Mixing
Is $S^2 times S^2$ diffeomorphic to $S^1 times S^3$?
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1
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I am trying to analyze whether $S^2 times S^2$ are diffeomorphic to $S^1 times S^3$. First of all, the dimension matches because they are all four-dimensional manifolds. Then I tried thinking about techniques to prove whether manifolds are diffeomorphic. The only thing that I could come up was to compute their de Rahm cohomology, but I do not have any background in that other than the definitions.
I found some random notes online hinting that one should think about the extension of smooth maps on the sphere to the closed ball. That does not make any sense to me though.
differential-topology
asked Nov 16 at 23:44
penny
545
add a comment |
up vote
1
down vote
favorite
I am trying to analyze whether $S^2 times S^2$ are diffeomorphic to $S^1 times S^3$. First of all, the dimension matches because they are all four-dimensional manifolds. Then I tried thinking about techniques to prove whether manifolds are diffeomorphic. The only thing that I could come up was to compute their de Rahm cohomology, but I do not have any background in that other than the definitions.
I found some random notes online hinting that one should think about the extension of smooth maps on the sphere to the closed ball. That does not make any sense to me though.
differential-topology
asked Nov 16 at 23:44
penny
545
The de Rham cohomology is also helpful.
– Laz
Nov 17 at 2:47
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am trying to analyze whether $S^2 times S^2$ are diffeomorphic to $S^1 times S^3$. First of all, the dimension matches because they are all four-dimensional manifolds. Then I tried thinking about techniques to prove whether manifolds are diffeomorphic. The only thing that I could come up was to compute their de Rahm cohomology, but I do not have any background in that other than the definitions.
I found some random notes online hinting that one should think about the extension of smooth maps on the sphere to the closed ball. That does not make any sense to me though.
differential-topology
asked Nov 16 at 23:44
penny
545
I am trying to analyze whether $S^2 times S^2$ are diffeomorphic to $S^1 times S^3$. First of all, the dimension matches because they are all four-dimensional manifolds. Then I tried thinking about techniques to prove whether manifolds are diffeomorphic. The only thing that I could come up was to compute their de Rahm cohomology, but I do not have any background in that other than the definitions.
I found some random notes online hinting that one should think about the extension of smooth maps on the sphere to the closed ball. That does not make any sense to me though.
differential-topology
differential-topology
asked Nov 16 at 23:44
penny
545
asked Nov 16 at 23:44
penny
545
edited Nov 17 at 1:53
asked Nov 16 at 23:44
penny
545
asked Nov 16 at 23:44
penny
545
asked Nov 16 at 23:44
penny
545
545
The de Rham cohomology is also helpful.
– Laz
Nov 17 at 2:47
add a comment |
The de Rham cohomology is also helpful.
– Laz
Nov 17 at 2:47
The de Rham cohomology is also helpful.
– Laz
Nov 17 at 2:47
The de Rham cohomology is also helpful.
– Laz
Nov 17 at 2:47
add a comment |
2 Answers
2
active
oldest
votes
up vote
4
down vote
They're not homeomorphic, so they won't be diffeomorphic. To see that they aren't homeomorphic, we compute their fundamental groups. Recalling that
$$ pi_1(S^1) = mathbb{Z} quad text{and} quad pi_1(S^2) = pi_1(S^3) = 1$$
and $pi_1(Xtimes Y) cong pi_1(X) oplus pi_1(Y)$, we find
$$pi_1(S^2 times S^2) = 1 oplus 1 = 1$$
while
$$pi_1(S^1 times S^3) = mathbb{Z} oplus 1 cong mathbb{Z} $$
Since their fundamental groups aren't isomorphic, they aren't homeomorphic, hence not diffeomorphic.
answered Nov 16 at 23:51
Hayden
13.7k12347
My apologies, but I am actually looking for a differential topological method since I am self-studying differential topology and do not have any background in algebraic topology. I know most of the basic techniques from differential topology except de Rham cohomology. If there are no any other answers, I will accept yours. My sincere apologies again and thank you for your answer.
