Geometric quantization: not understanding the curvature form and Weil's theorem
I am reading a bit about geometric quantization. The texts that I am following are N.M.J. Woodhouse's "Geometric Quantization" and an article from arXiv. I am having troubles understanding the prequantization step.
First, everybody cites 1958 Weil's theorem about the existence of Hermitian line bundles with prescribed curvature, but nobody provides its exact statement. Could anyone help me with it? In particular, do I prescribe just the curvature form, or a connection and its associated curvature form?
Second, one tries to produce Hermitian line bundles $L$ over a symplectic manifold $(M, omega)$ having $omega$ as the curvature form $R$ of $L$. I'm lost: how can we relate these two forms? $R$ takes values in $operatorname{End} _{Bbb C} (L)$, which $omega$ doesn't, so how can I identify them? Had $L$ been trivial, it would have been elementary, but how to do it in general? Or is this an abuse of terminology?
differential-geometry vector-bundles curvature quantum-mechanics symplectic-geometry
asked Apr 13 '16 at 17:54
Alex M.
28k103058
add a comment |
I am reading a bit about geometric quantization. The texts that I am following are N.M.J. Woodhouse's "Geometric Quantization" and an article from arXiv. I am having troubles understanding the prequantization step.
First, everybody cites 1958 Weil's theorem about the existence of Hermitian line bundles with prescribed curvature, but nobody provides its exact statement. Could anyone help me with it? In particular, do I prescribe just the curvature form, or a connection and its associated curvature form?
Second, one tries to produce Hermitian line bundles $L$ over a symplectic manifold $(M, omega)$ having $omega$ as the curvature form $R$ of $L$. I'm lost: how can we relate these two forms? $R$ takes values in $operatorname{End} _{Bbb C} (L)$, which $omega$ doesn't, so how can I identify them? Had $L$ been trivial, it would have been elementary, but how to do it in general? Or is this an abuse of terminology?
differential-geometry vector-bundles curvature quantum-mechanics symplectic-geometry
asked Apr 13 '16 at 17:54
Alex M.
28k103058
1
$text{End}_{Bbb C}(L)$ is canonically isomorphic to $L otimes_{Bbb C} L^*$, which is canonically isomorphic to the trivial line bundle by the evaluation map. For higher rank bundles, the endomorphism bundle might be topologically interesting.
– Mike Miller
Apr 13 '16 at 17:57
@MikeMiller: This is unexpected for me: do I understand correctly that $operatorname{End} _{Bbb C} (L)$ is trivial, even though $L$ might not be so?
– Alex M.
Apr 13 '16 at 18:08
@MikeMiller: Yes, $text{End}_{Bbb C}(L)$ is indeed trivial since $x mapsto operatorname{id}_x$ is a global section, nowhere zero. Ridiculous. Could you also help me with a pointer to an exact formulation of Weil's theorem? The internet was silent on this (or maybe I didn't know what terms to search for).
– Alex M.
Apr 13 '16 at 18:18
I'm not really aware of Weil's theorem, so I might run into the same troubles you are.
– Mike Miller
Apr 13 '16 at 18:44
I could write down a proof that if $L$ is a line bundle and $R in Omega^2$ a closed form with $[R] = c_1(L) in H^2_{dR}(M)$, then there's a connection on $L$ with $R$ as its curvature form - would that help?
– Mike Miller
Apr 13 '16 at 18:49
add a comment |
I am reading a bit about geometric quantization. The texts that I am following are N.M.J. Woodhouse's "Geometric Quantization" and an article from arXiv. I am having troubles understanding the prequantization step.
First, everybody cites 1958 Weil's theorem about the existence of Hermitian line bundles with prescribed curvature, but nobody provides its exact statement. Could anyone help me with it? In particular, do I prescribe just the curvature form, or a connection and its associated curvature form?
Second, one tries to produce Hermitian line bundles $L$ over a symplectic manifold $(M, omega)$ having $omega$ as the curvature form $R$ of $L$. I'm lost: how can we relate these two forms? $R$ takes values in $operatorname{End} _{Bbb C} (L)$, which $omega$ doesn't, so how can I identify them? Had $L$ been trivial, it would have been elementary, but how to do it in general? Or is this an abuse of terminology?
differential-geometry vector-bundles curvature quantum-mechanics symplectic-geometry
asked Apr 13 '16 at 17:54
Alex M.
