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Some basic questions regarding varieties in biprojective space (product of two $mathbbP^m$'s)
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I am just learning about product of projective spaces and I have some basic questions I would like to figure out. I will be working with $mathbbP^m times mathbbP^m$. And by bihomogeneous form I mean $F(mathbfx, mathbfy)$ such that $F(amathbfx, bmathbfy) = a^d_1b^d_2F(mathbfx, mathbfy)$ for some $d_1$ and $d_2 geq 0$. (coefficients in $mathbbC$)
Suppose I have a collection of bihomogeneous forms $S$.
1) What is the relation between the dimension of
affine variety
$$
(mathbfx, mathbfy) in mathbbC^2m+2: F = 0 (F in S)
$$
and the dimension (as a biprojective(?) variety)
$$
(mathbfx, mathbfy) in mathbbP^m times mathbbP^m: F = 0 (F in S)
$$? (My apologies for abusing the notation...)
2) Suppose I have a hyperplane
$$
V(L) = (mathbfx, mathbfy) in mathbbP^m times mathbbP^m: L(mathbfx) = 0 (F in S)
$$
where $L$ is a non-zero linear form in the $mathbfx$ variables. Does it then follow that
$$
dim (X cap V(L)) = dim X - 1
$$
as in the usual projective space?
Thank you very much!
Edit: I moved 1) of this question to mathovreflow https://mathoverflow.net/questions/307464/dimensions-of-a-vareity-and-its-affine-cone-in-biprojective-spaces. So now I am just asking for 2).
algebraic-geometry projective-geometry projective-space
asked Aug 2 at 15:40
Johnny T.
4861413
add a comment |Â
up vote
1
down vote
favorite
I am just learning about product of projective spaces and I have some basic questions I would like to figure out. I will be working with $mathbbP^m times mathbbP^m$. And by bihomogeneous form I mean $F(mathbfx, mathbfy)$ such that $F(amathbfx, bmathbfy) = a^d_1b^d_2F(mathbfx, mathbfy)$ for some $d_1$ and $d_2 geq 0$. (coefficients in $mathbbC$)
Suppose I have a collection of bihomogeneous forms $S$.
1) What is the relation between the dimension of
affine variety
$$
(mathbfx, mathbfy) in mathbbC^2m+2: F = 0 (F in S)
$$
and the dimension (as a biprojective(?) variety)
$$
(mathbfx, mathbfy) in mathbbP^m times mathbbP^m: F = 0 (F in S)
$$? (My apologies for abusing the notation...)
2) Suppose I have a hyperplane
$$
V(L) = (mathbfx, mathbfy) in mathbbP^m times mathbbP^m: L(mathbfx) = 0 (F in S)
$$
where $L$ is a non-zero linear form in the $mathbfx$ variables. Does it then follow that
$$
dim (X cap V(L)) = dim X - 1
$$
as in the usual projective space?
Thank you very much!
Edit: I moved 1) of this question to mathovreflow https://mathoverflow.net/questions/307464/dimensions-of-a-vareity-and-its-affine-cone-in-biprojective-spaces. So now I am just asking for 2).
algebraic-geometry projective-geometry projective-space
asked Aug 2 at 15:40
Johnny T.
4861413
Hint: do things one at a time in each $Bbb P^n$ then figure out how the results go together.
– KReiser
Aug 2 at 21:20
@KReiser I have made an honest attempt but I don't seem to be getting it... I would greatly appreciate proof (or more details). Thank you very much.
– Johnny T.
Aug 3 at 8:56
1
Please include some detail about your attempts in the question - it will help you get better answers. Additionally, cross-posting to MO inside of 24 hours is generally frowned upon.
– KReiser
Aug 3 at 15:44
I see. Let me delete the MO question in that case. And thank you for your answer. It is very helpful and it is greatly appreciated!
– Johnny T.
Aug 4 at 11:54
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am just learning about product of projective spaces and I have some basic questions I would like to figure out. I will be working with $mathbbP^m times mathbbP^m$. And by bihomogeneous form I mean $F(mathbfx, mathbfy)$ such that $F(amathbfx, bmathbfy) = a^d_1b^d_2F(mathbfx, mathbfy)$ for some $d_1$ and $d_2 geq 0$. (coefficients in $mathbbC$)
Suppose I have a collection of bihomogeneous forms $S$.