– penny
Nov 17 at 1:50
1
@penny That's understandable. A differential topology proof that comes to mind is that $S^2 times S^2$ has no smooth field of non-zero tangent vectors, while $S^1times S^3$ does, but I'll admit my differential topology is rusty.
– Hayden
Nov 17 at 2:04
add a comment |
up vote
1
down vote
I would have used the fundamental group as @Hayden, but you want a differential argument, so here it goes.
Since the tangent bundles $Tmathbb{S}^1$, $Tmathbb{S}^3$ are trivial (you can check this in Spivak's Volume 1, e.g, but the fact that $Tmathbb{S}^1$ is trivial is clear from the looks, and $mathbb{S}^3$ is a Lie group, so it's parallelizable), then $T(mathbb{S}^1timesmathbb{S}^3)cong Tmathbb{S}^1times Tmathbb{S}^3$ is again trivial.
This is not the case for $mathbb{S}^2timesmathbb{S}^2$.
To see this, observe that since $mathbb{S}^2timesmathbb{S}^2$ is a $CW$-complex, we can use any of its $CW$ structures to compute its Euler characteristic. For example, use the one for $mathbb{S}^2$ with one $0$-cell and one $2$-cell, to get one in $mathbb{S}^2timesmathbb{S}^2$ with one $0$-cell, two $2$-cells and one $4$-cell. Then its Euler characteristic is 4, now use the Poincaré-Hopf Index Theorem to prove that any vector field on $mathbb{S}^2timesmathbb{S}^2$ must have a zero, contradicting a possible triviallity of its tangent bundle.
$textbf{Credit}:$ to Mike Miller for the final argument using the Euler characteristic.
answered Nov 17 at 2:45
Laz
79749
It is not clear to me how you intend the argument in the final line to go. I do not know how to prove that if $E$ and $F$ are nontrivial vector bundles, their Cartesian product is nontrivial. I am not sure I believe that it's true. Perhaps what you wanted to point out is that $chi(S^2 times S^2) = 4$, and so it carries no nonvanishing field of tangent vectors.
– Mike Miller
2 days ago
Thanks @MikeMiller, you are totally right about my sloppyness. I just happened to know the non-triviallity of $T(mathbb{S}^2timesmathbb{S}^2)$ but I was going for an argument that was going to take me way to long to prove it. I corrected it using your argument which is a thousand times finer. I just want to point out that in this case $E$, $F$ are not any, are the same sphere, of even dimension.
– Laz
2 days ago
I don't need credit, I was just curious if there was a different argument that worked. Thank you though!
– Mike Miller
2 days ago
You totally got it, thanks again!
– Laz
2 days ago
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
They're not homeomorphic, so they won't be diffeomorphic. To see that they aren't homeomorphic, we compute their fundamental groups. Recalling that
$$ pi_1(S^1) = mathbb{Z} quad text{and} quad pi_1(S^2) = pi_1(S^3) = 1$$
and $pi_1(Xtimes Y) cong pi_1(X) oplus pi_1(Y)$, we find
$$pi_1(S^2 times S^2) = 1 oplus 1 = 1$$
while
$$pi_1(S^1 times S^3) = mathbb{Z} oplus 1 cong mathbb{Z} $$
Since their fundamental groups aren't isomorphic, they aren't homeomorphic, hence not diffeomorphic.
answered Nov 16 at 23:51
Hayden
13.7k12347
My apologies, but I am actually looking for a differential topological method since I am self-studying differential topology and do not have any background in algebraic topology. I know most of the basic techniques from differential topology except de Rham cohomology. If there are no any other answers, I will accept yours. My sincere apologies again and thank you for your answer.
– penny
Nov 17 at 1:50
1
@penny That's understandable. A differential topology proof that comes to mind is that $S^2 times S^2$ has no smooth field of non-zero tangent vectors, while $S^1times S^3$ does, but I'll admit my differential topology is rusty.
