# Tag Info

## Hot answers tagged lie-algebras

1

I think, the answer is negative. Taking $\mathfrak{sl}_3(K)$ and $(e,f,h)$ given by $$e=\begin{pmatrix} 0 & 1 & 0 \cr 0 & 0 & 0 \cr 0 & 0 & 0\end{pmatrix},\; f=\begin{pmatrix} 0 & 0 & 0 \cr 1 & 0 & 0 \cr 0 & 0 & 0\end{pmatrix},\; h=\begin{pmatrix} 1 & 0 & 0 \cr 0 & -1 & 0 \cr 0 & 0 & 0\... 1 Okay I figured it out, turns out this is some special notation the author defines way earlier in the paper that I missed. Let G be a Lie group and let \mathfrak{g} be its Lie algebra. If S\subseteq\mathfrak{g} then we define$$ {}^GS:=\{{}^gs:=\mathrm{Ad}(g)s:g\in G,\,\text{and}\,s\in S\}  Where $\mathrm{Ad}(g)\cdot h:=ghg^{-1}$. Following the ...

1

I only know the case where $\Phi,\Phi'$ are assumed to be irreducible root systems. Recall that for irreducible systems, roots of the same length are conjugate under the Weyl group $\mathcal{W}$. (These can be found in Humphreys' Introduction to Lie Algebras and Representation Theory, Section 10.4) Since and isomorphism of roots system comes from an ...

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