Help to correct answer: Product of orientable manifolds is orientable












1












$begingroup$


In the first reply to this post, I made a comment, but nobody answered me yet. Link: Orientability of a product of smooth manifolds implies orientability of each factor



My problem is in the return of the affirmation $(Leftarrow)$.



Problem statement:




Given two smooth manifolds $M$ and $N$, show that the product manifold
$M times N$ is orientable if and only if $M$ and $N$ are orientable.




First reply from rmdmc89:




$1)$ If $M^m,N^n$ are both orientable, there are volume forms (i.e., non-zero top-forms) $omegainOmega^m(M)$ and $sigmainOmega^n(N)$. Define $etainOmega^{n+m}(M)$ as:
$$eta(X_1,...,X_n,Y_1,...,Y_m):=omega(X_1,...,X_n)sigma(Y_1,...,Y_m)$$



which is a volume form on $Mtimes N$, so $Mtimes N$ is orientable.



$2)$ Conversely, if $Mtimes N$ is orientable, there is a volume form $etainOmega^{n+m}(Mtimes N)$. Fix a point $qin N$ and a basis $left{left.frac{partial}{partial y_1}right|_q, ...,left.frac{partial}{partial y_n}right|_qright}$ a for $T_qN$. Define $omegainOmega^m(M)$ as:
$$w(X_1,...,X_n):=eta_{(cdot,q)}left(X_1,...,X_n,left.frac{partial}{partial y_1}right|_q,...,left.frac{partial}{partial y_m}right|_qright)$$



which is a non-zero top-form on $M$, so $M$ is orientable. We can make a similar construction for $N$.




My comment on the first answer:



"I think it's wrong at the beginning of "Define $η∈Ω^{n+m}(M)$" must be "Define $η∈Ω^{n+m}(Mtimes N)$". And even then, I think the product is not well defined. In $Λ^{m+n}(M×N)$ has more vectors than ωσ can actually receive. Am I right? Think of the case $m=n=1$ you will see. I have a problem similar to posted, my idea was to do as you but I found this problem. Am I correct?"



How to correct?






differential-geometry manifolds riemannian-geometry smooth-manifolds orientation







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$endgroup$












  • $begingroup$
    You may want to rewrite the question here, so that the post is self-contained as it should be. In principle, since it hasn't been accepted, the answer you've commented on may be deleted any moment.
    $endgroup$
    – Saucy O'Path
    Jun 10 '18 at 20:00












  • $begingroup$
    @SaucyO'Path Yes!
    $endgroup$
    – Mancala
    Jun 10 '18 at 20:01
















1












$begingroup$


In the first reply to this post, I made a comment, but nobody answered me yet. Link: Orientability of a product of smooth manifolds implies orientability of each factor



My problem is in the return of the affirmation $(Leftarrow)$.



Problem statement:




Given two smooth manifolds $M$ and $N$, show that the product manifold
$M times N$ is orientable if and only if $M$ and $N$ are orientable.




First reply from rmdmc89:




$1)$ If $M^m,N^n$ are both orientable, there are volume forms (i.e., non-zero top-forms) $omegainOmega^m(M)$ and $sigmainOmega^n(N)$. Define $etainOmega^{n+m}(M)$ as:
$$eta(X_1,...,X_n,Y_1,...,Y_m):=omega(X_1,...,X_n)sigma(Y_1,...,Y_m)$$



which is a volume form on $Mtimes N$, so $Mtimes N$ is orientable.



$2)$ Conversely, if $Mtimes N$ is orientable, there is a volume form $etainOmega^{n+m}(Mtimes N)$. Fix a point $qin N$ and a basis $left{left.frac{partial}{partial y_1}right|_q, ...,left.frac{partial}{partial y_n}right|_qright}$ a for $T_qN$. Define $omegainOmega^m(M)$ as:
$$w(X_1,...,X_n):=eta_{(cdot,q)}left(X_1,...,X_n,left.frac{partial}{partial y_1}right|_q,...,left.frac{partial}{partial y_m}right|_qright)$$



which is a non-zero top-form on $M$, so $M$ is orientable. We can make a similar construction for $N$.




