Show that $mathbb{S}^{n+m}$ is not homeomorphic to a product of orientable manifolds
$begingroup$
I want to prove that the sphere $mathbb{S}^{n+m}$ is not homeomorphic to the product of N and M, orientable manifolfs with $textit{dim};N=n$ and $textit{dim};M=m$. I know that I have to use the de Rham cohomology of the sphere to prove it, but I don´t know how to do this. Any hint?
differential-geometry orientation de-rham-cohomology
asked Dec 4 '18 at 19:26
davidivadfuldavidivadful
1189
$endgroup$
add a comment |
$begingroup$
I want to prove that the sphere $mathbb{S}^{n+m}$ is not homeomorphic to the product of N and M, orientable manifolfs with $textit{dim};N=n$ and $textit{dim};M=m$. I know that I have to use the de Rham cohomology of the sphere to prove it, but I don´t know how to do this. Any hint?
differential-geometry orientation de-rham-cohomology
asked Dec 4 '18 at 19:26
davidivadfuldavidivadful
1189
$endgroup$
$begingroup$
Do you know Poincare duality?
$endgroup$
– Jason DeVito
Dec 4 '18 at 19:38
$begingroup$
Yes, but I don't know how to apply it here
$endgroup$
– davidivadful
Dec 4 '18 at 19:39
1
$begingroup$
Perhaps more pointed: do you know the Kunneth theorem?
$endgroup$
– Mike Miller
Dec 4 '18 at 19:40
$begingroup$
I don't know Kunneth theorem or any theorem about products
$endgroup$
– davidivadful
Dec 4 '18 at 19:41
add a comment |
$begingroup$
I want to prove that the sphere $mathbb{S}^{n+m}$ is not homeomorphic to the product of N and M, orientable manifolfs with $textit{dim};N=n$ and $textit{dim};M=m$. I know that I have to use the de Rham cohomology of the sphere to prove it, but I don´t know how to do this. Any hint?
differential-geometry orientation de-rham-cohomology
asked Dec 4 '18 at 19:26
davidivadfuldavidivadful
1189
$endgroup$
I want to prove that the sphere $mathbb{S}^{n+m}$ is not homeomorphic to the product of N and M, orientable manifolfs with $textit{dim};N=n$ and $textit{dim};M=m$. I know that I have to use the de Rham cohomology of the sphere to prove it, but I don´t know how to do this. Any hint?
differential-geometry orientation de-rham-cohomology
differential-geometry orientation de-rham-cohomology
asked Dec 4 '18 at 19:26
davidivadfuldavidivadful
1189
asked Dec 4 '18 at 19:26
davidivadfuldavidivadful
1189
asked Dec 4 '18 at 19:26
davidivadfuldavidivadful
1189
asked Dec 4 '18 at 19:26
davidivadfuldavidivadful
1189
asked Dec 4 '18 at 19:26
davidivadfuldavidivadful
1189
1189
$begingroup$
Do you know Poincare duality?
$endgroup$
– Jason DeVito
Dec 4 '18 at 19:38
$begingroup$
Yes, but I don't know how to apply it here
$endgroup$
– davidivadful
Dec 4 '18 at 19:39
1
$begingroup$
Perhaps more pointed: do you know the Kunneth theorem?
$endgroup$
– Mike Miller
Dec 4 '18 at 19:40
$begingroup$
I don't know Kunneth theorem or any theorem about products
$endgroup$
– davidivadful
Dec 4 '18 at 19:41
add a comment |
$begingroup$
Do you know Poincare duality?
$endgroup$
– Jason DeVito
Dec 4 '18 at 19:38
$begingroup$
Yes, but I don't know how to apply it here
$endgroup$
– davidivadful
Dec 4 '18 at 19:39
1
$begingroup$
Perhaps more pointed: do you know the Kunneth theorem?
$endgroup$
– Mike Miller
Dec 4 '18 at 19:40
$begingroup$
I don't know Kunneth theorem or any theorem about products
$endgroup$
– davidivadful
Dec 4 '18 at 19:41
$begingroup$
Do you know Poincare duality?
