Definition of connected sum and orientation problem
I am reading Kosinski's book. To define the connected sum of $M_1^n$and $M_2^n$ (oriented and closed manifolds) we choose two embeddings of the disk $h_i:mathbb{D}^nto M_i$ such that $h_1$ preserves the orientation and $h_2$ reverse it then we can construct the quotient manifold
$$frac{M_1setminus h_1(0) sqcup M_2setminus h_2(0)} sim $$
where $xin h_1(mathbb{D}^n) sim h_2(frac{h_1^{-1}(x)}{||h_1^{-1}(x)||^2})in h_2(mathbb{D}^n) $ .
Now Kosinski shows that this construction doesn't depend on the choice of the embeddings because we know that all the embeddings of $mathbb{D}^n$ that preserve the orientation are isotopic.
From this though it follows that we can remove the assumption that $h_i$ should preserve(or not) the orientation in the definition of connected sum that shouldn't pose restrictions on the positivity of the diffeomorphism.
I explain why. If say, $h_1$ doesn't preserve the orientation,we just change the orientation on $M_1$ and we get an orientation preserving embedding. Then
if $(-M_1)sharp M_2$ is diffeomorphic to $M_1sharp M_2$
we have shown that for the definition it is not necessary to consider orientation.
Am I right?
algebraic-topology differential-topology geometric-topology low-dimensional-topology
asked Nov 27 '18 at 19:08
Warlock of Firetop Mountain
2,269815
add a comment |
I am reading Kosinski's book. To define the connected sum of $M_1^n$and $M_2^n$ (oriented and closed manifolds) we choose two embeddings of the disk $h_i:mathbb{D}^nto M_i$ such that $h_1$ preserves the orientation and $h_2$ reverse it then we can construct the quotient manifold
$$frac{M_1setminus h_1(0) sqcup M_2setminus h_2(0)} sim $$
where $xin h_1(mathbb{D}^n) sim h_2(frac{h_1^{-1}(x)}{||h_1^{-1}(x)||^2})in h_2(mathbb{D}^n) $ .
Now Kosinski shows that this construction doesn't depend on the choice of the embeddings because we know that all the embeddings of $mathbb{D}^n$ that preserve the orientation are isotopic.
From this though it follows that we can remove the assumption that $h_i$ should preserve(or not) the orientation in the definition of connected sum that shouldn't pose restrictions on the positivity of the diffeomorphism.
I explain why. If say, $h_1$ doesn't preserve the orientation,we just change the orientation on $M_1$ and we get an orientation preserving embedding. Then
if $(-M_1)sharp M_2$ is diffeomorphic to $M_1sharp M_2$
we have shown that for the definition it is not necessary to consider orientation.
Am I right?
algebraic-topology differential-topology geometric-topology low-dimensional-topology
asked Nov 27 '18 at 19:08
Warlock of Firetop Mountain
2,269815
add a comment |
I am reading Kosinski's book. To define the connected sum of $M_1^n$and $M_2^n$ (oriented and closed manifolds) we choose two embeddings of the disk $h_i:mathbb{D}^nto M_i$ such that $h_1$ preserves the orientation and $h_2$ reverse it then we can construct the quotient manifold
$$frac{M_1setminus h_1(0) sqcup M_2setminus h_2(0)} sim $$
where $xin h_1(mathbb{D}^n) sim h_2(frac{h_1^{-1}(x)}{||h_1^{-1}(x)||^2})in h_2(mathbb{D}^n) $ .
Now Kosinski shows that this construction doesn't depend on the choice of the embeddings because we know that all the embeddings of $mathbb{D}^n$ that preserve the orientation are isotopic.
From this though it follows that we can remove the assumption that $h_i$ should preserve(or not) the orientation in the definition of connected sum that shouldn't pose restrictions on the positivity of the diffeomorphism.
I explain why. If say, $h_1$ doesn't preserve the orientation,we just change the orientation on $M_1$ and we get an orientation preserving embedding. Then
if $(-M_1)sharp M_2$ is diffeomorphic to $M_1sharp M_2$
we have shown that for the definition it is not necessary to consider orientation.
