$omega$ a closed 2-form and $bigwedge_{i=1}^n omega ne 0$ on a compact orientable smooth $2n$-manifold w/o...
$begingroup$
Suppose $M$ is a compact orientable smooth $2n$-manifold without boundary, and let $omega$ be a closed $2$-form such that $bigwedge_{i=1}^n omega_p ne 0$ at every point $p$. Show that $H^2_{dR}(M) ne 0$.
That is, $(M, omega)$ is a symplectic manifold.
I'm not sure what to try with this. Any ideas on how to show this?
differential-geometry differential-topology smooth-manifolds exterior-algebra de-rham-cohomology
asked Dec 19 '18 at 3:17
Al JebrAl Jebr
4,38943378
$endgroup$
|
show 4 more comments
$begingroup$
Suppose $M$ is a compact orientable smooth $2n$-manifold without boundary, and let $omega$ be a closed $2$-form such that $bigwedge_{i=1}^n omega_p ne 0$ at every point $p$. Show that $H^2_{dR}(M) ne 0$.
That is, $(M, omega)$ is a symplectic manifold.
I'm not sure what to try with this. Any ideas on how to show this?
differential-geometry differential-topology smooth-manifolds exterior-algebra de-rham-cohomology
asked Dec 19 '18 at 3:17
Al JebrAl Jebr
4,38943378
$endgroup$
$begingroup$
Fyi, this is telling you something important about symplectic manifolds: their second cohomology can’t vanish.
$endgroup$
– symplectomorphic
Dec 19 '18 at 3:39
$begingroup$
what is $Lambda_{i=1}^n$ do you mean $Lambda^n$ ?
$endgroup$
– Tsemo Aristide
Dec 19 '18 at 3:45
1
$begingroup$
Be careful! Do you want $omega^nneq 0$ happening in $Omega^{2n}(M)$ or in $H^{2n}(M)$? The latter is easy, the former is not true.
$endgroup$
– user10354138
Dec 19 '18 at 13:43
1
$begingroup$
@user10354138 I think your comment is incorrect; the definition of a symplectic form is a closed 2-form so that $omega^n$ is pointwise nonzero. Such 2-forms are popular, both at conferences and at parties.
$endgroup$
– user98602
Dec 19 '18 at 21:23
2
$begingroup$
Please be careful with your notation. You mean to say that $omega^n$ is nowhere $0$, I believe, not that it is not identically $0$.
$endgroup$
– Ted Shifrin
Dec 19 '18 at 22:57
|
show 4 more comments
$begingroup$
Suppose $M$ is a compact orientable smooth $2n$-manifold without boundary, and let $omega$ be a closed $2$-form such that $bigwedge_{i=1}^n omega_p ne 0$ at every point $p$. Show that $H^2_{dR}(M) ne 0$.
That is, $(M, omega)$ is a symplectic manifold.
I'm not sure what to try with this. Any ideas on how to show this?
differential-geometry differential-topology smooth-manifolds exterior-algebra de-rham-cohomology
asked Dec 19 '18 at 3:17
Al JebrAl Jebr
4,38943378
$endgroup$
Suppose $M$ is a compact orientable smooth $2n$-manifold without boundary, and let $omega$ be a closed $2$-form such that $bigwedge_{i=1}^n omega_p ne 0$ at every point $p$. Show that $H^2_{dR}(M) ne 0$.
That is, $(M, omega)$ is a symplectic manifold.
I'm not sure what to try with this. Any ideas on how to show this?
differential-geometry differential-topology smooth-manifolds exterior-algebra de-rham-cohomology
differential-geometry differential-topology smooth-manifolds exterior-algebra de-rham-cohomology
asked Dec 19 '18 at 3:17
Al JebrAl Jebr
4,38943378
asked Dec 19 '18 at 3:17
Al JebrAl Jebr
4,38943378
edited Dec 20 '18 at 15:44
user98602
asked Dec 19 '18 at 3:17
Al JebrAl Jebr
4,38943378
asked Dec 19 '18 at 3:17
Al JebrAl Jebr
4,38943378
asked Dec 19 '18 at 3:17
Al JebrAl Jebr
4,38943378
4,38943378
$begingroup$
Fyi, this is telling you something important about symplectic manifolds: their second cohomology can’t vanish.
