Mixing
Exact sequences of sheaves
I don't understand something : in the notes by Justin Campell "Some examples of the Riemann-Hilbert correspondence", it is stated that there is an exact sequence $$ 0 to delta_{infty} to j_!IC_{Bbb A^1} to IC_{Bbb P^1}to 0$$ where $Bbb P^1$ is stratified as $Bbb A^1 sqcup {infty}$ and $delta_{infty} $ is the skyscraper sheaf with constant stalk $Bbb Q$.
However, taking the stalk at infinity gives $0 to Bbb Q to 0 to Bbb Q to 0 $ which is clearly a contradiction. As far as I understand, that this sequence was obtained by taking the Verdier dual of the sequence $$ 0 to IC_{Bbb P^1} to j_*IC_{Bbb A^1} to delta_{infty} to 0 $$ which seems plausible but where is my mistake ?
algebraic-geometry algebraic-topology sheaf-cohomology
asked Nov 20 '18 at 18:01
student
1078
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show 1 more comment
I don't understand something : in the notes by Justin Campell "Some examples of the Riemann-Hilbert correspondence", it is stated that there is an exact sequence $$ 0 to delta_{infty} to j_!IC_{Bbb A^1} to IC_{Bbb P^1}to 0$$ where $Bbb P^1$ is stratified as $Bbb A^1 sqcup {infty}$ and $delta_{infty} $ is the skyscraper sheaf with constant stalk $Bbb Q$.
However, taking the stalk at infinity gives $0 to Bbb Q to 0 to Bbb Q to 0 $ which is clearly a contradiction. As far as I understand, that this sequence was obtained by taking the Verdier dual of the sequence $$ 0 to IC_{Bbb P^1} to j_*IC_{Bbb A^1} to delta_{infty} to 0 $$ which seems plausible but where is my mistake ?
algebraic-geometry algebraic-topology sheaf-cohomology
asked Nov 20 '18 at 18:01
student
1078
I think the author meant exact sequence of perverse sheaves.
– Roland
Nov 20 '18 at 18:25
@Roland : I see, so is there a "concrete" description of what is an exact sequence of perverse sheaves ? Is there a way I can think of it in term of classical exact sequence of sheaves ? I find it really confusing that perverse epimorphism does not induce surjective map on stalks.
– student
Nov 20 '18 at 18:30
Think in term of derived category. For a sequence $0to Ato Bto Cto 0$ in a triangulated category $mathcal{T}$ to be exact for a $t$-structure, it is enough that they all lie in the heart and that $Ato Bto Cto A[1]$ is distinguished (Just take the long exact sequence of cohomology). I'm not sure about the converse though. But here, you have the triangle $delta_{infty}to j_! IC_{mathbb{A}^1}to IC_{mathbb{P}^1}todelta_{infty}[1]$ (which is just a rotation of a triangle of sheaves), and they all are perverse sheaves. So this is an exact sequence of perverse sheaves.
– Roland
Nov 20 '18 at 18:42
But yes, the stalk functor is not exact in the category of perverse sheaves.
– Roland
Nov 20 '18 at 18:43
@Roland : thanks a lot, your explanations are wonderful as always !
– student
Nov 20 '18 at 19:54
|
show 1 more comment
I don't understand something : in the notes by Justin Campell "Some examples of the Riemann-Hilbert correspondence", it is stated that there is an exact sequence $$ 0 to delta_{infty} to j_!IC_{Bbb A^1} to IC_{Bbb P^1}to 0$$ where $Bbb P^1$ is stratified as $Bbb A^1 sqcup {infty}$ and $delta_{infty} $ is the skyscraper sheaf with constant stalk $Bbb Q$.