28k103058
I am reading a bit about geometric quantization. The texts that I am following are N.M.J. Woodhouse's "Geometric Quantization" and an article from arXiv. I am having troubles understanding the prequantization step.
First, everybody cites 1958 Weil's theorem about the existence of Hermitian line bundles with prescribed curvature, but nobody provides its exact statement. Could anyone help me with it? In particular, do I prescribe just the curvature form, or a connection and its associated curvature form?
Second, one tries to produce Hermitian line bundles $L$ over a symplectic manifold $(M, omega)$ having $omega$ as the curvature form $R$ of $L$. I'm lost: how can we relate these two forms? $R$ takes values in $operatorname{End} _{Bbb C} (L)$, which $omega$ doesn't, so how can I identify them? Had $L$ been trivial, it would have been elementary, but how to do it in general? Or is this an abuse of terminology?
differential-geometry vector-bundles curvature quantum-mechanics symplectic-geometry
differential-geometry vector-bundles curvature quantum-mechanics symplectic-geometry
asked Apr 13 '16 at 17:54
Alex M.
28k103058
asked Apr 13 '16 at 17:54
Alex M.
28k103058
asked Apr 13 '16 at 17:54
Alex M.
28k103058
asked Apr 13 '16 at 17:54
Alex M.
28k103058
asked Apr 13 '16 at 17:54
Alex M.
28k103058
28k103058
1
$text{End}_{Bbb C}(L)$ is canonically isomorphic to $L otimes_{Bbb C} L^*$, which is canonically isomorphic to the trivial line bundle by the evaluation map. For higher rank bundles, the endomorphism bundle might be topologically interesting.
– Mike Miller
Apr 13 '16 at 17:57
@MikeMiller: This is unexpected for me: do I understand correctly that $operatorname{End} _{Bbb C} (L)$ is trivial, even though $L$ might not be so?
– Alex M.
Apr 13 '16 at 18:08
@MikeMiller: Yes, $text{End}_{Bbb C}(L)$ is indeed trivial since $x mapsto operatorname{id}_x$ is a global section, nowhere zero. Ridiculous. Could you also help me with a pointer to an exact formulation of Weil's theorem? The internet was silent on this (or maybe I didn't know what terms to search for).
– Alex M.
Apr 13 '16 at 18:18
I'm not really aware of Weil's theorem, so I might run into the same troubles you are.
– Mike Miller
Apr 13 '16 at 18:44
I could write down a proof that if $L$ is a line bundle and $R in Omega^2$ a closed form with $[R] = c_1(L) in H^2_{dR}(M)$, then there's a connection on $L$ with $R$ as its curvature form - would that help?
– Mike Miller
Apr 13 '16 at 18:49
add a comment |
1
$text{End}_{Bbb C}(L)$ is canonically isomorphic to $L otimes_{Bbb C} L^*$, which is canonically isomorphic to the trivial line bundle by the evaluation map. For higher rank bundles, the endomorphism bundle might be topologically interesting.
– Mike Miller
Apr 13 '16 at 17:57
@MikeMiller: This is unexpected for me: do I understand correctly that $operatorname{End} _{Bbb C} (L)$ is trivial, even though $L$ might not be so?
– Alex M.
Apr 13 '16 at 18:08
@MikeMiller: Yes, $text{End}_{Bbb C}(L)$ is indeed trivial since $x mapsto operatorname{id}_x$ is a global section, nowhere zero. Ridiculous. Could you also help me with a pointer to an exact formulation of Weil's theorem? The internet was silent on this (or maybe I didn't know what terms to search for).
– Alex M.
Apr 13 '16 at 18:18
I'm not really aware of Weil's theorem, so I might run into the same troubles you are.
– Mike Miller
Apr 13 '16 at 18:44
I could write down a proof that if $L$ is a line bundle and $R in Omega^2$ a closed form with $[R] = c_1(L) in H^2_{dR}(M)$, then there's a connection on $L$ with $R$ as its curvature form - would that help?
– Mike Miller
Apr 13 '16 at 18:49
1
1
$text{End}_{Bbb C}(L)$ is canonically isomorphic to $L otimes_{Bbb C} L^*$, which is canonically isomorphic to the trivial line bundle by the evaluation map. For higher rank bundles, the endomorphism bundle might be topologically interesting.
– Mike Miller
Apr 13 '16 at 17:57
$text{End}_{Bbb C}(L)$ is canonically isomorphic to $L otimes_{Bbb C} L^*$, which is canonically isomorphic to the trivial line bundle by the evaluation map. For higher rank bundles, the endomorphism bundle might be topologically interesting.
