This is somewhat extension of question in why does Lie bracket of two coordinate vector fields always vanish? Now i want to understand the meaning of vanishing Lie bracket. $i.e$, For vector field $X$, $Y$ If \begin{align} [X,Y]=0 \end{align} for all $Y$ on $M$, Of course i know if $X, Y$ are coordinate basis, then […]

Let $g$ be a simple Lie algebra. Let $(g \wedge g)^g = \{a \wedge b \in g \wedge g: x.(a \wedge b) = [x,a] \wedge b + a \wedge [x,b] = 0\}$ be the set of $g$ invariants under the adjoint action. Do we have $(g \wedge g)^g = 0$? If $g$ is a semisimple […]

Let $G$ and $H$ be two Lie groups and $\rho: G \to H$ be a homomorphism. How to differentiate $\rho$ to obtain a Lie algebra homomorphism $d\rho_e: T_eG \to T_eH$?

Let $L$ be a finite dimensional semisimple Lie algebra. Let $\lambda_1$, …..,$\lambda_l$ be the fundamental dominant weights for the root system $R$ of $L$. Show how to construct an arbitrary irreducible $L$ module of highest weight $\lambda$ denoted by $V(\lambda)$ where $\lambda$ is an integral dominant weight as a direct summand in a suitable tensor […]

I know that $\langle U,V \rangle = \langle dR_{x_{t(e)}}U, dR_{x_{t(e)}}V \rangle$ and $\langle U,V \rangle = \langle dL_{x_{t(e)}}U, dL_{x_{t(e)}}V \rangle$ because it is bi-invariant. How do I proceed?

Let $L$ be a Lie algebra. why if $L$ be supersolvable then $L’=[L,L]$ (derived algebra) is nilpotent.

Fact : $\phi : L_1\rightarrow L_2$ is $surjective$ Lie algebra homomorphism. If $h\in L_1$ and ${\rm ad}_h$ is diagonalizable then ${\rm ad}_{\phi(h)}$ is diagonalizable Defn $x\in {\rm gl}\ ({\bf C}^n)$ has Jordan decomposition if $$ x= d+n $$ where $[d,n]=0$, $d$ is diagonalizable and $n$ is nilpotent. EXE (cf. 90page in Erdmann and Wildon’s book) […]

First, what I know is that given the basis: $$e = \left(\begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array}\right),f = \left(\begin{array}{cc} 0 & 0 \\ 1 & 0 \end{array}\right),h = \left(\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right)$$ I want to find the ‘structure constants’, but furthermore that the adjoint representation of $sl(2,F)$, […]

Can anyone explain why a semi-simple finite dimensional Lie algebra $\mathfrak{g}$ has to be perfect ? The natural way to prove something like that would be to look to the algebra generated by the Lie brackets, which when $\mathfrak{g}$ is not perfect would be expected to be solvable. But it doesn’t seem to work.

We can define a Lie algebra letting $\mathbb{R}$ be the vector space and also the field. We can then have $[x,y]=xy-yx=0$ for all $x,y$. Is there a one-dimensional Lie algebra such that $[x,y]$ is not identically zero?

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