Relations between homogeneous polynomials The Next CEO of Stack OverflowIs this stronger Knaster-Kuratowski-Mazurkiewicz Lemma true?Hypersurfaces containing a general chain of linesVarieties with an ample vector bundle mapping to their tangent bundleRestriction of the Canonical Divisor $K_X$ to a general fiberSmoothness of a (given) global section of a vector bundle over G(2,6)Kodaira fibration and moduli space of Riemann surfacesA Special Case of Maximal Rank Conjecturesurjectivity of double dual map for weil divisors on normal varietiesNorm in an order of a number field and Hilbert polynomialNumbers of solutions equal on every finite commutative ring
Relations between homogeneous polynomials
The Next CEO of Stack OverflowIs this stronger Knaster-Kuratowski-Mazurkiewicz Lemma true?Hypersurfaces containing a general chain of linesVarieties with an ample vector bundle mapping to their tangent bundleRestriction of the Canonical Divisor $K_X$ to a general fiberSmoothness of a (given) global section of a vector bundle over G(2,6)Kodaira fibration and moduli space of Riemann surfacesA Special Case of Maximal Rank Conjecturesurjectivity of double dual map for weil divisors on normal varietiesNorm in an order of a number field and Hilbert polynomialNumbers of solutions equal on every finite commutative ring
$begingroup$
Edit: The formulation of my question was incorrect, for several reasons. Here is what I hope to be the correct formulation:
Let $mathbbP$ be a projective space, and $V$ a general linear subspace of $H^0(mathcalO_mathbbP(d))$ (that is, a general point in the corresponding Grassmannian). Then for $p<d$ the multiplication map
$$H^0(mathcalO_mathbbP(p))otimes Vrightarrow H^0(mathcalO_mathbbP(p+d))$$
is of maximal rank, i.e. either injective or surjective.
Is this true? Known? Sasha's answer shows that it is true when $dim V leq dim BbbP+1$.
ag.algebraic-geometry ac.commutative-algebra
asked Mar 19 at 20:17
abxabx
23.9k34885
$endgroup$
|
show 2 more comments
$begingroup$
Edit: The formulation of my question was incorrect, for several reasons. Here is what I hope to be the correct formulation:
Let $mathbbP$ be a projective space, and $V$ a general linear subspace of $H^0(mathcalO_mathbbP(d))$ (that is, a general point in the corresponding Grassmannian). Then for $p<d$ the multiplication map
$$H^0(mathcalO_mathbbP(p))otimes Vrightarrow H^0(mathcalO_mathbbP(p+d))$$
is of maximal rank, i.e. either injective or surjective.
Is this true? Known? Sasha's answer shows that it is true when $dim V leq dim BbbP+1$.
ag.algebraic-geometry ac.commutative-algebra
asked Mar 19 at 20:17
abxabx
23.9k34885
$endgroup$
$begingroup$
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
Mar 19 at 20:57
$begingroup$
Sure. Is that different from what I wrote?
$endgroup$
– abx
Mar 19 at 21:00
$begingroup$
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
Mar 19 at 21:04
1
$begingroup$
Oh, right. $ellleq n$ would be fine for me, though this can be certainly weakened.
$endgroup$
– abx
Mar 19 at 21:43
3
$begingroup$
$ell leq n$ seems far too strong (as in, it makes the question trivial), as then $F_i = x_i^d$ is clearly independent.
$endgroup$
– user44191
Mar 19 at 21:48
|
show 2 more comments
$begingroup$
Edit: The formulation of my question was incorrect, for several reasons. Here is what I hope to be the correct formulation:
Let $mathbbP$ be a projective space, and $V$ a general linear subspace of $H^0(mathcalO_mathbbP(d))$ (that is, a general point in the corresponding Grassmannian). Then for $p<d$ the multiplication map
$$H^0(mathcalO_mathbbP(p))otimes Vrightarrow H^0(mathcalO_mathbbP(p+d))$$
is of maximal rank, i.e. either injective or surjective.