However, taking the stalk at infinity gives $0 to Bbb Q to 0 to Bbb Q to 0 $ which is clearly a contradiction. As far as I understand, that this sequence was obtained by taking the Verdier dual of the sequence $$ 0 to IC_{Bbb P^1} to j_*IC_{Bbb A^1} to delta_{infty} to 0 $$ which seems plausible but where is my mistake ?
algebraic-geometry algebraic-topology sheaf-cohomology
asked Nov 20 '18 at 18:01
student
1078
I don't understand something : in the notes by Justin Campell "Some examples of the Riemann-Hilbert correspondence", it is stated that there is an exact sequence $$ 0 to delta_{infty} to j_!IC_{Bbb A^1} to IC_{Bbb P^1}to 0$$ where $Bbb P^1$ is stratified as $Bbb A^1 sqcup {infty}$ and $delta_{infty} $ is the skyscraper sheaf with constant stalk $Bbb Q$.
However, taking the stalk at infinity gives $0 to Bbb Q to 0 to Bbb Q to 0 $ which is clearly a contradiction. As far as I understand, that this sequence was obtained by taking the Verdier dual of the sequence $$ 0 to IC_{Bbb P^1} to j_*IC_{Bbb A^1} to delta_{infty} to 0 $$ which seems plausible but where is my mistake ?
algebraic-geometry algebraic-topology sheaf-cohomology
algebraic-geometry algebraic-topology sheaf-cohomology
asked Nov 20 '18 at 18:01
student
1078
asked Nov 20 '18 at 18:01
student
1078
asked Nov 20 '18 at 18:01
student
1078
asked Nov 20 '18 at 18:01
student
1078
asked Nov 20 '18 at 18:01
student
1078
1078
I think the author meant exact sequence of perverse sheaves.
– Roland
Nov 20 '18 at 18:25
@Roland : I see, so is there a "concrete" description of what is an exact sequence of perverse sheaves ? Is there a way I can think of it in term of classical exact sequence of sheaves ? I find it really confusing that perverse epimorphism does not induce surjective map on stalks.
– student
Nov 20 '18 at 18:30
Think in term of derived category. For a sequence $0to Ato Bto Cto 0$ in a triangulated category $mathcal{T}$ to be exact for a $t$-structure, it is enough that they all lie in the heart and that $Ato Bto Cto A[1]$ is distinguished (Just take the long exact sequence of cohomology). I'm not sure about the converse though. But here, you have the triangle $delta_{infty}to j_! IC_{mathbb{A}^1}to IC_{mathbb{P}^1}todelta_{infty}[1]$ (which is just a rotation of a triangle of sheaves), and they all are perverse sheaves. So this is an exact sequence of perverse sheaves.
– Roland
Nov 20 '18 at 18:42
But yes, the stalk functor is not exact in the category of perverse sheaves.
– Roland
Nov 20 '18 at 18:43
@Roland : thanks a lot, your explanations are wonderful as always !
– student
Nov 20 '18 at 19:54
|
show 1 more comment
I think the author meant exact sequence of perverse sheaves.
– Roland
Nov 20 '18 at 18:25
@Roland : I see, so is there a "concrete" description of what is an exact sequence of perverse sheaves ? Is there a way I can think of it in term of classical exact sequence of sheaves ? I find it really confusing that perverse epimorphism does not induce surjective map on stalks.
– student
Nov 20 '18 at 18:30
Think in term of derived category. For a sequence $0to Ato Bto Cto 0$ in a triangulated category $mathcal{T}$ to be exact for a $t$-structure, it is enough that they all lie in the heart and that $Ato Bto Cto A[1]$ is distinguished (Just take the long exact sequence of cohomology). I'm not sure about the converse though. But here, you have the triangle $delta_{infty}to j_! IC_{mathbb{A}^1}to IC_{mathbb{P}^1}todelta_{infty}[1]$ (which is just a rotation of a triangle of sheaves), and they all are perverse sheaves. So this is an exact sequence of perverse sheaves.
– Roland
Nov 20 '18 at 18:42
But yes, the stalk functor is not exact in the category of perverse sheaves.
