Skip to main content
Link and `\operatorname` while this is on the front page
Source Link
LSpice
  • 12.9k
  • 4
  • 45
  • 69

A relevant paper of Slodowy is LNM 815LNM 815, Simple singularities and simple algebraic groups. On p29 he calculates the unipotent variety of $PGL_2$$\operatorname{PGL}_2$ in characteristic $2$ as being given by the three equations \begin{align*}X^2+YZ&=0\\ Y(X+1)&=S^2\\ Z(X+1)&=T^2\end{align*} in the five variables.

But (again in characteristic $2$) the nilpotent variety in $\mathfrak{pgl_2}$ is, I believe, just the $2$-dimensional vector space spanned by the root spaces, on which the toral subalgebra acts by derivations. So they are not isomorphic.

Yet another question is when they are isomorphic as schemes. Slodowy points out that the equations above define a non-reduced scheme. But of course the nilpotent variety of $\mathfrak{pgl}_2$ is reduced.

By contrast, I think the unipotent and nilpotent varieties of $\mathrm{SL}_2$ are isomorphic (and are both reduced). It would be reasonable to conjecture that the answer is 'yes' when $G$ is simply-connected, and 'no' when the covering map from the simply-connected cover is not smooth.

A relevant paper of Slodowy is LNM 815. On p29 he calculates the unipotent variety of $PGL_2$ in characteristic $2$ as being given by the three equations \begin{align*}X^2+YZ&=0\\ Y(X+1)&=S^2\\ Z(X+1)&=T^2\end{align*} in the five variables.

But (again in characteristic $2$) the nilpotent variety in $\mathfrak{pgl_2}$ is, I believe, just the $2$-dimensional vector space spanned by the root spaces, on which the toral subalgebra acts by derivations. So they are not isomorphic.

Yet another question is when they are isomorphic as schemes. Slodowy points out that the equations above define a non-reduced scheme. But of course the nilpotent variety of $\mathfrak{pgl}_2$ is reduced.

By contrast, I think the unipotent and nilpotent varieties of $\mathrm{SL}_2$ are isomorphic (and are both reduced). It would be reasonable to conjecture that the answer is 'yes' when $G$ is simply-connected, and 'no' when the covering map from the simply-connected cover is not smooth.

A relevant paper of Slodowy is LNM 815, Simple singularities and simple algebraic groups. On p29 he calculates the unipotent variety of $\operatorname{PGL}_2$ in characteristic $2$ as being given by the three equations \begin{align*}X^2+YZ&=0\\ Y(X+1)&=S^2\\ Z(X+1)&=T^2\end{align*} in the five variables.

But (again in characteristic $2$) the nilpotent variety in $\mathfrak{pgl_2}$ is, I believe, just the $2$-dimensional vector space spanned by the root spaces, on which the toral subalgebra acts by derivations. So they are not isomorphic.

Yet another question is when they are isomorphic as schemes. Slodowy points out that the equations above define a non-reduced scheme. But of course the nilpotent variety of $\mathfrak{pgl}_2$ is reduced.

By contrast, I think the unipotent and nilpotent varieties of $\mathrm{SL}_2$ are isomorphic (and are both reduced). It would be reasonable to conjecture that the answer is 'yes' when $G$ is simply-connected, and 'no' when the covering map from the simply-connected cover is not smooth.

Source Link
David Stewart
  • 631
  • 1
  • 5
  • 11

A relevant paper of Slodowy is LNM 815. On p29 he calculates the unipotent variety of $PGL_2$ in characteristic $2$ as being given by the three equations \begin{align*}X^2+YZ&=0\\ Y(X+1)&=S^2\\ Z(X+1)&=T^2\end{align*} in the five variables.

But (again in characteristic $2$) the nilpotent variety in $\mathfrak{pgl_2}$ is, I believe, just the $2$-dimensional vector space spanned by the root spaces, on which the toral subalgebra acts by derivations. So they are not isomorphic.

Yet another question is when they are isomorphic as schemes. Slodowy points out that the equations above define a non-reduced scheme. But of course the nilpotent variety of $\mathfrak{pgl}_2$ is reduced.

By contrast, I think the unipotent and nilpotent varieties of $\mathrm{SL}_2$ are isomorphic (and are both reduced). It would be reasonable to conjecture that the answer is 'yes' when $G$ is simply-connected, and 'no' when the covering map from the simply-connected cover is not smooth.