1) What is the relation between the dimension of
affine variety
$$
(mathbfx, mathbfy) in mathbbC^2m+2: F = 0 (F in S)
$$
and the dimension (as a biprojective(?) variety)
$$
(mathbfx, mathbfy) in mathbbP^m times mathbbP^m: F = 0 (F in S)
$$? (My apologies for abusing the notation...)
2) Suppose I have a hyperplane
$$
V(L) = (mathbfx, mathbfy) in mathbbP^m times mathbbP^m: L(mathbfx) = 0 (F in S)
$$
where $L$ is a non-zero linear form in the $mathbfx$ variables. Does it then follow that
$$
dim (X cap V(L)) = dim X - 1
$$
as in the usual projective space?
Thank you very much!
Edit: I moved 1) of this question to mathovreflow https://mathoverflow.net/questions/307464/dimensions-of-a-vareity-and-its-affine-cone-in-biprojective-spaces. So now I am just asking for 2).
algebraic-geometry projective-geometry projective-space
asked Aug 2 at 15:40
Johnny T.
4861413
I am just learning about product of projective spaces and I have some basic questions I would like to figure out. I will be working with $mathbbP^m times mathbbP^m$. And by bihomogeneous form I mean $F(mathbfx, mathbfy)$ such that $F(amathbfx, bmathbfy) = a^d_1b^d_2F(mathbfx, mathbfy)$ for some $d_1$ and $d_2 geq 0$. (coefficients in $mathbbC$)
Suppose I have a collection of bihomogeneous forms $S$.
1) What is the relation between the dimension of
affine variety
$$
(mathbfx, mathbfy) in mathbbC^2m+2: F = 0 (F in S)
$$
and the dimension (as a biprojective(?) variety)
$$
(mathbfx, mathbfy) in mathbbP^m times mathbbP^m: F = 0 (F in S)
$$? (My apologies for abusing the notation...)
2) Suppose I have a hyperplane
$$
V(L) = (mathbfx, mathbfy) in mathbbP^m times mathbbP^m: L(mathbfx) = 0 (F in S)
$$
where $L$ is a non-zero linear form in the $mathbfx$ variables. Does it then follow that
$$
dim (X cap V(L)) = dim X - 1
$$
as in the usual projective space?
Thank you very much!
Edit: I moved 1) of this question to mathovreflow https://mathoverflow.net/questions/307464/dimensions-of-a-vareity-and-its-affine-cone-in-biprojective-spaces. So now I am just asking for 2).
algebraic-geometry projective-geometry projective-space
asked Aug 2 at 15:40
Johnny T.
4861413
edited Aug 3 at 10:15
asked Aug 2 at 15:40
Johnny T.
4861413
asked Aug 2 at 15:40
Johnny T.
4861413
asked Aug 2 at 15:40
Johnny T.
4861413
4861413
Hint: do things one at a time in each $Bbb P^n$ then figure out how the results go together.
– KReiser
Aug 2 at 21:20
@KReiser I have made an honest attempt but I don't seem to be getting it... I would greatly appreciate proof (or more details). Thank you very much.
– Johnny T.
Aug 3 at 8:56
1
Please include some detail about your attempts in the question - it will help you get better answers. Additionally, cross-posting to MO inside of 24 hours is generally frowned upon.
– KReiser
Aug 3 at 15:44
I see. Let me delete the MO question in that case. And thank you for your answer. It is very helpful and it is greatly appreciated!
– Johnny T.
Aug 4 at 11:54
add a comment |Â
Hint: do things one at a time in each $Bbb P^n$ then figure out how the results go together.
– KReiser
Aug 2 at 21:20
@KReiser I have made an honest attempt but I don't seem to be getting it... I would greatly appreciate proof (or more details). Thank you very much.
– Johnny T.
Aug 3 at 8:56
1
Please include some detail about your attempts in the question - it will help you get better answers. Additionally, cross-posting to MO inside of 24 hours is generally frowned upon.
– KReiser
Aug 3 at 15:44
I see. Let me delete the MO question in that case. And thank you for your answer. It is very helpful and it is greatly appreciated!
– Johnny T.
Aug 4 at 11:54
Hint: do things one at a time in each $Bbb P^n$ then figure out how the results go together.
– KReiser
Aug 2 at 21:20
Hint: do things one at a time in each $Bbb P^n$ then figure out how the results go together.