– Roland
Nov 20 '18 at 18:43
@Roland : thanks a lot, your explanations are wonderful as always !
– student
Nov 20 '18 at 19:54
I think the author meant exact sequence of perverse sheaves.
– Roland
Nov 20 '18 at 18:25
I think the author meant exact sequence of perverse sheaves.
– Roland
Nov 20 '18 at 18:25
@Roland : I see, so is there a "concrete" description of what is an exact sequence of perverse sheaves ? Is there a way I can think of it in term of classical exact sequence of sheaves ? I find it really confusing that perverse epimorphism does not induce surjective map on stalks.
– student
Nov 20 '18 at 18:30
@Roland : I see, so is there a "concrete" description of what is an exact sequence of perverse sheaves ? Is there a way I can think of it in term of classical exact sequence of sheaves ? I find it really confusing that perverse epimorphism does not induce surjective map on stalks.
– student
Nov 20 '18 at 18:30
Think in term of derived category. For a sequence $0to Ato Bto Cto 0$ in a triangulated category $mathcal{T}$ to be exact for a $t$-structure, it is enough that they all lie in the heart and that $Ato Bto Cto A[1]$ is distinguished (Just take the long exact sequence of cohomology). I'm not sure about the converse though. But here, you have the triangle $delta_{infty}to j_! IC_{mathbb{A}^1}to IC_{mathbb{P}^1}todelta_{infty}[1]$ (which is just a rotation of a triangle of sheaves), and they all are perverse sheaves. So this is an exact sequence of perverse sheaves.
– Roland
Nov 20 '18 at 18:42
Think in term of derived category. For a sequence $0to Ato Bto Cto 0$ in a triangulated category $mathcal{T}$ to be exact for a $t$-structure, it is enough that they all lie in the heart and that $Ato Bto Cto A[1]$ is distinguished (Just take the long exact sequence of cohomology). I'm not sure about the converse though. But here, you have the triangle $delta_{infty}to j_! IC_{mathbb{A}^1}to IC_{mathbb{P}^1}todelta_{infty}[1]$ (which is just a rotation of a triangle of sheaves), and they all are perverse sheaves. So this is an exact sequence of perverse sheaves.
– Roland
Nov 20 '18 at 18:42
But yes, the stalk functor is not exact in the category of perverse sheaves.
– Roland
Nov 20 '18 at 18:43
But yes, the stalk functor is not exact in the category of perverse sheaves.
– Roland
Nov 20 '18 at 18:43
@Roland : thanks a lot, your explanations are wonderful as always !
– student
Nov 20 '18 at 19:54
@Roland : thanks a lot, your explanations are wonderful as always !
– student
Nov 20 '18 at 19:54
|
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I think the author meant exact sequence of perverse sheaves.
– Roland
Nov 20 '18 at 18:25
@Roland : I see, so is there a "concrete" description of what is an exact sequence of perverse sheaves ? Is there a way I can think of it in term of classical exact sequence of sheaves ? I find it really confusing that perverse epimorphism does not induce surjective map on stalks.
– student
Nov 20 '18 at 18:30
Think in term of derived category. For a sequence $0to Ato Bto Cto 0$ in a triangulated category $mathcal{T}$ to be exact for a $t$-structure, it is enough that they all lie in the heart and that $Ato Bto Cto A[1]$ is distinguished (Just take the long exact sequence of cohomology). I'm not sure about the converse though. But here, you have the triangle $delta_{infty}to j_! IC_{mathbb{A}^1}to IC_{mathbb{P}^1}todelta_{infty}[1]$ (which is just a rotation of a triangle of sheaves), and they all are perverse sheaves. So this is an exact sequence of perverse sheaves.
– Roland
Nov 20 '18 at 18:42
But yes, the stalk functor is not exact in the category of perverse sheaves.
– Roland
Nov 20 '18 at 18:43
@Roland : thanks a lot, your explanations are wonderful as always !
– student
Nov 20 '18 at 19:54