Is this true? Known? Sasha's answer shows that it is true when $dim V leq dim BbbP+1$.
ag.algebraic-geometry ac.commutative-algebra
asked Mar 19 at 20:17
abxabx
23.9k34885
$endgroup$
Edit: The formulation of my question was incorrect, for several reasons. Here is what I hope to be the correct formulation:
Let $mathbbP$ be a projective space, and $V$ a general linear subspace of $H^0(mathcalO_mathbbP(d))$ (that is, a general point in the corresponding Grassmannian). Then for $p<d$ the multiplication map
$$H^0(mathcalO_mathbbP(p))otimes Vrightarrow H^0(mathcalO_mathbbP(p+d))$$
is of maximal rank, i.e. either injective or surjective.
Is this true? Known? Sasha's answer shows that it is true when $dim V leq dim BbbP+1$.
ag.algebraic-geometry ac.commutative-algebra
ag.algebraic-geometry ac.commutative-algebra
asked Mar 19 at 20:17
abxabx
23.9k34885
asked Mar 19 at 20:17
abxabx
23.9k34885
edited Mar 20 at 8:17
abx
asked Mar 19 at 20:17
abxabx
23.9k34885
asked Mar 19 at 20:17
abxabx
23.9k34885
asked Mar 19 at 20:17
abxabx
23.9k34885
23.9k34885
$begingroup$
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
Mar 19 at 20:57
$begingroup$
Sure. Is that different from what I wrote?
$endgroup$
– abx
Mar 19 at 21:00
$begingroup$
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
Mar 19 at 21:04
1
$begingroup$
Oh, right. $ellleq n$ would be fine for me, though this can be certainly weakened.
$endgroup$
– abx
Mar 19 at 21:43
3
$begingroup$
$ell leq n$ seems far too strong (as in, it makes the question trivial), as then $F_i = x_i^d$ is clearly independent.
$endgroup$
– user44191
Mar 19 at 21:48
|
show 2 more comments
$begingroup$
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
Mar 19 at 20:57
$begingroup$
Sure. Is that different from what I wrote?
$endgroup$
– abx
Mar 19 at 21:00
$begingroup$
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
Mar 19 at 21:04
1
$begingroup$
Oh, right. $ellleq n$ would be fine for me, though this can be certainly weakened.
$endgroup$
– abx
Mar 19 at 21:43
3
$begingroup$
$ell leq n$ seems far too strong (as in, it makes the question trivial), as then $F_i = x_i^d$ is clearly independent.
$endgroup$
– user44191
Mar 19 at 21:48
$begingroup$
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
Mar 19 at 20:57
$begingroup$
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
Mar 19 at 20:57
$begingroup$
Sure. Is that different from what I wrote?
$endgroup$
– abx
Mar 19 at 21:00
$begingroup$
Sure. Is that different from what I wrote?
$endgroup$
– abx
Mar 19 at 21:00
$begingroup$
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
Mar 19 at 21:04
$begingroup$
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
Mar 19 at 21:04
1
1
$begingroup$
Oh, right. $ellleq n$ would be fine for me, though this can be certainly weakened.
$endgroup$
– abx
Mar 19 at 21:43
$begingroup$
Oh, right. $ellleq n$ would be fine for me, though this can be certainly weakened.
$endgroup$
– abx
Mar 19 at 21:43
3
3
$begingroup$
$ell leq n$ seems far too strong (as in, it makes the question trivial), as then $F_i = x_i^d$ is clearly independent.
$endgroup$
– user44191
Mar 19 at 21:48
$begingroup$
$ell leq n$ seems far too strong (as in, it makes the question trivial), as then $F_i = x_i^d$ is clearly independent.
$endgroup$
– user44191
Mar 19 at 21:48
|
show 2 more comments
2 Answers
2
active
oldest
votes
$begingroup$
I am posting this as an answer since the comment thread is already long. The question is a special case of Fröberg's Conjecture.
MR0813632 (87f:13022)
Fröberg, Ralf(S-STOC)
An inequality for Hilbert series of graded algebras.
Math. Scand. 56 (1985), no. 2, 117–144.
13H15 (13D03 13H10)
This special case is mostly solved by work of Gleb Nenashev.
