Relations between homogeneous polynomialsIs 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


Is 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













9












$begingroup$


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 $dim V leq dim BbbP+1$.










share|cite|improve this question











$endgroup$











  • $begingroup$
    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
















9












$begingroup$


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 $dim V leq dim BbbP+1$.










share|cite|improve this question











$endgroup$











  • $begingroup$
    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














9












9








9


1



$begingroup$


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 $dim V leq dim BbbP+1$.










share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 20 at 8:17







abx

















asked Mar 19 at 20:17









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
    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











  • $begingroup$
    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











2 Answers
2






active

oldest

votes


















7












$begingroup$

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
$$textdim H^0(mathbbP,mathcalO_mathbbP(p))otimes V geq textdim H^0(mathbbP,mathcalO_mathbbP(p+d)) + textdim H^0(mathbbP,mathcalO_mathbbP(p))^2.$$






share|cite|improve this answer











$endgroup$












  • $begingroup$
    Great, thanks a lot!
    $endgroup$
    – abx
    Mar 20 at 10:35










  • $begingroup$
    You are welcome.
    $endgroup$
    – Jason Starr
    Mar 20 at 10:57


















9












$begingroup$

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.






share|cite|improve this answer











$endgroup$








  • 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










Your Answer





StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "504"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);













draft saved

draft discarded


















StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f325793%2frelations-between-homogeneous-polynomials%23new-answer', 'question_page');

);

Post as a guest















Required, but never shown

























2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









7












$begingroup$

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
$$textdim H^0(mathbbP,mathcalO_mathbbP(p))otimes V geq textdim H^0(mathbbP,mathcalO_mathbbP(p+d)) + textdim H^0(mathbbP,mathcalO_mathbbP(p))^2.$$






share|cite|improve this answer











$endgroup$












  • $begingroup$
    Great, thanks a lot!
    $endgroup$
    – abx
    Mar 20 at 10:35










  • $begingroup$
    You are welcome.
    $endgroup$
    – Jason Starr
    Mar 20 at 10:57















7












$begingroup$

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
$$textdim H^0(mathbbP,mathcalO_mathbbP(p))otimes V geq textdim H^0(mathbbP,mathcalO_mathbbP(p+d)) + textdim H^0(mathbbP,mathcalO_mathbbP(p))^2.$$






share|cite|improve this answer











$endgroup$












  • $begingroup$
    Great, thanks a lot!
    $endgroup$
    – abx
    Mar 20 at 10:35










  • $begingroup$
    You are welcome.
    $endgroup$
    – Jason Starr
    Mar 20 at 10:57













7












7








7





$begingroup$

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
$$textdim H^0(mathbbP,mathcalO_mathbbP(p))otimes V geq textdim H^0(mathbbP,mathcalO_mathbbP(p+d)) + textdim H^0(mathbbP,mathcalO_mathbbP(p))^2.$$






share|cite|improve this answer











$endgroup$



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
$$textdim H^0(mathbbP,mathcalO_mathbbP(p))otimes V geq textdim H^0(mathbbP,mathcalO_mathbbP(p+d)) + textdim H^0(mathbbP,mathcalO_mathbbP(p))^2.$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








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
















  • $begingroup$
    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











9












$begingroup$

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.






share|cite|improve this answer











$endgroup$








  • 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















9












$begingroup$

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.






share|cite|improve this answer











$endgroup$








  • 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













9












9








9





$begingroup$

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.






share|cite|improve this answer











$endgroup$



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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Mar 20 at 6:35

























answered Mar 19 at 20:46









SashaSasha

21.1k22755




21.1k22755







  • 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




    $begingroup$
    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

















draft saved

draft discarded
















































Thanks for contributing an answer to MathOverflow!


  • Please be sure to answer the question. Provide details and share your research!

But avoid


  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.

Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f325793%2frelations-between-homogeneous-polynomials%23new-answer', 'question_page');

);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







Popular posts from this blog

Færeyskur hestur Heimild | Tengill | Tilvísanir | LeiðsagnarvalRossið - síða um færeyska hrossið á færeyskuGott ár hjá færeyska hestinum

He _____ here since 1970 . Answer needed [closed]What does “since he was so high” mean?Meaning of “catch birds for”?How do I ensure “since” takes the meaning I want?“Who cares here” meaningWhat does “right round toward” mean?the time tense (had now been detected)What does the phrase “ring around the roses” mean here?Correct usage of “visited upon”Meaning of “foiled rail sabotage bid”It was the third time I had gone to Rome or It is the third time I had been to Rome

Slayer Innehåll Historia | Stil, komposition och lyrik | Bandets betydelse och framgångar | Sidoprojekt och samarbeten | Kontroverser | Medlemmar | Utmärkelser och nomineringar | Turnéer och festivaler | Diskografi | Referenser | Externa länkar | Navigeringsmenywww.slayer.net”Metal Massacre vol. 1””Metal Massacre vol. 3””Metal Massacre Volume III””Show No Mercy””Haunting the Chapel””Live Undead””Hell Awaits””Reign in Blood””Reign in Blood””Gold & Platinum – Reign in Blood””Golden Gods Awards Winners”originalet”Kerrang! Hall Of Fame””Slayer Looks Back On 37-Year Career In New Video Series: Part Two””South of Heaven””Gold & Platinum – South of Heaven””Seasons in the Abyss””Gold & Platinum - Seasons in the Abyss””Divine Intervention””Divine Intervention - Release group by Slayer””Gold & Platinum - Divine Intervention””Live Intrusion””Undisputed Attitude””Abolish Government/Superficial Love””Release “Slatanic Slaughter: A Tribute to Slayer” by Various Artists””Diabolus in Musica””Soundtrack to the Apocalypse””God Hates Us All””Systematic - Relationships””War at the Warfield””Gold & Platinum - War at the Warfield””Soundtrack to the Apocalypse””Gold & Platinum - Still Reigning””Metallica, Slayer, Iron Mauden Among Winners At Metal Hammer Awards””Eternal Pyre””Eternal Pyre - Slayer release group””Eternal Pyre””Metal Storm Awards 2006””Kerrang! Hall Of Fame””Slayer Wins 'Best Metal' Grammy Award””Slayer Guitarist Jeff Hanneman Dies””Bullet-For My Valentine booed at Metal Hammer Golden Gods Awards””Unholy Aliance””The End Of Slayer?””Slayer: We Could Thrash Out Two More Albums If We're Fast Enough...””'The Unholy Alliance: Chapter III' UK Dates Added”originalet”Megadeth And Slayer To Co-Headline 'Canadian Carnage' Trek”originalet”World Painted Blood””Release “World Painted Blood” by Slayer””Metallica Heading To Cinemas””Slayer, Megadeth To Join Forces For 'European Carnage' Tour - Dec. 18, 2010”originalet”Slayer's Hanneman Contracts Acute Infection; Band To Bring In Guest Guitarist””Cannibal Corpse's Pat O'Brien Will Step In As Slayer's Guest Guitarist”originalet”Slayer’s Jeff Hanneman Dead at 49””Dave Lombardo Says He Made Only $67,000 In 2011 While Touring With Slayer””Slayer: We Do Not Agree With Dave Lombardo's Substance Or Timeline Of Events””Slayer Welcomes Drummer Paul Bostaph Back To The Fold””Slayer Hope to Unveil Never-Before-Heard Jeff Hanneman Material on Next Album””Slayer Debut New Song 'Implode' During Surprise Golden Gods Appearance””Release group Repentless by Slayer””Repentless - Slayer - Credits””Slayer””Metal Storm Awards 2015””Slayer - to release comic book "Repentless #1"””Slayer To Release 'Repentless' 6.66" Vinyl Box Set””BREAKING NEWS: Slayer Announce Farewell Tour””Slayer Recruit Lamb of God, Anthrax, Behemoth + Testament for Final Tour””Slayer lägger ner efter 37 år””Slayer Announces Second North American Leg Of 'Final' Tour””Final World Tour””Slayer Announces Final European Tour With Lamb of God, Anthrax And Obituary””Slayer To Tour Europe With Lamb of God, Anthrax And Obituary””Slayer To Play 'Last French Show Ever' At Next Year's Hellfst””Slayer's Final World Tour Will Extend Into 2019””Death Angel's Rob Cavestany On Slayer's 'Farewell' Tour: 'Some Of Us Could See This Coming'””Testament Has No Plans To Retire Anytime Soon, Says Chuck Billy””Anthrax's Scott Ian On Slayer's 'Farewell' Tour Plans: 'I Was Surprised And I Wasn't Surprised'””Slayer””Slayer's Morbid Schlock””Review/Rock; For Slayer, the Mania Is the Message””Slayer - Biography””Slayer - Reign In Blood”originalet”Dave Lombardo””An exclusive oral history of Slayer”originalet”Exclusive! Interview With Slayer Guitarist Jeff Hanneman”originalet”Thinking Out Loud: Slayer's Kerry King on hair metal, Satan and being polite””Slayer Lyrics””Slayer - Biography””Most influential artists for extreme metal music””Slayer - Reign in Blood””Slayer guitarist Jeff Hanneman dies aged 49””Slatanic Slaughter: A Tribute to Slayer””Gateway to Hell: A Tribute to Slayer””Covered In Blood””Slayer: The Origins of Thrash in San Francisco, CA.””Why They Rule - #6 Slayer”originalet”Guitar World's 100 Greatest Heavy Metal Guitarists Of All Time”originalet”The fans have spoken: Slayer comes out on top in readers' polls”originalet”Tribute to Jeff Hanneman (1964-2013)””Lamb Of God Frontman: We Sound Like A Slayer Rip-Off””BEHEMOTH Frontman Pays Tribute To SLAYER's JEFF HANNEMAN””Slayer, Hatebreed Doing Double Duty On This Year's Ozzfest””System of a Down””Lacuna Coil’s Andrea Ferro Talks Influences, Skateboarding, Band Origins + More””Slayer - Reign in Blood””Into The Lungs of Hell””Slayer rules - en utställning om fans””Slayer and Their Fans Slashed Through a No-Holds-Barred Night at Gas Monkey””Home””Slayer””Gold & Platinum - The Big 4 Live from Sofia, Bulgaria””Exclusive! Interview With Slayer Guitarist Kerry King””2008-02-23: Wiltern, Los Angeles, CA, USA””Slayer's Kerry King To Perform With Megadeth Tonight! - Oct. 21, 2010”originalet”Dave Lombardo - Biography”Slayer Case DismissedArkiveradUltimate Classic Rock: Slayer guitarist Jeff Hanneman dead at 49.”Slayer: "We could never do any thing like Some Kind Of Monster..."””Cannibal Corpse'S Pat O'Brien Will Step In As Slayer'S Guest Guitarist | The Official Slayer Site”originalet”Slayer Wins 'Best Metal' Grammy Award””Slayer Guitarist Jeff Hanneman Dies””Kerrang! Awards 2006 Blog: Kerrang! Hall Of Fame””Kerrang! Awards 2013: Kerrang! Legend”originalet”Metallica, Slayer, Iron Maien Among Winners At Metal Hammer Awards””Metal Hammer Golden Gods Awards””Bullet For My Valentine Booed At Metal Hammer Golden Gods Awards””Metal Storm Awards 2006””Metal Storm Awards 2015””Slayer's Concert History””Slayer - Relationships””Slayer - Releases”Slayers officiella webbplatsSlayer på MusicBrainzOfficiell webbplatsSlayerSlayerr1373445760000 0001 1540 47353068615-5086262726cb13906545x(data)6033143kn20030215029