Relations between homogeneous polynomials
$begingroup$
Let $F_1,ldots ,F_ellin mathbb{C}[X_1,ldots , X_n]$ be general homogeneous polynomials of degree $d$. For $p<d$, it seems likely that there is no nontrivial relation $sum F_iG_i=0$ with $G_1,ldots ,G_ell$ homogeneous of degree $p$. Is this known? If yes, what would be a reference (or a proof)?
ag.algebraic-geometry ac.commutative-algebra
asked 8 hours ago
abxabx
23.8k34885
$endgroup$
|
show 1 more comment
$begingroup$
Let $F_1,ldots ,F_ellin mathbb{C}[X_1,ldots , X_n]$ be general homogeneous polynomials of degree $d$. For $p<d$, it seems likely that there is no nontrivial relation $sum F_iG_i=0$ with $G_1,ldots ,G_ell$ homogeneous of degree $p$. Is this known? If yes, what would be a reference (or a proof)?
ag.algebraic-geometry ac.commutative-algebra
asked 8 hours ago
abxabx
23.8k34885
$endgroup$
$begingroup$
I think this question needs some clarification. I think you may mean something along the lines of: let $X$ be the moduli space of $n$-tuples of homogeneous polynomials of degree $d$; within $X$, there is a subspace $Y$ consisting of homogeneous polynomials that satisfy a nontrivial relation $sum F_i G_i = 0$ for some $G_i$ homogeneous of degree $p < d$ (i.e. the union of the subspaces determined by each tuple $G$). Is it true that $Y subsetneq X$?
$endgroup$
– user44191
7 hours ago
$begingroup$
Sure. Is that different from what I wrote?
$endgroup$
– abx
7 hours ago
$begingroup$
I found what you wrote ambiguous or underdetermined, e.g. with what you mean by "general"; I thought what I wrote made it more explicitly clear (though it may be overkill as written).
$endgroup$
– user44191
7 hours ago
$begingroup$
This seem trivially false if I haven't misunderstood the question. Take $n = 1, p = 0, d = 1, ell = 2$. Then $F_1 = ax, F_2 = bx$ for some $a, b$, so $b F_1 - a F_2 = 0$. I think both $n, ell$ play some role.
$endgroup$
– user44191
7 hours ago
$begingroup$
Oh, right. $ellleq n$ would be fine for me, though this can be certainly weakened.
$endgroup$
– abx
7 hours ago
|
show 1 more comment
$begingroup$
Let $F_1,ldots ,F_ellin mathbb{C}[X_1,ldots , X_n]$ be general homogeneous polynomials of degree $d$. For $p<d$, it seems likely that there is no nontrivial relation $sum F_iG_i=0$ with $G_1,ldots ,G_ell$ homogeneous of degree $p$. Is this known? If yes, what would be a reference (or a proof)?
ag.algebraic-geometry ac.commutative-algebra
asked 8 hours ago
abxabx
23.8k34885
$endgroup$
Let $F_1,ldots ,F_ellin mathbb{C}[X_1,ldots , X_n]$ be general homogeneous polynomials of degree $d$. For $p<d$, it seems likely that there is no nontrivial relation $sum F_iG_i=0$ with $G_1,ldots ,G_ell$ homogeneous of degree $p$. Is this known? If yes, what would be a reference (or a proof)?
ag.algebraic-geometry ac.commutative-algebra
ag.algebraic-geometry ac.commutative-algebra
asked 8 hours ago
abxabx
23.8k34885
asked 8 hours ago
abxabx
23.8k34885
edited 8 hours ago
abx
asked 8 hours ago
abxabx
23.8k34885
asked 8 hours ago
abxabx
23.8k34885
asked 8 hours ago
abxabx
23.8k34885
23.8k34885
$begingroup$
I think this question needs some clarification. I think you may mean something along the lines of: let $X$ be the moduli space of $n$-tuples of homogeneous polynomials of degree $d$; within $X$, there is a subspace $Y$ consisting of homogeneous polynomials that satisfy a nontrivial relation $sum F_i G_i = 0$ for some $G_i$ homogeneous of degree $p < d$ (i.e. the union of the subspaces determined by each tuple $G$). Is it true that $Y subsetneq X$?