MR3621254
Nenashev, Gleb(S-STOC)
A note on Fröberg's conjecture for forms of equal degrees.
C. R. Math. Acad. Sci. Paris 355 (2017), no. 3, 272–276.
13D40
https://arxiv.org/pdf/1512.04324.pdf
Theorem 1 proves the maximal rank conjecture for these maps except for a few values of $p$ near the "changeover" from injectivity to surjectivity. In particular, Nenashev proves injectivity whenever $$textdim H^0(mathbbP,mathcalO_mathbbP(p))otimes V leq textdim H^0(mathbbP,mathcalO_mathbbP(p+d)) - textdim H^0(mathbbP,mathcalO_mathbbP(p))^2,$$
and surjectivity whenever
$$textdim H^0(mathbbP,mathcalO_mathbbP(p))otimes V geq textdim H^0(mathbbP,mathcalO_mathbbP(p+d)) + textdim H^0(mathbbP,mathcalO_mathbbP(p))^2.$$
$endgroup$
$begingroup$
Great, thanks a lot!
$endgroup$
– abx
Mar 20 at 10:35
$begingroup$
You are welcome.
$endgroup$
– Jason Starr
Mar 20 at 10:57
add a comment |
$begingroup$
I think this can be controlled as follows. Let $Z subset mathbbP^n-1$ be the complete intersection defined by $F_i$. Then there is a Koszul resolution
$$
dots to mathcalO(-2d)^binomell2 to mathcalO(-d)^ell to mathcalO to mathcalO_Z to 0.
$$
Twisting it by $mathcalO(d+p)$ we obtain
$$
dots to mathcalO(p-d)^binomell2 to mathcalO(p)^ell to mathcalO(d+p) to mathcalO_Z(d+p) to 0.tag*
$$
Your question is equivalent to injectivity of the induced map
$$
H^0(mathcalO(p)^ell) to H^0(mathcalO(d+p)).
$$
If $n ge ell$ the cohomology spectral sequence of the twisted Koszul complex proves this. Is that enough for your purposes?
EDIT (the spectral sequence argument). The hypercohomology spectral sequence of $(*)$ has first term
$$
E_1^i,j = H^jleft(mathcalO(d+p+id)^binomell-iright),qquad i le 0
$$
and converges to $E_infty^k = H^k(mathcalO_Z(d+p))$. Since a line bundle on a projective space can have only $H^0$ or $H^n-1$, the nonzero terms are only in the rows 0 and $n-1$. The leftmost term of the top row is
$$
E_1^-ell,n-1 = H^n-1left(mathcalO(d+p-ell d)right)
$$
is in the total grading $-ell + n - 1 ge -1$, hence all differentials from it go to terms of total grading $ge 0$. The same of course is true for the other terms in the top row. On the other hand, the leftmost term in the bottom row is
$$
E_1^-1,0 = H^0left(mathcalO(p)^ellright)
$$
is in the total degree $-1$. Thus, no differentials go to this term. Therefore, if the kernel of the differential
$$
d_1^-1,0 colon H^0(mathcalO(p)^ell) to H^0(mathcalO(d+p)).
$$
is nonzero, it survives in the spectral sequence and gives a contribution to $E_infty^-1 = H^-1(mathcalO_Z(d+p)) = 0$, which is absurd.
answered Mar 19 at 20:46
SashaSasha
21.2k22756
$endgroup$
1
$begingroup$
Thanks, but could you explain why the Koszul complex gives that?
$endgroup$
– abx
Mar 19 at 21:03
1
$begingroup$
Let $E$ the cokernel of $ldots rightarrow mathcalO(p-d)^binoml2$. 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
Mar 19 at 22:20
1
$begingroup$
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
Mar 19 at 22:22
$begingroup$
@abx: I added an explanation of the spectral sequence argument.
$endgroup$
– Sasha
Mar 20 at 6:35
$begingroup$
Thanks very much @Sasha, unfortunately I need a little more. I have changed to a more intrinsic formulation.