– Hayden
Nov 17 at 2:04
add a comment |
up vote
4
down vote
They're not homeomorphic, so they won't be diffeomorphic. To see that they aren't homeomorphic, we compute their fundamental groups. Recalling that
$$ pi_1(S^1) = mathbb{Z} quad text{and} quad pi_1(S^2) = pi_1(S^3) = 1$$
and $pi_1(Xtimes Y) cong pi_1(X) oplus pi_1(Y)$, we find
$$pi_1(S^2 times S^2) = 1 oplus 1 = 1$$
while
$$pi_1(S^1 times S^3) = mathbb{Z} oplus 1 cong mathbb{Z} $$
Since their fundamental groups aren't isomorphic, they aren't homeomorphic, hence not diffeomorphic.
answered Nov 16 at 23:51
Hayden
13.7k12347
My apologies, but I am actually looking for a differential topological method since I am self-studying differential topology and do not have any background in algebraic topology. I know most of the basic techniques from differential topology except de Rham cohomology. If there are no any other answers, I will accept yours. My sincere apologies again and thank you for your answer.
– penny
Nov 17 at 1:50
1
@penny That's understandable. A differential topology proof that comes to mind is that $S^2 times S^2$ has no smooth field of non-zero tangent vectors, while $S^1times S^3$ does, but I'll admit my differential topology is rusty.
– Hayden
Nov 17 at 2:04
add a comment |
up vote
4
down vote
up vote
4
down vote
They're not homeomorphic, so they won't be diffeomorphic. To see that they aren't homeomorphic, we compute their fundamental groups. Recalling that
$$ pi_1(S^1) = mathbb{Z} quad text{and} quad pi_1(S^2) = pi_1(S^3) = 1$$
and $pi_1(Xtimes Y) cong pi_1(X) oplus pi_1(Y)$, we find
$$pi_1(S^2 times S^2) = 1 oplus 1 = 1$$
while
$$pi_1(S^1 times S^3) = mathbb{Z} oplus 1 cong mathbb{Z} $$
Since their fundamental groups aren't isomorphic, they aren't homeomorphic, hence not diffeomorphic.
answered Nov 16 at 23:51
Hayden
13.7k12347
They're not homeomorphic, so they won't be diffeomorphic. To see that they aren't homeomorphic, we compute their fundamental groups. Recalling that
$$ pi_1(S^1) = mathbb{Z} quad text{and} quad pi_1(S^2) = pi_1(S^3) = 1$$
and $pi_1(Xtimes Y) cong pi_1(X) oplus pi_1(Y)$, we find
$$pi_1(S^2 times S^2) = 1 oplus 1 = 1$$
while
$$pi_1(S^1 times S^3) = mathbb{Z} oplus 1 cong mathbb{Z} $$
Since their fundamental groups aren't isomorphic, they aren't homeomorphic, hence not diffeomorphic.
answered Nov 16 at 23:51
Hayden
13.7k12347
answered Nov 16 at 23:51
Hayden
13.7k12347
answered Nov 16 at 23:51
Hayden
13.7k12347
answered Nov 16 at 23:51
Hayden
13.7k12347
13.7k12347
My apologies, but I am actually looking for a differential topological method since I am self-studying differential topology and do not have any background in algebraic topology. I know most of the basic techniques from differential topology except de Rham cohomology. If there are no any other answers, I will accept yours. My sincere apologies again and thank you for your answer.
– penny
Nov 17 at 1:50
1
@penny That's understandable. A differential topology proof that comes to mind is that $S^2 times S^2$ has no smooth field of non-zero tangent vectors, while $S^1times S^3$ does, but I'll admit my differential topology is rusty.
– Hayden
Nov 17 at 2:04
add a comment |
My apologies, but I am actually looking for a differential topological method since I am self-studying differential topology and do not have any background in algebraic topology. I know most of the basic techniques from differential topology except de Rham cohomology. If there are no any other answers, I will accept yours. My sincere apologies again and thank you for your answer.
– penny
Nov 17 at 1:50
1
@penny That's understandable. A differential topology proof that comes to mind is that $S^2 times S^2$ has no smooth field of non-zero tangent vectors, while $S^1times S^3$ does, but I'll admit my differential topology is rusty.
– Hayden
Nov 17 at 2:04
My apologies, but I am actually looking for a differential topological method since I am self-studying differential topology and do not have any background in algebraic topology. I know most of the basic techniques from differential topology except de Rham cohomology. If there are no any other answers, I will accept yours. My sincere apologies again and thank you for your answer.
