Hot answers tagged

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 $\mathbb{F}_q^n$ is 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 ...



Only top voted, non community-wiki answers of a minimum length are eligible