

First, I would let $\Phi$ be a base for an indecomposable root system, and choose labels for the elements of $\Phi$ according to the following algorithm: Suppose you have chosen $\Phi_k:=\{\alpha_1,\ldots,\alpha_k\}$ such that the corresponding root system $\Delta_k$ is indecomposable (when $k=1$ this is automatic). Then, there exists $\alpha_{k+1}\in\Phi$ ...

2

Here are counterexamples for the identity $(*)$, with $$A=\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix},\; B=\begin{pmatrix} 1 & 0 \\ 1 & 0 \end{pmatrix},\; C=\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix},\; D=\begin{pmatrix} d_1 & 1 \\ 0 & d_1 \end{pmatrix},\;$$ for all $d_1\in K$, where $K$ is a field of characteristic zero. ...

2

This kind of formula only works when you have the action of the lie group $G = SO(3)$ on some manifold $M$, in this case, the euclidean space $\mathbb{R}^3$. Let then $G$ act on $M$ by $(g,x) \mapsto gx$. A left-invariant vector field $V$ in $G$ induces a vector field on $M$ such that the flow in $M$ is: $${\gamma(t,x) = \exp(tV)x,}$$ where $\exp: ... 1 I don't remember the exact reference, but this question is dealt in Meinolf Geck's book An Introduction to Algebraic Geometry and Algebraic Groups. The problem is that in the Bruhat decomposition$\displaystyle G = \coprod_{w \in W} BwB$, if you write$g = b_1wb_2$for some element$g \in G$, the elements$b_1$and$b_2$are not unique in general. Thus, ... 1 Yes, consider for example the finite group$G=GL(n,q)$. The order$o(n,q)$of this group and the number$f(n,q)=(1+q)(1+q+q^2)\cdots (1+q+q^2+\cdots q^{n-1})$of complete flags in$\mathbb{F}_q^nis related by $$o(n,q)=q^{\binom{n}{2}}(q-1)^nf(n,q),$$ and the combinatorial explanation uses the Bruhat decomposition $$GL(n,q)=\bigcup_{w\in ... 1 Since \Gamma_{0,1} is contained in \Lambda^2V\subset V\otimes V, you see that \Gamma_{a,b}\subset Sym^a V\otimes Sym^b \Gamma_{0,1}\subset \otimes^a V\otimes \otimes^b(\otimes^2 V)\subset\otimes^{a+2b}V. 1 If U is in the Lie algebra of G_A, you have exp(tU)Aexp(-tU)=A. If you differentiate this, you obtain: UA-AU=0. 1 By definition, a vector space is a Lie algebra, if the Lie bracket is "closed under commutation relation", i.e., [x,y]\in L for all x,y\in L, and satisfies skew-symmetry and the Jacobi identity. This holds for finite-dimensional and infinite-dimensional vector spaces. 1 (promoting my comment to an answer) You can use the fact that L_+ and L_- are each others adjoints, IOW$$\langle L_+x|y\rangle=\langle x|L_-y\rangle$$for all x,y. Applying this to y=L_+v, x=v gives$$ \begin{aligned} \Vert L_+v\Vert^2&=\langle L_+v\mid L_+v\rangle\\ &=\langle v\mid L_-L_+ v\rangle\\ &=\langle v\mid ... 1 Sorry to resurrect such an old post... The matrix you wrote is not in\text{SO}(5)$, as it not an orthogonal matrix. Only for simply connected Lie groups can a representation of the Lie algebra be lifted to a representation of the Lie group.$\text{SO}(5)$is not simply connected. So not every representation of its Lie algebra$(\text{B}_2)$can be lifted ... 1 To each lie algebra$\mathfrak{g}$there is a unique simply connected Lie group$G$having$\mathfrak{g}$as its lie algebra, and furthermore any other Lie group$H$having lie algebra$\mathfrak{g}$is covered by the universal one$G$, in other words is a quotinet of$G$by some discrete central subgroup$K\$. (In fact, covering space theory goes on further ...

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