Existence of Harder-Narasimhan filtration The Next CEO of Stack OverflowThe Harder-Narasimhan filtration with inverse slopes.Rank of a coherent sheaf in terms of coefficients of the Hilbert polynomialDeformation of a point on a Quot schemeDefinition of degree of a coherent sheafDo finite groups act admissibly on separated schemes of finite type over kA characterization of pure sheafOne-dimensional Sheaves and Pushforwards of Vector BundlesOn purity of structure sheaf of a closed subschemeSome basic question on torsion filtration .On torsion sheaf of a coherent sheaf of $dim X$
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Existence of Harder-Narasimhan filtration
The Next CEO of Stack OverflowThe Harder-Narasimhan filtration with inverse slopes.Rank of a coherent sheaf in terms of coefficients of the Hilbert polynomialDeformation of a point on a Quot schemeDefinition of degree of a coherent sheafDo finite groups act admissibly on separated schemes of finite type over kA characterization of pure sheafOne-dimensional Sheaves and Pushforwards of Vector BundlesOn purity of structure sheaf of a closed subschemeSome basic question on torsion filtration .On torsion sheaf of a coherent sheaf of $dim X$
$begingroup$
I am trying to understand the proof of the existence of Harder-Narasimhan filtration from Huybrechts and Lehn.
Let $X$ be a projective scheme with a fixed ample line bundle. Then the theorem says that every pure sheaf has a unique Harder-Narasimhan filtration.
The book first proves the following lemma : let $E$ be a purely $d$-dimensional sheaf. Then there is a subsheaf $Fsubset E$ such that for all subsheaves $Gsubset E$ one has $p(F)geq p(G) $ and in case of equality $Gsubset F$. Moreover $F$ is uniquely determined and semistable. $F$ is the maximal destabilizing subsheaf.
My doubt is as follows. Once we establish the existence of such an $F$, the book says by induction we can assume that $E/F$ has a Harder Narasimhan filtration.
What are we inducting on? My guess is the dimension of the sheaf $E$. But if so I am not able to see why dimension of $E/F$ is strictly less than dimension of $ E$. Any help will be appreciated!
algebraic-geometry schemes vector-bundles
asked Mar 6 '15 at 9:46
gradstudentgradstudent
1,313720
$endgroup$
add a comment |
$begingroup$
I am trying to understand the proof of the existence of Harder-Narasimhan filtration from Huybrechts and Lehn.
Let $X$ be a projective scheme with a fixed ample line bundle. Then the theorem says that every pure sheaf has a unique Harder-Narasimhan filtration.
The book first proves the following lemma : let $E$ be a purely $d$-dimensional sheaf. Then there is a subsheaf $Fsubset E$ such that for all subsheaves $Gsubset E$ one has $p(F)geq p(G) $ and in case of equality $Gsubset F$. Moreover $F$ is uniquely determined and semistable. $F$ is the maximal destabilizing subsheaf.
My doubt is as follows. Once we establish the existence of such an $F$, the book says by induction we can assume that $E/F$ has a Harder Narasimhan filtration.
What are we inducting on? My guess is the dimension of the sheaf $E$. But if so I am not able to see why dimension of $E/F$ is strictly less than dimension of $ E$. Any help will be appreciated!
algebraic-geometry schemes vector-bundles
asked Mar 6 '15 at 9:46
gradstudentgradstudent
1,313720
$endgroup$
2
$begingroup$
I remember this proof being terrible. I do think that one is inducting on dimension, since one then passes from $E$ to $E/mathrmHN_1(E)$ which has strictly smaller dimension, and then proceeds from there.
$endgroup$
– Alex Youcis
Mar 6 '15 at 9:59
$begingroup$
Thanks @Alex! But can you tell me why it has strictly smaller dimension?
$endgroup$
– gradstudent
Mar 6 '15 at 10:04
add a comment |
$begingroup$
I am trying to understand the proof of the existence of Harder-Narasimhan filtration from Huybrechts and Lehn.
Let $X$ be a projective scheme with a fixed ample line bundle. Then the theorem says that every pure sheaf has a unique Harder-Narasimhan filtration.
The book first proves the following lemma : let $E$ be a purely $d$-dimensional sheaf. Then there is a subsheaf $Fsubset E$ such that for all subsheaves $Gsubset E$ one has $p(F)geq p(G) $ and in case of equality $Gsubset F$. Moreover $F$ is uniquely determined and semistable. $F$ is the maximal destabilizing subsheaf.
My doubt is as follows. Once we establish the existence of such an $F$, the book says by induction we can assume that $E/F$ has a Harder Narasimhan filtration.
What are we inducting on? My guess is the dimension of the sheaf $E$. But if so I am not able to see why dimension of $E/F$ is strictly less than dimension of $ E$. Any help will be appreciated!
algebraic-geometry schemes vector-bundles
asked Mar 6 '15 at 9:46
gradstudentgradstudent
1,313720
$endgroup$
I am trying to understand the proof of the existence of Harder-Narasimhan filtration from Huybrechts and Lehn.
