Skip to main content
MathOverflow

Return to Answer

deleted 3 characters in body
Source Link
Alexander Premet
Alexander Premet
  • 2.6k
  • 21
  • 18

To each unipotent element $u\in \frak{g}$$u\in G$ one assigns its weighted Dynkin diagram which is basically a map $\Delta\colon\, \Pi\rightarrow \{0,1,2\}$ where $\Pi$ is a basis of simple roots of the root system of $G$. From $\Delta$ one can read off the characters on ${\mathfrak g}={\rm Lie}(G)$ of a maximal torus $T_0$ of an ${\rm SL}_2$ subgroup containing $u$ (all such subgroups are $G$-conjugate in characteristic $0$ by a classical result of Kostant, as David Stewart pointed out in his comment). If $\Delta(\alpha)=1$ for some $\alpha\in \Pi$ then $\mathfrak g$ has a faithful composition factor for the adjoint action of ${\rm SL}_2$. If $\Delta(\Pi)\subseteq \{0,2\}$ then all weights of $T_0$ on $\mathfrak g$ are even which implies that $-1\in{\rm SL}_2$ acts trivially on $\mathfrak g$. So the case of ${\rm PSL}_2$ occurs if and only if $\Delta(\Pi)\subseteq \{0,2\}$. In this case $u$ is sometimes referred to as even. All one has to do now (for $G$ exceptional) is to open Carter's book on groups of Lie type and examine the list of Dynkin labels on pp. 401--406. For $G$ classical it is explained in Carter's book how to construct $\Delta$ from the partition associated with $u$. So it fairly straightforward to figure out which unipotent elements are even for $G$ classical. Although any even unipotent element of $G$ has to be Richardson, there are plenty of them around. For example, all distinguished unipotent elements of $G$ are even.

To each unipotent element $u\in \frak{g}$ one assigns its weighted Dynkin diagram which is basically a map $\Delta\colon\, \Pi\rightarrow \{0,1,2\}$ where $\Pi$ is a basis of simple roots of the root system of $G$. From $\Delta$ one can read off the characters on ${\mathfrak g}={\rm Lie}(G)$ of a maximal torus $T_0$ of an ${\rm SL}_2$ subgroup containing $u$ (all such subgroups are $G$-conjugate in characteristic $0$ by a classical result of Kostant, as David Stewart pointed out in his comment). If $\Delta(\alpha)=1$ for some $\alpha\in \Pi$ then $\mathfrak g$ has a faithful composition factor for the adjoint action of ${\rm SL}_2$. If $\Delta(\Pi)\subseteq \{0,2\}$ then all weights of $T_0$ on $\mathfrak g$ are even which implies that $-1\in{\rm SL}_2$ acts trivially on $\mathfrak g$. So the case of ${\rm PSL}_2$ occurs if and only if $\Delta(\Pi)\subseteq \{0,2\}$. In this case $u$ is sometimes referred to as even. All one has to do now (for $G$ exceptional) is to open Carter's book on groups of Lie type and examine the list of Dynkin labels on pp. 401--406. For $G$ classical it is explained in Carter's book how to construct $\Delta$ from the partition associated with $u$. So it fairly straightforward to figure out which unipotent elements are even for $G$ classical. Although any even unipotent element of $G$ has to be Richardson, there are plenty of them around. For example, all distinguished unipotent elements of are even.

To each unipotent element $u\in G$ one assigns its weighted Dynkin diagram which is basically a map $\Delta\colon\, \Pi\rightarrow \{0,1,2\}$ where $\Pi$ is a basis of simple roots of the root system of $G$. From $\Delta$ one can read off the characters on ${\mathfrak g}={\rm Lie}(G)$ of a maximal torus $T_0$ of an ${\rm SL}_2$ subgroup containing $u$ (all such subgroups are $G$-conjugate in characteristic $0$ by a classical result of Kostant, as David Stewart pointed out in his comment). If $\Delta(\alpha)=1$ for some $\alpha\in \Pi$ then $\mathfrak g$ has a faithful composition factor for the adjoint action of ${\rm SL}_2$. If $\Delta(\Pi)\subseteq \{0,2\}$ then all weights of $T_0$ on $\mathfrak g$ are even which implies that $-1\in{\rm SL}_2$ acts trivially on $\mathfrak g$. So the case of ${\rm PSL}_2$ occurs if and only if $\Delta(\Pi)\subseteq \{0,2\}$. In this case $u$ is sometimes referred to as even. All one has to do now (for $G$ exceptional) is to open Carter's book on groups of Lie type and examine the list of Dynkin labels on pp. 401--406. For $G$ classical it is explained in Carter's book how to construct $\Delta$ from the partition associated with $u$. So it fairly straightforward to figure out which unipotent elements are even for $G$ classical. Although any even unipotent element of $G$ has to be Richardson, there are plenty of them around. For example, all distinguished unipotent elements of $G$ are even.

Source Link
Alexander Premet
Alexander Premet
  • 2.6k
  • 21
  • 18

To each unipotent element $u\in \frak{g}$ one assigns its weighted Dynkin diagram which is basically a map $\Delta\colon\, \Pi\rightarrow \{0,1,2\}$ where $\Pi$ is a basis of simple roots of the root system of $G$. From $\Delta$ one can read off the characters on ${\mathfrak g}={\rm Lie}(G)$ of a maximal torus $T_0$ of an ${\rm SL}_2$ subgroup containing $u$ (all such subgroups are $G$-conjugate in characteristic $0$ by a classical result of Kostant, as David Stewart pointed out in his comment). If $\Delta(\alpha)=1$ for some $\alpha\in \Pi$ then $\mathfrak g$ has a faithful composition factor for the adjoint action of ${\rm SL}_2$. If $\Delta(\Pi)\subseteq \{0,2\}$ then all weights of $T_0$ on $\mathfrak g$ are even which implies that $-1\in{\rm SL}_2$ acts trivially on $\mathfrak g$. So the case of ${\rm PSL}_2$ occurs if and only if $\Delta(\Pi)\subseteq \{0,2\}$. In this case $u$ is sometimes referred to as even. All one has to do now (for $G$ exceptional) is to open Carter's book on groups of Lie type and examine the list of Dynkin labels on pp. 401--406. For $G$ classical it is explained in Carter's book how to construct $\Delta$ from the partition associated with $u$. So it fairly straightforward to figure out which unipotent elements are even for $G$ classical. Although any even unipotent element of $G$ has to be Richardson, there are plenty of them around. For example, all distinguished unipotent elements of are even.