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4

Let $x\in L$ and consider the adjoint operator $ad(x)$ for $x\in L$. Its eigenvectors $y\neq 0$ are given by $[x,y]=\lambda y$. Suppose that $\lambda\neq 0$. Then $\langle x,y\rangle $ is a $2$-dimensional non-abelian subalgebra, a contradiction to the assumption. Hence $\lambda=0$ and all adjoint operators have only $\lambda=0$ as eigenvalue. Hence they are ...


3

The image of the exponential map from the Lie algebra of a Lie group to the group is contained in the connected component of the identity element. In the case of $O(n)$, which is not connected, this means that not all elements of $O(n)$ are the exponential of something in the Lie algebra.


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If $G$ is a real Lie group, so in particular it is a real manifold, then its tangent spaces will be real vector spaces. In particular its Lie algebra will be a real vector space. If, on the other hand, $G$ is a complex Lie group, then its Lie algebra will be a complex vector space.


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Is this are you looking for? (using Einstein convention) $$\left[T^a,T^b\right]^i_j=(T^aT^b )_{ij}-(T^bT^a )_{ij}$$ $$=(T^a)^i_p(T^b)^p_j-(T^b)^i_q(T^a)^q_j=\epsilon_{aip}\epsilon_{bpj}-\epsilon_{biq}\epsilon_{aqj}.$$ And then whatever you need to do you probably will need to use the following identity: ...


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The idea here is that we want to find an invariant subspace, that is, a subspace $W$ of $V$ such that if $w \in W$ we have that $gw \in W$ for all $g \in S_3$. As Tobias mentioned, we want to pick a subspace where no matter how we permute the basis vectors, we get the same subspace. The subspace $W = <(1,1,1)>$ is invariant, and it's complement is a ...


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For the computation of the Killing form of the classical Lie algebras, the crucial thing is to note that a bilinear form $\rho: {\mathfrak g}\times {\mathfrak g}\to {\mathbb C}$ which is non-degenerate and associative - i.e. $\rho([X,Y],Z)=\rho(X,[Y,Z])$ - is unique up to scalar. The Killing form is associative in general and non-degenerate for semisimple ...


1

The conditions on a linear operator to be nilpotent, respectively singular, are polynomial in whatever coordinates you introduce on your space(s). Indeed, if $T$ is a linear operator on a space of dimension $N$, then $T$ is nilpotent if and only if $T^N = 0$ - and $T^N$ entries (matrix elements if you will) are polynomial in the matrix elements of $T$; and ...


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Unfortunately, I'm not very familiar with Lie brackets, so I only have a partial answer via the properties from their wikipedia page. Hopefully someone else can do the final step. Recall the Jacobi Identity $$[X,[Y,Z]]+[Z,[X,Y]]+[Y,[Z,X]]=0$$ As $[X,Y]=Y$, this gives us that $$[X,[Y,Z]]+[Z,Y]+[Y,[Z,X]]=0$$ Also, recall that the Lie bracket is ...


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This is not true, any simply-connected $n$-dimensional nilpotent Lie group is homeomorphic to $R^n$, but its Lie algebra is not always commutative. Thus is not isomorphic to the Lie algebra of the $n$-dimensional simply connected commutative Lie group $R^n$.



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