– KReiser
Aug 2 at 21:20
@KReiser I have made an honest attempt but I don't seem to be getting it... I would greatly appreciate proof (or more details). Thank you very much.
– Johnny T.
Aug 3 at 8:56
@KReiser I have made an honest attempt but I don't seem to be getting it... I would greatly appreciate proof (or more details). Thank you very much.
– Johnny T.
Aug 3 at 8:56
1
1
Please include some detail about your attempts in the question - it will help you get better answers. Additionally, cross-posting to MO inside of 24 hours is generally frowned upon.
– KReiser
Aug 3 at 15:44
Please include some detail about your attempts in the question - it will help you get better answers. Additionally, cross-posting to MO inside of 24 hours is generally frowned upon.
– KReiser
Aug 3 at 15:44
I see. Let me delete the MO question in that case. And thank you for your answer. It is very helpful and it is greatly appreciated!
– Johnny T.
Aug 4 at 11:54
I see. Let me delete the MO question in that case. And thank you for your answer. It is very helpful and it is greatly appreciated!
– Johnny T.
Aug 4 at 11:54
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
1) Assume $F$ is nonzero. $V(F)subset Bbb C^2m+2$ is codimension 1 (or dimension $2m+1$), so it remains to determine what $dim V(F)subset Bbb P^mtimesBbb P^m$ is. If $d_1$ or $d_2$ is zero, then $V(F)subset Bbb P^mtimes Bbb P^m$ may be empty, so now we assume that both are strictly positive.
Embed $Bbb P^mtimes Bbb P^m$ into $Bbb P^m^2+2m$ by the Segre embedding. WLOG, $d_1leq d_2$. Pick $l$ a homogeneous linear polynomial in $Bbb C[x_0,cdots,x_m]$ so that no irreducible component of $V(F)cap Bbb P^m$ is contained in $V(l)$. Then $G=l^d_2-d_1F$ is a homogeneous polynomial on $Bbb P^m^2+2m$ using the Segre coordinates so that $V(G)cap (Bbb P^mtimes Bbb P^m) = V(F)subset (Bbb P^mtimes Bbb P^m)$ on the open set $D(l)subset Bbb P^mtimes Bbb P^m$. Since no irreducible component of $V(F)$ is contained in $V(l)$, for each irreducible component $X_isubset V(F)$, $dim X_i = dim X_icap D(l)$. But we may compute this final quantity in $V(G)cap (Bbb P^mtimes Bbb P^m)$, and since codimensions add, we have that $dim X_i= 2m-1$.
The intersection theory of multiple such forms $F$ will follow the same pattern: whatever the dimensions of each irreducible component of $V(F)subset Bbb C^2m+2$, the dimension of the corresponding irreducible component in $Bbb P^mtimesBbb P^m$ will be 2 less, and then you can intersect to your heart's content.
2) No, it may be possible that $Xcap V(L)$ is empty: consider $m=1$ with $X=V(x_0)$ and $L=x_1$. In fact, you can embed fairly large varieties which do not intersect: $V(x_0,cdots,x_k)timesBbb P^m$ and $V(x_k+1,cdots,x_m)timesBbb P^m$ don't intersect inside $Bbb P^mtimesBbb P^m$.
answered Aug 3 at 22:54
KReiser
7,44011230
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
1) Assume $F$ is nonzero. $V(F)subset Bbb C^2m+2$ is codimension 1 (or dimension $2m+1$), so it remains to determine what $dim V(F)subset Bbb P^mtimesBbb P^m$ is. If $d_1$ or $d_2$ is zero, then $V(F)subset Bbb P^mtimes Bbb P^m$ may be empty, so now we assume that both are strictly positive.
Embed $Bbb P^mtimes Bbb P^m$ into $Bbb P^m^2+2m$ by the Segre embedding. WLOG, $d_1leq d_2$. Pick $l$ a homogeneous linear polynomial in $Bbb C[x_0,cdots,x_m]$ so that no irreducible component of $V(F)cap Bbb P^m$ is contained in $V(l)$. Then $G=l^d_2-d_1F$ is a homogeneous polynomial on $Bbb P^m^2+2m$ using the Segre coordinates so that $V(G)cap (Bbb P^mtimes Bbb P^m) = V(F)subset (Bbb P^mtimes Bbb P^m)$ on the open set $D(l)subset Bbb P^mtimes Bbb P^m$. Since no irreducible component of $V(F)$ is contained in $V(l)$, for each irreducible component $X_isubset V(F)$, $dim X_i = dim X_icap D(l)$. But we may compute this final quantity in $V(G)cap (Bbb P^mtimes Bbb P^m)$, and since codimensions add, we have that $dim X_i= 2m-1$.
