How to show that a hypersurface is a diagonal intersected with hyperplanes?
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Suppose I have a hypersurface $V(F) = mathbfx in k^n: F(mathbfx) = 0 $, where $F$ is a homogeneous form of degree $d > 1$. I would like to show that there exists some diagonal form $D(y_1, ldots, y_N) = A_1 y_1^d + ldots + A_N y^d in k[y_1, ldots, y_N]$ and some linear forms $L_i(mathbfy) in k[y_1, ldots, y_N]$, say $1 leq i leq T$, such that
$$
V(F) = V(D) cap V(L_1) cap ldots cap V(L_T).
$$
Here $k = mathbbQ, mathbbR$ or $mathbbC$. I seem to recall someone telling me this follows fairly easily from some algebraic geometry (possibly using Veronese embedding) but I couldn't figure it out. Any comments would be appreciated.
ag.algebraic-geometry nt.number-theory
edited Aug 19 at 16:06
Zach Teitler
2,30411024
asked Aug 19 at 12:00
Johnny T.
862518
add a comment |Â
up vote
5
down vote
favorite
Suppose I have a hypersurface $V(F) = mathbfx in k^n: F(mathbfx) = 0 $, where $F$ is a homogeneous form of degree $d > 1$. I would like to show that there exists some diagonal form $D(y_1, ldots, y_N) = A_1 y_1^d + ldots + A_N y^d in k[y_1, ldots, y_N]$ and some linear forms $L_i(mathbfy) in k[y_1, ldots, y_N]$, say $1 leq i leq T$, such that
$$
V(F) = V(D) cap V(L_1) cap ldots cap V(L_T).
$$
Here $k = mathbbQ, mathbbR$ or $mathbbC$. I seem to recall someone telling me this follows fairly easily from some algebraic geometry (possibly using Veronese embedding) but I couldn't figure it out. Any comments would be appreciated.
ag.algebraic-geometry nt.number-theory
edited Aug 19 at 16:06
Zach Teitler
2,30411024
asked Aug 19 at 12:00
Johnny T.
862518
1
It doesn't seem clear what you are asking. Are $n$ and $N$ the same? If so then $V$ has codimension one in $mathbbP_mathbbC^n-1$, and for the desired inequality to occur then $F$ has to have the same vanishing locus of the diagonal form. If $n < N$ then one would have to specify some embedding of $mathbbP_mathbbC^n-1$ into $mathbbP_mathbbC^N-1$. This is possibly where you might use Vernoese embedding. You can also look at this paper (projecteuclid.org/euclid.dmj/1143936000)
– Stanley Yao Xiao
Aug 19 at 12:07
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
Suppose I have a hypersurface $V(F) = mathbfx in k^n: F(mathbfx) = 0 $, where $F$ is a homogeneous form of degree $d > 1$. I would like to show that there exists some diagonal form $D(y_1, ldots, y_N) = A_1 y_1^d + ldots + A_N y^d in k[y_1, ldots, y_N]$ and some linear forms $L_i(mathbfy) in k[y_1, ldots, y_N]$, say $1 leq i leq T$, such that
$$
V(F) = V(D) cap V(L_1) cap ldots cap V(L_T).
$$
Here $k = mathbbQ, mathbbR$ or $mathbbC$. I seem to recall someone telling me this follows fairly easily from some algebraic geometry (possibly using Veronese embedding) but I couldn't figure it out. Any comments would be appreciated.
ag.algebraic-geometry nt.number-theory
edited Aug 19 at 16:06
Zach Teitler
2,30411024
asked Aug 19 at 12:00
Johnny T.