$endgroup$
– symplectomorphic
Dec 19 '18 at 3:39
$begingroup$
what is $Lambda_{i=1}^n$ do you mean $Lambda^n$ ?
$endgroup$
– Tsemo Aristide
Dec 19 '18 at 3:45
1
$begingroup$
Be careful! Do you want $omega^nneq 0$ happening in $Omega^{2n}(M)$ or in $H^{2n}(M)$? The latter is easy, the former is not true.
$endgroup$
– user10354138
Dec 19 '18 at 13:43
1
$begingroup$
@user10354138 I think your comment is incorrect; the definition of a symplectic form is a closed 2-form so that $omega^n$ is pointwise nonzero. Such 2-forms are popular, both at conferences and at parties.
$endgroup$
– user98602
Dec 19 '18 at 21:23
2
$begingroup$
Please be careful with your notation. You mean to say that $omega^n$ is nowhere $0$, I believe, not that it is not identically $0$.
$endgroup$
– Ted Shifrin
Dec 19 '18 at 22:57
|
show 4 more comments
$begingroup$
Fyi, this is telling you something important about symplectic manifolds: their second cohomology can’t vanish.
$endgroup$
– symplectomorphic
Dec 19 '18 at 3:39
$begingroup$
what is $Lambda_{i=1}^n$ do you mean $Lambda^n$ ?
$endgroup$
– Tsemo Aristide
Dec 19 '18 at 3:45
1
$begingroup$
Be careful! Do you want $omega^nneq 0$ happening in $Omega^{2n}(M)$ or in $H^{2n}(M)$? The latter is easy, the former is not true.
$endgroup$
– user10354138
Dec 19 '18 at 13:43
1
$begingroup$
@user10354138 I think your comment is incorrect; the definition of a symplectic form is a closed 2-form so that $omega^n$ is pointwise nonzero. Such 2-forms are popular, both at conferences and at parties.
$endgroup$
– user98602
Dec 19 '18 at 21:23
2
$begingroup$
Please be careful with your notation. You mean to say that $omega^n$ is nowhere $0$, I believe, not that it is not identically $0$.
$endgroup$
– Ted Shifrin
Dec 19 '18 at 22:57
$begingroup$
Fyi, this is telling you something important about symplectic manifolds: their second cohomology can’t vanish.
$endgroup$
– symplectomorphic
Dec 19 '18 at 3:39
$begingroup$
Fyi, this is telling you something important about symplectic manifolds: their second cohomology can’t vanish.
$endgroup$
– symplectomorphic
Dec 19 '18 at 3:39
$begingroup$
what is $Lambda_{i=1}^n$ do you mean $Lambda^n$ ?
$endgroup$
– Tsemo Aristide
Dec 19 '18 at 3:45
$begingroup$
what is $Lambda_{i=1}^n$ do you mean $Lambda^n$ ?
$endgroup$
– Tsemo Aristide
Dec 19 '18 at 3:45
1
1
$begingroup$
Be careful! Do you want $omega^nneq 0$ happening in $Omega^{2n}(M)$ or in $H^{2n}(M)$? The latter is easy, the former is not true.
$endgroup$
– user10354138
Dec 19 '18 at 13:43
$begingroup$
Be careful! Do you want $omega^nneq 0$ happening in $Omega^{2n}(M)$ or in $H^{2n}(M)$? The latter is easy, the former is not true.
$endgroup$
– user10354138
Dec 19 '18 at 13:43
1
1
$begingroup$
@user10354138 I think your comment is incorrect; the definition of a symplectic form is a closed 2-form so that $omega^n$ is pointwise nonzero. Such 2-forms are popular, both at conferences and at parties.
$endgroup$
– user98602
Dec 19 '18 at 21:23
$begingroup$
@user10354138 I think your comment is incorrect; the definition of a symplectic form is a closed 2-form so that $omega^n$ is pointwise nonzero. Such 2-forms are popular, both at conferences and at parties.
