How does this example from Spivak that $H_c^n(mathbb R^n) ne 0$?
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I am not sure how this integral that is being calculated using Stoke's theorem shows that the $n$th de Rham cohomology group with compact supports of $mathbb R^n$ is not trivial.
How does the fact that the integral of $omega$ over $mathbb R^n$ is zero show that $H_c^n(mathbb R^n) ne 0$?
differential-geometry proof-explanation differential-topology homology-cohomology de-rham-cohomology
asked Dec 10 '18 at 23:28
Al JebrAl Jebr
4,37443378
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add a comment |
$begingroup$
I am not sure how this integral that is being calculated using Stoke's theorem shows that the $n$th de Rham cohomology group with compact supports of $mathbb R^n$ is not trivial.
How does the fact that the integral of $omega$ over $mathbb R^n$ is zero show that $H_c^n(mathbb R^n) ne 0$?
differential-geometry proof-explanation differential-topology homology-cohomology de-rham-cohomology
asked Dec 10 '18 at 23:28
Al JebrAl Jebr
4,37443378
$endgroup$
add a comment |
$begingroup$
I am not sure how this integral that is being calculated using Stoke's theorem shows that the $n$th de Rham cohomology group with compact supports of $mathbb R^n$ is not trivial.
How does the fact that the integral of $omega$ over $mathbb R^n$ is zero show that $H_c^n(mathbb R^n) ne 0$?
differential-geometry proof-explanation differential-topology homology-cohomology de-rham-cohomology
asked Dec 10 '18 at 23:28
Al JebrAl Jebr
4,37443378
$endgroup$
I am not sure how this integral that is being calculated using Stoke's theorem shows that the $n$th de Rham cohomology group with compact supports of $mathbb R^n$ is not trivial.
How does the fact that the integral of $omega$ over $mathbb R^n$ is zero show that $H_c^n(mathbb R^n) ne 0$?
differential-geometry proof-explanation differential-topology homology-cohomology de-rham-cohomology
differential-geometry proof-explanation differential-topology homology-cohomology de-rham-cohomology
asked Dec 10 '18 at 23:28
Al JebrAl Jebr
4,37443378
asked Dec 10 '18 at 23:28
Al JebrAl Jebr
4,37443378
asked Dec 10 '18 at 23:28
Al JebrAl Jebr
4,37443378
asked Dec 10 '18 at 23:28
Al JebrAl Jebr
4,37443378
asked Dec 10 '18 at 23:28
Al JebrAl Jebr
4,37443378
4,37443378
add a comment |
add a comment |
1 Answer
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Since $mathbf R^n$ has dimension $n$ and $omega$ is an $n$-form, we have that $omega$ has to be closed (as there are no nontrivial $(n+1)$-forms on $mathbf R^n$). Hence $omegain Z_c^n(mathbf R^n)$. But $omeganotin B_c^n(mathbf R^n)$ because we'd would have $eta$ with compact support such that $omega=deta$, and that would imply that $int_{mathbf R^n} omega =0$ as Spivak claims. This is a contradiction because $omega$ is non-negative and positive at some point (hence on a neighborhood of that point by continuity). Hence $omega$ is a nonzero element of $H_c^n(mathbf R^n)$.
answered Dec 10 '18 at 23:40
ZirchtZircht
1,623913
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$begingroup$
Why is $int_{partial mathbb R^n} eta =0$?
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– Al Jebr
Dec 11 '18 at 0:34
1
$begingroup$
@AlJebr It's because $partial mathbf R^n = emptyset$.
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– Zircht
Dec 11 '18 at 0:41
add a comment |
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1 Answer
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1 Answer
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oldest
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active
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votes
$begingroup$
Since $mathbf R^n$ has dimension $n$ and $omega$ is an $n$-form, we have that $omega$ has to be closed (as there are no nontrivial $(n+1)$-forms on $mathbf R^n$). Hence $omegain Z_c^n(mathbf R^n)$. But $omeganotin B_c^n(mathbf R^n)$ because we'd would have $eta$ with compact support such that $omega=deta$, and that would imply that $int_{mathbf R^n} omega =0$ as Spivak claims. This is a contradiction because $omega$ is non-negative and positive at some point (hence on a neighborhood of that point by continuity). Hence $omega$ is a nonzero element of $H_c^n(mathbf R^n)$.
answered Dec 10 '18 at 23:40
ZirchtZircht
1,623913
$endgroup$
$begingroup$
Why is $int_{partial mathbb R^n} eta =0$?
