Using Algebraic Topology, show that $M times N$ is orientable if and only if $M$ and $N$ are both orientable.
$begingroup$
This is an exercise 3.3.5 in Hatcher's Algebraic Topology book.
3.3.5: Show that $M times N$ is orientable if and only if $M$ and $N$ are both orientable.
Notation: For $x in M$, $H_n(M|x):=H_n(M,M-x)$.
Definition: An orientation of an $m$-dimensional manifold $M$ is a function $x to mu_x$ assigning to each $x in M$ a local orientation of $mu_x in H_n(M|x)$ satisfying the local consistency condition that each $x in M$ has a neighbourhood $U approx mathbb R^m subset M$ containing an open ball $B$ of finite radius about $x$ such that all the local orientations
$mu_y$ at points $y in B$ are the images of one generator $mu_B$ of $H_n(M|B) approx H_n(mathbb R^m |B)$ under the natural maps $H_n(M|B) to H_n(B|y)$ induced by inclusion. If an orientation exists for $M$, then $M$ is called orientable.
Attempt: Let $x to mu_x, y to mu_y$ be respective functions assigning to $xin M$ and $yin N$ local orientations as described above.
By Section 3B of Hatcher, there is a cross product homomorphism $$H_i(M,M-x) otimes H_j(N, N-y) to H_{i+j}(Mtimes N, (M-x)times (N-y)).$$
Since $M-xsubset M$ and $N-y subset N$, we have $(M-x)times (N-y) subset M times N$. At this point, can we say that $$(M-x)times (N-y) subset M times N-(x,y)?$$
If this holds, then the function $(x,y) to mu_x otimes mu_y$ assigning to $(x,y) in Mtimes N$ local orientation in the above manner gives the global orientation for $Mtimes N$.
Any suggestion will be appreciated.
algebraic-topology
asked Dec 22 '18 at 17:15
M. Giovanni LucarettiM. Giovanni Lucaretti
797
$endgroup$
add a comment |
$begingroup$
This is an exercise 3.3.5 in Hatcher's Algebraic Topology book.
3.3.5: Show that $M times N$ is orientable if and only if $M$ and $N$ are both orientable.
Notation: For $x in M$, $H_n(M|x):=H_n(M,M-x)$.
Definition: An orientation of an $m$-dimensional manifold $M$ is a function $x to mu_x$ assigning to each $x in M$ a local orientation of $mu_x in H_n(M|x)$ satisfying the local consistency condition that each $x in M$ has a neighbourhood $U approx mathbb R^m subset M$ containing an open ball $B$ of finite radius about $x$ such that all the local orientations
$mu_y$ at points $y in B$ are the images of one generator $mu_B$ of $H_n(M|B) approx H_n(mathbb R^m |B)$ under the natural maps $H_n(M|B) to H_n(B|y)$ induced by inclusion. If an orientation exists for $M$, then $M$ is called orientable.
Attempt: Let $x to mu_x, y to mu_y$ be respective functions assigning to $xin M$ and $yin N$ local orientations as described above.
By Section 3B of Hatcher, there is a cross product homomorphism $$H_i(M,M-x) otimes H_j(N, N-y) to H_{i+j}(Mtimes N, (M-x)times (N-y)).$$
Since $M-xsubset M$ and $N-y subset N$, we have $(M-x)times (N-y) subset M times N$. At this point, can we say that $$(M-x)times (N-y) subset M times N-(x,y)?$$
If this holds, then the function $(x,y) to mu_x otimes mu_y$ assigning to $(x,y) in Mtimes N$ local orientation in the above manner gives the global orientation for $Mtimes N$.
Any suggestion will be appreciated.
algebraic-topology
asked Dec 22 '18 at 17:15
M. Giovanni LucarettiM. Giovanni Lucaretti
797
$endgroup$
add a comment |
$begingroup$
This is an exercise 3.3.5 in Hatcher's Algebraic Topology book.
3.3.5: Show that $M times N$ is orientable if and only if $M$ and $N$ are both orientable.
Notation: For $x in M$, $H_n(M|x):=H_n(M,M-x)$.
Definition: An orientation of an $m$-dimensional manifold $M$ is a function $x to mu_x$ assigning to each $x in M$ a local orientation of $mu_x in H_n(M|x)$ satisfying the local consistency condition that each $x in M$ has a neighbourhood $U approx mathbb R^m subset M$ containing an open ball $B$ of finite radius about $x$ such that all the local orientations
$mu_y$ at points $y in B$ are the images of one generator $mu_B$ of $H_n(M|B) approx H_n(mathbb R^m |B)$ under the natural maps $H_n(M|B) to H_n(B|y)$ induced by inclusion. If an orientation exists for $M$, then $M$ is called orientable.