Let $X$ be a projective scheme with a fixed ample line bundle. Then the theorem says that every pure sheaf has a unique Harder-Narasimhan filtration.
The book first proves the following lemma : let $E$ be a purely $d$-dimensional sheaf. Then there is a subsheaf $Fsubset E$ such that for all subsheaves $Gsubset E$ one has $p(F)geq p(G) $ and in case of equality $Gsubset F$. Moreover $F$ is uniquely determined and semistable. $F$ is the maximal destabilizing subsheaf.
My doubt is as follows. Once we establish the existence of such an $F$, the book says by induction we can assume that $E/F$ has a Harder Narasimhan filtration.
What are we inducting on? My guess is the dimension of the sheaf $E$. But if so I am not able to see why dimension of $E/F$ is strictly less than dimension of $ E$. Any help will be appreciated!
algebraic-geometry schemes vector-bundles
algebraic-geometry schemes vector-bundles
asked Mar 6 '15 at 9:46
gradstudentgradstudent
1,313720
asked Mar 6 '15 at 9:46
gradstudentgradstudent
1,313720
edited Mar 6 '15 at 17:43
gradstudent
asked Mar 6 '15 at 9:46
gradstudentgradstudent
1,313720
asked Mar 6 '15 at 9:46
gradstudentgradstudent
1,313720
asked Mar 6 '15 at 9:46
gradstudentgradstudent
1,313720
1,313720
2
$begingroup$
I remember this proof being terrible. I do think that one is inducting on dimension, since one then passes from $E$ to $E/mathrmHN_1(E)$ which has strictly smaller dimension, and then proceeds from there.
$endgroup$
– Alex Youcis
Mar 6 '15 at 9:59
$begingroup$
Thanks @Alex! But can you tell me why it has strictly smaller dimension?
$endgroup$
– gradstudent
Mar 6 '15 at 10:04
add a comment |
2
$begingroup$
I remember this proof being terrible. I do think that one is inducting on dimension, since one then passes from $E$ to $E/mathrmHN_1(E)$ which has strictly smaller dimension, and then proceeds from there.
$endgroup$
– Alex Youcis
Mar 6 '15 at 9:59
$begingroup$
Thanks @Alex! But can you tell me why it has strictly smaller dimension?
$endgroup$
– gradstudent
Mar 6 '15 at 10:04
2
2
$begingroup$
I remember this proof being terrible. I do think that one is inducting on dimension, since one then passes from $E$ to $E/mathrmHN_1(E)$ which has strictly smaller dimension, and then proceeds from there.
$endgroup$
– Alex Youcis
Mar 6 '15 at 9:59
$begingroup$
I remember this proof being terrible. I do think that one is inducting on dimension, since one then passes from $E$ to $E/mathrmHN_1(E)$ which has strictly smaller dimension, and then proceeds from there.
$endgroup$
– Alex Youcis
Mar 6 '15 at 9:59
$begingroup$
Thanks @Alex! But can you tell me why it has strictly smaller dimension?
$endgroup$
– gradstudent
Mar 6 '15 at 10:04
$begingroup$
Thanks @Alex! But can you tell me why it has strictly smaller dimension?
$endgroup$
– gradstudent
Mar 6 '15 at 10:04
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I think we can induct on the rank of $E$. Look at the sequence $0 to F to E to E/F to 0$. Since $rk(E) = rk(F) + rk(E/F)$, and $F$ is a proper nonzero subsheaf of $E$, so $rk(E/F) < rk(E)$, and so we can proceed via induction.
answered Nov 20 '15 at 16:25
nujranujra
1
$endgroup$
add a comment |
$begingroup$
I was thinking about induction on $alpha_d(E)$, where $alpha_d(E)/d!$ is the leading coefficient of the Hilbert polynomial of $E$, with the same notation as Huybrechts & Lehn. It is additive on exact sequences so we should have $alpha_d(E/F)<alpha_d(E)$. Moreover the base case $alpha_d(E)=0$ should be trivial since it implies $E=0$ (by a previous remark after the definition of the Hilbert polynomial).
