Action of torus on an affine toric variety
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Let $V$ be an affine toric variety (where the base field is $mathbb C$), i.e., $V$ is an irreducible affine variety containing a torus $T$ as a Zariski open subset and $T$ has an action on $V$.
The action if $T$ on $V$ is given by a morphism $Ttimes Vrightarrow V$. Let $finmathbb C[V]$, a regular function on $V$. For $tin T$, we define $tcdot f$ which is given by $pmapsto f(t^-1cdot p)$.
My question is why $tcdot finmathbb C[V]$.
This is from the book 'Toric varieties' by Cox, Little, Schenck; page 19, Theorem 1.1.17
Thank you.
algebraic-geometry commutative-algebra affine-varieties
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up vote
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Let $V$ be an affine toric variety (where the base field is $mathbb C$), i.e., $V$ is an irreducible affine variety containing a torus $T$ as a Zariski open subset and $T$ has an action on $V$.
The action if $T$ on $V$ is given by a morphism $Ttimes Vrightarrow V$. Let $finmathbb C[V]$, a regular function on $V$. For $tin T$, we define $tcdot f$ which is given by $pmapsto f(t^-1cdot p)$.
My question is why $tcdot finmathbb C[V]$.
This is from the book 'Toric varieties' by Cox, Little, Schenck; page 19, Theorem 1.1.17
Thank you.
algebraic-geometry commutative-algebra affine-varieties
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $V$ be an affine toric variety (where the base field is $mathbb C$), i.e., $V$ is an irreducible affine variety containing a torus $T$ as a Zariski open subset and $T$ has an action on $V$.
The action if $T$ on $V$ is given by a morphism $Ttimes Vrightarrow V$. Let $finmathbb C[V]$, a regular function on $V$. For $tin T$, we define $tcdot f$ which is given by $pmapsto f(t^-1cdot p)$.
My question is why $tcdot finmathbb C[V]$.
This is from the book 'Toric varieties' by Cox, Little, Schenck; page 19, Theorem 1.1.17
Thank you.
algebraic-geometry commutative-algebra affine-varieties
Let $V$ be an affine toric variety (where the base field is $mathbb C$), i.e., $V$ is an irreducible affine variety containing a torus $T$ as a Zariski open subset and $T$ has an action on $V$.
The action if $T$ on $V$ is given by a morphism $Ttimes Vrightarrow V$. Let $finmathbb C[V]$, a regular function on $V$. For $tin T$, we define $tcdot f$ which is given by $pmapsto f(t^-1cdot p)$.
My question is why $tcdot finmathbb C[V]$.
This is from the book 'Toric varieties' by Cox, Little, Schenck; page 19, Theorem 1.1.17
Thank you.
algebraic-geometry commutative-algebra affine-varieties
asked Jul 16 at 6:29


2015
1,2881521
1,2881521
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1 Answer
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Recall that a function in $Bbb C[V]$ is the same as a regular map $Vto Bbb A^1_Bbb C$, and that the composition of regular maps is a regular map.
The map $i:Tto T$ given by $tmapsto t^-1$ is a regular map, and the map $m:Ttimes V$ given by $(t,p)mapsto tcdot p$ is a regular map. Thus $Vstackrelt_0times idto Ttimes V stackrelitimes idto Ttimes V stackrelmto V$ for a fixed $t_0$ is given by $vmapsto t_0^-1cdot v$. Further, post-composing with $f: Vto Bbb A^1_Bbb C$ gives us that the composite map is $pmapsto f(t_0^-1cdot p)$ and since this is a composition of regular maps, it is a regular map from $Vto Bbb A^1_Bbb C$ and thus an element of $Bbb C[V]$.
I understand your argument and I will accept the answer. I have one additional question : if $f$ is given by some polynomial then can we say anything about the polynomial of $t_0cdot f$?
– 2015
Jul 16 at 6:55
I can't think of anything simpler than "translate the input by $t_0^-1$". The degree will remain the same, the roots of $f$ will be the roots of $f$ translated, etc. It's analogous to taking a polynomial $f(x)=a_nx^n+cdots+a_1x+a_0$ and asking what happens when we shift by $x=x+1$.
– KReiser
Jul 16 at 7:08
Thank you for your answer.
– 2015
Jul 16 at 7:09
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Recall that a function in $Bbb C[V]$ is the same as a regular map $Vto Bbb A^1_Bbb C$, and that the composition of regular maps is a regular map.
The map $i:Tto T$ given by $tmapsto t^-1$ is a regular map, and the map $m:Ttimes V$ given by $(t,p)mapsto tcdot p$ is a regular map. Thus $Vstackrelt_0times idto Ttimes V stackrelitimes idto Ttimes V stackrelmto V$ for a fixed $t_0$ is given by $vmapsto t_0^-1cdot v$. Further, post-composing with $f: Vto Bbb A^1_Bbb C$ gives us that the composite map is $pmapsto f(t_0^-1cdot p)$ and since this is a composition of regular maps, it is a regular map from $Vto Bbb A^1_Bbb C$ and thus an element of $Bbb C[V]$.
I understand your argument and I will accept the answer. I have one additional question : if $f$ is given by some polynomial then can we say anything about the polynomial of $t_0cdot f$?
– 2015
Jul 16 at 6:55
I can't think of anything simpler than "translate the input by $t_0^-1$". The degree will remain the same, the roots of $f$ will be the roots of $f$ translated, etc. It's analogous to taking a polynomial $f(x)=a_nx^n+cdots+a_1x+a_0$ and asking what happens when we shift by $x=x+1$.
– KReiser
Jul 16 at 7:08
Thank you for your answer.