$endgroup$
– Al Jebr
Dec 11 '18 at 0:34
1
$begingroup$
@AlJebr It's because $partial mathbf R^n = emptyset$.
$endgroup$
– Zircht
Dec 11 '18 at 0:41
add a comment |
$begingroup$
Since $mathbf R^n$ has dimension $n$ and $omega$ is an $n$-form, we have that $omega$ has to be closed (as there are no nontrivial $(n+1)$-forms on $mathbf R^n$). Hence $omegain Z_c^n(mathbf R^n)$. But $omeganotin B_c^n(mathbf R^n)$ because we'd would have $eta$ with compact support such that $omega=deta$, and that would imply that $int_{mathbf R^n} omega =0$ as Spivak claims. This is a contradiction because $omega$ is non-negative and positive at some point (hence on a neighborhood of that point by continuity). Hence $omega$ is a nonzero element of $H_c^n(mathbf R^n)$.
answered Dec 10 '18 at 23:40
ZirchtZircht
1,623913
$endgroup$
$begingroup$
Why is $int_{partial mathbb R^n} eta =0$?
$endgroup$
– Al Jebr
Dec 11 '18 at 0:34
1
$begingroup$
@AlJebr It's because $partial mathbf R^n = emptyset$.
$endgroup$
– Zircht
Dec 11 '18 at 0:41
add a comment |
$begingroup$
Since $mathbf R^n$ has dimension $n$ and $omega$ is an $n$-form, we have that $omega$ has to be closed (as there are no nontrivial $(n+1)$-forms on $mathbf R^n$). Hence $omegain Z_c^n(mathbf R^n)$. But $omeganotin B_c^n(mathbf R^n)$ because we'd would have $eta$ with compact support such that $omega=deta$, and that would imply that $int_{mathbf R^n} omega =0$ as Spivak claims. This is a contradiction because $omega$ is non-negative and positive at some point (hence on a neighborhood of that point by continuity). Hence $omega$ is a nonzero element of $H_c^n(mathbf R^n)$.
answered Dec 10 '18 at 23:40
ZirchtZircht
1,623913
$endgroup$
Since $mathbf R^n$ has dimension $n$ and $omega$ is an $n$-form, we have that $omega$ has to be closed (as there are no nontrivial $(n+1)$-forms on $mathbf R^n$). Hence $omegain Z_c^n(mathbf R^n)$. But $omeganotin B_c^n(mathbf R^n)$ because we'd would have $eta$ with compact support such that $omega=deta$, and that would imply that $int_{mathbf R^n} omega =0$ as Spivak claims. This is a contradiction because $omega$ is non-negative and positive at some point (hence on a neighborhood of that point by continuity). Hence $omega$ is a nonzero element of $H_c^n(mathbf R^n)$.
answered Dec 10 '18 at 23:40
ZirchtZircht
1,623913
answered Dec 10 '18 at 23:40
ZirchtZircht
1,623913
answered Dec 10 '18 at 23:40
ZirchtZircht
1,623913
answered Dec 10 '18 at 23:40
ZirchtZircht
1,623913
1,623913
$begingroup$
Why is $int_{partial mathbb R^n} eta =0$?
$endgroup$
– Al Jebr
Dec 11 '18 at 0:34
1
$begingroup$
@AlJebr It's because $partial mathbf R^n = emptyset$.
$endgroup$
– Zircht
Dec 11 '18 at 0:41
add a comment |
$begingroup$
Why is $int_{partial mathbb R^n} eta =0$?
$endgroup$
– Al Jebr
Dec 11 '18 at 0:34
1
$begingroup$
@AlJebr It's because $partial mathbf R^n = emptyset$.
$endgroup$
– Zircht
Dec 11 '18 at 0:41
$begingroup$
Why is $int_{partial mathbb R^n} eta =0$?
$endgroup$
– Al Jebr
Dec 11 '18 at 0:34
$begingroup$
Why is $int_{partial mathbb R^n} eta =0$?
$endgroup$
– Al Jebr
Dec 11 '18 at 0:34
1
1
$begingroup$
@AlJebr It's because $partial mathbf R^n = emptyset$.
$endgroup$
– Zircht
Dec 11 '18 at 0:41
$begingroup$
@AlJebr It's because $partial mathbf R^n = emptyset$.
$endgroup$
– Zircht
Dec 11 '18 at 0:41
add a comment |
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