– penny
Nov 17 at 1:50
My apologies, but I am actually looking for a differential topological method since I am self-studying differential topology and do not have any background in algebraic topology. I know most of the basic techniques from differential topology except de Rham cohomology. If there are no any other answers, I will accept yours. My sincere apologies again and thank you for your answer.
– penny
Nov 17 at 1:50
1
1
@penny That's understandable. A differential topology proof that comes to mind is that $S^2 times S^2$ has no smooth field of non-zero tangent vectors, while $S^1times S^3$ does, but I'll admit my differential topology is rusty.
– Hayden
Nov 17 at 2:04
@penny That's understandable. A differential topology proof that comes to mind is that $S^2 times S^2$ has no smooth field of non-zero tangent vectors, while $S^1times S^3$ does, but I'll admit my differential topology is rusty.
– Hayden
Nov 17 at 2:04
add a comment |
up vote
1
down vote
I would have used the fundamental group as @Hayden, but you want a differential argument, so here it goes.
Since the tangent bundles $Tmathbb{S}^1$, $Tmathbb{S}^3$ are trivial (you can check this in Spivak's Volume 1, e.g, but the fact that $Tmathbb{S}^1$ is trivial is clear from the looks, and $mathbb{S}^3$ is a Lie group, so it's parallelizable), then $T(mathbb{S}^1timesmathbb{S}^3)cong Tmathbb{S}^1times Tmathbb{S}^3$ is again trivial.
This is not the case for $mathbb{S}^2timesmathbb{S}^2$.
To see this, observe that since $mathbb{S}^2timesmathbb{S}^2$ is a $CW$-complex, we can use any of its $CW$ structures to compute its Euler characteristic. For example, use the one for $mathbb{S}^2$ with one $0$-cell and one $2$-cell, to get one in $mathbb{S}^2timesmathbb{S}^2$ with one $0$-cell, two $2$-cells and one $4$-cell. Then its Euler characteristic is 4, now use the Poincaré-Hopf Index Theorem to prove that any vector field on $mathbb{S}^2timesmathbb{S}^2$ must have a zero, contradicting a possible triviallity of its tangent bundle.
$textbf{Credit}:$ to Mike Miller for the final argument using the Euler characteristic.
answered Nov 17 at 2:45
Laz
79749
It is not clear to me how you intend the argument in the final line to go. I do not know how to prove that if $E$ and $F$ are nontrivial vector bundles, their Cartesian product is nontrivial. I am not sure I believe that it's true. Perhaps what you wanted to point out is that $chi(S^2 times S^2) = 4$, and so it carries no nonvanishing field of tangent vectors.
– Mike Miller
2 days ago
Thanks @MikeMiller, you are totally right about my sloppyness. I just happened to know the non-triviallity of $T(mathbb{S}^2timesmathbb{S}^2)$ but I was going for an argument that was going to take me way to long to prove it. I corrected it using your argument which is a thousand times finer. I just want to point out that in this case $E$, $F$ are not any, are the same sphere, of even dimension.
– Laz
2 days ago
I don't need credit, I was just curious if there was a different argument that worked. Thank you though!
– Mike Miller
2 days ago
You totally got it, thanks again!
– Laz
2 days ago
add a comment |
up vote
1
down vote
I would have used the fundamental group as @Hayden, but you want a differential argument, so here it goes.
Since the tangent bundles $Tmathbb{S}^1$, $Tmathbb{S}^3$ are trivial (you can check this in Spivak's Volume 1, e.g, but the fact that $Tmathbb{S}^1$ is trivial is clear from the looks, and $mathbb{S}^3$ is a Lie group, so it's parallelizable), then $T(mathbb{S}^1timesmathbb{S}^3)cong Tmathbb{S}^1times Tmathbb{S}^3$ is again trivial.
This is not the case for $mathbb{S}^2timesmathbb{S}^2$.