$endgroup$
– Jason DeVito
Dec 4 '18 at 19:38
$begingroup$
Do you know Poincare duality?
$endgroup$
– Jason DeVito
Dec 4 '18 at 19:38
$begingroup$
Yes, but I don't know how to apply it here
$endgroup$
– davidivadful
Dec 4 '18 at 19:39
$begingroup$
Yes, but I don't know how to apply it here
$endgroup$
– davidivadful
Dec 4 '18 at 19:39
1
1
$begingroup$
Perhaps more pointed: do you know the Kunneth theorem?
$endgroup$
– Mike Miller
Dec 4 '18 at 19:40
$begingroup$
Perhaps more pointed: do you know the Kunneth theorem?
$endgroup$
– Mike Miller
Dec 4 '18 at 19:40
$begingroup$
I don't know Kunneth theorem or any theorem about products
$endgroup$
– davidivadful
Dec 4 '18 at 19:41
$begingroup$
I don't know Kunneth theorem or any theorem about products
$endgroup$
– davidivadful
Dec 4 '18 at 19:41
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We want to find two closed forms $omega_1$ and $omega_2$ on $M times N$ so that $omega_1 wedge omega_2$ is a closed form of top degree with $int omega_1 wedge omega_2 neq 0$. Therefore, $[omega_1] wedge [omega_2]$ is nonzero in cohomology, and therefore each $[omega_i]$ must have been as well.
Write $omega_M$ for a volume form on $M$ - a nonvanishing top-dimensional form - which necessarily has $int_M omega_M neq 0$. Similarly for $omega_N$. Our desired forms are $pi_M^* omega_M$ and $pi_N^* omega_N$.
Now the Fubini theorem on integrals of functions on $Bbb R^{n + m}$ of the form $f cdot g$, where $f$ is a compactly supported function on $Bbb R^n$ (respectively, $Bbb R^m$) states that $$int_{Bbb R^{n+m}} fg = int_{Bbb R^n} f cdot int_{Bbb R^m} g.$$
The usual argument to take a theorem about compactly supported integrals in $Bbb R^n$ and make it into a theorem about integrals of differential forms is to apply a partition of unity to your atlas and them sum up the result. Nothing changes here (you should use two partitions of unity: one for $M$ and one for $N$), and you find that $$int_{Mtimes N} pi_M^*omega_M wedge pi_N^* omega_N = int_M omega_M cdot int_N omega_N neq 0,$$ as desired.
answered Dec 4 '18 at 19:47
Mike MillerMike Miller
36.8k470137
$endgroup$
$begingroup$
But aren't M and N are too general to find a closed form?
$endgroup$
– davidivadful
Dec 4 '18 at 19:55
1
$begingroup$
@davidivadful Nope! It might not be completely explicit, but that doesn't matter. There is a crucial assumption on $M$ and $N$ you have to use. :)
$endgroup$
– Mike Miller
Dec 4 '18 at 19:57
$begingroup$
But a form in $M$ gives me a form in $Mtimes N$? And closed?
$endgroup$
– davidivadful
Dec 4 '18 at 20:07
1
$begingroup$
@davidivadful You'll have to think about how to do that part: it only uses simple operations on forms that you already know, and their properties.
$endgroup$
– Mike Miller
Dec 4 '18 at 20:09
1
$begingroup$
You want to consider $pi_M^* omega_M wedge pi_N^* omega_N$. By Fubini's theorem, the integral is $int_M omega_M int_N omega_N$.
$endgroup$
– Mike Miller
Dec 5 '18 at 13:54
|
show 5 more comments
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1 Answer
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$begingroup$
We want to find two closed forms $omega_1$ and $omega_2$ on $M times N$ so that $omega_1 wedge omega_2$ is a closed form of top degree with $int omega_1 wedge omega_2 neq 0$. Therefore, $[omega_1] wedge [omega_2]$ is nonzero in cohomology, and therefore each $[omega_i]$ must have been as well.
Write $omega_M$ for a volume form on $M$ - a nonvanishing top-dimensional form - which necessarily has $int_M omega_M neq 0$. Similarly for $omega_N$. Our desired forms are $pi_M^* omega_M$ and $pi_N^* omega_N$.