862518
Suppose I have a hypersurface $V(F) = mathbfx in k^n: F(mathbfx) = 0 $, where $F$ is a homogeneous form of degree $d > 1$. I would like to show that there exists some diagonal form $D(y_1, ldots, y_N) = A_1 y_1^d + ldots + A_N y^d in k[y_1, ldots, y_N]$ and some linear forms $L_i(mathbfy) in k[y_1, ldots, y_N]$, say $1 leq i leq T$, such that
$$
V(F) = V(D) cap V(L_1) cap ldots cap V(L_T).
$$
Here $k = mathbbQ, mathbbR$ or $mathbbC$. I seem to recall someone telling me this follows fairly easily from some algebraic geometry (possibly using Veronese embedding) but I couldn't figure it out. Any comments would be appreciated.
ag.algebraic-geometry nt.number-theory
ag.algebraic-geometry nt.number-theory
edited Aug 19 at 16:06
Zach Teitler
2,30411024
asked Aug 19 at 12:00
Johnny T.
862518
edited Aug 19 at 16:06
Zach Teitler
2,30411024
asked Aug 19 at 12:00
Johnny T.
862518
edited Aug 19 at 16:06
Zach Teitler
2,30411024
edited Aug 19 at 16:06
Zach Teitler
2,30411024
edited Aug 19 at 16:06
Zach Teitler
2,30411024
2,30411024
asked Aug 19 at 12:00
Johnny T.
862518
asked Aug 19 at 12:00
Johnny T.
862518
asked Aug 19 at 12:00
Johnny T.
862518
862518
1
It doesn't seem clear what you are asking. Are $n$ and $N$ the same? If so then $V$ has codimension one in $mathbbP_mathbbC^n-1$, and for the desired inequality to occur then $F$ has to have the same vanishing locus of the diagonal form. If $n < N$ then one would have to specify some embedding of $mathbbP_mathbbC^n-1$ into $mathbbP_mathbbC^N-1$. This is possibly where you might use Vernoese embedding. You can also look at this paper (projecteuclid.org/euclid.dmj/1143936000)
– Stanley Yao Xiao
Aug 19 at 12:07
add a comment |Â
1
It doesn't seem clear what you are asking. Are $n$ and $N$ the same? If so then $V$ has codimension one in $mathbbP_mathbbC^n-1$, and for the desired inequality to occur then $F$ has to have the same vanishing locus of the diagonal form. If $n < N$ then one would have to specify some embedding of $mathbbP_mathbbC^n-1$ into $mathbbP_mathbbC^N-1$. This is possibly where you might use Vernoese embedding. You can also look at this paper (projecteuclid.org/euclid.dmj/1143936000)
– Stanley Yao Xiao
Aug 19 at 12:07
1
1
It doesn't seem clear what you are asking. Are $n$ and $N$ the same? If so then $V$ has codimension one in $mathbbP_mathbbC^n-1$, and for the desired inequality to occur then $F$ has to have the same vanishing locus of the diagonal form. If $n < N$ then one would have to specify some embedding of $mathbbP_mathbbC^n-1$ into $mathbbP_mathbbC^N-1$. This is possibly where you might use Vernoese embedding. You can also look at this paper (projecteuclid.org/euclid.dmj/1143936000)
– Stanley Yao Xiao
Aug 19 at 12:07
It doesn't seem clear what you are asking. Are $n$ and $N$ the same? If so then $V$ has codimension one in $mathbbP_mathbbC^n-1$, and for the desired inequality to occur then $F$ has to have the same vanishing locus of the diagonal form. If $n < N$ then one would have to specify some embedding of $mathbbP_mathbbC^n-1$ into $mathbbP_mathbbC^N-1$. This is possibly where you might use Vernoese embedding. You can also look at this paper (projecteuclid.org/euclid.dmj/1143936000)
– Stanley Yao Xiao
Aug 19 at 12:07
add a comment |Â
1 Answer
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The map from $V(L_1) cap dots cap V(L_T)$ to $mathbb P^N-1$ is a linear map. So an equivalent way of stating this is that there are $N$ linear forms $y_1,dots,y_N$ in $x_1,dots,x_n$ such that $sum_i=1^N A_i y_i^d = F$.