$endgroup$
– user98602
Dec 19 '18 at 21:23
2
2
$begingroup$
Please be careful with your notation. You mean to say that $omega^n$ is nowhere $0$, I believe, not that it is not identically $0$.
$endgroup$
– Ted Shifrin
Dec 19 '18 at 22:57
$begingroup$
Please be careful with your notation. You mean to say that $omega^n$ is nowhere $0$, I believe, not that it is not identically $0$.
$endgroup$
– Ted Shifrin
Dec 19 '18 at 22:57
|
show 4 more comments
2 Answers
2
active
oldest
votes
$begingroup$
If the second de Rham cohomology were trivial, $omega$ would be $dalpha$ for some 1-form $alpha$.
Then use the non-boundary condition and Stokes via
$$int_Momega^n=int_Md(alphawedgeomega^{n-1})=int_{partial M}cdots=cdots$$
to derive a contradiction with $omega^nneq0$.
answered Dec 19 '18 at 3:34
symplectomorphicsymplectomorphic
12.6k22041
$endgroup$
$begingroup$
How is $omega^n= d(alpha wedge omega^{n-1})$? I'm getting $$d(alpha wedge omega^{n-1})=dalpha wedge omega^{n-1} -alpha wedge domega^{n-1}=omega^n-alpha wedge domega^{n-1}.$$
$endgroup$
– Al Jebr
Dec 19 '18 at 3:45
$begingroup$
@AlJebr Show that if a form $omega$ is closed, so is $omega^k$ for all $k$, by showing that the wedge product of any two closed forms is closed.
$endgroup$
– user98602
Dec 19 '18 at 12:02
$begingroup$
Why is $int_M omega^n =0$ a contradiction?
$endgroup$
– Al Jebr
Dec 19 '18 at 16:26
$begingroup$
@AlJebr What do you know about $omega^n$?
$endgroup$
– user98602
Dec 19 '18 at 21:21
$begingroup$
@MikeMiller You only know that there is a point $pin M$ such that $omega^n(p)neq 0$. You are not given this for all $pin M$ in the question statement by one of the possible interpretation of $bigwedge_{i=1}^nomega$.
$endgroup$
– user10354138
Dec 20 '18 at 8:22
|
show 1 more comment
$begingroup$
This is a long comment. To emphasize that you must interpret the hypothesis as being that $omega^n (p) ne 0$ for all $pin M$, let me give a counterexample with the other interpretation (@user10354138). Let $M=S^4times S^4$, with obvious projection maps $pi_i$, $i=1,2$, to the $i$th factor. Choose any nonzero exact $2$-form $eta$ on $S^4$ with the property that $eta^2 = etawedgeeta$ is not identically $0$. For example, take a compactly-supported exact $2$-form on $Bbb R^4$, which extends naturally to be a form on $S^4$. [To be explicit, take $rho_1$ to be a smooth function that is $1$ on the unit ball in $Bbb R^3$ and $0$ outside the ball of radius $2$, and let $rho_2$ be $1$ on the ball of radius $1$ centered at $(1,0,0,0)$ and $0$ outside the ball of radius $2$ centered at that point. Let $eta = d(rho_1,dx-rho_2,dy)$. Note that $eta^2 = drho_1wedge drho_2wedge dxwedge dy$ will be nonzero at certain points of the intersection of the two balls of radius $2$.] Now let $omega = pi_1^*eta + pi_2^*eta$. Then $omega^4 = 6pi_1^*(eta^2)wedgepi_2^*(eta^2)$ will be not identically zero on $S^4times S^4$, and yet $H^2(S^4times S^4) = 0$.
answered Dec 20 '18 at 17:48
Ted ShifrinTed Shifrin
64.7k44692
$endgroup$
add a comment |
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%2f3045955%2fomega-a-closed-2-form-and-bigwedge-i-1n-omega-ne-0-on-a-compact-orien%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
$begingroup$
If the second de Rham cohomology were trivial, $omega$ would be $dalpha$ for some 1-form $alpha$.