To see this, observe that since $mathbb{S}^2timesmathbb{S}^2$ is a $CW$-complex, we can use any of its $CW$ structures to compute its Euler characteristic. For example, use the one for $mathbb{S}^2$ with one $0$-cell and one $2$-cell, to get one in $mathbb{S}^2timesmathbb{S}^2$ with one $0$-cell, two $2$-cells and one $4$-cell. Then its Euler characteristic is 4, now use the Poincaré-Hopf Index Theorem to prove that any vector field on $mathbb{S}^2timesmathbb{S}^2$ must have a zero, contradicting a possible triviallity of its tangent bundle.
$textbf{Credit}:$ to Mike Miller for the final argument using the Euler characteristic.
answered Nov 17 at 2:45
Laz
79749
It is not clear to me how you intend the argument in the final line to go. I do not know how to prove that if $E$ and $F$ are nontrivial vector bundles, their Cartesian product is nontrivial. I am not sure I believe that it's true. Perhaps what you wanted to point out is that $chi(S^2 times S^2) = 4$, and so it carries no nonvanishing field of tangent vectors.
– Mike Miller
2 days ago
Thanks @MikeMiller, you are totally right about my sloppyness. I just happened to know the non-triviallity of $T(mathbb{S}^2timesmathbb{S}^2)$ but I was going for an argument that was going to take me way to long to prove it. I corrected it using your argument which is a thousand times finer. I just want to point out that in this case $E$, $F$ are not any, are the same sphere, of even dimension.
– Laz
2 days ago
I don't need credit, I was just curious if there was a different argument that worked. Thank you though!
– Mike Miller
2 days ago
You totally got it, thanks again!
– Laz
2 days ago
add a comment |
up vote
1
down vote
up vote
1
down vote
I would have used the fundamental group as @Hayden, but you want a differential argument, so here it goes.
Since the tangent bundles $Tmathbb{S}^1$, $Tmathbb{S}^3$ are trivial (you can check this in Spivak's Volume 1, e.g, but the fact that $Tmathbb{S}^1$ is trivial is clear from the looks, and $mathbb{S}^3$ is a Lie group, so it's parallelizable), then $T(mathbb{S}^1timesmathbb{S}^3)cong Tmathbb{S}^1times Tmathbb{S}^3$ is again trivial.
This is not the case for $mathbb{S}^2timesmathbb{S}^2$.
To see this, observe that since $mathbb{S}^2timesmathbb{S}^2$ is a $CW$-complex, we can use any of its $CW$ structures to compute its Euler characteristic. For example, use the one for $mathbb{S}^2$ with one $0$-cell and one $2$-cell, to get one in $mathbb{S}^2timesmathbb{S}^2$ with one $0$-cell, two $2$-cells and one $4$-cell. Then its Euler characteristic is 4, now use the Poincaré-Hopf Index Theorem to prove that any vector field on $mathbb{S}^2timesmathbb{S}^2$ must have a zero, contradicting a possible triviallity of its tangent bundle.
$textbf{Credit}:$ to Mike Miller for the final argument using the Euler characteristic.
answered Nov 17 at 2:45
Laz
79749
I would have used the fundamental group as @Hayden, but you want a differential argument, so here it goes.
Since the tangent bundles $Tmathbb{S}^1$, $Tmathbb{S}^3$ are trivial (you can check this in Spivak's Volume 1, e.g, but the fact that $Tmathbb{S}^1$ is trivial is clear from the looks, and $mathbb{S}^3$ is a Lie group, so it's parallelizable), then $T(mathbb{S}^1timesmathbb{S}^3)cong Tmathbb{S}^1times Tmathbb{S}^3$ is again trivial.
This is not the case for $mathbb{S}^2timesmathbb{S}^2$.
To see this, observe that since $mathbb{S}^2timesmathbb{S}^2$ is a $CW$-complex, we can use any of its $CW$ structures to compute its Euler characteristic. For example, use the one for $mathbb{S}^2$ with one $0$-cell and one $2$-cell, to get one in $mathbb{S}^2timesmathbb{S}^2$ with one $0$-cell, two $2$-cells and one $4$-cell. Then its Euler characteristic is 4, now use the Poincaré-Hopf Index Theorem to prove that any vector field on $mathbb{S}^2timesmathbb{S}^2$ must have a zero, contradicting a possible triviallity of its tangent bundle.
