The dimension of a closed set of $mathbb{C}P^ntimesmathbb{C}P^ntimes mathbb{C}P^n$












0












$begingroup$


Let $Xsubset mathbb{C}P^ntimesmathbb{C}P^ntimes mathbb{C}P^n$ be the set of points ${(x_1,x_2,x_3)}$ s.t. some $x_i=x_j$, then why is the codimension of $X$ $n$?



Now $X=Z(x_1=x_2)cup Z(x_3=x_2)cup Z(x_1=x_3)$ and each union component has codimension $n$. Why is $X$ also with codimension $n$?






algebraic-geometry







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$endgroup$












  • $begingroup$
    Consider writing it as the union of multiple varieties you can express easily. What varieties can you write it as a union of in a simple case, like $n=1$ or $n=2$?
    $endgroup$
    – KReiser
    Dec 6 '18 at 4:08












  • $begingroup$
    @KReiser it's the union of $x_1=x_2$, $x_1=x_3$, $x_2=x_3$, each of them has codimension $1$, then why is the union still with codimension $1$?
    $endgroup$
    – 6666
    Dec 6 '18 at 4:21










  • $begingroup$
    @KReiser I mean $n=1$
    $endgroup$
    – 6666
    Dec 6 '18 at 4:31


















0












$begingroup$


Let $Xsubset mathbb{C}P^ntimesmathbb{C}P^ntimes mathbb{C}P^n$ be the set of points ${(x_1,x_2,x_3)}$ s.t. some $x_i=x_j$, then why is the codimension of $X$ $n$?



Now $X=Z(x_1=x_2)cup Z(x_3=x_2)cup Z(x_1=x_3)$ and each union component has codimension $n$. Why is $X$ also with codimension $n$?






algebraic-geometry







share|cite|improve this question











$endgroup$












  • $begingroup$
    Consider writing it as the union of multiple varieties you can express easily. What varieties can you write it as a union of in a simple case, like $n=1$ or $n=2$?
    $endgroup$
    – KReiser
    Dec 6 '18 at 4:08












  • $begingroup$
    @KReiser it's the union of $x_1=x_2$, $x_1=x_3$, $x_2=x_3$, each of them has codimension $1$, then why is the union still with codimension $1$?
    $endgroup$
    – 6666
    Dec 6 '18 at 4:21










  • $begingroup$
    @KReiser I mean $n=1$
    $endgroup$
    – 6666
    Dec 6 '18 at 4:31
















0












0








0





$begingroup$


Let $Xsubset mathbb{C}P^ntimesmathbb{C}P^ntimes mathbb{C}P^n$ be the set of points ${(x_1,x_2,x_3)}$ s.t. some $x_i=x_j$, then why is the codimension of $X$ $n$?



Now $X=Z(x_1=x_2)cup Z(x_3=x_2)cup Z(x_1=x_3)$ and each union component has codimension $n$. Why is $X$ also with codimension $n$?






algebraic-geometry







share|cite|improve this question











$endgroup$




Let $Xsubset mathbb{C}P^ntimesmathbb{C}P^ntimes mathbb{C}P^n$ be the set of points ${(x_1,x_2,x_3)}$ s.t. some $x_i=x_j$, then why is the codimension of $X$ $n$?



Now $X=Z(x_1=x_2)cup Z(x_3=x_2)cup Z(x_1=x_3)$ and each union component has codimension $n$. Why is $X$ also with codimension $n$?






algebraic-geometry




algebraic-geometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 7 '18 at 5:35







6666

















asked Dec 6 '18 at 3:56









66666666

1,294620




1,294620












  • $begingroup$
    Consider writing it as the union of multiple varieties you can express easily. What varieties can you write it as a union of in a simple case, like $n=1$ or $n=2$?
    $endgroup$
    – KReiser
    Dec 6 '18 at 4:08












  • $begingroup$
    @KReiser it's the union of $x_1=x_2$, $x_1=x_3$, $x_2=x_3$, each of them has codimension $1$, then why is the union still with codimension $1$?
    $endgroup$
    – 6666
    Dec 6 '18 at 4:21










  • $begingroup$
    @KReiser I mean $n=1$
    $endgroup$
    – 6666
    Dec 6 '18 at 4:31




















  • $begingroup$
    Consider writing it as the union of multiple varieties you can express easily. What varieties can you write it as a union of in a simple case, like $n=1$ or $n=2$?
    $endgroup$
    – KReiser
    Dec 6 '18 at 4:08












  • $begingroup$
    @KReiser it's the union of $x_1=x_2$, $x_1=x_3$, $x_2=x_3$, each of them has codimension $1$, then why is the union still with codimension $1$?
    $endgroup$
    – 6666
    Dec 6 '18 at 4:21