$endgroup$
– user44191
7 hours ago
$begingroup$
Sure. Is that different from what I wrote?
$endgroup$
– abx
7 hours ago
$begingroup$
I found what you wrote ambiguous or underdetermined, e.g. with what you mean by "general"; I thought what I wrote made it more explicitly clear (though it may be overkill as written).
$endgroup$
– user44191
7 hours ago
$begingroup$
This seem trivially false if I haven't misunderstood the question. Take $n = 1, p = 0, d = 1, ell = 2$. Then $F_1 = ax, F_2 = bx$ for some $a, b$, so $b F_1 - a F_2 = 0$. I think both $n, ell$ play some role.
$endgroup$
– user44191
7 hours ago
$begingroup$
Oh, right. $ellleq n$ would be fine for me, though this can be certainly weakened.
$endgroup$
– abx
7 hours ago
|
show 1 more comment
$begingroup$
I think this question needs some clarification. I think you may mean something along the lines of: let $X$ be the moduli space of $n$-tuples of homogeneous polynomials of degree $d$; within $X$, there is a subspace $Y$ consisting of homogeneous polynomials that satisfy a nontrivial relation $sum F_i G_i = 0$ for some $G_i$ homogeneous of degree $p < d$ (i.e. the union of the subspaces determined by each tuple $G$). Is it true that $Y subsetneq X$?
$endgroup$
– user44191
7 hours ago
$begingroup$
Sure. Is that different from what I wrote?
$endgroup$
– abx
7 hours ago
$begingroup$
I found what you wrote ambiguous or underdetermined, e.g. with what you mean by "general"; I thought what I wrote made it more explicitly clear (though it may be overkill as written).
$endgroup$
– user44191
7 hours ago
$begingroup$
This seem trivially false if I haven't misunderstood the question. Take $n = 1, p = 0, d = 1, ell = 2$. Then $F_1 = ax, F_2 = bx$ for some $a, b$, so $b F_1 - a F_2 = 0$. I think both $n, ell$ play some role.
$endgroup$
– user44191
7 hours ago
$begingroup$
Oh, right. $ellleq n$ would be fine for me, though this can be certainly weakened.
$endgroup$
– abx
7 hours ago
$begingroup$
I think this question needs some clarification. I think you may mean something along the lines of: let $X$ be the moduli space of $n$-tuples of homogeneous polynomials of degree $d$; within $X$, there is a subspace $Y$ consisting of homogeneous polynomials that satisfy a nontrivial relation $sum F_i G_i = 0$ for some $G_i$ homogeneous of degree $p < d$ (i.e. the union of the subspaces determined by each tuple $G$). Is it true that $Y subsetneq X$?
$endgroup$
– user44191
7 hours ago
$begingroup$
I think this question needs some clarification. I think you may mean something along the lines of: let $X$ be the moduli space of $n$-tuples of homogeneous polynomials of degree $d$; within $X$, there is a subspace $Y$ consisting of homogeneous polynomials that satisfy a nontrivial relation $sum F_i G_i = 0$ for some $G_i$ homogeneous of degree $p < d$ (i.e. the union of the subspaces determined by each tuple $G$). Is it true that $Y subsetneq X$?
$endgroup$
– user44191
7 hours ago
$begingroup$
Sure. Is that different from what I wrote?
$endgroup$
– abx
7 hours ago
$begingroup$
Sure. Is that different from what I wrote?
$endgroup$
– abx
7 hours ago
$begingroup$
I found what you wrote ambiguous or underdetermined, e.g. with what you mean by "general"; I thought what I wrote made it more explicitly clear (though it may be overkill as written).
$endgroup$
– user44191
7 hours ago
$begingroup$
I found what you wrote ambiguous or underdetermined, e.g. with what you mean by "general"; I thought what I wrote made it more explicitly clear (though it may be overkill as written).
$endgroup$
– user44191
7 hours ago
$begingroup$
This seem trivially false if I haven't misunderstood the question. Take $n = 1, p = 0, d = 1, ell = 2$. Then $F_1 = ax, F_2 = bx$ for some $a, b$, so $b F_1 - a F_2 = 0$. I think both $n, ell$ play some role.