– Mike Miller
Apr 13 '16 at 17:57
@MikeMiller: This is unexpected for me: do I understand correctly that $operatorname{End} _{Bbb C} (L)$ is trivial, even though $L$ might not be so?
– Alex M.
Apr 13 '16 at 18:08
@MikeMiller: This is unexpected for me: do I understand correctly that $operatorname{End} _{Bbb C} (L)$ is trivial, even though $L$ might not be so?
– Alex M.
Apr 13 '16 at 18:08
@MikeMiller: Yes, $text{End}_{Bbb C}(L)$ is indeed trivial since $x mapsto operatorname{id}_x$ is a global section, nowhere zero. Ridiculous. Could you also help me with a pointer to an exact formulation of Weil's theorem? The internet was silent on this (or maybe I didn't know what terms to search for).
– Alex M.
Apr 13 '16 at 18:18
@MikeMiller: Yes, $text{End}_{Bbb C}(L)$ is indeed trivial since $x mapsto operatorname{id}_x$ is a global section, nowhere zero. Ridiculous. Could you also help me with a pointer to an exact formulation of Weil's theorem? The internet was silent on this (or maybe I didn't know what terms to search for).
– Alex M.
Apr 13 '16 at 18:18
I'm not really aware of Weil's theorem, so I might run into the same troubles you are.
– Mike Miller
Apr 13 '16 at 18:44
I'm not really aware of Weil's theorem, so I might run into the same troubles you are.
– Mike Miller
Apr 13 '16 at 18:44
I could write down a proof that if $L$ is a line bundle and $R in Omega^2$ a closed form with $[R] = c_1(L) in H^2_{dR}(M)$, then there's a connection on $L$ with $R$ as its curvature form - would that help?
– Mike Miller
Apr 13 '16 at 18:49
I could write down a proof that if $L$ is a line bundle and $R in Omega^2$ a closed form with $[R] = c_1(L) in H^2_{dR}(M)$, then there's a connection on $L$ with $R$ as its curvature form - would that help?
– Mike Miller
Apr 13 '16 at 18:49
add a comment |
1 Answer
1
active
oldest
votes
First, $text{End}(L)$ is trivial by the canonical isomorphisms $L otimes L^* to text{End}(L)$ and $L otimes L^* to Bbb C$. So we may canonically identify $Omega^2(text{End}(L)) cong Omega^2$ (where here I'm using complex-valued forms).
Then based on looking at your first reference, it seems to me the desired theorem is the following.
Let $L$ be a complex line bundle over some smooth manifold $M$, and $R in Omega^2(M) otimes Bbb C$. Then if $R$ is closed and satisfies $[R] = c_1(L) in H^2_{dR}(M)$, then $R$ is the curvature form of some connection on $L$.
You could make everything here Hermitian but I'm not going to bother.
1) First, note that if $A$ is a connection on $L$, the induced connection on $text{End}(L) cong Bbb C$ is the trivial connection.
2) The Bianchi identity is that $d_A F(A) = 0$ if $F(A)$ is the curvature of $A$. By 1), this says that $F(A)$ is closed.
3) You may as well define $c_1(L)$ to be the homology class of the curvature of any connection on $L$. This is well-defined because of the formula $F(A+a) = F(A) + d_Aa + a wedge a$; in the case of line bundles this simply degenerates to $F(A+a) = F(A) + da$. Weil proved that the set of possible $c_1(L)$ is precisely $2pi H^2(M;Bbb Z) subset H^2_{dR}(M)$, one for each line bundle (modulo torsion classes; these correspond to flat complex line bundles).
4) Now to write down a proof of the theorem. Let $R$ be your desired curvature form, and $F(A)$ the curvature of some random connection. We want to solve $R = F(A+a) = F(A)+da$. This simplifies to $R-F(A) = da$; since we assumed $[R]=[F(A)]$ in cohomology, this is solvable, as desired.
answered Apr 13 '16 at 19:39
Mike Miller
36.4k470137
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%2f1741098%2fgeometric-quantization-not-understanding-the-curvature-form-and-weils-theorem%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
First, $text{End}(L)$ is trivial by the canonical isomorphisms $L otimes L^* to text{End}(L)$ and $L otimes L^* to Bbb C$. So we may canonically identify $Omega^2(text{End}(L)) cong Omega^2$ (where here I'm using complex-valued forms).
