Hilbert scheme of n points and moduli space of ideal sheaves
Let $X$ be a smooth projective variety. I have seen people identify the Hilbert scheme of $n$ points on $X$ with the moduli space of stable sheaves with Chern character $(1,0,....,-n)$ (say on $mathbb{P}^2$).
On the other hand, this identification clearly cannot work for all $X$, say $X=mathbb{P}^1$, so we may need to assume dimension of $X$ is at least $2$.
My question: 1. what are the conditions needed for us to make the above identification?
- Let $Y_1,Y_2$ be two distinct length $n$ subschemes on $X$, how to we prove the two ideal sheaves $I_{Y_1},I_{Y_2}$ are not isomorphic?
Thanks for the help!
algebraic-geometry moduli-space
asked Dec 4 '18 at 20:56
Rust QRust Q
1278
add a comment |
Let $X$ be a smooth projective variety. I have seen people identify the Hilbert scheme of $n$ points on $X$ with the moduli space of stable sheaves with Chern character $(1,0,....,-n)$ (say on $mathbb{P}^2$).
On the other hand, this identification clearly cannot work for all $X$, say $X=mathbb{P}^1$, so we may need to assume dimension of $X$ is at least $2$.
My question: 1. what are the conditions needed for us to make the above identification?
- Let $Y_1,Y_2$ be two distinct length $n$ subschemes on $X$, how to we prove the two ideal sheaves $I_{Y_1},I_{Y_2}$ are not isomorphic?
Thanks for the help!
algebraic-geometry moduli-space
asked Dec 4 '18 at 20:56
Rust QRust Q
1278
add a comment |
Let $X$ be a smooth projective variety. I have seen people identify the Hilbert scheme of $n$ points on $X$ with the moduli space of stable sheaves with Chern character $(1,0,....,-n)$ (say on $mathbb{P}^2$).
On the other hand, this identification clearly cannot work for all $X$, say $X=mathbb{P}^1$, so we may need to assume dimension of $X$ is at least $2$.
My question: 1. what are the conditions needed for us to make the above identification?
- Let $Y_1,Y_2$ be two distinct length $n$ subschemes on $X$, how to we prove the two ideal sheaves $I_{Y_1},I_{Y_2}$ are not isomorphic?
Thanks for the help!
algebraic-geometry moduli-space
asked Dec 4 '18 at 20:56
Rust QRust Q
1278
Let $X$ be a smooth projective variety. I have seen people identify the Hilbert scheme of $n$ points on $X$ with the moduli space of stable sheaves with Chern character $(1,0,....,-n)$ (say on $mathbb{P}^2$).
On the other hand, this identification clearly cannot work for all $X$, say $X=mathbb{P}^1$, so we may need to assume dimension of $X$ is at least $2$.
My question: 1. what are the conditions needed for us to make the above identification?
- Let $Y_1,Y_2$ be two distinct length $n$ subschemes on $X$, how to we prove the two ideal sheaves $I_{Y_1},I_{Y_2}$ are not isomorphic?
Thanks for the help!
algebraic-geometry moduli-space
algebraic-geometry moduli-space
asked Dec 4 '18 at 20:56
Rust QRust Q
1278
asked Dec 4 '18 at 20:56
Rust QRust Q
1278
edited Dec 4 '18 at 22:42
Rust Q
asked Dec 4 '18 at 20:56
Rust QRust Q
1278
asked Dec 4 '18 at 20:56
Rust QRust Q
1278
asked Dec 4 '18 at 20:56
Rust QRust Q
1278
1278
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
1) The first condition is $dim(X) ge 2$, the second is $Pic^0(X) = 0$. For a proof see Lemma B.5.6 in https://arxiv.org/abs/1605.02010
2) If $I = I_Y$ then $Y$ is the support of the cokernel of the natural map $I to I^{veevee}$. The proof of (1) is just the same argument applied to a family of sheaves.
answered Dec 5 '18 at 4:20
SashaSasha
4,418139
Just to clarify: to prove (2), an isom between $I_1$ and $I_2$ would induced an isom between the double duals which is compatible with the natural map $I_i$ to double dual. Thus gives an isom of the cokernel which leads to a contradiction by looking at the support.
– Rust Q
Dec 5 '18 at 5:33
Right. But my point was that the map $I mapsto supp(Coker(I to I^{veevee}))$ is the inverse of the map $Y mapsto I_Y$. And definitely, if a map has inverse then it is injective. Your argument is just an implementation of a general principle in this particular situation.
– Sasha
Dec 5 '18 at 6:19
I understand now. Thank you very much!
– Rust Q
Dec 5 '18 at 16:14
add a comment |
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: "69"
};
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
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3026159%2fhilbert-scheme-of-n-points-and-moduli-space-of-ideal-sheaves%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
1) The first condition is $dim(X) ge 2$, the second is $Pic^0(X) = 0$. For a proof see Lemma B.5.6 in https://arxiv.org/abs/1605.02010
2) If $I = I_Y$ then $Y$ is the support of the cokernel of the natural map $I to I^{veevee}$. The proof of (1) is just the same argument applied to a family of sheaves.
answered Dec 5 '18 at 4:20
SashaSasha
4,418139
Just to clarify: to prove (2), an isom between $I_1$ and $I_2$ would induced an isom between the double duals which is compatible with the natural map $I_i$ to double dual. Thus gives an isom of the cokernel which leads to a contradiction by looking at the support.