$endgroup$
– user44191
7 hours ago
$begingroup$
This seem trivially false if I haven't misunderstood the question. Take $n = 1, p = 0, d = 1, ell = 2$. Then $F_1 = ax, F_2 = bx$ for some $a, b$, so $b F_1 - a F_2 = 0$. I think both $n, ell$ play some role.
$endgroup$
– user44191
7 hours ago
$begingroup$
Oh, right. $ellleq n$ would be fine for me, though this can be certainly weakened.
$endgroup$
– abx
7 hours ago
$begingroup$
Oh, right. $ellleq n$ would be fine for me, though this can be certainly weakened.
$endgroup$
– abx
7 hours ago
|
show 1 more comment
1 Answer
1
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oldest
votes
$begingroup$
I think this can be controlled as follows. Let $Z subset mathbb{P}^{n-1}$ be the complete intersection defined by $F_i$. Then there is a Koszul resolution
$$
dots to mathcal{O}(-2d)^{binom{ell}{2}} to mathcal{O}(-d)^ell to mathcal{O} to mathcal{O}_Z to 0.
$$
Twisting it by $mathcal{O}(d+p)$ we obtain
$$
dots to mathcal{O}(p-d)^{binom{ell}{2}} to mathcal{O}(p)^ell to mathcal{O}(d+p) to mathcal{O}_Z(d+p) to 0.
$$
Your question is equivalent to injectivity of the induced map
$$
H^0(mathcal{O}(p)^ell) to H^0(mathcal{O}(d+p)).
$$
If $n ge ell$ the cohomology spectral sequence of the twisted Koszul complex proves this. Is that enough for your purposes?
answered 8 hours ago
SashaSasha
21k22755
$endgroup$
1
$begingroup$
Thanks, but could you explain why the Koszul complex gives that?
$endgroup$
– abx
7 hours ago
$begingroup$
Let $E$ the cokernel of $ldots rightarrow mathcal{O}(p-d)^{binom{l}{2}}$. This truncated comples has length $l-1$. Hence, the dimensions of cohomology groups which play a role in order to compute $H^0(E)$ are less than $l-1$.
$endgroup$
– Libli
6 hours ago
$begingroup$
Hence, if you assume that $l leq n$ and noting that all terms of this truncated complex are acyclic in degree strictly less than $n$, you get that $H^0(E)= 0$ and the injectivity claimed by Sasha.
$endgroup$
– Libli
6 hours ago
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
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votes
$begingroup$
I think this can be controlled as follows. Let $Z subset mathbb{P}^{n-1}$ be the complete intersection defined by $F_i$. Then there is a Koszul resolution
$$
dots to mathcal{O}(-2d)^{binom{ell}{2}} to mathcal{O}(-d)^ell to mathcal{O} to mathcal{O}_Z to 0.
$$
Twisting it by $mathcal{O}(d+p)$ we obtain
$$
dots to mathcal{O}(p-d)^{binom{ell}{2}} to mathcal{O}(p)^ell to mathcal{O}(d+p) to mathcal{O}_Z(d+p) to 0.
$$
Your question is equivalent to injectivity of the induced map
$$
H^0(mathcal{O}(p)^ell) to H^0(mathcal{O}(d+p)).
$$
If $n ge ell$ the cohomology spectral sequence of the twisted Koszul complex proves this. Is that enough for your purposes?
answered 8 hours ago
SashaSasha
21k22755
$endgroup$
1
$begingroup$
Thanks, but could you explain why the Koszul complex gives that?
$endgroup$
– abx
7 hours ago
$begingroup$
Let $E$ the cokernel of $ldots rightarrow mathcal{O}(p-d)^{binom{l}{2}}$. This truncated comples has length $l-1$. Hence, the dimensions of cohomology groups which play a role in order to compute $H^0(E)$ are less than $l-1$.