My comment on the first answer:



"I think it's wrong at the beginning of "Define $η∈Ω^{n+m}(M)$" must be "Define $η∈Ω^{n+m}(Mtimes N)$". And even then, I think the product is not well defined. In $Λ^{m+n}(M×N)$ has more vectors than ωσ can actually receive. Am I right? Think of the case $m=n=1$ you will see. I have a problem similar to posted, my idea was to do as you but I found this problem. Am I correct?"



How to correct?






differential-geometry manifolds riemannian-geometry smooth-manifolds orientation







share|cite|improve this question











$endgroup$












  • $begingroup$
    You may want to rewrite the question here, so that the post is self-contained as it should be. In principle, since it hasn't been accepted, the answer you've commented on may be deleted any moment.
    $endgroup$
    – Saucy O'Path
    Jun 10 '18 at 20:00












  • $begingroup$
    @SaucyO'Path Yes!
    $endgroup$
    – Mancala
    Jun 10 '18 at 20:01














1












1








1


1



$begingroup$


In the first reply to this post, I made a comment, but nobody answered me yet. Link: Orientability of a product of smooth manifolds implies orientability of each factor



My problem is in the return of the affirmation $(Leftarrow)$.



Problem statement:




Given two smooth manifolds $M$ and $N$, show that the product manifold
$M times N$ is orientable if and only if $M$ and $N$ are orientable.




First reply from rmdmc89:




$1)$ If $M^m,N^n$ are both orientable, there are volume forms (i.e., non-zero top-forms) $omegainOmega^m(M)$ and $sigmainOmega^n(N)$. Define $etainOmega^{n+m}(M)$ as:
$$eta(X_1,...,X_n,Y_1,...,Y_m):=omega(X_1,...,X_n)sigma(Y_1,...,Y_m)$$



which is a volume form on $Mtimes N$, so $Mtimes N$ is orientable.



$2)$ Conversely, if $Mtimes N$ is orientable, there is a volume form $etainOmega^{n+m}(Mtimes N)$. Fix a point $qin N$ and a basis $left{left.frac{partial}{partial y_1}right|_q, ...,left.frac{partial}{partial y_n}right|_qright}$ a for $T_qN$. Define $omegainOmega^m(M)$ as:
$$w(X_1,...,X_n):=eta_{(cdot,q)}left(X_1,...,X_n,left.frac{partial}{partial y_1}right|_q,...,left.frac{partial}{partial y_m}right|_qright)$$



which is a non-zero top-form on $M$, so $M$ is orientable. We can make a similar construction for $N$.




My comment on the first answer:



"I think it's wrong at the beginning of "Define $η∈Ω^{n+m}(M)$" must be "Define $η∈Ω^{n+m}(Mtimes N)$". And even then, I think the product is not well defined. In $Λ^{m+n}(M×N)$ has more vectors than ωσ can actually receive. Am I right? Think of the case $m=n=1$ you will see. I have a problem similar to posted, my idea was to do as you but I found this problem. Am I correct?"



How to correct?






differential-geometry manifolds riemannian-geometry smooth-manifolds orientation







share|cite|improve this question











$endgroup$




In the first reply to this post, I made a comment, but nobody answered me yet. Link: Orientability of a product of smooth manifolds implies orientability of each factor



My problem is in the return of the affirmation $(Leftarrow)$.



Problem statement:




Given two smooth manifolds $M$ and $N$, show that the product manifold
$M times N$ is orientable if and only if $M$ and $N$ are orientable.




First reply from rmdmc89:




$1)$ If $M^m,N^n$ are both orientable, there are volume forms (i.e., non-zero top-forms) $omegainOmega^m(M)$ and $sigmainOmega^n(N)$. Define $etainOmega^{n+m}(M)$ as:
$$eta(X_1,...,X_n,Y_1,...,Y_m):=omega(X_1,...,X_n)sigma(Y_1,...,Y_m)$$



which is a volume form on $Mtimes N$, so $Mtimes N$ is orientable.