Then based on looking at your first reference, it seems to me the desired theorem is the following.
Let $L$ be a complex line bundle over some smooth manifold $M$, and $R in Omega^2(M) otimes Bbb C$. Then if $R$ is closed and satisfies $[R] = c_1(L) in H^2_{dR}(M)$, then $R$ is the curvature form of some connection on $L$.
You could make everything here Hermitian but I'm not going to bother.
1) First, note that if $A$ is a connection on $L$, the induced connection on $text{End}(L) cong Bbb C$ is the trivial connection.
2) The Bianchi identity is that $d_A F(A) = 0$ if $F(A)$ is the curvature of $A$. By 1), this says that $F(A)$ is closed.
3) You may as well define $c_1(L)$ to be the homology class of the curvature of any connection on $L$. This is well-defined because of the formula $F(A+a) = F(A) + d_Aa + a wedge a$; in the case of line bundles this simply degenerates to $F(A+a) = F(A) + da$. Weil proved that the set of possible $c_1(L)$ is precisely $2pi H^2(M;Bbb Z) subset H^2_{dR}(M)$, one for each line bundle (modulo torsion classes; these correspond to flat complex line bundles).
4) Now to write down a proof of the theorem. Let $R$ be your desired curvature form, and $F(A)$ the curvature of some random connection. We want to solve $R = F(A+a) = F(A)+da$. This simplifies to $R-F(A) = da$; since we assumed $[R]=[F(A)]$ in cohomology, this is solvable, as desired.
answered Apr 13 '16 at 19:39
Mike Miller
36.4k470137
add a comment |
First, $text{End}(L)$ is trivial by the canonical isomorphisms $L otimes L^* to text{End}(L)$ and $L otimes L^* to Bbb C$. So we may canonically identify $Omega^2(text{End}(L)) cong Omega^2$ (where here I'm using complex-valued forms).
Then based on looking at your first reference, it seems to me the desired theorem is the following.
Let $L$ be a complex line bundle over some smooth manifold $M$, and $R in Omega^2(M) otimes Bbb C$. Then if $R$ is closed and satisfies $[R] = c_1(L) in H^2_{dR}(M)$, then $R$ is the curvature form of some connection on $L$.
You could make everything here Hermitian but I'm not going to bother.
1) First, note that if $A$ is a connection on $L$, the induced connection on $text{End}(L) cong Bbb C$ is the trivial connection.
2) The Bianchi identity is that $d_A F(A) = 0$ if $F(A)$ is the curvature of $A$. By 1), this says that $F(A)$ is closed.
3) You may as well define $c_1(L)$ to be the homology class of the curvature of any connection on $L$. This is well-defined because of the formula $F(A+a) = F(A) + d_Aa + a wedge a$; in the case of line bundles this simply degenerates to $F(A+a) = F(A) + da$. Weil proved that the set of possible $c_1(L)$ is precisely $2pi H^2(M;Bbb Z) subset H^2_{dR}(M)$, one for each line bundle (modulo torsion classes; these correspond to flat complex line bundles).
4) Now to write down a proof of the theorem. Let $R$ be your desired curvature form, and $F(A)$ the curvature of some random connection. We want to solve $R = F(A+a) = F(A)+da$. This simplifies to $R-F(A) = da$; since we assumed $[R]=[F(A)]$ in cohomology, this is solvable, as desired.
answered Apr 13 '16 at 19:39
Mike Miller
36.4k470137
add a comment |
First, $text{End}(L)$ is trivial by the canonical isomorphisms $L otimes L^* to text{End}(L)$ and $L otimes L^* to Bbb C$. So we may canonically identify $Omega^2(text{End}(L)) cong Omega^2$ (where here I'm using complex-valued forms).
Then based on looking at your first reference, it seems to me the desired theorem is the following.
Let $L$ be a complex line bundle over some smooth manifold $M$, and $R in Omega^2(M) otimes Bbb C$. Then if $R$ is closed and satisfies $[R] = c_1(L) in H^2_{dR}(M)$, then $R$ is the curvature form of some connection on $L$.
You could make everything here Hermitian but I'm not going to bother.
1) First, note that if $A$ is a connection on $L$, the induced connection on $text{End}(L) cong Bbb C$ is the trivial connection.
2) The Bianchi identity is that $d_A F(A) = 0$ if $F(A)$ is the curvature of $A$. By 1), this says that $F(A)$ is closed.
