$G$-variety maps send closed orbits to closed orbits if $G$ is linear reductive
$begingroup$
Is it true that if $G$ is a linear reductive group and $X$ and $Y$ are two $G$-varieties with a surjective $G$-map $f:Xrightarrow Y$ then $f$ sends a closed orbit to a closed orbit? If not, is there an easy condition to guarantee this property?
I tried to use the fact that we have a commutative diagram
$$
newcommand{ra}[1]{!!!!!!!!!!!!xrightarrow{quad#1quad}!!!!!!!!}
newcommand{da}[1]{leftdownarrow{scriptstyle#1}vphantom{displaystyleint_0^1}right.}
%
begin{array}{llllllllllll}
X & ra{f} &Y\
da{q_1} & & da{q_2}\
Z & ra{g} & T \
end{array}
$$
where $Z,T$ are the quotients $X/G,Y/G$ respectively but I could not figure out anything out of it.
Thanks in advance.
group-theory algebraic-geometry
asked Jan 9 at 15:53
LeventLevent
2,729925
$endgroup$
add a comment |
$begingroup$
Is it true that if $G$ is a linear reductive group and $X$ and $Y$ are two $G$-varieties with a surjective $G$-map $f:Xrightarrow Y$ then $f$ sends a closed orbit to a closed orbit? If not, is there an easy condition to guarantee this property?
I tried to use the fact that we have a commutative diagram
$$
newcommand{ra}[1]{!!!!!!!!!!!!xrightarrow{quad#1quad}!!!!!!!!}
newcommand{da}[1]{leftdownarrow{scriptstyle#1}vphantom{displaystyleint_0^1}right.}
%
begin{array}{llllllllllll}
X & ra{f} &Y\
da{q_1} & & da{q_2}\
Z & ra{g} & T \
end{array}
$$
where $Z,T$ are the quotients $X/G,Y/G$ respectively but I could not figure out anything out of it.
Thanks in advance.
group-theory algebraic-geometry
asked Jan 9 at 15:53
LeventLevent
2,729925
$endgroup$
add a comment |
$begingroup$
Is it true that if $G$ is a linear reductive group and $X$ and $Y$ are two $G$-varieties with a surjective $G$-map $f:Xrightarrow Y$ then $f$ sends a closed orbit to a closed orbit? If not, is there an easy condition to guarantee this property?
I tried to use the fact that we have a commutative diagram
$$
newcommand{ra}[1]{!!!!!!!!!!!!xrightarrow{quad#1quad}!!!!!!!!}
newcommand{da}[1]{leftdownarrow{scriptstyle#1}vphantom{displaystyleint_0^1}right.}
%
begin{array}{llllllllllll}
X & ra{f} &Y\
da{q_1} & & da{q_2}\
Z & ra{g} & T \
end{array}
$$
where $Z,T$ are the quotients $X/G,Y/G$ respectively but I could not figure out anything out of it.
Thanks in advance.
group-theory algebraic-geometry
asked Jan 9 at 15:53
LeventLevent
2,729925
$endgroup$
Is it true that if $G$ is a linear reductive group and $X$ and $Y$ are two $G$-varieties with a surjective $G$-map $f:Xrightarrow Y$ then $f$ sends a closed orbit to a closed orbit? If not, is there an easy condition to guarantee this property?
I tried to use the fact that we have a commutative diagram
$$
newcommand{ra}[1]{!!!!!!!!!!!!xrightarrow{quad#1quad}!!!!!!!!}
newcommand{da}[1]{leftdownarrow{scriptstyle#1}vphantom{displaystyleint_0^1}right.}
%
begin{array}{llllllllllll}
X & ra{f} &Y\
da{q_1} & & da{q_2}\
Z & ra{g} & T \
end{array}
$$
where $Z,T$ are the quotients $X/G,Y/G$ respectively but I could not figure out anything out of it.