– Rust Q
Dec 5 '18 at 5:33
Right. But my point was that the map $I mapsto supp(Coker(I to I^{veevee}))$ is the inverse of the map $Y mapsto I_Y$. And definitely, if a map has inverse then it is injective. Your argument is just an implementation of a general principle in this particular situation.
– Sasha
Dec 5 '18 at 6:19
I understand now. Thank you very much!
– Rust Q
Dec 5 '18 at 16:14
add a comment |
1) The first condition is $dim(X) ge 2$, the second is $Pic^0(X) = 0$. For a proof see Lemma B.5.6 in https://arxiv.org/abs/1605.02010
2) If $I = I_Y$ then $Y$ is the support of the cokernel of the natural map $I to I^{veevee}$. The proof of (1) is just the same argument applied to a family of sheaves.
answered Dec 5 '18 at 4:20
SashaSasha
4,418139
Just to clarify: to prove (2), an isom between $I_1$ and $I_2$ would induced an isom between the double duals which is compatible with the natural map $I_i$ to double dual. Thus gives an isom of the cokernel which leads to a contradiction by looking at the support.
– Rust Q
Dec 5 '18 at 5:33
Right. But my point was that the map $I mapsto supp(Coker(I to I^{veevee}))$ is the inverse of the map $Y mapsto I_Y$. And definitely, if a map has inverse then it is injective. Your argument is just an implementation of a general principle in this particular situation.
– Sasha
Dec 5 '18 at 6:19
I understand now. Thank you very much!
– Rust Q
Dec 5 '18 at 16:14
add a comment |
1) The first condition is $dim(X) ge 2$, the second is $Pic^0(X) = 0$. For a proof see Lemma B.5.6 in https://arxiv.org/abs/1605.02010
2) If $I = I_Y$ then $Y$ is the support of the cokernel of the natural map $I to I^{veevee}$. The proof of (1) is just the same argument applied to a family of sheaves.
answered Dec 5 '18 at 4:20
SashaSasha
4,418139
1) The first condition is $dim(X) ge 2$, the second is $Pic^0(X) = 0$. For a proof see Lemma B.5.6 in https://arxiv.org/abs/1605.02010
2) If $I = I_Y$ then $Y$ is the support of the cokernel of the natural map $I to I^{veevee}$. The proof of (1) is just the same argument applied to a family of sheaves.
answered Dec 5 '18 at 4:20
SashaSasha
4,418139
answered Dec 5 '18 at 4:20
SashaSasha
4,418139
answered Dec 5 '18 at 4:20
SashaSasha
4,418139
answered Dec 5 '18 at 4:20
SashaSasha
4,418139
4,418139
Just to clarify: to prove (2), an isom between $I_1$ and $I_2$ would induced an isom between the double duals which is compatible with the natural map $I_i$ to double dual. Thus gives an isom of the cokernel which leads to a contradiction by looking at the support.
– Rust Q
Dec 5 '18 at 5:33
Right. But my point was that the map $I mapsto supp(Coker(I to I^{veevee}))$ is the inverse of the map $Y mapsto I_Y$. And definitely, if a map has inverse then it is injective. Your argument is just an implementation of a general principle in this particular situation.
– Sasha
Dec 5 '18 at 6:19
I understand now. Thank you very much!
– Rust Q
Dec 5 '18 at 16:14
add a comment |
Just to clarify: to prove (2), an isom between $I_1$ and $I_2$ would induced an isom between the double duals which is compatible with the natural map $I_i$ to double dual. Thus gives an isom of the cokernel which leads to a contradiction by looking at the support.
– Rust Q
Dec 5 '18 at 5:33
Right. But my point was that the map $I mapsto supp(Coker(I to I^{veevee}))$ is the inverse of the map $Y mapsto I_Y$. And definitely, if a map has inverse then it is injective. Your argument is just an implementation of a general principle in this particular situation.
– Sasha
Dec 5 '18 at 6:19
I understand now. Thank you very much!
– Rust Q
Dec 5 '18 at 16:14
Just to clarify: to prove (2), an isom between $I_1$ and $I_2$ would induced an isom between the double duals which is compatible with the natural map $I_i$ to double dual. Thus gives an isom of the cokernel which leads to a contradiction by looking at the support.
– Rust Q
Dec 5 '18 at 5:33
Just to clarify: to prove (2), an isom between $I_1$ and $I_2$ would induced an isom between the double duals which is compatible with the natural map $I_i$ to double dual. Thus gives an isom of the cokernel which leads to a contradiction by looking at the support.
– Rust Q
Dec 5 '18 at 5:33
Right. But my point was that the map $I mapsto supp(Coker(I to I^{veevee}))$ is the inverse of the map $Y mapsto I_Y$. And definitely, if a map has inverse then it is injective. Your argument is just an implementation of a general principle in this particular situation.
– Sasha
Dec 5 '18 at 6:19
Right. But my point was that the map $I mapsto supp(Coker(I to I^{veevee}))$ is the inverse of the map $Y mapsto I_Y$. And definitely, if a map has inverse then it is injective. Your argument is just an implementation of a general principle in this particular situation.
– Sasha
Dec 5 '18 at 6:19
I understand now. Thank you very much!
– Rust Q
Dec 5 '18 at 16:14
I understand now. Thank you very much!
– Rust Q
Dec 5 '18 at 16:14
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- 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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3026159%2fhilbert-scheme-of-n-points-and-moduli-space-of-ideal-sheaves%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