One way to think about this is to consider, for each $N$, the locus in the space of degree $d$ homogeneous forms that can be written as $sum_i=1^N A_i y_i^d$. If the dimension of this locus for some $N$ is equal to the dimension for $N+1$, it follows that the sum of a generic element of this locus with a generic linear form is a generic element of the locus, so the sum of two generic elements of the locus is a generic element of the locus, so the locus is a dense subset of a linear subspace. However, the space of linear forms raised to the $d$th power is not contained in any linear subspace (this can be expressed as a statement of the Veronese embedding, but can also be proved directly algebraically), so for this $N$ the locus is in fact dense in the whole space. Hence for $2N$ the locus is the whole space. (And their must be such an $N$ because the dimension is bounded.)
answered Aug 19 at 13:06
Will Sawin
64.7k6128271
2
I think the essential point of the problem is exactly to prove that this subvariety is not contained in a linear subspace, which is not at all immediately clear from the statement. One can either interpret it as the image of the Veronese embedding, after which it becomes immediately clear (it's amazing the psychological effect dualizing has) or do something explicit as in mathoverflow.net/questions/98714/….
– dhy
Aug 19 at 13:14
2
@dhy Here's how I would express it: If you write $y_i sum_i=1^n z_i x_i$, you can take any linear form on this space and write it as a degree $d$ homogeneous polynomial in the $z_i$. If the form is nonzero on some monomial in the $x_i$, the corresponding polynomial will contain a nonzero coefficient of that monomial in the $z_i$ and hence be nonzero. (this uses characteristic $>d$, it is false without that) But every nonzero polynomial is nonzero on some point.
– Will Sawin
Aug 19 at 13:42
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
The map from $V(L_1) cap dots cap V(L_T)$ to $mathbb P^N-1$ is a linear map. So an equivalent way of stating this is that there are $N$ linear forms $y_1,dots,y_N$ in $x_1,dots,x_n$ such that $sum_i=1^N A_i y_i^d = F$.
One way to think about this is to consider, for each $N$, the locus in the space of degree $d$ homogeneous forms that can be written as $sum_i=1^N A_i y_i^d$. If the dimension of this locus for some $N$ is equal to the dimension for $N+1$, it follows that the sum of a generic element of this locus with a generic linear form is a generic element of the locus, so the sum of two generic elements of the locus is a generic element of the locus, so the locus is a dense subset of a linear subspace. However, the space of linear forms raised to the $d$th power is not contained in any linear subspace (this can be expressed as a statement of the Veronese embedding, but can also be proved directly algebraically), so for this $N$ the locus is in fact dense in the whole space. Hence for $2N$ the locus is the whole space. (And their must be such an $N$ because the dimension is bounded.)
answered Aug 19 at 13:06
Will Sawin
64.7k6128271
2
I think the essential point of the problem is exactly to prove that this subvariety is not contained in a linear subspace, which is not at all immediately clear from the statement. One can either interpret it as the image of the Veronese embedding, after which it becomes immediately clear (it's amazing the psychological effect dualizing has) or do something explicit as in mathoverflow.net/questions/98714/….
– dhy
Aug 19 at 13:14
2
@dhy Here's how I would express it: If you write $y_i sum_i=1^n z_i x_i$, you can take any linear form on this space and write it as a degree $d$ homogeneous polynomial in the $z_i$. If the form is nonzero on some monomial in the $x_i$, the corresponding polynomial will contain a nonzero coefficient of that monomial in the $z_i$ and hence be nonzero. (this uses characteristic $>d$, it is false without that) But every nonzero polynomial is nonzero on some point.
– Will Sawin
Aug 19 at 13:42
add a comment |Â
up vote
5
down vote
The map from $V(L_1) cap dots cap V(L_T)$ to $mathbb P^N-1$ is a linear map. So an equivalent way of stating this is that there are $N$ linear forms $y_1,dots,y_N$ in $x_1,dots,x_n$ such that $sum_i=1^N A_i y_i^d = F$.