$endgroup$
– abx
Mar 20 at 8:14
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I am posting this as an answer since the comment thread is already long. The question is a special case of Fröberg's Conjecture.
MR0813632 (87f:13022)
Fröberg, Ralf(S-STOC)
An inequality for Hilbert series of graded algebras.
Math. Scand. 56 (1985), no. 2, 117–144.
13H15 (13D03 13H10)
This special case is mostly solved by work of Gleb Nenashev.
MR3621254
Nenashev, Gleb(S-STOC)
A note on Fröberg's conjecture for forms of equal degrees.
C. R. Math. Acad. Sci. Paris 355 (2017), no. 3, 272–276.
13D40
https://arxiv.org/pdf/1512.04324.pdf
Theorem 1 proves the maximal rank conjecture for these maps except for a few values of $p$ near the "changeover" from injectivity to surjectivity. In particular, Nenashev proves injectivity whenever $$textdim H^0(mathbbP,mathcalO_mathbbP(p))otimes V leq textdim H^0(mathbbP,mathcalO_mathbbP(p+d)) - textdim H^0(mathbbP,mathcalO_mathbbP(p))^2,$$
and surjectivity whenever
$$textdim H^0(mathbbP,mathcalO_mathbbP(p))otimes V geq textdim H^0(mathbbP,mathcalO_mathbbP(p+d)) + textdim H^0(mathbbP,mathcalO_mathbbP(p))^2.$$
$endgroup$
$begingroup$
Great, thanks a lot!
$endgroup$
– abx
Mar 20 at 10:35
$begingroup$
You are welcome.
$endgroup$
– Jason Starr
Mar 20 at 10:57
add a comment |
$begingroup$
I am posting this as an answer since the comment thread is already long. The question is a special case of Fröberg's Conjecture.
MR0813632 (87f:13022)
Fröberg, Ralf(S-STOC)
An inequality for Hilbert series of graded algebras.
Math. Scand. 56 (1985), no. 2, 117–144.
13H15 (13D03 13H10)
This special case is mostly solved by work of Gleb Nenashev.
MR3621254
Nenashev, Gleb(S-STOC)
A note on Fröberg's conjecture for forms of equal degrees.
C. R. Math. Acad. Sci. Paris 355 (2017), no. 3, 272–276.
13D40
https://arxiv.org/pdf/1512.04324.pdf
Theorem 1 proves the maximal rank conjecture for these maps except for a few values of $p$ near the "changeover" from injectivity to surjectivity. In particular, Nenashev proves injectivity whenever $$textdim H^0(mathbbP,mathcalO_mathbbP(p))otimes V leq textdim H^0(mathbbP,mathcalO_mathbbP(p+d)) - textdim H^0(mathbbP,mathcalO_mathbbP(p))^2,$$
and surjectivity whenever
$$textdim H^0(mathbbP,mathcalO_mathbbP(p))otimes V geq textdim H^0(mathbbP,mathcalO_mathbbP(p+d)) + textdim H^0(mathbbP,mathcalO_mathbbP(p))^2.$$
$endgroup$
$begingroup$
Great, thanks a lot!
$endgroup$
– abx
Mar 20 at 10:35
$begingroup$
You are welcome.
$endgroup$
– Jason Starr
Mar 20 at 10:57
add a comment |
$begingroup$
I am posting this as an answer since the comment thread is already long. The question is a special case of Fröberg's Conjecture.
MR0813632 (87f:13022)
Fröberg, Ralf(S-STOC)
An inequality for Hilbert series of graded algebras.
Math. Scand. 56 (1985), no. 2, 117–144.
13H15 (13D03 13H10)
This special case is mostly solved by work of Gleb Nenashev.
MR3621254
Nenashev, Gleb(S-STOC)
A note on Fröberg's conjecture for forms of equal degrees.