$endgroup$
– Libli
6 hours ago
$begingroup$
Hence, if you assume that $l leq n$ and noting that all terms of this truncated complex are acyclic in degree strictly less than $n$, you get that $H^0(E)= 0$ and the injectivity claimed by Sasha.
$endgroup$
– Libli
6 hours ago
add a comment |
$begingroup$
I think this can be controlled as follows. Let $Z subset mathbb{P}^{n-1}$ be the complete intersection defined by $F_i$. Then there is a Koszul resolution
$$
dots to mathcal{O}(-2d)^{binom{ell}{2}} to mathcal{O}(-d)^ell to mathcal{O} to mathcal{O}_Z to 0.
$$
Twisting it by $mathcal{O}(d+p)$ we obtain
$$
dots to mathcal{O}(p-d)^{binom{ell}{2}} to mathcal{O}(p)^ell to mathcal{O}(d+p) to mathcal{O}_Z(d+p) to 0.
$$
Your question is equivalent to injectivity of the induced map
$$
H^0(mathcal{O}(p)^ell) to H^0(mathcal{O}(d+p)).
$$
If $n ge ell$ the cohomology spectral sequence of the twisted Koszul complex proves this. Is that enough for your purposes?
answered 8 hours ago
SashaSasha
21k22755
$endgroup$
1
$begingroup$
Thanks, but could you explain why the Koszul complex gives that?
$endgroup$
– abx
7 hours ago
$begingroup$
Let $E$ the cokernel of $ldots rightarrow mathcal{O}(p-d)^{binom{l}{2}}$. This truncated comples has length $l-1$. Hence, the dimensions of cohomology groups which play a role in order to compute $H^0(E)$ are less than $l-1$.
$endgroup$
– Libli
6 hours ago
$begingroup$
Hence, if you assume that $l leq n$ and noting that all terms of this truncated complex are acyclic in degree strictly less than $n$, you get that $H^0(E)= 0$ and the injectivity claimed by Sasha.
$endgroup$
– Libli
6 hours ago
add a comment |
$begingroup$
I think this can be controlled as follows. Let $Z subset mathbb{P}^{n-1}$ be the complete intersection defined by $F_i$. Then there is a Koszul resolution
$$
dots to mathcal{O}(-2d)^{binom{ell}{2}} to mathcal{O}(-d)^ell to mathcal{O} to mathcal{O}_Z to 0.
$$
Twisting it by $mathcal{O}(d+p)$ we obtain
$$
dots to mathcal{O}(p-d)^{binom{ell}{2}} to mathcal{O}(p)^ell to mathcal{O}(d+p) to mathcal{O}_Z(d+p) to 0.
$$
Your question is equivalent to injectivity of the induced map
$$
H^0(mathcal{O}(p)^ell) to H^0(mathcal{O}(d+p)).
$$
If $n ge ell$ the cohomology spectral sequence of the twisted Koszul complex proves this. Is that enough for your purposes?
answered 8 hours ago
SashaSasha
21k22755
$endgroup$
I think this can be controlled as follows. Let $Z subset mathbb{P}^{n-1}$ be the complete intersection defined by $F_i$. Then there is a Koszul resolution
$$
dots to mathcal{O}(-2d)^{binom{ell}{2}} to mathcal{O}(-d)^ell to mathcal{O} to mathcal{O}_Z to 0.
$$
Twisting it by $mathcal{O}(d+p)$ we obtain
$$
dots to mathcal{O}(p-d)^{binom{ell}{2}} to mathcal{O}(p)^ell to mathcal{O}(d+p) to mathcal{O}_Z(d+p) to 0.
$$
Your question is equivalent to injectivity of the induced map
$$
H^0(mathcal{O}(p)^ell) to H^0(mathcal{O}(d+p)).
$$
If $n ge ell$ the cohomology spectral sequence of the twisted Koszul complex proves this. Is that enough for your purposes?
answered 8 hours ago
SashaSasha
21k22755
answered 8 hours ago
SashaSasha
21k22755
answered 8 hours ago
SashaSasha
21k22755
answered 8 hours ago
SashaSasha
21k22755
21k22755
1
$begingroup$
Thanks, but could you explain why the Koszul complex gives that?