  • $begingroup$
    @KReiser I mean $n=1$
    $endgroup$
    – 6666
    Dec 6 '18 at 4:31


















$begingroup$
Consider writing it as the union of multiple varieties you can express easily. What varieties can you write it as a union of in a simple case, like $n=1$ or $n=2$?
$endgroup$
– KReiser
Dec 6 '18 at 4:08






$begingroup$
Consider writing it as the union of multiple varieties you can express easily. What varieties can you write it as a union of in a simple case, like $n=1$ or $n=2$?
$endgroup$
– KReiser
Dec 6 '18 at 4:08














$begingroup$
@KReiser it's the union of $x_1=x_2$, $x_1=x_3$, $x_2=x_3$, each of them has codimension $1$, then why is the union still with codimension $1$?
$endgroup$
– 6666
Dec 6 '18 at 4:21




$begingroup$
@KReiser it's the union of $x_1=x_2$, $x_1=x_3$, $x_2=x_3$, each of them has codimension $1$, then why is the union still with codimension $1$?
$endgroup$
– 6666
Dec 6 '18 at 4:21












$begingroup$
@KReiser I mean $n=1$
$endgroup$
– 6666
Dec 6 '18 at 4:31






$begingroup$
@KReiser I mean $n=1$
$endgroup$
– 6666
Dec 6 '18 at 4:31












1 Answer
1






active

oldest

votes


















1












$begingroup$

Maybe there is confusion that the $x_i$ are like coordinates? But they are not, each $x_i$ is rather a line which itself has $n+1$ homogeneous coordinates.



So, lets change notation from $(x_1,x_2,x_3)$ to $(x,y,z)$. Then $(x,y,z) in mathbb CP^ntimesmathbb C P^n times mathbb C P^n$, and each variable can be written in homogeneous coordinates like so: $$x = [x_0:cdots:x_n]$$
$$y = [y_0:cdots:y_n]$$
$$z = [z_0:cdots:z_n]$$



Setting for example $x = y$ means that $[x_0:cdots:x_n] = [y_0:cdots:y_n]$. When $x_0 = y_0 = 1$, this leaves us with $n$ separate equations to impose, $x_1=y_1,x_2=y_2,ldots x_n=y_n$. Therefore $Z(x=y)$ is of codimension $n$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Sorry, I feel confused. I agree $Z(x=y)$ is of codimension $n$, my question is why $Z(x=y)cup Z(z=y)cup Z(z=x)$ has codimension $n$?
    $endgroup$
    – 6666
    Dec 7 '18 at 6:37






  • 1




    $begingroup$
    The dimension of $X$ is by definition the maximum of the dimension of its components, in this case they are all the same. So the codimension is the same as any of its components too.
    $endgroup$
    – Ben
    Dec 7 '18 at 6:40











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Maybe there is confusion that the $x_i$ are like coordinates? But they are not, each $x_i$ is rather a line which itself has $n+1$ homogeneous coordinates.



So, lets change notation from $(x_1,x_2,x_3)$ to $(x,y,z)$. Then $(x,y,z) in mathbb CP^ntimesmathbb C P^n times mathbb C P^n$, and each variable can be written in homogeneous coordinates like so: $$x = [x_0:cdots:x_n]$$
$$y = [y_0:cdots:y_n]$$
$$z = [z_0:cdots:z_n]$$



Setting for example $x = y$ means that $[x_0:cdots:x_n] = [y_0:cdots:y_n]$. When $x_0 = y_0 = 1$, this leaves us with $n$ separate equations to impose, $x_1=y_1,x_2=y_2,ldots x_n=y_n$. Therefore $Z(x=y)$ is of codimension $n$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Sorry, I feel confused. I agree $Z(x=y)$ is of codimension $n$, my question is why $Z(x=y)cup Z(z=y)cup Z(z=x)$ has codimension $n$?
    $endgroup$
    – 6666
    Dec 7 '18 at 6:37






  • 1




    $begingroup$
    The dimension of $X$ is by definition the maximum of the dimension of its components, in this case they are all the same. So the codimension is the same as any of its components too.
    $endgroup$
    – Ben
    Dec 7 '18 at 6:40
















1












$begingroup$

Maybe there is confusion that the $x_i$ are like coordinates? But they are not, each $x_i$ is rather a line which itself has $n+1$ homogeneous coordinates.