Now the Fubini theorem on integrals of functions on $Bbb R^{n + m}$ of the form $f cdot g$, where $f$ is a compactly supported function on $Bbb R^n$ (respectively, $Bbb R^m$) states that $$int_{Bbb R^{n+m}} fg = int_{Bbb R^n} f cdot int_{Bbb R^m} g.$$
The usual argument to take a theorem about compactly supported integrals in $Bbb R^n$ and make it into a theorem about integrals of differential forms is to apply a partition of unity to your atlas and them sum up the result. Nothing changes here (you should use two partitions of unity: one for $M$ and one for $N$), and you find that $$int_{Mtimes N} pi_M^*omega_M wedge pi_N^* omega_N = int_M omega_M cdot int_N omega_N neq 0,$$ as desired.
answered Dec 4 '18 at 19:47
Mike MillerMike Miller
36.8k470137
$endgroup$
$begingroup$
But aren't M and N are too general to find a closed form?
$endgroup$
– davidivadful
Dec 4 '18 at 19:55
1
$begingroup$
@davidivadful Nope! It might not be completely explicit, but that doesn't matter. There is a crucial assumption on $M$ and $N$ you have to use. :)
$endgroup$
– Mike Miller
Dec 4 '18 at 19:57
$begingroup$
But a form in $M$ gives me a form in $Mtimes N$? And closed?
$endgroup$
– davidivadful
Dec 4 '18 at 20:07
1
$begingroup$
@davidivadful You'll have to think about how to do that part: it only uses simple operations on forms that you already know, and their properties.
$endgroup$
– Mike Miller
Dec 4 '18 at 20:09
1
$begingroup$
You want to consider $pi_M^* omega_M wedge pi_N^* omega_N$. By Fubini's theorem, the integral is $int_M omega_M int_N omega_N$.
$endgroup$
– Mike Miller
Dec 5 '18 at 13:54
|
show 5 more comments
$begingroup$
We want to find two closed forms $omega_1$ and $omega_2$ on $M times N$ so that $omega_1 wedge omega_2$ is a closed form of top degree with $int omega_1 wedge omega_2 neq 0$. Therefore, $[omega_1] wedge [omega_2]$ is nonzero in cohomology, and therefore each $[omega_i]$ must have been as well.
Write $omega_M$ for a volume form on $M$ - a nonvanishing top-dimensional form - which necessarily has $int_M omega_M neq 0$. Similarly for $omega_N$. Our desired forms are $pi_M^* omega_M$ and $pi_N^* omega_N$.
Now the Fubini theorem on integrals of functions on $Bbb R^{n + m}$ of the form $f cdot g$, where $f$ is a compactly supported function on $Bbb R^n$ (respectively, $Bbb R^m$) states that $$int_{Bbb R^{n+m}} fg = int_{Bbb R^n} f cdot int_{Bbb R^m} g.$$
The usual argument to take a theorem about compactly supported integrals in $Bbb R^n$ and make it into a theorem about integrals of differential forms is to apply a partition of unity to your atlas and them sum up the result. Nothing changes here (you should use two partitions of unity: one for $M$ and one for $N$), and you find that $$int_{Mtimes N} pi_M^*omega_M wedge pi_N^* omega_N = int_M omega_M cdot int_N omega_N neq 0,$$ as desired.
answered Dec 4 '18 at 19:47
Mike MillerMike Miller
36.8k470137
$endgroup$
$begingroup$
But aren't M and N are too general to find a closed form?
$endgroup$
– davidivadful
Dec 4 '18 at 19:55
1
$begingroup$
@davidivadful Nope! It might not be completely explicit, but that doesn't matter. There is a crucial assumption on $M$ and $N$ you have to use. :)
$endgroup$
– Mike Miller
Dec 4 '18 at 19:57
$begingroup$
But a form in $M$ gives me a form in $Mtimes N$? And closed?
$endgroup$
– davidivadful
Dec 4 '18 at 20:07
1
$begingroup$
@davidivadful You'll have to think about how to do that part: it only uses simple operations on forms that you already know, and their properties.
