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If $\pi:G\to H$ is a covering of Lie groups (such as $\operatorname{Spin}(2n,\Bbb C)\to\operatorname{SO}(2n,\Bbb C)$), then the derivative at $1$ of $\pi$ gives an isomorphism $$\operatorname{Lie}(G)\cong\operatorname{Lie}(H).$$ Since the root system and the Weyl group are defined intrinsically in the Lie algebra, this says that $G$ and $H$ have the same ...


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The result is false if $H=A$ so I assume $H\neq A$. Here is a proof, which works in an arbitrary nilpotent algebra $A$, that every proper subalgebra $H$ is strictly contained in its 2-sided normalizer $N(H)=\{v:vH\cup Hv\subset H\}$. (Under this generality $N(H)$ is not necessarily a subalgebra; it's contained in the left normalizer, which, in the Leibniz ...


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An irreducible representation $\phi:\mathfrak{g}_\mathbb{C} \rightarrow GL(\mathbb{C,n})$ is a Lie algebra homomorphism by definition. That is, it's linear. So, to consider your example, we see that for $x,y \in \mathfrak{so}(3)$, \begin{align} \pi(x+iy) = \pi(x)+i\pi(y) \end{align} which shows that working with \begin{align} \pi: \mathfrak{so}(3) \...



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