Thanks in advance.
group-theory algebraic-geometry
group-theory algebraic-geometry
asked Jan 9 at 15:53
LeventLevent
2,729925
asked Jan 9 at 15:53
LeventLevent
2,729925
edited Jan 11 at 13:06
Levent
asked Jan 9 at 15:53
LeventLevent
2,729925
asked Jan 9 at 15:53
LeventLevent
2,729925
asked Jan 9 at 15:53
LeventLevent
2,729925
2,729925
add a comment |
add a comment |
1 Answer
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No. Take $G = mathbb{C}^{ast}$, $X = mathbb{C}^2$ and $Y = mathbb{C}$, with $t in G$ acting by $(x,y) mapsto (tx, t^{-1} y)$ and $z mapsto tz$. Let the map $X to Y$ be projection on the first coordinate. Then the closed orbit $xy=1$ is sent to the non-closed (in fact, open) orbit $z neq 0$.
As a more sophisticated example, to show that replacing "reductive" with "simple" doesn't help, let $G = SL_2$, let $X$ be $2 times 2$ matrices and let $Y = mathbb{C}^2$. We take $G$ to act by left multiplication in each case, and map $X to Y$ by sending a matrix to its first column. Then the closed orbit $det left[ begin{smallmatrix} u& v \ w & x end{smallmatrix} right]= 1$ is sent to the non-closed (in fact open) orbit $left[ begin{smallmatrix} u \ w end{smallmatrix} right] neq left[ begin{smallmatrix} 0 \ 0 end{smallmatrix} right]$.
answered Jan 18 at 16:48
David E SpeyerDavid E Speyer
46k4126210
$endgroup$
$begingroup$
Thank you for your answer and for your very simple examples. Are aware of a condition to guarantee that this is true? In my case, $G=SL_3$.
$endgroup$
– Levent
Jan 20 at 0:42
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
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active
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$begingroup$
No. Take $G = mathbb{C}^{ast}$, $X = mathbb{C}^2$ and $Y = mathbb{C}$, with $t in G$ acting by $(x,y) mapsto (tx, t^{-1} y)$ and $z mapsto tz$. Let the map $X to Y$ be projection on the first coordinate. Then the closed orbit $xy=1$ is sent to the non-closed (in fact, open) orbit $z neq 0$.
As a more sophisticated example, to show that replacing "reductive" with "simple" doesn't help, let $G = SL_2$, let $X$ be $2 times 2$ matrices and let $Y = mathbb{C}^2$. We take $G$ to act by left multiplication in each case, and map $X to Y$ by sending a matrix to its first column. Then the closed orbit $det left[ begin{smallmatrix} u& v \ w & x end{smallmatrix} right]= 1$ is sent to the non-closed (in fact open) orbit $left[ begin{smallmatrix} u \ w end{smallmatrix} right] neq left[ begin{smallmatrix} 0 \ 0 end{smallmatrix} right]$.
answered Jan 18 at 16:48
David E SpeyerDavid E Speyer
46k4126210
$endgroup$
$begingroup$
Thank you for your answer and for your very simple examples. Are aware of a condition to guarantee that this is true? In my case, $G=SL_3$.
$endgroup$
– Levent
Jan 20 at 0:42
add a comment |
$begingroup$
No. Take $G = mathbb{C}^{ast}$, $X = mathbb{C}^2$ and $Y = mathbb{C}$, with $t in G$ acting by $(x,y) mapsto (tx, t^{-1} y)$ and $z mapsto tz$. Let the map $X to Y$ be projection on the first coordinate. Then the closed orbit $xy=1$ is sent to the non-closed (in fact, open) orbit $z neq 0$.