$endgroup$
– abx
7 hours ago
$begingroup$
Let $E$ the cokernel of $ldots rightarrow mathcal{O}(p-d)^{binom{l}{2}}$. This truncated comples has length $l-1$. Hence, the dimensions of cohomology groups which play a role in order to compute $H^0(E)$ are less than $l-1$.
$endgroup$
– Libli
6 hours ago
$begingroup$
Hence, if you assume that $l leq n$ and noting that all terms of this truncated complex are acyclic in degree strictly less than $n$, you get that $H^0(E)= 0$ and the injectivity claimed by Sasha.
$endgroup$
– Libli
6 hours ago
add a comment |
1
$begingroup$
Thanks, but could you explain why the Koszul complex gives that?
$endgroup$
– abx
7 hours ago
$begingroup$
Let $E$ the cokernel of $ldots rightarrow mathcal{O}(p-d)^{binom{l}{2}}$. This truncated comples has length $l-1$. Hence, the dimensions of cohomology groups which play a role in order to compute $H^0(E)$ are less than $l-1$.
$endgroup$
– Libli
6 hours ago
$begingroup$
Hence, if you assume that $l leq n$ and noting that all terms of this truncated complex are acyclic in degree strictly less than $n$, you get that $H^0(E)= 0$ and the injectivity claimed by Sasha.
$endgroup$
– Libli
6 hours ago
1
1
$begingroup$
Thanks, but could you explain why the Koszul complex gives that?
$endgroup$
– abx
7 hours ago
$begingroup$
Thanks, but could you explain why the Koszul complex gives that?
$endgroup$
– abx
7 hours ago
$begingroup$
Let $E$ the cokernel of $ldots rightarrow mathcal{O}(p-d)^{binom{l}{2}}$. This truncated comples has length $l-1$. Hence, the dimensions of cohomology groups which play a role in order to compute $H^0(E)$ are less than $l-1$.
$endgroup$
– Libli
6 hours ago
$begingroup$
Let $E$ the cokernel of $ldots rightarrow mathcal{O}(p-d)^{binom{l}{2}}$. This truncated comples has length $l-1$. Hence, the dimensions of cohomology groups which play a role in order to compute $H^0(E)$ are less than $l-1$.
$endgroup$
– Libli
6 hours ago
$begingroup$
Hence, if you assume that $l leq n$ and noting that all terms of this truncated complex are acyclic in degree strictly less than $n$, you get that $H^0(E)= 0$ and the injectivity claimed by Sasha.
$endgroup$
– Libli
6 hours ago
$begingroup$
Hence, if you assume that $l leq n$ and noting that all terms of this truncated complex are acyclic in degree strictly less than $n$, you get that $H^0(E)= 0$ and the injectivity claimed by Sasha.
$endgroup$
– Libli
6 hours ago
add a comment |
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$begingroup$
I think this question needs some clarification. I think you may mean something along the lines of: let $X$ be the moduli space of $n$-tuples of homogeneous polynomials of degree $d$; within $X$, there is a subspace $Y$ consisting of homogeneous polynomials that satisfy a nontrivial relation $sum F_i G_i = 0$ for some $G_i$ homogeneous of degree $p < d$ (i.e. the union of the subspaces determined by each tuple $G$). Is it true that $Y subsetneq X$?
$endgroup$
– user44191
7 hours ago
$begingroup$
Sure. Is that different from what I wrote?
$endgroup$
– abx
7 hours ago
$begingroup$
I found what you wrote ambiguous or underdetermined, e.g. with what you mean by "general"; I thought what I wrote made it more explicitly clear (though it may be overkill as written).
$endgroup$
– user44191
7 hours ago
$begingroup$
This seem trivially false if I haven't misunderstood the question. Take $n = 1, p = 0, d = 1, ell = 2$. Then $F_1 = ax, F_2 = bx$ for some $a, b$, so $b F_1 - a F_2 = 0$. I think both $n, ell$ play some role.
$endgroup$
– user44191
7 hours ago
$begingroup$
Oh, right. $ellleq n$ would be fine for me, though this can be certainly weakened.
$endgroup$
– abx
7 hours ago