Am I right?
algebraic-topology differential-topology geometric-topology low-dimensional-topology
asked Nov 27 '18 at 19:08
Warlock of Firetop Mountain
2,269815
I am reading Kosinski's book. To define the connected sum of $M_1^n$and $M_2^n$ (oriented and closed manifolds) we choose two embeddings of the disk $h_i:mathbb{D}^nto M_i$ such that $h_1$ preserves the orientation and $h_2$ reverse it then we can construct the quotient manifold
$$frac{M_1setminus h_1(0) sqcup M_2setminus h_2(0)} sim $$
where $xin h_1(mathbb{D}^n) sim h_2(frac{h_1^{-1}(x)}{||h_1^{-1}(x)||^2})in h_2(mathbb{D}^n) $ .
Now Kosinski shows that this construction doesn't depend on the choice of the embeddings because we know that all the embeddings of $mathbb{D}^n$ that preserve the orientation are isotopic.
From this though it follows that we can remove the assumption that $h_i$ should preserve(or not) the orientation in the definition of connected sum that shouldn't pose restrictions on the positivity of the diffeomorphism.
I explain why. If say, $h_1$ doesn't preserve the orientation,we just change the orientation on $M_1$ and we get an orientation preserving embedding. Then
if $(-M_1)sharp M_2$ is diffeomorphic to $M_1sharp M_2$
we have shown that for the definition it is not necessary to consider orientation.
Am I right?
algebraic-topology differential-topology geometric-topology low-dimensional-topology
algebraic-topology differential-topology geometric-topology low-dimensional-topology
asked Nov 27 '18 at 19:08
Warlock of Firetop Mountain
2,269815
asked Nov 27 '18 at 19:08
Warlock of Firetop Mountain
2,269815
edited Nov 27 '18 at 22:22
asked Nov 27 '18 at 19:08
Warlock of Firetop Mountain
2,269815
asked Nov 27 '18 at 19:08
Warlock of Firetop Mountain
2,269815
asked Nov 27 '18 at 19:08
Warlock of Firetop Mountain
2,269815
2,269815
add a comment |
add a comment |
1 Answer
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No, those two manifolds are not always diffeomorphic, or even homotopy equivalent. The simplest counterexample is usually given as $Bbb{CP}^2 # overline{Bbb{CP}}^2$ and $Bbb{CP}^2 # Bbb{CP}^2$. One has signature 2, the other has signature 0.
If one of the manifolds is not orientable, then there is only one embedding of the disc up to isotopy, and the choice of embedding of the disc in the other manifold doesn't matter.
It is a fluke of luck that you can ignore this for surfaces, where every surface admits an orientation reversing self-diffeomorphism.
answered Nov 27 '18 at 20:12
Mike Miller
36.1k470137
Do you know of an example of closed orientable (simply connected?) $M$ and $N$ which do not admit orientation reversing diffeos, but yet $Msharp N$ and $Msharp overline{N}$ are diffeomorphic?
– Jason DeVito
Nov 27 '18 at 20:20
2
I wasn't aware of the result you just quoted. But here is an example. According to arxiv.org/pdf/1708.06582.pdf (remark 2.8), there is an $1240$-dimensional exotic sphere $Sigma^{1240}$ of order $7$ for which $mathbb{H}P^{310}sharp Sigma sharp ...sharp Sigma cong mathbb{H}P^{310}$ for any number of $Sigma$ summands. Now, an exotic sphere admits an orientation reversing diffeo iff it's order $2$, so $Sigma$ works for $N$. Further, $mathbb{H}P^k$ does not admit an orientation reversing diffeo for $k > 1$ (since $p_1$ is non-trivial), so this works for $M$.
– Jason DeVito
Nov 27 '18 at 20:46
2
And, as I'm sure you know, $overline{Sigma}cong underbrace{Sigma sharp ... sharp Sigma}_{6text times}$. Finally, this is the first time in my life that I've used an example whose dimension is in the thousands. Fun day!
– Jason DeVito
Nov 27 '18 at 20:49
1
@WarlockofFiretopMountain Once you know the definition of signature, it is immediate from knowing the cohomology ring of the two factors.
– Mike Miller
Nov 27 '18 at 21:42
1
@WarlockofFiretopMountain: Kosinski's book "Differential Manifolds" has a lot of this stuff about connect sums, as well as some of the facts I used about exotic spheres.
