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3

1

$begingroup$

Suppose $G$ is a connected complex algebraic group acting on a variety $X$. Write $mathfrak{g}$ for the Lie algebra of $G$. Then both $G$ and $mathfrak{g}$ act on $mathbb{C}(X)$, the ring of rational functions on $X$.

We clearly have an inclusion $mathbb{C}(X)^{G}subseteqmathbb{C}(X)^{mathfrak{g}}$, since $mathfrak{g}$ is acting by infinitesimal translations. My question is, do we also have the opposite inclusion, i.e. do we get $mathbb{C}(X)^{G}=mathbb{C}(X)^{mathfrak{g}}$?

lie-groups lie-algebras algebraic-groups

share|cite|improve this question

edited yesterday

freeRmodule

asked Feb 26 at 17:57

freeRmodulefreeRmodule

699310

$endgroup$

This question has an open bounty worth +50
reputation from freeRmodule ending in 3 days.

This question has not received enough attention.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Are you assuming that $G$ is irreducible as a variety? Otherwise, consider the order two group $G={pm I_2}$ (the center of $SL(2, {mathbb C})$) acting on $X={mathbb C}^2$ in the natural fashion (the generator of $G$ sends $(x,y)$ to $(-x, -y)$).
  $endgroup$
  – Moishe Kohan
  yesterday
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yeah, I want to assume $G$ is connected, let me add that assumption. Thanks!
  $endgroup$
  – freeRmodule
  yesterday
  

  
  
  
  

add a comment | 

3

1

$begingroup$

Suppose $G$ is a connected complex algebraic group acting on a variety $X$. Write $mathfrak{g}$ for the Lie algebra of $G$. Then both $G$ and $mathfrak{g}$ act on $mathbb{C}(X)$, the ring of rational functions on $X$.

We clearly have an inclusion $mathbb{C}(X)^{G}subseteqmathbb{C}(X)^{mathfrak{g}}$, since $mathfrak{g}$ is acting by infinitesimal translations. My question is, do we also have the opposite inclusion, i.e. do we get $mathbb{C}(X)^{G}=mathbb{C}(X)^{mathfrak{g}}$?

lie-groups lie-algebras algebraic-groups

share|cite|improve this question

edited yesterday

freeRmodule

asked Feb 26 at 17:57

freeRmodulefreeRmodule

699310

$endgroup$

This question has an open bounty worth +50
reputation from freeRmodule ending in 3 days.

This question has not received enough attention.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Are you assuming that $G$ is irreducible as a variety? Otherwise, consider the order two group $G={pm I_2}$ (the center of $SL(2, {mathbb C})$) acting on $X={mathbb C}^2$ in the natural fashion (the generator of $G$ sends $(x,y)$ to $(-x, -y)$).
  $endgroup$
  – Moishe Kohan
  yesterday
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yeah, I want to assume $G$ is connected, let me add that assumption. Thanks!
  $endgroup$
  – freeRmodule
  yesterday
  

  
  
  
  

add a comment | 

$begingroup$

Suppose $G$ is a connected complex algebraic group acting on a variety $X$. Write $mathfrak{g}$ for the Lie algebra of $G$. Then both $G$ and $mathfrak{g}$ act on $mathbb{C}(X)$, the ring of rational functions on $X$.

We clearly have an inclusion $mathbb{C}(X)^{G}subseteqmathbb{C}(X)^{mathfrak{g}}$, since $mathfrak{g}$ is acting by infinitesimal translations. My question is, do we also have the opposite inclusion, i.e. do we get $mathbb{C}(X)^{G}=mathbb{C}(X)^{mathfrak{g}}$?

lie-groups lie-algebras algebraic-groups

share|cite|improve this question

edited yesterday

freeRmodule

asked Feb 26 at 17:57

freeRmodulefreeRmodule

699310

$endgroup$

share|cite|improve this question

edited yesterday

freeRmodule

asked Feb 26 at 17:57

freeRmodulefreeRmodule

699310

share|cite|improve this question

edited yesterday

freeRmodule

asked Feb 26 at 17:57

freeRmodulefreeRmodule

699310

asked Feb 26 at 17:57

freeRmodulefreeRmodule

699310

asked Feb 26 at 17:57

freeRmodulefreeRmodule

699310