$2)$ Conversely, if $Mtimes N$ is orientable, there is a volume form $etainOmega^{n+m}(Mtimes N)$. Fix a point $qin N$ and a basis $left{left.frac{partial}{partial y_1}right|_q, ...,left.frac{partial}{partial y_n}right|_qright}$ a for $T_qN$. Define $omegainOmega^m(M)$ as:
$$w(X_1,...,X_n):=eta_{(cdot,q)}left(X_1,...,X_n,left.frac{partial}{partial y_1}right|_q,...,left.frac{partial}{partial y_m}right|_qright)$$



which is a non-zero top-form on $M$, so $M$ is orientable. We can make a similar construction for $N$.




My comment on the first answer:



"I think it's wrong at the beginning of "Define $η∈Ω^{n+m}(M)$" must be "Define $η∈Ω^{n+m}(Mtimes N)$". And even then, I think the product is not well defined. In $Λ^{m+n}(M×N)$ has more vectors than ωσ can actually receive. Am I right? Think of the case $m=n=1$ you will see. I have a problem similar to posted, my idea was to do as you but I found this problem. Am I correct?"



How to correct?






differential-geometry manifolds riemannian-geometry smooth-manifolds orientation




differential-geometry manifolds riemannian-geometry smooth-manifolds orientation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 9 '18 at 15:45









davidivadful

1289




1289










asked Jun 10 '18 at 19:55









MancalaMancala

371311




371311












  • $begingroup$
    You may want to rewrite the question here, so that the post is self-contained as it should be. In principle, since it hasn't been accepted, the answer you've commented on may be deleted any moment.
    $endgroup$
    – Saucy O'Path
    Jun 10 '18 at 20:00












  • $begingroup$
    @SaucyO'Path Yes!
    $endgroup$
    – Mancala
    Jun 10 '18 at 20:01


















  • $begingroup$
    You may want to rewrite the question here, so that the post is self-contained as it should be. In principle, since it hasn't been accepted, the answer you've commented on may be deleted any moment.
    $endgroup$
    – Saucy O'Path
    Jun 10 '18 at 20:00












  • $begingroup$
    @SaucyO'Path Yes!
    $endgroup$
    – Mancala
    Jun 10 '18 at 20:01
















$begingroup$
You may want to rewrite the question here, so that the post is self-contained as it should be. In principle, since it hasn't been accepted, the answer you've commented on may be deleted any moment.
$endgroup$
– Saucy O'Path
Jun 10 '18 at 20:00






$begingroup$
You may want to rewrite the question here, so that the post is self-contained as it should be. In principle, since it hasn't been accepted, the answer you've commented on may be deleted any moment.
$endgroup$
– Saucy O'Path
Jun 10 '18 at 20:00














$begingroup$
@SaucyO'Path Yes!
$endgroup$
– Mancala
Jun 10 '18 at 20:01




$begingroup$
@SaucyO'Path Yes!
$endgroup$
– Mancala
Jun 10 '18 at 20:01










1 Answer
1






active

oldest

votes


















1












$begingroup$

You’re right that $eta in Omega^{n+m} (M times N)$. The product is fine as defined . Maybe you’d like a different formulation. Let $omega in Omega^m(M)$ and $sigma in Omega^n(N)$ be two volume forms. Let $pi_M : Mtimes N to M$ and $pi_N : Mtimes N to N$ be the canonical projection maps. Then let $eta = pi_M^ast(omega) wedge pi_N^ast(sigma)$. This is a volume form on $M times N$ and is more or less the $eta$ defined in the original response.



