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typografical correction
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Salvatore Siciliano
Salvatore Siciliano
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Of course, the easiest case is when g$\mathfrak{g}$ has nilpotence class 2 (that is, $[[\mathfrak{g},\mathfrak{g}],\mathfrak{g}]=0$). Under this assumption one has trivially that $im(ad_{\xi})⊆Z(\mathfrak{g})$, and $im(ad_{\xi}$) contains a 1-dimensional ideal whenever $\xi$ is not in $Z(\mathfrak{g})$.

Of course, the easiest case is when g has nilpotence class 2 (that is, $[[\mathfrak{g},\mathfrak{g}],\mathfrak{g}]=0$). Under this assumption one has trivially that $im(ad_{\xi})⊆Z(\mathfrak{g})$, and $im(ad_{\xi}$) contains a 1-dimensional ideal whenever $\xi$ is not in $Z(\mathfrak{g})$.

Of course, the easiest case is when $\mathfrak{g}$ has nilpotence class 2 (that is, $[[\mathfrak{g},\mathfrak{g}],\mathfrak{g}]=0$). Under this assumption one has trivially that $im(ad_{\xi})⊆Z(\mathfrak{g})$, and $im(ad_{\xi}$) contains a 1-dimensional ideal whenever $\xi$ is not in $Z(\mathfrak{g})$.

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Salvatore Siciliano
Salvatore Siciliano
  • 5.9k
  • 26
  • 41

Of course, the easiest case is when g has nilpotence class 2 (that is, $[[\mathfrak{g},\mathfrak{g}],\mathfrak{g}]=0$). Under this assumption one has trivially that $im(ad_{\xi})⊆Z(\mathfrak{g})$, and $im(ad_{\xi}$) contains a 1-dimensional ideal whenever $\xi$ is not in $Z(\mathfrak{g})$.