C. R. Math. Acad. Sci. Paris 355 (2017), no. 3, 272–276.
13D40
https://arxiv.org/pdf/1512.04324.pdf
Theorem 1 proves the maximal rank conjecture for these maps except for a few values of $p$ near the "changeover" from injectivity to surjectivity. In particular, Nenashev proves injectivity whenever $$textdim H^0(mathbbP,mathcalO_mathbbP(p))otimes V leq textdim H^0(mathbbP,mathcalO_mathbbP(p+d)) - textdim H^0(mathbbP,mathcalO_mathbbP(p))^2,$$
and surjectivity whenever
$$textdim H^0(mathbbP,mathcalO_mathbbP(p))otimes V geq textdim H^0(mathbbP,mathcalO_mathbbP(p+d)) + textdim H^0(mathbbP,mathcalO_mathbbP(p))^2.$$
$endgroup$
I am posting this as an answer since the comment thread is already long. The question is a special case of Fröberg's Conjecture.
MR0813632 (87f:13022)
Fröberg, Ralf(S-STOC)
An inequality for Hilbert series of graded algebras.
Math. Scand. 56 (1985), no. 2, 117–144.
13H15 (13D03 13H10)
This special case is mostly solved by work of Gleb Nenashev.
MR3621254
Nenashev, Gleb(S-STOC)
A note on Fröberg's conjecture for forms of equal degrees.
C. R. Math. Acad. Sci. Paris 355 (2017), no. 3, 272–276.
13D40
https://arxiv.org/pdf/1512.04324.pdf
Theorem 1 proves the maximal rank conjecture for these maps except for a few values of $p$ near the "changeover" from injectivity to surjectivity. In particular, Nenashev proves injectivity whenever $$textdim H^0(mathbbP,mathcalO_mathbbP(p))otimes V leq textdim H^0(mathbbP,mathcalO_mathbbP(p+d)) - textdim H^0(mathbbP,mathcalO_mathbbP(p))^2,$$
and surjectivity whenever
$$textdim H^0(mathbbP,mathcalO_mathbbP(p))otimes V geq textdim H^0(mathbbP,mathcalO_mathbbP(p+d)) + textdim H^0(mathbbP,mathcalO_mathbbP(p))^2.$$
answered Mar 20 at 8:45
community wiki
Jason Starr
$begingroup$
Great, thanks a lot!
$endgroup$
– abx
Mar 20 at 10:35
$begingroup$
You are welcome.
$endgroup$
– Jason Starr
Mar 20 at 10:57
add a comment |
$begingroup$
Great, thanks a lot!
$endgroup$
– abx
Mar 20 at 10:35
$begingroup$
You are welcome.
$endgroup$
– Jason Starr
Mar 20 at 10:57
$begingroup$
Great, thanks a lot!
$endgroup$
– abx
Mar 20 at 10:35
$begingroup$
Great, thanks a lot!
$endgroup$
– abx
Mar 20 at 10:35
$begingroup$
You are welcome.
$endgroup$
– Jason Starr
Mar 20 at 10:57
$begingroup$
You are welcome.
$endgroup$
– Jason Starr
Mar 20 at 10:57
add a comment |
$begingroup$
I think this can be controlled as follows. Let $Z subset mathbbP^n-1$ be the complete intersection defined by $F_i$. Then there is a Koszul resolution
$$
dots to mathcalO(-2d)^binomell2 to mathcalO(-d)^ell to mathcalO to mathcalO_Z to 0.
$$
Twisting it by $mathcalO(d+p)$ we obtain
$$
dots to mathcalO(p-d)^binomell2 to mathcalO(p)^ell to mathcalO(d+p) to mathcalO_Z(d+p) to 0.tag*
$$
Your question is equivalent to injectivity of the induced map
$$
H^0(mathcalO(p)^ell) to H^0(mathcalO(d+p)).
$$
If $n ge ell$ the cohomology spectral sequence of the twisted Koszul complex proves this. Is that enough for your purposes?