Edit: As Shifrin points out below, the form $eta$ in the original answer doesn’t make sense. I think my construction of a volume form is correct however.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks Osama Ghani. If the product is well defined, then I want is to understand where I am failing in the response. I have read the answer several times and I can not accept that it is correct :(
    $endgroup$
    – Mancala
    Jun 10 '18 at 20:30






  • 1




    $begingroup$
    Why do you think the product isn’t well defined? As for $bigwedge^{m+n}(M times N)$, this accepts $m+n$ vector (fields) as does $omega sigma$. Where do you see a mismatch?
    $endgroup$
    – Osama Ghani
    Jun 10 '18 at 20:32












  • $begingroup$
    I'm going to write what I do not understand:
    $endgroup$
    – Mancala
    Jun 10 '18 at 20:35










  • $begingroup$
    No, the original response is just plain wrong. One needs to take $m+n$ vector fields on $Mtimes N$. Thus, one needs to project $m$ of them to $M$ and $n$ of them to $N$, but one needs to take the (at least partial) skew-symmetrization involved with the wedge product.
    $endgroup$
    – Ted Shifrin
    Jun 10 '18 at 20:47










  • $begingroup$
    Ah of course, I don’t know how I overlooked that. My apologies @Mancala . I do believe my own answer is correct though. I’m not able to really justify it being a volume form but I think it has something to do with the Künneth formula.
    $endgroup$
    – Osama Ghani
    Jun 10 '18 at 21:04











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1 Answer
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active

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

You’re right that $eta in Omega^{n+m} (M times N)$. The product is fine as defined . Maybe you’d like a different formulation. Let $omega in Omega^m(M)$ and $sigma in Omega^n(N)$ be two volume forms. Let $pi_M : Mtimes N to M$ and $pi_N : Mtimes N to N$ be the canonical projection maps. Then let $eta = pi_M^ast(omega) wedge pi_N^ast(sigma)$. This is a volume form on $M times N$ and is more or less the $eta$ defined in the original response.



Edit: As Shifrin points out below, the form $eta$ in the original answer doesn’t make sense. I think my construction of a volume form is correct however.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks Osama Ghani. If the product is well defined, then I want is to understand where I am failing in the response. I have read the answer several times and I can not accept that it is correct :(
    $endgroup$
    – Mancala
    Jun 10 '18 at 20:30






  • 1




    $begingroup$
    Why do you think the product isn’t well defined? As for $bigwedge^{m+n}(M times N)$, this accepts $m+n$ vector (fields) as does $omega sigma$. Where do you see a mismatch?
    $endgroup$
    – Osama Ghani
    Jun 10 '18 at 20:32












  • $begingroup$
    I'm going to write what I do not understand:
    $endgroup$
    – Mancala
    Jun 10 '18 at 20:35










  • $begingroup$
    No, the original response is just plain wrong. One needs to take $m+n$ vector fields on $Mtimes N$. Thus, one needs to project $m$ of them to $M$ and $n$ of them to $N$, but one needs to take the (at least partial) skew-symmetrization involved with the wedge product.
    $endgroup$
    – Ted Shifrin
    Jun 10 '18 at 20:47










  • $begingroup$
    Ah of course, I don’t know how I overlooked that. My apologies @Mancala . I do believe my own answer is correct though. I’m not able to really justify it being a volume form but I think it has something to do with the Künneth formula.
    $endgroup$
    – Osama Ghani
    Jun 10 '18 at 21:04
















1












$begingroup$

You’re right that $eta in Omega^{n+m} (M times N)$. The product is fine as defined . Maybe you’d like a different formulation. Let $omega in Omega^m(M)$ and $sigma in Omega^n(N)$ be two volume forms. Let $pi_M : Mtimes N to M$ and $pi_N : Mtimes N to N$ be the canonical projection maps. Then let $eta = pi_M^ast(omega) wedge pi_N^ast(sigma)$. This is a volume form on $M times N$ and is more or less the $eta$ defined in the original response.