– Jason DeVito
Nov 27 '18 at 21:50
|
show 3 more comments
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No, those two manifolds are not always diffeomorphic, or even homotopy equivalent. The simplest counterexample is usually given as $Bbb{CP}^2 # overline{Bbb{CP}}^2$ and $Bbb{CP}^2 # Bbb{CP}^2$. One has signature 2, the other has signature 0.
If one of the manifolds is not orientable, then there is only one embedding of the disc up to isotopy, and the choice of embedding of the disc in the other manifold doesn't matter.
It is a fluke of luck that you can ignore this for surfaces, where every surface admits an orientation reversing self-diffeomorphism.
answered Nov 27 '18 at 20:12
Mike Miller
36.1k470137
Do you know of an example of closed orientable (simply connected?) $M$ and $N$ which do not admit orientation reversing diffeos, but yet $Msharp N$ and $Msharp overline{N}$ are diffeomorphic?
– Jason DeVito
Nov 27 '18 at 20:20
2
I wasn't aware of the result you just quoted. But here is an example. According to arxiv.org/pdf/1708.06582.pdf (remark 2.8), there is an $1240$-dimensional exotic sphere $Sigma^{1240}$ of order $7$ for which $mathbb{H}P^{310}sharp Sigma sharp ...sharp Sigma cong mathbb{H}P^{310}$ for any number of $Sigma$ summands. Now, an exotic sphere admits an orientation reversing diffeo iff it's order $2$, so $Sigma$ works for $N$. Further, $mathbb{H}P^k$ does not admit an orientation reversing diffeo for $k > 1$ (since $p_1$ is non-trivial), so this works for $M$.
– Jason DeVito
Nov 27 '18 at 20:46
2
And, as I'm sure you know, $overline{Sigma}cong underbrace{Sigma sharp ... sharp Sigma}_{6text times}$. Finally, this is the first time in my life that I've used an example whose dimension is in the thousands. Fun day!
– Jason DeVito
Nov 27 '18 at 20:49
1
@WarlockofFiretopMountain Once you know the definition of signature, it is immediate from knowing the cohomology ring of the two factors.
– Mike Miller
Nov 27 '18 at 21:42
1
@WarlockofFiretopMountain: Kosinski's book "Differential Manifolds" has a lot of this stuff about connect sums, as well as some of the facts I used about exotic spheres.
– Jason DeVito
Nov 27 '18 at 21:50
|
show 3 more comments
No, those two manifolds are not always diffeomorphic, or even homotopy equivalent. The simplest counterexample is usually given as $Bbb{CP}^2 # overline{Bbb{CP}}^2$ and $Bbb{CP}^2 # Bbb{CP}^2$. One has signature 2, the other has signature 0.
If one of the manifolds is not orientable, then there is only one embedding of the disc up to isotopy, and the choice of embedding of the disc in the other manifold doesn't matter.
It is a fluke of luck that you can ignore this for surfaces, where every surface admits an orientation reversing self-diffeomorphism.
answered Nov 27 '18 at 20:12
Mike Miller
36.1k470137
Do you know of an example of closed orientable (simply connected?) $M$ and $N$ which do not admit orientation reversing diffeos, but yet $Msharp N$ and $Msharp overline{N}$ are diffeomorphic?
– Jason DeVito
Nov 27 '18 at 20:20
2
I wasn't aware of the result you just quoted. But here is an example. According to arxiv.org/pdf/1708.06582.pdf (remark 2.8), there is an $1240$-dimensional exotic sphere $Sigma^{1240}$ of order $7$ for which $mathbb{H}P^{310}sharp Sigma sharp ...sharp Sigma cong mathbb{H}P^{310}$ for any number of $Sigma$ summands. Now, an exotic sphere admits an orientation reversing diffeo iff it's order $2$, so $Sigma$ works for $N$. Further, $mathbb{H}P^k$ does not admit an orientation reversing diffeo for $k > 1$ (since $p_1$ is non-trivial), so this works for $M$.
– Jason DeVito
Nov 27 '18 at 20:46
2
And, as I'm sure you know, $overline{Sigma}cong underbrace{Sigma sharp ... sharp Sigma}_{6text times}$. Finally, this is the first time in my life that I've used an example whose dimension is in the thousands. Fun day!
– Jason DeVito
Nov 27 '18 at 20:49
1
@WarlockofFiretopMountain Once you know the definition of signature, it is immediate from knowing the cohomology ring of the two factors.