Then use the non-boundary condition and Stokes via
$$int_Momega^n=int_Md(alphawedgeomega^{n-1})=int_{partial M}cdots=cdots$$
to derive a contradiction with $omega^nneq0$.
answered Dec 19 '18 at 3:34
symplectomorphicsymplectomorphic
12.6k22041
$endgroup$
$begingroup$
How is $omega^n= d(alpha wedge omega^{n-1})$? I'm getting $$d(alpha wedge omega^{n-1})=dalpha wedge omega^{n-1} -alpha wedge domega^{n-1}=omega^n-alpha wedge domega^{n-1}.$$
$endgroup$
– Al Jebr
Dec 19 '18 at 3:45
$begingroup$
@AlJebr Show that if a form $omega$ is closed, so is $omega^k$ for all $k$, by showing that the wedge product of any two closed forms is closed.
$endgroup$
– user98602
Dec 19 '18 at 12:02
$begingroup$
Why is $int_M omega^n =0$ a contradiction?
$endgroup$
– Al Jebr
Dec 19 '18 at 16:26
$begingroup$
@AlJebr What do you know about $omega^n$?
$endgroup$
– user98602
Dec 19 '18 at 21:21
$begingroup$
@MikeMiller You only know that there is a point $pin M$ such that $omega^n(p)neq 0$. You are not given this for all $pin M$ in the question statement by one of the possible interpretation of $bigwedge_{i=1}^nomega$.
$endgroup$
– user10354138
Dec 20 '18 at 8:22
|
show 1 more comment
$begingroup$
If the second de Rham cohomology were trivial, $omega$ would be $dalpha$ for some 1-form $alpha$.
Then use the non-boundary condition and Stokes via
$$int_Momega^n=int_Md(alphawedgeomega^{n-1})=int_{partial M}cdots=cdots$$
to derive a contradiction with $omega^nneq0$.
answered Dec 19 '18 at 3:34
symplectomorphicsymplectomorphic
12.6k22041
$endgroup$
$begingroup$
How is $omega^n= d(alpha wedge omega^{n-1})$? I'm getting $$d(alpha wedge omega^{n-1})=dalpha wedge omega^{n-1} -alpha wedge domega^{n-1}=omega^n-alpha wedge domega^{n-1}.$$
$endgroup$
– Al Jebr
Dec 19 '18 at 3:45
$begingroup$
@AlJebr Show that if a form $omega$ is closed, so is $omega^k$ for all $k$, by showing that the wedge product of any two closed forms is closed.
$endgroup$
– user98602
Dec 19 '18 at 12:02
$begingroup$
Why is $int_M omega^n =0$ a contradiction?
$endgroup$
– Al Jebr
Dec 19 '18 at 16:26
$begingroup$
@AlJebr What do you know about $omega^n$?
$endgroup$
– user98602
Dec 19 '18 at 21:21
$begingroup$
@MikeMiller You only know that there is a point $pin M$ such that $omega^n(p)neq 0$. You are not given this for all $pin M$ in the question statement by one of the possible interpretation of $bigwedge_{i=1}^nomega$.
$endgroup$
– user10354138
Dec 20 '18 at 8:22
|
show 1 more comment
$begingroup$
If the second de Rham cohomology were trivial, $omega$ would be $dalpha$ for some 1-form $alpha$.
Then use the non-boundary condition and Stokes via
$$int_Momega^n=int_Md(alphawedgeomega^{n-1})=int_{partial M}cdots=cdots$$
to derive a contradiction with $omega^nneq0$.
answered Dec 19 '18 at 3:34
symplectomorphicsymplectomorphic
12.6k22041
$endgroup$
If the second de Rham cohomology were trivial, $omega$ would be $dalpha$ for some 1-form $alpha$.