Edit: As Shifrin points out below, the form $eta$ in the original answer doesn’t make sense. I think my construction of a volume form is correct however.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks Osama Ghani. If the product is well defined, then I want is to understand where I am failing in the response. I have read the answer several times and I can not accept that it is correct :(
    $endgroup$
    – Mancala
    Jun 10 '18 at 20:30






  • 1




    $begingroup$
    Why do you think the product isn’t well defined? As for $bigwedge^{m+n}(M times N)$, this accepts $m+n$ vector (fields) as does $omega sigma$. Where do you see a mismatch?
    $endgroup$
    – Osama Ghani
    Jun 10 '18 at 20:32












  • $begingroup$
    I'm going to write what I do not understand:
    $endgroup$
    – Mancala
    Jun 10 '18 at 20:35










  • $begingroup$
    No, the original response is just plain wrong. One needs to take $m+n$ vector fields on $Mtimes N$. Thus, one needs to project $m$ of them to $M$ and $n$ of them to $N$, but one needs to take the (at least partial) skew-symmetrization involved with the wedge product.
    $endgroup$
    – Ted Shifrin
    Jun 10 '18 at 20:47










  • $begingroup$
    Ah of course, I don’t know how I overlooked that. My apologies @Mancala . I do believe my own answer is correct though. I’m not able to really justify it being a volume form but I think it has something to do with the Künneth formula.
    $endgroup$
    – Osama Ghani
    Jun 10 '18 at 21:04














1












1








1





$begingroup$

You’re right that $eta in Omega^{n+m} (M times N)$. The product is fine as defined . Maybe you’d like a different formulation. Let $omega in Omega^m(M)$ and $sigma in Omega^n(N)$ be two volume forms. Let $pi_M : Mtimes N to M$ and $pi_N : Mtimes N to N$ be the canonical projection maps. Then let $eta = pi_M^ast(omega) wedge pi_N^ast(sigma)$. This is a volume form on $M times N$ and is more or less the $eta$ defined in the original response.



Edit: As Shifrin points out below, the form $eta$ in the original answer doesn’t make sense. I think my construction of a volume form is correct however.






share|cite|improve this answer











$endgroup$



You’re right that $eta in Omega^{n+m} (M times N)$. The product is fine as defined . Maybe you’d like a different formulation. Let $omega in Omega^m(M)$ and $sigma in Omega^n(N)$ be two volume forms. Let $pi_M : Mtimes N to M$ and $pi_N : Mtimes N to N$ be the canonical projection maps. Then let $eta = pi_M^ast(omega) wedge pi_N^ast(sigma)$. This is a volume form on $M times N$ and is more or less the $eta$ defined in the original response.



Edit: As Shifrin points out below, the form $eta$ in the original answer doesn’t make sense. I think my construction of a volume form is correct however.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jun 10 '18 at 21:13

























answered Jun 10 '18 at 20:15









Osama GhaniOsama Ghani

1,172313




1,172313












  • $begingroup$
    Thanks Osama Ghani. If the product is well defined, then I want is to understand where I am failing in the response. I have read the answer several times and I can not accept that it is correct :(
    $endgroup$
    – Mancala
    Jun 10 '18 at 20:30






  • 1




    $begingroup$
    Why do you think the product isn’t well defined? As for $bigwedge^{m+n}(M times N)$, this accepts $m+n$ vector (fields) as does $omega sigma$. Where do you see a mismatch?
    $endgroup$
    – Osama Ghani
    Jun 10 '18 at 20:32












  • $begingroup$
    I'm going to write what I do not understand:
    $endgroup$
    – Mancala
    Jun 10 '18 at 20:35










  • $begingroup$
    No, the original response is just plain wrong. One needs to take $m+n$ vector fields on $Mtimes N$. Thus, one needs to project $m$ of them to $M$ and $n$ of them to $N$, but one needs to take the (at least partial) skew-symmetrization involved with the wedge product.
    $endgroup$
    – Ted Shifrin
    Jun 10 '18 at 20:47










  • $begingroup$
    Ah of course, I don’t know how I overlooked that. My apologies @Mancala . I do believe my own answer is correct though. I’m not able to really justify it being a volume form but I think it has something to do with the Künneth formula.
    $endgroup$
    – Osama Ghani
    Jun 10 '18 at 21:04


