The intersection theory of multiple such forms $F$ will follow the same pattern: whatever the dimensions of each irreducible component of $V(F)subset Bbb C^2m+2$, the dimension of the corresponding irreducible component in $Bbb P^mtimesBbb P^m$ will be 2 less, and then you can intersect to your heart's content.
2) No, it may be possible that $Xcap V(L)$ is empty: consider $m=1$ with $X=V(x_0)$ and $L=x_1$. In fact, you can embed fairly large varieties which do not intersect: $V(x_0,cdots,x_k)timesBbb P^m$ and $V(x_k+1,cdots,x_m)timesBbb P^m$ don't intersect inside $Bbb P^mtimesBbb P^m$.
answered Aug 3 at 22:54
KReiser
7,44011230
add a comment |Â
up vote
1
down vote
accepted
1) Assume $F$ is nonzero. $V(F)subset Bbb C^2m+2$ is codimension 1 (or dimension $2m+1$), so it remains to determine what $dim V(F)subset Bbb P^mtimesBbb P^m$ is. If $d_1$ or $d_2$ is zero, then $V(F)subset Bbb P^mtimes Bbb P^m$ may be empty, so now we assume that both are strictly positive.
Embed $Bbb P^mtimes Bbb P^m$ into $Bbb P^m^2+2m$ by the Segre embedding. WLOG, $d_1leq d_2$. Pick $l$ a homogeneous linear polynomial in $Bbb C[x_0,cdots,x_m]$ so that no irreducible component of $V(F)cap Bbb P^m$ is contained in $V(l)$. Then $G=l^d_2-d_1F$ is a homogeneous polynomial on $Bbb P^m^2+2m$ using the Segre coordinates so that $V(G)cap (Bbb P^mtimes Bbb P^m) = V(F)subset (Bbb P^mtimes Bbb P^m)$ on the open set $D(l)subset Bbb P^mtimes Bbb P^m$. Since no irreducible component of $V(F)$ is contained in $V(l)$, for each irreducible component $X_isubset V(F)$, $dim X_i = dim X_icap D(l)$. But we may compute this final quantity in $V(G)cap (Bbb P^mtimes Bbb P^m)$, and since codimensions add, we have that $dim X_i= 2m-1$.
The intersection theory of multiple such forms $F$ will follow the same pattern: whatever the dimensions of each irreducible component of $V(F)subset Bbb C^2m+2$, the dimension of the corresponding irreducible component in $Bbb P^mtimesBbb P^m$ will be 2 less, and then you can intersect to your heart's content.
2) No, it may be possible that $Xcap V(L)$ is empty: consider $m=1$ with $X=V(x_0)$ and $L=x_1$. In fact, you can embed fairly large varieties which do not intersect: $V(x_0,cdots,x_k)timesBbb P^m$ and $V(x_k+1,cdots,x_m)timesBbb P^m$ don't intersect inside $Bbb P^mtimesBbb P^m$.
answered Aug 3 at 22:54
KReiser
7,44011230
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
1) Assume $F$ is nonzero. $V(F)subset Bbb C^2m+2$ is codimension 1 (or dimension $2m+1$), so it remains to determine what $dim V(F)subset Bbb P^mtimesBbb P^m$ is. If $d_1$ or $d_2$ is zero, then $V(F)subset Bbb P^mtimes Bbb P^m$ may be empty, so now we assume that both are strictly positive.
Embed $Bbb P^mtimes Bbb P^m$ into $Bbb P^m^2+2m$ by the Segre embedding. WLOG, $d_1leq d_2$. Pick $l$ a homogeneous linear polynomial in $Bbb C[x_0,cdots,x_m]$ so that no irreducible component of $V(F)cap Bbb P^m$ is contained in $V(l)$. Then $G=l^d_2-d_1F$ is a homogeneous polynomial on $Bbb P^m^2+2m$ using the Segre coordinates so that $V(G)cap (Bbb P^mtimes Bbb P^m) = V(F)subset (Bbb P^mtimes Bbb P^m)$ on the open set $D(l)subset Bbb P^mtimes Bbb P^m$. Since no irreducible component of $V(F)$ is contained in $V(l)$, for each irreducible component $X_isubset V(F)$, $dim X_i = dim X_icap D(l)$. But we may compute this final quantity in $V(G)cap (Bbb P^mtimes Bbb P^m)$, and since codimensions add, we have that $dim X_i= 2m-1$.