Then use the non-boundary condition and Stokes via
$$int_Momega^n=int_Md(alphawedgeomega^{n-1})=int_{partial M}cdots=cdots$$
to derive a contradiction with $omega^nneq0$.
answered Dec 19 '18 at 3:34
symplectomorphicsymplectomorphic
12.6k22041
answered Dec 19 '18 at 3:34
symplectomorphicsymplectomorphic
12.6k22041
answered Dec 19 '18 at 3:34
symplectomorphicsymplectomorphic
12.6k22041
answered Dec 19 '18 at 3:34
symplectomorphicsymplectomorphic
12.6k22041
12.6k22041
$begingroup$
How is $omega^n= d(alpha wedge omega^{n-1})$? I'm getting $$d(alpha wedge omega^{n-1})=dalpha wedge omega^{n-1} -alpha wedge domega^{n-1}=omega^n-alpha wedge domega^{n-1}.$$
$endgroup$
– Al Jebr
Dec 19 '18 at 3:45
$begingroup$
@AlJebr Show that if a form $omega$ is closed, so is $omega^k$ for all $k$, by showing that the wedge product of any two closed forms is closed.
$endgroup$
– user98602
Dec 19 '18 at 12:02
$begingroup$
Why is $int_M omega^n =0$ a contradiction?
$endgroup$
– Al Jebr
Dec 19 '18 at 16:26
$begingroup$
@AlJebr What do you know about $omega^n$?
$endgroup$
– user98602
Dec 19 '18 at 21:21
$begingroup$
@MikeMiller You only know that there is a point $pin M$ such that $omega^n(p)neq 0$. You are not given this for all $pin M$ in the question statement by one of the possible interpretation of $bigwedge_{i=1}^nomega$.
$endgroup$
– user10354138
Dec 20 '18 at 8:22
|
show 1 more comment
$begingroup$
How is $omega^n= d(alpha wedge omega^{n-1})$? I'm getting $$d(alpha wedge omega^{n-1})=dalpha wedge omega^{n-1} -alpha wedge domega^{n-1}=omega^n-alpha wedge domega^{n-1}.$$
$endgroup$
– Al Jebr
Dec 19 '18 at 3:45
$begingroup$
@AlJebr Show that if a form $omega$ is closed, so is $omega^k$ for all $k$, by showing that the wedge product of any two closed forms is closed.
$endgroup$
– user98602
Dec 19 '18 at 12:02
$begingroup$
Why is $int_M omega^n =0$ a contradiction?
$endgroup$
– Al Jebr
Dec 19 '18 at 16:26
$begingroup$
@AlJebr What do you know about $omega^n$?
$endgroup$
– user98602
Dec 19 '18 at 21:21
$begingroup$
@MikeMiller You only know that there is a point $pin M$ such that $omega^n(p)neq 0$. You are not given this for all $pin M$ in the question statement by one of the possible interpretation of $bigwedge_{i=1}^nomega$.
$endgroup$
– user10354138
Dec 20 '18 at 8:22
$begingroup$
How is $omega^n= d(alpha wedge omega^{n-1})$? I'm getting $$d(alpha wedge omega^{n-1})=dalpha wedge omega^{n-1} -alpha wedge domega^{n-1}=omega^n-alpha wedge domega^{n-1}.$$
$endgroup$
– Al Jebr
Dec 19 '18 at 3:45
$begingroup$
How is $omega^n= d(alpha wedge omega^{n-1})$? I'm getting $$d(alpha wedge omega^{n-1})=dalpha wedge omega^{n-1} -alpha wedge domega^{n-1}=omega^n-alpha wedge domega^{n-1}.$$
$endgroup$
– Al Jebr
Dec 19 '18 at 3:45
$begingroup$
@AlJebr Show that if a form $omega$ is closed, so is $omega^k$ for all $k$, by showing that the wedge product of any two closed forms is closed.
$endgroup$
– user98602
Dec 19 '18 at 12:02
$begingroup$
@AlJebr Show that if a form $omega$ is closed, so is $omega^k$ for all $k$, by showing that the wedge product of any two closed forms is closed.
$endgroup$
– user98602
Dec 19 '18 at 12:02
$begingroup$
Why is $int_M omega^n =0$ a contradiction?
$endgroup$
– Al Jebr
Dec 19 '18 at 16:26
$begingroup$
Why is $int_M omega^n =0$ a contradiction?
$endgroup$
– Al Jebr
Dec 19 '18 at 16:26
$begingroup$
@AlJebr What do you know about $omega^n$?
$endgroup$
– user98602
Dec 19 '18 at 21:21
$begingroup$
@AlJebr What do you know about $omega^n$?