EDIT (the spectral sequence argument). The hypercohomology spectral sequence of $(*)$ has first term
$$
E_1^i,j = H^jleft(mathcalO(d+p+id)^binomell-iright),qquad i le 0
$$
and converges to $E_infty^k = H^k(mathcalO_Z(d+p))$. Since a line bundle on a projective space can have only $H^0$ or $H^n-1$, the nonzero terms are only in the rows 0 and $n-1$. The leftmost term of the top row is
$$
E_1^-ell,n-1 = H^n-1left(mathcalO(d+p-ell d)right)
$$
is in the total grading $-ell + n - 1 ge -1$, hence all differentials from it go to terms of total grading $ge 0$. The same of course is true for the other terms in the top row. On the other hand, the leftmost term in the bottom row is
$$
E_1^-1,0 = H^0left(mathcalO(p)^ellright)
$$
is in the total degree $-1$. Thus, no differentials go to this term. Therefore, if the kernel of the differential
$$
d_1^-1,0 colon H^0(mathcalO(p)^ell) to H^0(mathcalO(d+p)).
$$
is nonzero, it survives in the spectral sequence and gives a contribution to $E_infty^-1 = H^-1(mathcalO_Z(d+p)) = 0$, which is absurd.
answered Mar 19 at 20:46
SashaSasha
21.2k22756
$endgroup$
1
$begingroup$
Thanks, but could you explain why the Koszul complex gives that?
$endgroup$
– abx
Mar 19 at 21:03
1
$begingroup$
Let $E$ the cokernel of $ldots rightarrow mathcalO(p-d)^binoml2$. 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
Mar 19 at 22:20
1
$begingroup$
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
Mar 19 at 22:22
$begingroup$
@abx: I added an explanation of the spectral sequence argument.
$endgroup$
– Sasha
Mar 20 at 6:35
$begingroup$
Thanks very much @Sasha, unfortunately I need a little more. I have changed to a more intrinsic formulation.
$endgroup$
– abx
Mar 20 at 8:14
add a comment |
$begingroup$
I think this can be controlled as follows. Let $Z subset mathbbP^n-1$ be the complete intersection defined by $F_i$. Then there is a Koszul resolution
$$
dots to mathcalO(-2d)^binomell2 to mathcalO(-d)^ell to mathcalO to mathcalO_Z to 0.
$$
Twisting it by $mathcalO(d+p)$ we obtain
$$
dots to mathcalO(p-d)^binomell2 to mathcalO(p)^ell to mathcalO(d+p) to mathcalO_Z(d+p) to 0.tag*
$$
Your question is equivalent to injectivity of the induced map
$$
H^0(mathcalO(p)^ell) to H^0(mathcalO(d+p)).
$$
If $n ge ell$ the cohomology spectral sequence of the twisted Koszul complex proves this. Is that enough for your purposes?
EDIT (the spectral sequence argument). The hypercohomology spectral sequence of $(*)$ has first term
$$
E_1^i,j = H^jleft(mathcalO(d+p+id)^binomell-iright),qquad i le 0
$$
and converges to $E_infty^k = H^k(mathcalO_Z(d+p))$. Since a line bundle on a projective space can have only $H^0$ or $H^n-1$, the nonzero terms are only in the rows 0 and $n-1$. The leftmost term of the top row is
$$
E_1^-ell,n-1 = H^n-1left(mathcalO(d+p-ell d)right)
$$
is in the total grading $-ell + n - 1 ge -1$, hence all differentials from it go to terms of total grading $ge 0$. The same of course is true for the other terms in the top row. On the other hand, the leftmost term in the bottom row is
$$
E_1^-1,0 = H^0left(mathcalO(p)^ellright)
$$
is in the total degree $-1$. Thus, no differentials go to this term. Therefore, if the kernel of the differential
$$
d_1^-1,0 colon H^0(mathcalO(p)^ell) to H^0(mathcalO(d+p)).
$$
is nonzero, it survives in the spectral sequence and gives a contribution to $E_infty^-1 = H^-1(mathcalO_Z(d+p)) = 0$, which is absurd.
answered Mar 19 at 20:46
SashaSasha
21.2k22756
$endgroup$
1
$begingroup$
Thanks, but could you explain why the Koszul complex gives that?
$endgroup$
– abx
Mar 19 at 21:03
1
$begingroup$
Let $E$ the cokernel of $ldots rightarrow mathcalO(p-d)^binoml2$. 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
Mar 19 at 22:20
1
$begingroup$
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
Mar 19 at 22:22
$begingroup$
@abx: I added an explanation of the spectral sequence argument.