  • $begingroup$
    Thanks Osama Ghani. If the product is well defined, then I want is to understand where I am failing in the response. I have read the answer several times and I can not accept that it is correct :(
    $endgroup$
    – Mancala
    Jun 10 '18 at 20:30






  • 1




    $begingroup$
    Why do you think the product isn’t well defined? As for $bigwedge^{m+n}(M times N)$, this accepts $m+n$ vector (fields) as does $omega sigma$. Where do you see a mismatch?
    $endgroup$
    – Osama Ghani
    Jun 10 '18 at 20:32












  • $begingroup$
    I'm going to write what I do not understand:
    $endgroup$
    – Mancala
    Jun 10 '18 at 20:35










  • $begingroup$
    No, the original response is just plain wrong. One needs to take $m+n$ vector fields on $Mtimes N$. Thus, one needs to project $m$ of them to $M$ and $n$ of them to $N$, but one needs to take the (at least partial) skew-symmetrization involved with the wedge product.
    $endgroup$
    – Ted Shifrin
    Jun 10 '18 at 20:47










  • $begingroup$
    Ah of course, I don’t know how I overlooked that. My apologies @Mancala . I do believe my own answer is correct though. I’m not able to really justify it being a volume form but I think it has something to do with the Künneth formula.
    $endgroup$
    – Osama Ghani
    Jun 10 '18 at 21:04
















$begingroup$
Thanks Osama Ghani. If the product is well defined, then I want is to understand where I am failing in the response. I have read the answer several times and I can not accept that it is correct :(
$endgroup$
– Mancala
Jun 10 '18 at 20:30




$begingroup$
Thanks Osama Ghani. If the product is well defined, then I want is to understand where I am failing in the response. I have read the answer several times and I can not accept that it is correct :(
$endgroup$
– Mancala
Jun 10 '18 at 20:30




1




1




$begingroup$
Why do you think the product isn’t well defined? As for $bigwedge^{m+n}(M times N)$, this accepts $m+n$ vector (fields) as does $omega sigma$. Where do you see a mismatch?
$endgroup$
– Osama Ghani
Jun 10 '18 at 20:32






$begingroup$
Why do you think the product isn’t well defined? As for $bigwedge^{m+n}(M times N)$, this accepts $m+n$ vector (fields) as does $omega sigma$. Where do you see a mismatch?
$endgroup$
– Osama Ghani
Jun 10 '18 at 20:32














$begingroup$
I'm going to write what I do not understand:
$endgroup$
– Mancala
Jun 10 '18 at 20:35




$begingroup$
I'm going to write what I do not understand:
$endgroup$
– Mancala
Jun 10 '18 at 20:35












$begingroup$
No, the original response is just plain wrong. One needs to take $m+n$ vector fields on $Mtimes N$. Thus, one needs to project $m$ of them to $M$ and $n$ of them to $N$, but one needs to take the (at least partial) skew-symmetrization involved with the wedge product.
$endgroup$
– Ted Shifrin
Jun 10 '18 at 20:47




$begingroup$
No, the original response is just plain wrong. One needs to take $m+n$ vector fields on $Mtimes N$. Thus, one needs to project $m$ of them to $M$ and $n$ of them to $N$, but one needs to take the (at least partial) skew-symmetrization involved with the wedge product.
$endgroup$
– Ted Shifrin
Jun 10 '18 at 20:47












$begingroup$
Ah of course, I don’t know how I overlooked that. My apologies @Mancala . I do believe my own answer is correct though. I’m not able to really justify it being a volume form but I think it has something to do with the Künneth formula.
$endgroup$
– Osama Ghani
Jun 10 '18 at 21:04




$begingroup$
Ah of course, I don’t know how I overlooked that. My apologies @Mancala . I do believe my own answer is correct though. I’m not able to really justify it being a volume form but I think it has something to do with the Künneth formula.
$endgroup$
– Osama Ghani
Jun 10 '18 at 21:04


















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