$endgroup$
– user98602
Dec 19 '18 at 21:21
$begingroup$
@MikeMiller You only know that there is a point $pin M$ such that $omega^n(p)neq 0$. You are not given this for all $pin M$ in the question statement by one of the possible interpretation of $bigwedge_{i=1}^nomega$.
$endgroup$
– user10354138
Dec 20 '18 at 8:22
$begingroup$
@MikeMiller You only know that there is a point $pin M$ such that $omega^n(p)neq 0$. You are not given this for all $pin M$ in the question statement by one of the possible interpretation of $bigwedge_{i=1}^nomega$.
$endgroup$
– user10354138
Dec 20 '18 at 8:22
|
show 1 more comment
$begingroup$
This is a long comment. To emphasize that you must interpret the hypothesis as being that $omega^n (p) ne 0$ for all $pin M$, let me give a counterexample with the other interpretation (@user10354138). Let $M=S^4times S^4$, with obvious projection maps $pi_i$, $i=1,2$, to the $i$th factor. Choose any nonzero exact $2$-form $eta$ on $S^4$ with the property that $eta^2 = etawedgeeta$ is not identically $0$. For example, take a compactly-supported exact $2$-form on $Bbb R^4$, which extends naturally to be a form on $S^4$. [To be explicit, take $rho_1$ to be a smooth function that is $1$ on the unit ball in $Bbb R^3$ and $0$ outside the ball of radius $2$, and let $rho_2$ be $1$ on the ball of radius $1$ centered at $(1,0,0,0)$ and $0$ outside the ball of radius $2$ centered at that point. Let $eta = d(rho_1,dx-rho_2,dy)$. Note that $eta^2 = drho_1wedge drho_2wedge dxwedge dy$ will be nonzero at certain points of the intersection of the two balls of radius $2$.] Now let $omega = pi_1^*eta + pi_2^*eta$. Then $omega^4 = 6pi_1^*(eta^2)wedgepi_2^*(eta^2)$ will be not identically zero on $S^4times S^4$, and yet $H^2(S^4times S^4) = 0$.
answered Dec 20 '18 at 17:48
Ted ShifrinTed Shifrin
64.7k44692
$endgroup$
add a comment |
$begingroup$
This is a long comment. To emphasize that you must interpret the hypothesis as being that $omega^n (p) ne 0$ for all $pin M$, let me give a counterexample with the other interpretation (@user10354138). Let $M=S^4times S^4$, with obvious projection maps $pi_i$, $i=1,2$, to the $i$th factor. Choose any nonzero exact $2$-form $eta$ on $S^4$ with the property that $eta^2 = etawedgeeta$ is not identically $0$. For example, take a compactly-supported exact $2$-form on $Bbb R^4$, which extends naturally to be a form on $S^4$. [To be explicit, take $rho_1$ to be a smooth function that is $1$ on the unit ball in $Bbb R^3$ and $0$ outside the ball of radius $2$, and let $rho_2$ be $1$ on the ball of radius $1$ centered at $(1,0,0,0)$ and $0$ outside the ball of radius $2$ centered at that point. Let $eta = d(rho_1,dx-rho_2,dy)$. Note that $eta^2 = drho_1wedge drho_2wedge dxwedge dy$ will be nonzero at certain points of the intersection of the two balls of radius $2$.] Now let $omega = pi_1^*eta + pi_2^*eta$. Then $omega^4 = 6pi_1^*(eta^2)wedgepi_2^*(eta^2)$ will be not identically zero on $S^4times S^4$, and yet $H^2(S^4times S^4) = 0$.