The intersection theory of multiple such forms $F$ will follow the same pattern: whatever the dimensions of each irreducible component of $V(F)subset Bbb C^2m+2$, the dimension of the corresponding irreducible component in $Bbb P^mtimesBbb P^m$ will be 2 less, and then you can intersect to your heart's content.
2) No, it may be possible that $Xcap V(L)$ is empty: consider $m=1$ with $X=V(x_0)$ and $L=x_1$. In fact, you can embed fairly large varieties which do not intersect: $V(x_0,cdots,x_k)timesBbb P^m$ and $V(x_k+1,cdots,x_m)timesBbb P^m$ don't intersect inside $Bbb P^mtimesBbb P^m$.
answered Aug 3 at 22:54
KReiser
7,44011230
1) Assume $F$ is nonzero. $V(F)subset Bbb C^2m+2$ is codimension 1 (or dimension $2m+1$), so it remains to determine what $dim V(F)subset Bbb P^mtimesBbb P^m$ is. If $d_1$ or $d_2$ is zero, then $V(F)subset Bbb P^mtimes Bbb P^m$ may be empty, so now we assume that both are strictly positive.
Embed $Bbb P^mtimes Bbb P^m$ into $Bbb P^m^2+2m$ by the Segre embedding. WLOG, $d_1leq d_2$. Pick $l$ a homogeneous linear polynomial in $Bbb C[x_0,cdots,x_m]$ so that no irreducible component of $V(F)cap Bbb P^m$ is contained in $V(l)$. Then $G=l^d_2-d_1F$ is a homogeneous polynomial on $Bbb P^m^2+2m$ using the Segre coordinates so that $V(G)cap (Bbb P^mtimes Bbb P^m) = V(F)subset (Bbb P^mtimes Bbb P^m)$ on the open set $D(l)subset Bbb P^mtimes Bbb P^m$. Since no irreducible component of $V(F)$ is contained in $V(l)$, for each irreducible component $X_isubset V(F)$, $dim X_i = dim X_icap D(l)$. But we may compute this final quantity in $V(G)cap (Bbb P^mtimes Bbb P^m)$, and since codimensions add, we have that $dim X_i= 2m-1$.
The intersection theory of multiple such forms $F$ will follow the same pattern: whatever the dimensions of each irreducible component of $V(F)subset Bbb C^2m+2$, the dimension of the corresponding irreducible component in $Bbb P^mtimesBbb P^m$ will be 2 less, and then you can intersect to your heart's content.
2) No, it may be possible that $Xcap V(L)$ is empty: consider $m=1$ with $X=V(x_0)$ and $L=x_1$. In fact, you can embed fairly large varieties which do not intersect: $V(x_0,cdots,x_k)timesBbb P^m$ and $V(x_k+1,cdots,x_m)timesBbb P^m$ don't intersect inside $Bbb P^mtimesBbb P^m$.
answered Aug 3 at 22:54
KReiser
7,44011230
answered Aug 3 at 22:54
KReiser
7,44011230
answered Aug 3 at 22:54
KReiser
7,44011230
answered Aug 3 at 22:54
KReiser
7,44011230
7,44011230
add a comment |Â
add a comment |Â
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Hint: do things one at a time in each $Bbb P^n$ then figure out how the results go together.
– KReiser
Aug 2 at 21:20
@KReiser I have made an honest attempt but I don't seem to be getting it... I would greatly appreciate proof (or more details). Thank you very much.
– Johnny T.
Aug 3 at 8:56
1
Please include some detail about your attempts in the question - it will help you get better answers. Additionally, cross-posting to MO inside of 24 hours is generally frowned upon.
– KReiser
Aug 3 at 15:44
I see. Let me delete the MO question in that case. And thank you for your answer. It is very helpful and it is greatly appreciated!
– Johnny T.
Aug 4 at 11:54