3) You may as well define $c_1(L)$ to be the homology class of the curvature of any connection on $L$. This is well-defined because of the formula $F(A+a) = F(A) + d_Aa + a wedge a$; in the case of line bundles this simply degenerates to $F(A+a) = F(A) + da$. Weil proved that the set of possible $c_1(L)$ is precisely $2pi H^2(M;Bbb Z) subset H^2_{dR}(M)$, one for each line bundle (modulo torsion classes; these correspond to flat complex line bundles).
4) Now to write down a proof of the theorem. Let $R$ be your desired curvature form, and $F(A)$ the curvature of some random connection. We want to solve $R = F(A+a) = F(A)+da$. This simplifies to $R-F(A) = da$; since we assumed $[R]=[F(A)]$ in cohomology, this is solvable, as desired.
answered Apr 13 '16 at 19:39
Mike Miller
36.4k470137
First, $text{End}(L)$ is trivial by the canonical isomorphisms $L otimes L^* to text{End}(L)$ and $L otimes L^* to Bbb C$. So we may canonically identify $Omega^2(text{End}(L)) cong Omega^2$ (where here I'm using complex-valued forms).
Then based on looking at your first reference, it seems to me the desired theorem is the following.
Let $L$ be a complex line bundle over some smooth manifold $M$, and $R in Omega^2(M) otimes Bbb C$. Then if $R$ is closed and satisfies $[R] = c_1(L) in H^2_{dR}(M)$, then $R$ is the curvature form of some connection on $L$.
You could make everything here Hermitian but I'm not going to bother.
1) First, note that if $A$ is a connection on $L$, the induced connection on $text{End}(L) cong Bbb C$ is the trivial connection.
2) The Bianchi identity is that $d_A F(A) = 0$ if $F(A)$ is the curvature of $A$. By 1), this says that $F(A)$ is closed.
3) You may as well define $c_1(L)$ to be the homology class of the curvature of any connection on $L$. This is well-defined because of the formula $F(A+a) = F(A) + d_Aa + a wedge a$; in the case of line bundles this simply degenerates to $F(A+a) = F(A) + da$. Weil proved that the set of possible $c_1(L)$ is precisely $2pi H^2(M;Bbb Z) subset H^2_{dR}(M)$, one for each line bundle (modulo torsion classes; these correspond to flat complex line bundles).
4) Now to write down a proof of the theorem. Let $R$ be your desired curvature form, and $F(A)$ the curvature of some random connection. We want to solve $R = F(A+a) = F(A)+da$. This simplifies to $R-F(A) = da$; since we assumed $[R]=[F(A)]$ in cohomology, this is solvable, as desired.
answered Apr 13 '16 at 19:39
Mike Miller
36.4k470137
edited Dec 27 '18 at 17:16
answered Apr 13 '16 at 19:39
Mike Miller
36.4k470137
answered Apr 13 '16 at 19:39
Mike Miller
36.4k470137
answered Apr 13 '16 at 19:39
Mike Miller
36.4k470137
36.4k470137
add a comment |
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%2f1741098%2fgeometric-quantization-not-understanding-the-curvature-form-and-weils-theorem%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
1
$text{End}_{Bbb C}(L)$ is canonically isomorphic to $L otimes_{Bbb C} L^*$, which is canonically isomorphic to the trivial line bundle by the evaluation map. For higher rank bundles, the endomorphism bundle might be topologically interesting.
– Mike Miller
Apr 13 '16 at 17:57
@MikeMiller: This is unexpected for me: do I understand correctly that $operatorname{End} _{Bbb C} (L)$ is trivial, even though $L$ might not be so?
– Alex M.
Apr 13 '16 at 18:08
@MikeMiller: Yes, $text{End}_{Bbb C}(L)$ is indeed trivial since $x mapsto operatorname{id}_x$ is a global section, nowhere zero. Ridiculous. Could you also help me with a pointer to an exact formulation of Weil's theorem? The internet was silent on this (or maybe I didn't know what terms to search for).
– Alex M.
Apr 13 '16 at 18:18
I'm not really aware of Weil's theorem, so I might run into the same troubles you are.
– Mike Miller
Apr 13 '16 at 18:44
I could write down a proof that if $L$ is a line bundle and $R in Omega^2$ a closed form with $[R] = c_1(L) in H^2_{dR}(M)$, then there's a connection on $L$ with $R$ as its curvature form - would that help?
– Mike Miller
Apr 13 '16 at 18:49