answered Dec 20 '18 at 17:48
Ted ShifrinTed Shifrin
64.7k44692
$endgroup$
add a comment |
$begingroup$
This is a long comment. To emphasize that you must interpret the hypothesis as being that $omega^n (p) ne 0$ for all $pin M$, let me give a counterexample with the other interpretation (@user10354138). Let $M=S^4times S^4$, with obvious projection maps $pi_i$, $i=1,2$, to the $i$th factor. Choose any nonzero exact $2$-form $eta$ on $S^4$ with the property that $eta^2 = etawedgeeta$ is not identically $0$. For example, take a compactly-supported exact $2$-form on $Bbb R^4$, which extends naturally to be a form on $S^4$. [To be explicit, take $rho_1$ to be a smooth function that is $1$ on the unit ball in $Bbb R^3$ and $0$ outside the ball of radius $2$, and let $rho_2$ be $1$ on the ball of radius $1$ centered at $(1,0,0,0)$ and $0$ outside the ball of radius $2$ centered at that point. Let $eta = d(rho_1,dx-rho_2,dy)$. Note that $eta^2 = drho_1wedge drho_2wedge dxwedge dy$ will be nonzero at certain points of the intersection of the two balls of radius $2$.] Now let $omega = pi_1^*eta + pi_2^*eta$. Then $omega^4 = 6pi_1^*(eta^2)wedgepi_2^*(eta^2)$ will be not identically zero on $S^4times S^4$, and yet $H^2(S^4times S^4) = 0$.
answered Dec 20 '18 at 17:48
Ted ShifrinTed Shifrin
64.7k44692
$endgroup$
This is a long comment. To emphasize that you must interpret the hypothesis as being that $omega^n (p) ne 0$ for all $pin M$, let me give a counterexample with the other interpretation (@user10354138). Let $M=S^4times S^4$, with obvious projection maps $pi_i$, $i=1,2$, to the $i$th factor. Choose any nonzero exact $2$-form $eta$ on $S^4$ with the property that $eta^2 = etawedgeeta$ is not identically $0$. For example, take a compactly-supported exact $2$-form on $Bbb R^4$, which extends naturally to be a form on $S^4$. [To be explicit, take $rho_1$ to be a smooth function that is $1$ on the unit ball in $Bbb R^3$ and $0$ outside the ball of radius $2$, and let $rho_2$ be $1$ on the ball of radius $1$ centered at $(1,0,0,0)$ and $0$ outside the ball of radius $2$ centered at that point. Let $eta = d(rho_1,dx-rho_2,dy)$. Note that $eta^2 = drho_1wedge drho_2wedge dxwedge dy$ will be nonzero at certain points of the intersection of the two balls of radius $2$.] Now let $omega = pi_1^*eta + pi_2^*eta$. Then $omega^4 = 6pi_1^*(eta^2)wedgepi_2^*(eta^2)$ will be not identically zero on $S^4times S^4$, and yet $H^2(S^4times S^4) = 0$.
answered Dec 20 '18 at 17:48
Ted ShifrinTed Shifrin
64.7k44692
edited Dec 20 '18 at 17:58
answered Dec 20 '18 at 17:48
Ted ShifrinTed Shifrin
64.7k44692
answered Dec 20 '18 at 17:48
Ted ShifrinTed Shifrin
64.7k44692
answered Dec 20 '18 at 17:48
Ted ShifrinTed Shifrin
64.7k44692
64.7k44692
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.
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%2f3045955%2fomega-a-closed-2-form-and-bigwedge-i-1n-omega-ne-0-on-a-compact-orien%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
$begingroup$
Fyi, this is telling you something important about symplectic manifolds: their second cohomology can’t vanish.
$endgroup$
– symplectomorphic
Dec 19 '18 at 3:39
$begingroup$
what is $Lambda_{i=1}^n$ do you mean $Lambda^n$ ?
$endgroup$
– Tsemo Aristide
Dec 19 '18 at 3:45
1
$begingroup$
Be careful! Do you want $omega^nneq 0$ happening in $Omega^{2n}(M)$ or in $H^{2n}(M)$? The latter is easy, the former is not true.
$endgroup$
– user10354138
Dec 19 '18 at 13:43
1
$begingroup$
@user10354138 I think your comment is incorrect; the definition of a symplectic form is a closed 2-form so that $omega^n$ is pointwise nonzero. Such 2-forms are popular, both at conferences and at parties.
$endgroup$
– user98602
Dec 19 '18 at 21:23
2
$begingroup$
Please be careful with your notation. You mean to say that $omega^n$ is nowhere $0$, I believe, not that it is not identically $0$.
$endgroup$
– Ted Shifrin
Dec 19 '18 at 22:57