$endgroup$
– Mike Miller
Dec 4 '18 at 20:09
1
$begingroup$
You want to consider $pi_M^* omega_M wedge pi_N^* omega_N$. By Fubini's theorem, the integral is $int_M omega_M int_N omega_N$.
$endgroup$
– Mike Miller
Dec 5 '18 at 13:54
|
show 5 more comments
$begingroup$
We want to find two closed forms $omega_1$ and $omega_2$ on $M times N$ so that $omega_1 wedge omega_2$ is a closed form of top degree with $int omega_1 wedge omega_2 neq 0$. Therefore, $[omega_1] wedge [omega_2]$ is nonzero in cohomology, and therefore each $[omega_i]$ must have been as well.
Write $omega_M$ for a volume form on $M$ - a nonvanishing top-dimensional form - which necessarily has $int_M omega_M neq 0$. Similarly for $omega_N$. Our desired forms are $pi_M^* omega_M$ and $pi_N^* omega_N$.
Now the Fubini theorem on integrals of functions on $Bbb R^{n + m}$ of the form $f cdot g$, where $f$ is a compactly supported function on $Bbb R^n$ (respectively, $Bbb R^m$) states that $$int_{Bbb R^{n+m}} fg = int_{Bbb R^n} f cdot int_{Bbb R^m} g.$$
The usual argument to take a theorem about compactly supported integrals in $Bbb R^n$ and make it into a theorem about integrals of differential forms is to apply a partition of unity to your atlas and them sum up the result. Nothing changes here (you should use two partitions of unity: one for $M$ and one for $N$), and you find that $$int_{Mtimes N} pi_M^*omega_M wedge pi_N^* omega_N = int_M omega_M cdot int_N omega_N neq 0,$$ as desired.
answered Dec 4 '18 at 19:47
Mike MillerMike Miller
36.8k470137
$endgroup$
We want to find two closed forms $omega_1$ and $omega_2$ on $M times N$ so that $omega_1 wedge omega_2$ is a closed form of top degree with $int omega_1 wedge omega_2 neq 0$. Therefore, $[omega_1] wedge [omega_2]$ is nonzero in cohomology, and therefore each $[omega_i]$ must have been as well.
Write $omega_M$ for a volume form on $M$ - a nonvanishing top-dimensional form - which necessarily has $int_M omega_M neq 0$. Similarly for $omega_N$. Our desired forms are $pi_M^* omega_M$ and $pi_N^* omega_N$.
Now the Fubini theorem on integrals of functions on $Bbb R^{n + m}$ of the form $f cdot g$, where $f$ is a compactly supported function on $Bbb R^n$ (respectively, $Bbb R^m$) states that $$int_{Bbb R^{n+m}} fg = int_{Bbb R^n} f cdot int_{Bbb R^m} g.$$
The usual argument to take a theorem about compactly supported integrals in $Bbb R^n$ and make it into a theorem about integrals of differential forms is to apply a partition of unity to your atlas and them sum up the result. Nothing changes here (you should use two partitions of unity: one for $M$ and one for $N$), and you find that $$int_{Mtimes N} pi_M^*omega_M wedge pi_N^* omega_N = int_M omega_M cdot int_N omega_N neq 0,$$ as desired.
answered Dec 4 '18 at 19:47
Mike MillerMike Miller
36.8k470137
edited Dec 9 '18 at 3:58
answered Dec 4 '18 at 19:47
Mike MillerMike Miller
36.8k470137
answered Dec 4 '18 at 19:47
Mike MillerMike Miller
36.8k470137
answered Dec 4 '18 at 19:47
Mike MillerMike Miller
36.8k470137
36.8k470137
$begingroup$
But aren't M and N are too general to find a closed form?
$endgroup$
– davidivadful
Dec 4 '18 at 19:55
1
$begingroup$
@davidivadful Nope! It might not be completely explicit, but that doesn't matter. There is a crucial assumption on $M$ and $N$ you have to use. :)
$endgroup$
– Mike Miller
Dec 4 '18 at 19:57
$begingroup$
But a form in $M$ gives me a form in $Mtimes N$? And closed?