Attempt: Let $x to mu_x, y to mu_y$ be respective functions assigning to $xin M$ and $yin N$ local orientations as described above.
By Section 3B of Hatcher, there is a cross product homomorphism $$H_i(M,M-x) otimes H_j(N, N-y) to H_{i+j}(Mtimes N, (M-x)times (N-y)).$$
Since $M-xsubset M$ and $N-y subset N$, we have $(M-x)times (N-y) subset M times N$. At this point, can we say that $$(M-x)times (N-y) subset M times N-(x,y)?$$
If this holds, then the function $(x,y) to mu_x otimes mu_y$ assigning to $(x,y) in Mtimes N$ local orientation in the above manner gives the global orientation for $Mtimes N$.
Any suggestion will be appreciated.
algebraic-topology
asked Dec 22 '18 at 17:15
M. Giovanni LucarettiM. Giovanni Lucaretti
797
$endgroup$
This is an exercise 3.3.5 in Hatcher's Algebraic Topology book.
3.3.5: Show that $M times N$ is orientable if and only if $M$ and $N$ are both orientable.
Notation: For $x in M$, $H_n(M|x):=H_n(M,M-x)$.
Definition: An orientation of an $m$-dimensional manifold $M$ is a function $x to mu_x$ assigning to each $x in M$ a local orientation of $mu_x in H_n(M|x)$ satisfying the local consistency condition that each $x in M$ has a neighbourhood $U approx mathbb R^m subset M$ containing an open ball $B$ of finite radius about $x$ such that all the local orientations
$mu_y$ at points $y in B$ are the images of one generator $mu_B$ of $H_n(M|B) approx H_n(mathbb R^m |B)$ under the natural maps $H_n(M|B) to H_n(B|y)$ induced by inclusion. If an orientation exists for $M$, then $M$ is called orientable.
Attempt: Let $x to mu_x, y to mu_y$ be respective functions assigning to $xin M$ and $yin N$ local orientations as described above.
By Section 3B of Hatcher, there is a cross product homomorphism $$H_i(M,M-x) otimes H_j(N, N-y) to H_{i+j}(Mtimes N, (M-x)times (N-y)).$$
Since $M-xsubset M$ and $N-y subset N$, we have $(M-x)times (N-y) subset M times N$. At this point, can we say that $$(M-x)times (N-y) subset M times N-(x,y)?$$
If this holds, then the function $(x,y) to mu_x otimes mu_y$ assigning to $(x,y) in Mtimes N$ local orientation in the above manner gives the global orientation for $Mtimes N$.
Any suggestion will be appreciated.
algebraic-topology
algebraic-topology
asked Dec 22 '18 at 17:15
M. Giovanni LucarettiM. Giovanni Lucaretti
797
asked Dec 22 '18 at 17:15
M. Giovanni LucarettiM. Giovanni Lucaretti
797
asked Dec 22 '18 at 17:15
M. Giovanni LucarettiM. Giovanni Lucaretti
797
asked Dec 22 '18 at 17:15
M. Giovanni LucarettiM. Giovanni Lucaretti
797
asked Dec 22 '18 at 17:15
M. Giovanni LucarettiM. Giovanni Lucaretti
797
797
add a comment |
add a comment |
1 Answer
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$begingroup$
Checking whether $(M-x) times (N-y) subset (M times N) - (x,y)$ is purely a matter of set theory. However it's not true in general because $(M-x) times (N-y)$ also excludes ${x} times N$ and $M times {y}$.
If you look at page $276$ of Hatcher, you see that $$C_ast(M)/C_ast(M-x) otimes C_ast(N)/C_ast(N-y) cong C_ast(M times N)/C_ast([(M-x) times N ]cup [M times (N-y)])$$ The latter term is the same as $C_ast(Mtimes N)/C_ast(Mtimes N - (x,y))$ and should help give the isomorphism in homology you're looking for.
edited Dec 22 '18 at 17:51
M. Giovanni Lucaretti
797
answered Dec 22 '18 at 17:42
Osama GhaniOsama Ghani
1,172313
$endgroup$
$begingroup$
The latter term is the same as $C_ast(Mtimes N)/C_ast(Mtimes N - (x,y))$ due to a set theoretical identity?