$endgroup$
– Sasha
Mar 20 at 6:35
$begingroup$
Thanks very much @Sasha, unfortunately I need a little more. I have changed to a more intrinsic formulation.
$endgroup$
– abx
Mar 20 at 8:14
add a comment |
$begingroup$
I think this can be controlled as follows. Let $Z subset mathbbP^n-1$ be the complete intersection defined by $F_i$. Then there is a Koszul resolution
$$
dots to mathcalO(-2d)^binomell2 to mathcalO(-d)^ell to mathcalO to mathcalO_Z to 0.
$$
Twisting it by $mathcalO(d+p)$ we obtain
$$
dots to mathcalO(p-d)^binomell2 to mathcalO(p)^ell to mathcalO(d+p) to mathcalO_Z(d+p) to 0.tag*
$$
Your question is equivalent to injectivity of the induced map
$$
H^0(mathcalO(p)^ell) to H^0(mathcalO(d+p)).
$$
If $n ge ell$ the cohomology spectral sequence of the twisted Koszul complex proves this. Is that enough for your purposes?
EDIT (the spectral sequence argument). The hypercohomology spectral sequence of $(*)$ has first term
$$
E_1^i,j = H^jleft(mathcalO(d+p+id)^binomell-iright),qquad i le 0
$$
and converges to $E_infty^k = H^k(mathcalO_Z(d+p))$. Since a line bundle on a projective space can have only $H^0$ or $H^n-1$, the nonzero terms are only in the rows 0 and $n-1$. The leftmost term of the top row is
$$
E_1^-ell,n-1 = H^n-1left(mathcalO(d+p-ell d)right)
$$
is in the total grading $-ell + n - 1 ge -1$, hence all differentials from it go to terms of total grading $ge 0$. The same of course is true for the other terms in the top row. On the other hand, the leftmost term in the bottom row is
$$
E_1^-1,0 = H^0left(mathcalO(p)^ellright)
$$
is in the total degree $-1$. Thus, no differentials go to this term. Therefore, if the kernel of the differential
$$
d_1^-1,0 colon H^0(mathcalO(p)^ell) to H^0(mathcalO(d+p)).
$$
is nonzero, it survives in the spectral sequence and gives a contribution to $E_infty^-1 = H^-1(mathcalO_Z(d+p)) = 0$, which is absurd.
answered Mar 19 at 20:46
SashaSasha
21.2k22756
$endgroup$
I think this can be controlled as follows. Let $Z subset mathbbP^n-1$ be the complete intersection defined by $F_i$. Then there is a Koszul resolution
$$
dots to mathcalO(-2d)^binomell2 to mathcalO(-d)^ell to mathcalO to mathcalO_Z to 0.
$$
Twisting it by $mathcalO(d+p)$ we obtain
$$
dots to mathcalO(p-d)^binomell2 to mathcalO(p)^ell to mathcalO(d+p) to mathcalO_Z(d+p) to 0.tag*
$$
Your question is equivalent to injectivity of the induced map
$$
H^0(mathcalO(p)^ell) to H^0(mathcalO(d+p)).
$$
If $n ge ell$ the cohomology spectral sequence of the twisted Koszul complex proves this. Is that enough for your purposes?
EDIT (the spectral sequence argument). The hypercohomology spectral sequence of $(*)$ has first term
$$
E_1^i,j = H^jleft(mathcalO(d+p+id)^binomell-iright),qquad i le 0
$$
and converges to $E_infty^k = H^k(mathcalO_Z(d+p))$. Since a line bundle on a projective space can have only $H^0$ or $H^n-1$, the nonzero terms are only in the rows 0 and $n-1$. The leftmost term of the top row is
$$
E_1^-ell,n-1 = H^n-1left(mathcalO(d+p-ell d)right)
$$
is in the total grading $-ell + n - 1 ge -1$, hence all differentials from it go to terms of total grading $ge 0$. The same of course is true for the other terms in the top row. On the other hand, the leftmost term in the bottom row is
$$
E_1^-1,0 = H^0left(mathcalO(p)^ellright)
$$
is in the total degree $-1$. Thus, no differentials go to this term. Therefore, if the kernel of the differential
$$
d_1^-1,0 colon H^0(mathcalO(p)^ell) to H^0(mathcalO(d+p)).