$endgroup$
– davidivadful
Dec 4 '18 at 20:07
1
$begingroup$
@davidivadful You'll have to think about how to do that part: it only uses simple operations on forms that you already know, and their properties.
$endgroup$
– Mike Miller
Dec 4 '18 at 20:09
1
$begingroup$
You want to consider $pi_M^* omega_M wedge pi_N^* omega_N$. By Fubini's theorem, the integral is $int_M omega_M int_N omega_N$.
$endgroup$
– Mike Miller
Dec 5 '18 at 13:54
|
show 5 more comments
$begingroup$
But aren't M and N are too general to find a closed form?
$endgroup$
– davidivadful
Dec 4 '18 at 19:55
1
$begingroup$
@davidivadful Nope! It might not be completely explicit, but that doesn't matter. There is a crucial assumption on $M$ and $N$ you have to use. :)
$endgroup$
– Mike Miller
Dec 4 '18 at 19:57
$begingroup$
But a form in $M$ gives me a form in $Mtimes N$? And closed?
$endgroup$
– davidivadful
Dec 4 '18 at 20:07
1
$begingroup$
@davidivadful You'll have to think about how to do that part: it only uses simple operations on forms that you already know, and their properties.
$endgroup$
– Mike Miller
Dec 4 '18 at 20:09
1
$begingroup$
You want to consider $pi_M^* omega_M wedge pi_N^* omega_N$. By Fubini's theorem, the integral is $int_M omega_M int_N omega_N$.
$endgroup$
– Mike Miller
Dec 5 '18 at 13:54
$begingroup$
But aren't M and N are too general to find a closed form?
$endgroup$
– davidivadful
Dec 4 '18 at 19:55
$begingroup$
But aren't M and N are too general to find a closed form?
$endgroup$
– davidivadful
Dec 4 '18 at 19:55
1
1
$begingroup$
@davidivadful Nope! It might not be completely explicit, but that doesn't matter. There is a crucial assumption on $M$ and $N$ you have to use. :)
$endgroup$
– Mike Miller
Dec 4 '18 at 19:57
$begingroup$
@davidivadful Nope! It might not be completely explicit, but that doesn't matter. There is a crucial assumption on $M$ and $N$ you have to use. :)
$endgroup$
– Mike Miller
Dec 4 '18 at 19:57
$begingroup$
But a form in $M$ gives me a form in $Mtimes N$? And closed?
$endgroup$
– davidivadful
Dec 4 '18 at 20:07
$begingroup$
But a form in $M$ gives me a form in $Mtimes N$? And closed?
$endgroup$
– davidivadful
Dec 4 '18 at 20:07
1
1
$begingroup$
@davidivadful You'll have to think about how to do that part: it only uses simple operations on forms that you already know, and their properties.
$endgroup$
– Mike Miller
Dec 4 '18 at 20:09
$begingroup$
@davidivadful You'll have to think about how to do that part: it only uses simple operations on forms that you already know, and their properties.
$endgroup$
– Mike Miller
Dec 4 '18 at 20:09
1
1
$begingroup$
You want to consider $pi_M^* omega_M wedge pi_N^* omega_N$. By Fubini's theorem, the integral is $int_M omega_M int_N omega_N$.
$endgroup$
– Mike Miller
Dec 5 '18 at 13:54
$begingroup$
You want to consider $pi_M^* omega_M wedge pi_N^* omega_N$. By Fubini's theorem, the integral is $int_M omega_M int_N omega_N$.
$endgroup$
– Mike Miller
Dec 5 '18 at 13:54
|
show 5 more comments
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$begingroup$
Do you know Poincare duality?
$endgroup$
– Jason DeVito
Dec 4 '18 at 19:38
$begingroup$
Yes, but I don't know how to apply it here
$endgroup$
– davidivadful
Dec 4 '18 at 19:39
1
$begingroup$
Perhaps more pointed: do you know the Kunneth theorem?
$endgroup$
– Mike Miller
Dec 4 '18 at 19:40
$begingroup$
I don't know Kunneth theorem or any theorem about products
$endgroup$
– davidivadful
Dec 4 '18 at 19:41