$endgroup$
– M. Giovanni Lucaretti
Dec 22 '18 at 17:52
$begingroup$
Yes! Hopefully it's clear that $(M times N) - (x,y) subset [(M-x) times N] cup [M times (N-y)]$. Any point either has a non-$x$ first coordinate or a non-$y$ second coordinate. The reverse inclusion can be seen by using De Morgan's law and realizing the complement is $(x times N) cap (M times y) = (x,y)$. (After writing this I realized the entire thing can be proven using De Morgan).
$endgroup$
– Osama Ghani
Dec 22 '18 at 17:57
add a comment |
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$begingroup$
Checking whether $(M-x) times (N-y) subset (M times N) - (x,y)$ is purely a matter of set theory. However it's not true in general because $(M-x) times (N-y)$ also excludes ${x} times N$ and $M times {y}$.
If you look at page $276$ of Hatcher, you see that $$C_ast(M)/C_ast(M-x) otimes C_ast(N)/C_ast(N-y) cong C_ast(M times N)/C_ast([(M-x) times N ]cup [M times (N-y)])$$ The latter term is the same as $C_ast(Mtimes N)/C_ast(Mtimes N - (x,y))$ and should help give the isomorphism in homology you're looking for.
edited Dec 22 '18 at 17:51
M. Giovanni Lucaretti
797
answered Dec 22 '18 at 17:42
Osama GhaniOsama Ghani
1,172313
$endgroup$
$begingroup$
The latter term is the same as $C_ast(Mtimes N)/C_ast(Mtimes N - (x,y))$ due to a set theoretical identity?
$endgroup$
– M. Giovanni Lucaretti
Dec 22 '18 at 17:52
$begingroup$
Yes! Hopefully it's clear that $(M times N) - (x,y) subset [(M-x) times N] cup [M times (N-y)]$. Any point either has a non-$x$ first coordinate or a non-$y$ second coordinate. The reverse inclusion can be seen by using De Morgan's law and realizing the complement is $(x times N) cap (M times y) = (x,y)$. (After writing this I realized the entire thing can be proven using De Morgan).
$endgroup$
– Osama Ghani
Dec 22 '18 at 17:57
add a comment |
$begingroup$
Checking whether $(M-x) times (N-y) subset (M times N) - (x,y)$ is purely a matter of set theory. However it's not true in general because $(M-x) times (N-y)$ also excludes ${x} times N$ and $M times {y}$.
If you look at page $276$ of Hatcher, you see that $$C_ast(M)/C_ast(M-x) otimes C_ast(N)/C_ast(N-y) cong C_ast(M times N)/C_ast([(M-x) times N ]cup [M times (N-y)])$$ The latter term is the same as $C_ast(Mtimes N)/C_ast(Mtimes N - (x,y))$ and should help give the isomorphism in homology you're looking for.
edited Dec 22 '18 at 17:51
M. Giovanni Lucaretti
797
answered Dec 22 '18 at 17:42
Osama GhaniOsama Ghani
1,172313
$endgroup$
$begingroup$
The latter term is the same as $C_ast(Mtimes N)/C_ast(Mtimes N - (x,y))$ due to a set theoretical identity?
$endgroup$
– M. Giovanni Lucaretti
Dec 22 '18 at 17:52
$begingroup$
Yes! Hopefully it's clear that $(M times N) - (x,y) subset [(M-x) times N] cup [M times (N-y)]$. Any point either has a non-$x$ first coordinate or a non-$y$ second coordinate. The reverse inclusion can be seen by using De Morgan's law and realizing the complement is $(x times N) cap (M times y) = (x,y)$. (After writing this I realized the entire thing can be proven using De Morgan).
$endgroup$
– Osama Ghani
Dec 22 '18 at 17:57
add a comment |
$begingroup$
Checking whether $(M-x) times (N-y) subset (M times N) - (x,y)$ is purely a matter of set theory. However it's not true in general because $(M-x) times (N-y)$ also excludes ${x} times N$ and $M times {y}$.
If you look at page $276$ of Hatcher, you see that $$C_ast(M)/C_ast(M-x) otimes C_ast(N)/C_ast(N-y) cong C_ast(M times N)/C_ast([(M-x) times N ]cup [M times (N-y)])$$ The latter term is the same as $C_ast(Mtimes N)/C_ast(Mtimes N - (x,y))$ and should help give the isomorphism in homology you're looking for.
edited Dec 22 '18 at 17:51
M. Giovanni Lucaretti
797
answered Dec 22 '18 at 17:42
Osama GhaniOsama Ghani
1,172313
$endgroup$
Checking whether $(M-x) times (N-y) subset (M times N) - (x,y)$ is purely a matter of set theory. However it's not true in general because $(M-x) times (N-y)$ also excludes ${x} times N$ and $M times {y}$.