So, lets change notation from $(x_1,x_2,x_3)$ to $(x,y,z)$. Then $(x,y,z) in mathbb CP^ntimesmathbb C P^n times mathbb C P^n$, and each variable can be written in homogeneous coordinates like so: $$x = [x_0:cdots:x_n]$$
$$y = [y_0:cdots:y_n]$$
$$z = [z_0:cdots:z_n]$$



Setting for example $x = y$ means that $[x_0:cdots:x_n] = [y_0:cdots:y_n]$. When $x_0 = y_0 = 1$, this leaves us with $n$ separate equations to impose, $x_1=y_1,x_2=y_2,ldots x_n=y_n$. Therefore $Z(x=y)$ is of codimension $n$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Sorry, I feel confused. I agree $Z(x=y)$ is of codimension $n$, my question is why $Z(x=y)cup Z(z=y)cup Z(z=x)$ has codimension $n$?
    $endgroup$
    – 6666
    Dec 7 '18 at 6:37






  • 1




    $begingroup$
    The dimension of $X$ is by definition the maximum of the dimension of its components, in this case they are all the same. So the codimension is the same as any of its components too.
    $endgroup$
    – Ben
    Dec 7 '18 at 6:40














1












1








1





$begingroup$

Maybe there is confusion that the $x_i$ are like coordinates? But they are not, each $x_i$ is rather a line which itself has $n+1$ homogeneous coordinates.



So, lets change notation from $(x_1,x_2,x_3)$ to $(x,y,z)$. Then $(x,y,z) in mathbb CP^ntimesmathbb C P^n times mathbb C P^n$, and each variable can be written in homogeneous coordinates like so: $$x = [x_0:cdots:x_n]$$
$$y = [y_0:cdots:y_n]$$
$$z = [z_0:cdots:z_n]$$



Setting for example $x = y$ means that $[x_0:cdots:x_n] = [y_0:cdots:y_n]$. When $x_0 = y_0 = 1$, this leaves us with $n$ separate equations to impose, $x_1=y_1,x_2=y_2,ldots x_n=y_n$. Therefore $Z(x=y)$ is of codimension $n$.






share|cite|improve this answer









$endgroup$



Maybe there is confusion that the $x_i$ are like coordinates? But they are not, each $x_i$ is rather a line which itself has $n+1$ homogeneous coordinates.



So, lets change notation from $(x_1,x_2,x_3)$ to $(x,y,z)$. Then $(x,y,z) in mathbb CP^ntimesmathbb C P^n times mathbb C P^n$, and each variable can be written in homogeneous coordinates like so: $$x = [x_0:cdots:x_n]$$
$$y = [y_0:cdots:y_n]$$
$$z = [z_0:cdots:z_n]$$



Setting for example $x = y$ means that $[x_0:cdots:x_n] = [y_0:cdots:y_n]$. When $x_0 = y_0 = 1$, this leaves us with $n$ separate equations to impose, $x_1=y_1,x_2=y_2,ldots x_n=y_n$. Therefore $Z(x=y)$ is of codimension $n$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 7 '18 at 6:28









BenBen

3,058616




3,058616












  • $begingroup$
    Sorry, I feel confused. I agree $Z(x=y)$ is of codimension $n$, my question is why $Z(x=y)cup Z(z=y)cup Z(z=x)$ has codimension $n$?
    $endgroup$
    – 6666
    Dec 7 '18 at 6:37






  • 1




    $begingroup$
    The dimension of $X$ is by definition the maximum of the dimension of its components, in this case they are all the same. So the codimension is the same as any of its components too.
    $endgroup$
    – Ben
    Dec 7 '18 at 6:40


















  • $begingroup$
    Sorry, I feel confused. I agree $Z(x=y)$ is of codimension $n$, my question is why $Z(x=y)cup Z(z=y)cup Z(z=x)$ has codimension $n$?
    $endgroup$
    – 6666
    Dec 7 '18 at 6:37






  • 1




    $begingroup$
    The dimension of $X$ is by definition the maximum of the dimension of its components, in this case they are all the same. So the codimension is the same as any of its components too.
    $endgroup$
    – Ben
    Dec 7 '18 at 6:40
















$begingroup$
Sorry, I feel confused. I agree $Z(x=y)$ is of codimension $n$, my question is why $Z(x=y)cup Z(z=y)cup Z(z=x)$ has codimension $n$?
$endgroup$
– 6666
Dec 7 '18 at 6:37




$begingroup$
Sorry, I feel confused. I agree $Z(x=y)$ is of codimension $n$, my question is why $Z(x=y)cup Z(z=y)cup Z(z=x)$ has codimension $n$?
$endgroup$
– 6666
Dec 7 '18 at 6:37




1




1




$begingroup$
The dimension of $X$ is by definition the maximum of the dimension of its components, in this case they are all the same. So the codimension is the same as any of its components too.
$endgroup$
– Ben
Dec 7 '18 at 6:40




$begingroup$
The dimension of $X$ is by definition the maximum of the dimension of its components, in this case they are all the same. So the codimension is the same as any of its components too.
$endgroup$
– Ben
Dec 7 '18 at 6:40


















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