$$
is nonzero, it survives in the spectral sequence and gives a contribution to $E_infty^-1 = H^-1(mathcalO_Z(d+p)) = 0$, which is absurd.
answered Mar 19 at 20:46
SashaSasha
21.2k22756
edited Mar 20 at 6:35
answered Mar 19 at 20:46
SashaSasha
21.2k22756
answered Mar 19 at 20:46
SashaSasha
21.2k22756
answered Mar 19 at 20:46
SashaSasha
21.2k22756
21.2k22756
1
$begingroup$
Thanks, but could you explain why the Koszul complex gives that?
$endgroup$
– abx
Mar 19 at 21:03
1
$begingroup$
Let $E$ the cokernel of $ldots rightarrow mathcalO(p-d)^binoml2$. 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
Mar 19 at 22:20
1
$begingroup$
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
Mar 19 at 22:22
$begingroup$
@abx: I added an explanation of the spectral sequence argument.
$endgroup$
– Sasha
Mar 20 at 6:35
$begingroup$
Thanks very much @Sasha, unfortunately I need a little more. I have changed to a more intrinsic formulation.
$endgroup$
– abx
Mar 20 at 8:14
add a comment |
1
$begingroup$
Thanks, but could you explain why the Koszul complex gives that?
$endgroup$
– abx
Mar 19 at 21:03
1
$begingroup$
Let $E$ the cokernel of $ldots rightarrow mathcalO(p-d)^binoml2$. 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
Mar 19 at 22:20
1
$begingroup$
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
Mar 19 at 22:22
$begingroup$
@abx: I added an explanation of the spectral sequence argument.
$endgroup$
– Sasha
Mar 20 at 6:35
$begingroup$
Thanks very much @Sasha, unfortunately I need a little more. I have changed to a more intrinsic formulation.
$endgroup$
– abx
Mar 20 at 8:14
1
1
$begingroup$
Thanks, but could you explain why the Koszul complex gives that?
$endgroup$
– abx
Mar 19 at 21:03
$begingroup$
Thanks, but could you explain why the Koszul complex gives that?
$endgroup$
– abx
Mar 19 at 21:03
1
1
$begingroup$
Let $E$ the cokernel of $ldots rightarrow mathcalO(p-d)^binoml2$. 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
Mar 19 at 22:20
$begingroup$
Let $E$ the cokernel of $ldots rightarrow mathcalO(p-d)^binoml2$. 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
Mar 19 at 22:20
1
1
$begingroup$
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
Mar 19 at 22:22
$begingroup$
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
Mar 19 at 22:22
$begingroup$
@abx: I added an explanation of the spectral sequence argument.
$endgroup$
– Sasha
Mar 20 at 6:35
$begingroup$
@abx: I added an explanation of the spectral sequence argument.
$endgroup$
– Sasha
Mar 20 at 6:35
$begingroup$
Thanks very much @Sasha, unfortunately I need a little more. I have changed to a more intrinsic formulation.
$endgroup$
– abx
Mar 20 at 8:14
$begingroup$
Thanks very much @Sasha, unfortunately I need a little more. I have changed to a more intrinsic formulation.
$endgroup$
– abx
Mar 20 at 8:14
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
Mar 19 at 20:57
$begingroup$
Sure. Is that different from what I wrote?
$endgroup$
– abx
Mar 19 at 21:00
$begingroup$
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
Mar 19 at 21:04
1
$begingroup$
Oh, right. $ellleq n$ would be fine for me, though this can be certainly weakened.
$endgroup$
– abx
Mar 19 at 21:43
3
$begingroup$
$ell leq n$ seems far too strong (as in, it makes the question trivial), as then $F_i = x_i^d$ is clearly independent.
$endgroup$
– user44191
Mar 19 at 21:48