If you look at page $276$ of Hatcher, you see that $$C_ast(M)/C_ast(M-x) otimes C_ast(N)/C_ast(N-y) cong C_ast(M times N)/C_ast([(M-x) times N ]cup [M times (N-y)])$$ The latter term is the same as $C_ast(Mtimes N)/C_ast(Mtimes N - (x,y))$ and should help give the isomorphism in homology you're looking for.
edited Dec 22 '18 at 17:51
M. Giovanni Lucaretti
797
answered Dec 22 '18 at 17:42
Osama GhaniOsama Ghani
1,172313
edited Dec 22 '18 at 17:51
M. Giovanni Lucaretti
797
edited Dec 22 '18 at 17:51
M. Giovanni Lucaretti
797
edited Dec 22 '18 at 17:51
M. Giovanni Lucaretti
797
797
answered Dec 22 '18 at 17:42
Osama GhaniOsama Ghani
1,172313
answered Dec 22 '18 at 17:42
Osama GhaniOsama Ghani
1,172313
answered Dec 22 '18 at 17:42
Osama GhaniOsama Ghani
1,172313
1,172313
$begingroup$
The latter term is the same as $C_ast(Mtimes N)/C_ast(Mtimes N - (x,y))$ due to a set theoretical identity?
$endgroup$
– M. Giovanni Lucaretti
Dec 22 '18 at 17:52
$begingroup$
Yes! Hopefully it's clear that $(M times N) - (x,y) subset [(M-x) times N] cup [M times (N-y)]$. Any point either has a non-$x$ first coordinate or a non-$y$ second coordinate. The reverse inclusion can be seen by using De Morgan's law and realizing the complement is $(x times N) cap (M times y) = (x,y)$. (After writing this I realized the entire thing can be proven using De Morgan).
$endgroup$
– Osama Ghani
Dec 22 '18 at 17:57
add a comment |
$begingroup$
The latter term is the same as $C_ast(Mtimes N)/C_ast(Mtimes N - (x,y))$ due to a set theoretical identity?
$endgroup$
– M. Giovanni Lucaretti
Dec 22 '18 at 17:52
$begingroup$
Yes! Hopefully it's clear that $(M times N) - (x,y) subset [(M-x) times N] cup [M times (N-y)]$. Any point either has a non-$x$ first coordinate or a non-$y$ second coordinate. The reverse inclusion can be seen by using De Morgan's law and realizing the complement is $(x times N) cap (M times y) = (x,y)$. (After writing this I realized the entire thing can be proven using De Morgan).
$endgroup$
– Osama Ghani
Dec 22 '18 at 17:57
$begingroup$
The latter term is the same as $C_ast(Mtimes N)/C_ast(Mtimes N - (x,y))$ due to a set theoretical identity?
$endgroup$
– M. Giovanni Lucaretti
Dec 22 '18 at 17:52
$begingroup$
The latter term is the same as $C_ast(Mtimes N)/C_ast(Mtimes N - (x,y))$ due to a set theoretical identity?
$endgroup$
– M. Giovanni Lucaretti
Dec 22 '18 at 17:52
$begingroup$
Yes! Hopefully it's clear that $(M times N) - (x,y) subset [(M-x) times N] cup [M times (N-y)]$. Any point either has a non-$x$ first coordinate or a non-$y$ second coordinate. The reverse inclusion can be seen by using De Morgan's law and realizing the complement is $(x times N) cap (M times y) = (x,y)$. (After writing this I realized the entire thing can be proven using De Morgan).
$endgroup$
– Osama Ghani
Dec 22 '18 at 17:57
$begingroup$
Yes! Hopefully it's clear that $(M times N) - (x,y) subset [(M-x) times N] cup [M times (N-y)]$. Any point either has a non-$x$ first coordinate or a non-$y$ second coordinate. The reverse inclusion can be seen by using De Morgan's law and realizing the complement is $(x times N) cap (M times y) = (x,y)$. (After writing this I realized the entire thing can be proven using De Morgan).
$endgroup$
– Osama Ghani
Dec 22 '18 at 17:57
add a comment |
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