One way to think about this is to consider, for each $N$, the locus in the space of degree $d$ homogeneous forms that can be written as $sum_i=1^N A_i y_i^d$. If the dimension of this locus for some $N$ is equal to the dimension for $N+1$, it follows that the sum of a generic element of this locus with a generic linear form is a generic element of the locus, so the sum of two generic elements of the locus is a generic element of the locus, so the locus is a dense subset of a linear subspace. However, the space of linear forms raised to the $d$th power is not contained in any linear subspace (this can be expressed as a statement of the Veronese embedding, but can also be proved directly algebraically), so for this $N$ the locus is in fact dense in the whole space. Hence for $2N$ the locus is the whole space. (And their must be such an $N$ because the dimension is bounded.)
answered Aug 19 at 13:06
Will Sawin
64.7k6128271
2
I think the essential point of the problem is exactly to prove that this subvariety is not contained in a linear subspace, which is not at all immediately clear from the statement. One can either interpret it as the image of the Veronese embedding, after which it becomes immediately clear (it's amazing the psychological effect dualizing has) or do something explicit as in mathoverflow.net/questions/98714/….
– dhy
Aug 19 at 13:14
2
@dhy Here's how I would express it: If you write $y_i sum_i=1^n z_i x_i$, you can take any linear form on this space and write it as a degree $d$ homogeneous polynomial in the $z_i$. If the form is nonzero on some monomial in the $x_i$, the corresponding polynomial will contain a nonzero coefficient of that monomial in the $z_i$ and hence be nonzero. (this uses characteristic $>d$, it is false without that) But every nonzero polynomial is nonzero on some point.
– Will Sawin
Aug 19 at 13:42
add a comment |Â
up vote
5
down vote
up vote
5
down vote
The map from $V(L_1) cap dots cap V(L_T)$ to $mathbb P^N-1$ is a linear map. So an equivalent way of stating this is that there are $N$ linear forms $y_1,dots,y_N$ in $x_1,dots,x_n$ such that $sum_i=1^N A_i y_i^d = F$.
One way to think about this is to consider, for each $N$, the locus in the space of degree $d$ homogeneous forms that can be written as $sum_i=1^N A_i y_i^d$. If the dimension of this locus for some $N$ is equal to the dimension for $N+1$, it follows that the sum of a generic element of this locus with a generic linear form is a generic element of the locus, so the sum of two generic elements of the locus is a generic element of the locus, so the locus is a dense subset of a linear subspace. However, the space of linear forms raised to the $d$th power is not contained in any linear subspace (this can be expressed as a statement of the Veronese embedding, but can also be proved directly algebraically), so for this $N$ the locus is in fact dense in the whole space. Hence for $2N$ the locus is the whole space. (And their must be such an $N$ because the dimension is bounded.)
answered Aug 19 at 13:06
Will Sawin
64.7k6128271
The map from $V(L_1) cap dots cap V(L_T)$ to $mathbb P^N-1$ is a linear map. So an equivalent way of stating this is that there are $N$ linear forms $y_1,dots,y_N$ in $x_1,dots,x_n$ such that $sum_i=1^N A_i y_i^d = F$.
One way to think about this is to consider, for each $N$, the locus in the space of degree $d$ homogeneous forms that can be written as $sum_i=1^N A_i y_i^d$. If the dimension of this locus for some $N$ is equal to the dimension for $N+1$, it follows that the sum of a generic element of this locus with a generic linear form is a generic element of the locus, so the sum of two generic elements of the locus is a generic element of the locus, so the locus is a dense subset of a linear subspace. However, the space of linear forms raised to the $d$th power is not contained in any linear subspace (this can be expressed as a statement of the Veronese embedding, but can also be proved directly algebraically), so for this $N$ the locus is in fact dense in the whole space. Hence for $2N$ the locus is the whole space. (And their must be such an $N$ because the dimension is bounded.)
answered Aug 19 at 13:06
Will Sawin
64.7k6128271
answered Aug 19 at 13:06
Will Sawin
64.7k6128271
answered Aug 19 at 13:06
Will Sawin
64.7k6128271
answered Aug 19 at 13:06
Will Sawin
64.7k6128271
64.7k6128271
2
I think the essential point of the problem is exactly to prove that this subvariety is not contained in a linear subspace, which is not at all immediately clear from the statement. One can either interpret it as the image of the Veronese embedding, after which it becomes immediately clear (it's amazing the psychological effect dualizing has) or do something explicit as in mathoverflow.net/questions/98714/….