As a more sophisticated example, to show that replacing "reductive" with "simple" doesn't help, let $G = SL_2$, let $X$ be $2 times 2$ matrices and let $Y = mathbb{C}^2$. We take $G$ to act by left multiplication in each case, and map $X to Y$ by sending a matrix to its first column. Then the closed orbit $det left[ begin{smallmatrix} u& v \ w & x end{smallmatrix} right]= 1$ is sent to the non-closed (in fact open) orbit $left[ begin{smallmatrix} u \ w end{smallmatrix} right] neq left[ begin{smallmatrix} 0 \ 0 end{smallmatrix} right]$.
answered Jan 18 at 16:48
David E SpeyerDavid E Speyer
46k4126210
$endgroup$
$begingroup$
Thank you for your answer and for your very simple examples. Are aware of a condition to guarantee that this is true? In my case, $G=SL_3$.
$endgroup$
– Levent
Jan 20 at 0:42
add a comment |
$begingroup$
No. Take $G = mathbb{C}^{ast}$, $X = mathbb{C}^2$ and $Y = mathbb{C}$, with $t in G$ acting by $(x,y) mapsto (tx, t^{-1} y)$ and $z mapsto tz$. Let the map $X to Y$ be projection on the first coordinate. Then the closed orbit $xy=1$ is sent to the non-closed (in fact, open) orbit $z neq 0$.
As a more sophisticated example, to show that replacing "reductive" with "simple" doesn't help, let $G = SL_2$, let $X$ be $2 times 2$ matrices and let $Y = mathbb{C}^2$. We take $G$ to act by left multiplication in each case, and map $X to Y$ by sending a matrix to its first column. Then the closed orbit $det left[ begin{smallmatrix} u& v \ w & x end{smallmatrix} right]= 1$ is sent to the non-closed (in fact open) orbit $left[ begin{smallmatrix} u \ w end{smallmatrix} right] neq left[ begin{smallmatrix} 0 \ 0 end{smallmatrix} right]$.
answered Jan 18 at 16:48
David E SpeyerDavid E Speyer
46k4126210
$endgroup$
No. Take $G = mathbb{C}^{ast}$, $X = mathbb{C}^2$ and $Y = mathbb{C}$, with $t in G$ acting by $(x,y) mapsto (tx, t^{-1} y)$ and $z mapsto tz$. Let the map $X to Y$ be projection on the first coordinate. Then the closed orbit $xy=1$ is sent to the non-closed (in fact, open) orbit $z neq 0$.
As a more sophisticated example, to show that replacing "reductive" with "simple" doesn't help, let $G = SL_2$, let $X$ be $2 times 2$ matrices and let $Y = mathbb{C}^2$. We take $G$ to act by left multiplication in each case, and map $X to Y$ by sending a matrix to its first column. Then the closed orbit $det left[ begin{smallmatrix} u& v \ w & x end{smallmatrix} right]= 1$ is sent to the non-closed (in fact open) orbit $left[ begin{smallmatrix} u \ w end{smallmatrix} right] neq left[ begin{smallmatrix} 0 \ 0 end{smallmatrix} right]$.
answered Jan 18 at 16:48
David E SpeyerDavid E Speyer
46k4126210
edited Jan 18 at 16:54
answered Jan 18 at 16:48
David E SpeyerDavid E Speyer
46k4126210
answered Jan 18 at 16:48
David E SpeyerDavid E Speyer
46k4126210
answered Jan 18 at 16:48
David E SpeyerDavid E Speyer
46k4126210
46k4126210
$begingroup$
Thank you for your answer and for your very simple examples. Are aware of a condition to guarantee that this is true? In my case, $G=SL_3$.
$endgroup$
– Levent
Jan 20 at 0:42
add a comment |
$begingroup$
Thank you for your answer and for your very simple examples. Are aware of a condition to guarantee that this is true? In my case, $G=SL_3$.
$endgroup$
– Levent
Jan 20 at 0:42
$begingroup$
Thank you for your answer and for your very simple examples. Are aware of a condition to guarantee that this is true? In my case, $G=SL_3$.
$endgroup$
– Levent
Jan 20 at 0:42
$begingroup$
Thank you for your answer and for your very simple examples. Are aware of a condition to guarantee that this is true? In my case, $G=SL_3$.
$endgroup$
– Levent
Jan 20 at 0:42
add a comment |
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