In fact thinking again about it, it is quite the same as inducting on the rank, since $textrk(E)=alpha_d(E)/alpha_d(O_X)$; the only (apparent) problem with rank is that it is not an integer in general.
answered Mar 28 at 15:25
OromisOromis
404412
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
I think we can induct on the rank of $E$. Look at the sequence $0 to F to E to E/F to 0$. Since $rk(E) = rk(F) + rk(E/F)$, and $F$ is a proper nonzero subsheaf of $E$, so $rk(E/F) < rk(E)$, and so we can proceed via induction.
answered Nov 20 '15 at 16:25
nujranujra
1
$endgroup$
add a comment |
$begingroup$
I think we can induct on the rank of $E$. Look at the sequence $0 to F to E to E/F to 0$. Since $rk(E) = rk(F) + rk(E/F)$, and $F$ is a proper nonzero subsheaf of $E$, so $rk(E/F) < rk(E)$, and so we can proceed via induction.
answered Nov 20 '15 at 16:25
nujranujra
1
$endgroup$
add a comment |
$begingroup$
I think we can induct on the rank of $E$. Look at the sequence $0 to F to E to E/F to 0$. Since $rk(E) = rk(F) + rk(E/F)$, and $F$ is a proper nonzero subsheaf of $E$, so $rk(E/F) < rk(E)$, and so we can proceed via induction.
answered Nov 20 '15 at 16:25
nujranujra
1
$endgroup$
I think we can induct on the rank of $E$. Look at the sequence $0 to F to E to E/F to 0$. Since $rk(E) = rk(F) + rk(E/F)$, and $F$ is a proper nonzero subsheaf of $E$, so $rk(E/F) < rk(E)$, and so we can proceed via induction.
answered Nov 20 '15 at 16:25
nujranujra
1
answered Nov 20 '15 at 16:25
nujranujra
1
answered Nov 20 '15 at 16:25
nujranujra
1
answered Nov 20 '15 at 16:25
nujranujra
1
1
add a comment |
add a comment |
$begingroup$
I was thinking about induction on $alpha_d(E)$, where $alpha_d(E)/d!$ is the leading coefficient of the Hilbert polynomial of $E$, with the same notation as Huybrechts & Lehn. It is additive on exact sequences so we should have $alpha_d(E/F)<alpha_d(E)$. Moreover the base case $alpha_d(E)=0$ should be trivial since it implies $E=0$ (by a previous remark after the definition of the Hilbert polynomial).
In fact thinking again about it, it is quite the same as inducting on the rank, since $textrk(E)=alpha_d(E)/alpha_d(O_X)$; the only (apparent) problem with rank is that it is not an integer in general.
answered Mar 28 at 15:25
OromisOromis
404412
$endgroup$
add a comment |
$begingroup$
I was thinking about induction on $alpha_d(E)$, where $alpha_d(E)/d!$ is the leading coefficient of the Hilbert polynomial of $E$, with the same notation as Huybrechts & Lehn. It is additive on exact sequences so we should have $alpha_d(E/F)<alpha_d(E)$. Moreover the base case $alpha_d(E)=0$ should be trivial since it implies $E=0$ (by a previous remark after the definition of the Hilbert polynomial).
In fact thinking again about it, it is quite the same as inducting on the rank, since $textrk(E)=alpha_d(E)/alpha_d(O_X)$; the only (apparent) problem with rank is that it is not an integer in general.
answered Mar 28 at 15:25
OromisOromis
404412
$endgroup$
add a comment |
$begingroup$
I was thinking about induction on $alpha_d(E)$, where $alpha_d(E)/d!$ is the leading coefficient of the Hilbert polynomial of $E$, with the same notation as Huybrechts & Lehn. It is additive on exact sequences so we should have $alpha_d(E/F)<alpha_d(E)$. Moreover the base case $alpha_d(E)=0$ should be trivial since it implies $E=0$ (by a previous remark after the definition of the Hilbert polynomial).
In fact thinking again about it, it is quite the same as inducting on the rank, since $textrk(E)=alpha_d(E)/alpha_d(O_X)$; the only (apparent) problem with rank is that it is not an integer in general.
answered Mar 28 at 15:25
OromisOromis
404412
$endgroup$
I was thinking about induction on $alpha_d(E)$, where $alpha_d(E)/d!$ is the leading coefficient of the Hilbert polynomial of $E$, with the same notation as Huybrechts & Lehn. It is additive on exact sequences so we should have $alpha_d(E/F)<alpha_d(E)$. Moreover the base case $alpha_d(E)=0$ should be trivial since it implies $E=0$ (by a previous remark after the definition of the Hilbert polynomial).
In fact thinking again about it, it is quite the same as inducting on the rank, since $textrk(E)=alpha_d(E)/alpha_d(O_X)$; the only (apparent) problem with rank is that it is not an integer in general.
answered Mar 28 at 15:25
OromisOromis
404412
answered Mar 28 at 15:25
OromisOromis
404412
answered Mar 28 at 15:25
OromisOromis
404412
answered Mar 28 at 15:25
OromisOromis
404412
404412
add a comment |
add a comment |
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I remember this proof being terrible. I do think that one is inducting on dimension, since one then passes from $E$ to $E/mathrmHN_1(E)$ which has strictly smaller dimension, and then proceeds from there.
$endgroup$
– Alex Youcis
Mar 6 '15 at 9:59
$begingroup$
Thanks @Alex! But can you tell me why it has strictly smaller dimension?
$endgroup$
– gradstudent
Mar 6 '15 at 10:04