– 2015
Jul 16 at 7:09
add a comment |Â
up vote
4
down vote
accepted
Recall that a function in $Bbb C[V]$ is the same as a regular map $Vto Bbb A^1_Bbb C$, and that the composition of regular maps is a regular map.
The map $i:Tto T$ given by $tmapsto t^-1$ is a regular map, and the map $m:Ttimes V$ given by $(t,p)mapsto tcdot p$ is a regular map. Thus $Vstackrelt_0times idto Ttimes V stackrelitimes idto Ttimes V stackrelmto V$ for a fixed $t_0$ is given by $vmapsto t_0^-1cdot v$. Further, post-composing with $f: Vto Bbb A^1_Bbb C$ gives us that the composite map is $pmapsto f(t_0^-1cdot p)$ and since this is a composition of regular maps, it is a regular map from $Vto Bbb A^1_Bbb C$ and thus an element of $Bbb C[V]$.
I understand your argument and I will accept the answer. I have one additional question : if $f$ is given by some polynomial then can we say anything about the polynomial of $t_0cdot f$?
– 2015
Jul 16 at 6:55
I can't think of anything simpler than "translate the input by $t_0^-1$". The degree will remain the same, the roots of $f$ will be the roots of $f$ translated, etc. It's analogous to taking a polynomial $f(x)=a_nx^n+cdots+a_1x+a_0$ and asking what happens when we shift by $x=x+1$.
– KReiser
Jul 16 at 7:08
Thank you for your answer.
– 2015
Jul 16 at 7:09
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Recall that a function in $Bbb C[V]$ is the same as a regular map $Vto Bbb A^1_Bbb C$, and that the composition of regular maps is a regular map.
The map $i:Tto T$ given by $tmapsto t^-1$ is a regular map, and the map $m:Ttimes V$ given by $(t,p)mapsto tcdot p$ is a regular map. Thus $Vstackrelt_0times idto Ttimes V stackrelitimes idto Ttimes V stackrelmto V$ for a fixed $t_0$ is given by $vmapsto t_0^-1cdot v$. Further, post-composing with $f: Vto Bbb A^1_Bbb C$ gives us that the composite map is $pmapsto f(t_0^-1cdot p)$ and since this is a composition of regular maps, it is a regular map from $Vto Bbb A^1_Bbb C$ and thus an element of $Bbb C[V]$.
Recall that a function in $Bbb C[V]$ is the same as a regular map $Vto Bbb A^1_Bbb C$, and that the composition of regular maps is a regular map.
The map $i:Tto T$ given by $tmapsto t^-1$ is a regular map, and the map $m:Ttimes V$ given by $(t,p)mapsto tcdot p$ is a regular map. Thus $Vstackrelt_0times idto Ttimes V stackrelitimes idto Ttimes V stackrelmto V$ for a fixed $t_0$ is given by $vmapsto t_0^-1cdot v$. Further, post-composing with $f: Vto Bbb A^1_Bbb C$ gives us that the composite map is $pmapsto f(t_0^-1cdot p)$ and since this is a composition of regular maps, it is a regular map from $Vto Bbb A^1_Bbb C$ and thus an element of $Bbb C[V]$.
answered Jul 16 at 6:43
KReiser
7,62511230
7,62511230
I understand your argument and I will accept the answer. I have one additional question : if $f$ is given by some polynomial then can we say anything about the polynomial of $t_0cdot f$?
– 2015
Jul 16 at 6:55
I can't think of anything simpler than "translate the input by $t_0^-1$". The degree will remain the same, the roots of $f$ will be the roots of $f$ translated, etc. It's analogous to taking a polynomial $f(x)=a_nx^n+cdots+a_1x+a_0$ and asking what happens when we shift by $x=x+1$.
– KReiser
Jul 16 at 7:08
Thank you for your answer.
– 2015
Jul 16 at 7:09
add a comment |Â
I understand your argument and I will accept the answer. I have one additional question : if $f$ is given by some polynomial then can we say anything about the polynomial of $t_0cdot f$?
– 2015
Jul 16 at 6:55
I can't think of anything simpler than "translate the input by $t_0^-1$". The degree will remain the same, the roots of $f$ will be the roots of $f$ translated, etc. It's analogous to taking a polynomial $f(x)=a_nx^n+cdots+a_1x+a_0$ and asking what happens when we shift by $x=x+1$.
– KReiser
Jul 16 at 7:08
Thank you for your answer.
– 2015
Jul 16 at 7:09
I understand your argument and I will accept the answer. I have one additional question : if $f$ is given by some polynomial then can we say anything about the polynomial of $t_0cdot f$?
– 2015
Jul 16 at 6:55
I understand your argument and I will accept the answer. I have one additional question : if $f$ is given by some polynomial then can we say anything about the polynomial of $t_0cdot f$?
– 2015
Jul 16 at 6:55
I can't think of anything simpler than "translate the input by $t_0^-1$". The degree will remain the same, the roots of $f$ will be the roots of $f$ translated, etc. It's analogous to taking a polynomial $f(x)=a_nx^n+cdots+a_1x+a_0$ and asking what happens when we shift by $x=x+1$.
– KReiser
Jul 16 at 7:08
I can't think of anything simpler than "translate the input by $t_0^-1$". The degree will remain the same, the roots of $f$ will be the roots of $f$ translated, etc. It's analogous to taking a polynomial $f(x)=a_nx^n+cdots+a_1x+a_0$ and asking what happens when we shift by $x=x+1$.
– KReiser
Jul 16 at 7:08
Thank you for your answer.
– 2015
Jul 16 at 7:09
Thank you for your answer.
– 2015
Jul 16 at 7:09
add a comment |Â
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