The dimension of a closed set of $mathbb{C}P^ntimesmathbb{C}P^ntimes mathbb{C}P^n$
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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
asked Dec 6 '18 at 3:56
66666666
1,294620
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add a comment |
$begingroup$
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
asked Dec 6 '18 at 3:56
66666666
1,294620
$endgroup$
$begingroup$
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 $n=2$?
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– KReiser
Dec 6 '18 at 4:08
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@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
add a comment |
$begingroup$
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
asked Dec 6 '18 at 3:56
66666666
1,294620
$endgroup$
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
asked Dec 6 '18 at 3:56
66666666
1,294620
asked Dec 6 '18 at 3:56
66666666
1,294620
edited Dec 7 '18 at 5:35
6666
asked Dec 6 '18 at 3:56
66666666
1,294620
asked Dec 6 '18 at 3:56
66666666
1,294620
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
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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 $n$.
answered Dec 7 '18 at 6:28
BenBen
3,058616
$endgroup$
$begingroup$
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
add a comment |
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1 Answer
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1 Answer
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$begingroup$
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 $n$.
answered Dec 7 '18 at 6:28
BenBen
3,058616
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $n$.
answered Dec 7 '18 at 6:28
BenBen
3,058616
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $n$.
answered Dec 7 '18 at 6:28
BenBen
3,058616
$endgroup$
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 $n$.
answered Dec 7 '18 at 6:28
BenBen
3,058616
answered Dec 7 '18 at 6:28
BenBen
3,058616
answered Dec 7 '18 at 6:28
BenBen
3,058616
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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