$textbf{Credit}:$ to Mike Miller for the final argument using the Euler characteristic.
answered Nov 17 at 2:45
Laz
79749
edited 2 days ago
answered Nov 17 at 2:45
Laz
79749
answered Nov 17 at 2:45
Laz
79749
answered Nov 17 at 2:45
Laz
79749
79749
It is not clear to me how you intend the argument in the final line to go. I do not know how to prove that if $E$ and $F$ are nontrivial vector bundles, their Cartesian product is nontrivial. I am not sure I believe that it's true. Perhaps what you wanted to point out is that $chi(S^2 times S^2) = 4$, and so it carries no nonvanishing field of tangent vectors.
– Mike Miller
2 days ago
Thanks @MikeMiller, you are totally right about my sloppyness. I just happened to know the non-triviallity of $T(mathbb{S}^2timesmathbb{S}^2)$ but I was going for an argument that was going to take me way to long to prove it. I corrected it using your argument which is a thousand times finer. I just want to point out that in this case $E$, $F$ are not any, are the same sphere, of even dimension.
– Laz
2 days ago
I don't need credit, I was just curious if there was a different argument that worked. Thank you though!
– Mike Miller
2 days ago
You totally got it, thanks again!
– Laz
2 days ago
add a comment |
It is not clear to me how you intend the argument in the final line to go. I do not know how to prove that if $E$ and $F$ are nontrivial vector bundles, their Cartesian product is nontrivial. I am not sure I believe that it's true. Perhaps what you wanted to point out is that $chi(S^2 times S^2) = 4$, and so it carries no nonvanishing field of tangent vectors.
– Mike Miller
2 days ago
Thanks @MikeMiller, you are totally right about my sloppyness. I just happened to know the non-triviallity of $T(mathbb{S}^2timesmathbb{S}^2)$ but I was going for an argument that was going to take me way to long to prove it. I corrected it using your argument which is a thousand times finer. I just want to point out that in this case $E$, $F$ are not any, are the same sphere, of even dimension.
– Laz
2 days ago
I don't need credit, I was just curious if there was a different argument that worked. Thank you though!
– Mike Miller
2 days ago
You totally got it, thanks again!
– Laz
2 days ago
It is not clear to me how you intend the argument in the final line to go. I do not know how to prove that if $E$ and $F$ are nontrivial vector bundles, their Cartesian product is nontrivial. I am not sure I believe that it's true. Perhaps what you wanted to point out is that $chi(S^2 times S^2) = 4$, and so it carries no nonvanishing field of tangent vectors.
– Mike Miller
2 days ago
It is not clear to me how you intend the argument in the final line to go. I do not know how to prove that if $E$ and $F$ are nontrivial vector bundles, their Cartesian product is nontrivial. I am not sure I believe that it's true. Perhaps what you wanted to point out is that $chi(S^2 times S^2) = 4$, and so it carries no nonvanishing field of tangent vectors.
– Mike Miller
2 days ago
Thanks @MikeMiller, you are totally right about my sloppyness. I just happened to know the non-triviallity of $T(mathbb{S}^2timesmathbb{S}^2)$ but I was going for an argument that was going to take me way to long to prove it. I corrected it using your argument which is a thousand times finer. I just want to point out that in this case $E$, $F$ are not any, are the same sphere, of even dimension.
– Laz
2 days ago
Thanks @MikeMiller, you are totally right about my sloppyness. I just happened to know the non-triviallity of $T(mathbb{S}^2timesmathbb{S}^2)$ but I was going for an argument that was going to take me way to long to prove it. I corrected it using your argument which is a thousand times finer. I just want to point out that in this case $E$, $F$ are not any, are the same sphere, of even dimension.
– Laz
2 days ago
I don't need credit, I was just curious if there was a different argument that worked. Thank you though!
– Mike Miller
2 days ago
I don't need credit, I was just curious if there was a different argument that worked. Thank you though!
– Mike Miller
2 days ago
You totally got it, thanks again!
– Laz
2 days ago
You totally got it, thanks again!
– Laz
2 days ago
add a comment |
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The de Rham cohomology is also helpful.
– Laz
Nov 17 at 2:47