– dhy
Aug 19 at 13:14
2
@dhy Here's how I would express it: If you write $y_i sum_i=1^n z_i x_i$, you can take any linear form on this space and write it as a degree $d$ homogeneous polynomial in the $z_i$. If the form is nonzero on some monomial in the $x_i$, the corresponding polynomial will contain a nonzero coefficient of that monomial in the $z_i$ and hence be nonzero. (this uses characteristic $>d$, it is false without that) But every nonzero polynomial is nonzero on some point.
– Will Sawin
Aug 19 at 13:42
add a comment |Â
2
I think the essential point of the problem is exactly to prove that this subvariety is not contained in a linear subspace, which is not at all immediately clear from the statement. One can either interpret it as the image of the Veronese embedding, after which it becomes immediately clear (it's amazing the psychological effect dualizing has) or do something explicit as in mathoverflow.net/questions/98714/….
– dhy
Aug 19 at 13:14
2
@dhy Here's how I would express it: If you write $y_i sum_i=1^n z_i x_i$, you can take any linear form on this space and write it as a degree $d$ homogeneous polynomial in the $z_i$. If the form is nonzero on some monomial in the $x_i$, the corresponding polynomial will contain a nonzero coefficient of that monomial in the $z_i$ and hence be nonzero. (this uses characteristic $>d$, it is false without that) But every nonzero polynomial is nonzero on some point.
– Will Sawin
Aug 19 at 13:42
2
2
I think the essential point of the problem is exactly to prove that this subvariety is not contained in a linear subspace, which is not at all immediately clear from the statement. One can either interpret it as the image of the Veronese embedding, after which it becomes immediately clear (it's amazing the psychological effect dualizing has) or do something explicit as in mathoverflow.net/questions/98714/….
– dhy
Aug 19 at 13:14
I think the essential point of the problem is exactly to prove that this subvariety is not contained in a linear subspace, which is not at all immediately clear from the statement. One can either interpret it as the image of the Veronese embedding, after which it becomes immediately clear (it's amazing the psychological effect dualizing has) or do something explicit as in mathoverflow.net/questions/98714/….
– dhy
Aug 19 at 13:14
2
2
@dhy Here's how I would express it: If you write $y_i sum_i=1^n z_i x_i$, you can take any linear form on this space and write it as a degree $d$ homogeneous polynomial in the $z_i$. If the form is nonzero on some monomial in the $x_i$, the corresponding polynomial will contain a nonzero coefficient of that monomial in the $z_i$ and hence be nonzero. (this uses characteristic $>d$, it is false without that) But every nonzero polynomial is nonzero on some point.
– Will Sawin
Aug 19 at 13:42
@dhy Here's how I would express it: If you write $y_i sum_i=1^n z_i x_i$, you can take any linear form on this space and write it as a degree $d$ homogeneous polynomial in the $z_i$. If the form is nonzero on some monomial in the $x_i$, the corresponding polynomial will contain a nonzero coefficient of that monomial in the $z_i$ and hence be nonzero. (this uses characteristic $>d$, it is false without that) But every nonzero polynomial is nonzero on some point.
– Will Sawin
Aug 19 at 13:42
add a comment |Â
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1
It doesn't seem clear what you are asking. Are $n$ and $N$ the same? If so then $V$ has codimension one in $mathbbP_mathbbC^n-1$, and for the desired inequality to occur then $F$ has to have the same vanishing locus of the diagonal form. If $n < N$ then one would have to specify some embedding of $mathbbP_mathbbC^n-1$ into $mathbbP_mathbbC^N-1$. This is possibly where you might use Vernoese embedding. You can also look at this paper (projecteuclid.org/euclid.dmj/1143936000)
– Stanley Yao Xiao
Aug 19 at 12:07