– Mike Miller
Nov 27 '18 at 21:42
1
@WarlockofFiretopMountain: Kosinski's book "Differential Manifolds" has a lot of this stuff about connect sums, as well as some of the facts I used about exotic spheres.
– Jason DeVito
Nov 27 '18 at 21:50
|
show 3 more comments
No, those two manifolds are not always diffeomorphic, or even homotopy equivalent. The simplest counterexample is usually given as $Bbb{CP}^2 # overline{Bbb{CP}}^2$ and $Bbb{CP}^2 # Bbb{CP}^2$. One has signature 2, the other has signature 0.
If one of the manifolds is not orientable, then there is only one embedding of the disc up to isotopy, and the choice of embedding of the disc in the other manifold doesn't matter.
It is a fluke of luck that you can ignore this for surfaces, where every surface admits an orientation reversing self-diffeomorphism.
answered Nov 27 '18 at 20:12
Mike Miller
36.1k470137
No, those two manifolds are not always diffeomorphic, or even homotopy equivalent. The simplest counterexample is usually given as $Bbb{CP}^2 # overline{Bbb{CP}}^2$ and $Bbb{CP}^2 # Bbb{CP}^2$. One has signature 2, the other has signature 0.
If one of the manifolds is not orientable, then there is only one embedding of the disc up to isotopy, and the choice of embedding of the disc in the other manifold doesn't matter.
It is a fluke of luck that you can ignore this for surfaces, where every surface admits an orientation reversing self-diffeomorphism.
answered Nov 27 '18 at 20:12
Mike Miller
36.1k470137
answered Nov 27 '18 at 20:12
Mike Miller
36.1k470137
answered Nov 27 '18 at 20:12
Mike Miller
36.1k470137
answered Nov 27 '18 at 20:12
Mike Miller
36.1k470137
36.1k470137
Do you know of an example of closed orientable (simply connected?) $M$ and $N$ which do not admit orientation reversing diffeos, but yet $Msharp N$ and $Msharp overline{N}$ are diffeomorphic?
– Jason DeVito
Nov 27 '18 at 20:20
2
I wasn't aware of the result you just quoted. But here is an example. According to arxiv.org/pdf/1708.06582.pdf (remark 2.8), there is an $1240$-dimensional exotic sphere $Sigma^{1240}$ of order $7$ for which $mathbb{H}P^{310}sharp Sigma sharp ...sharp Sigma cong mathbb{H}P^{310}$ for any number of $Sigma$ summands. Now, an exotic sphere admits an orientation reversing diffeo iff it's order $2$, so $Sigma$ works for $N$. Further, $mathbb{H}P^k$ does not admit an orientation reversing diffeo for $k > 1$ (since $p_1$ is non-trivial), so this works for $M$.
– Jason DeVito
Nov 27 '18 at 20:46
2
And, as I'm sure you know, $overline{Sigma}cong underbrace{Sigma sharp ... sharp Sigma}_{6text times}$. Finally, this is the first time in my life that I've used an example whose dimension is in the thousands. Fun day!
– Jason DeVito
Nov 27 '18 at 20:49
1
@WarlockofFiretopMountain Once you know the definition of signature, it is immediate from knowing the cohomology ring of the two factors.
– Mike Miller
Nov 27 '18 at 21:42
1
@WarlockofFiretopMountain: Kosinski's book "Differential Manifolds" has a lot of this stuff about connect sums, as well as some of the facts I used about exotic spheres.
– Jason DeVito
Nov 27 '18 at 21:50
|
show 3 more comments
Do you know of an example of closed orientable (simply connected?) $M$ and $N$ which do not admit orientation reversing diffeos, but yet $Msharp N$ and $Msharp overline{N}$ are diffeomorphic?
– Jason DeVito
Nov 27 '18 at 20:20
2
I wasn't aware of the result you just quoted. But here is an example. According to arxiv.org/pdf/1708.06582.pdf (remark 2.8), there is an $1240$-dimensional exotic sphere $Sigma^{1240}$ of order $7$ for which $mathbb{H}P^{310}sharp Sigma sharp ...sharp Sigma cong mathbb{H}P^{310}$ for any number of $Sigma$ summands. Now, an exotic sphere admits an orientation reversing diffeo iff it's order $2$, so $Sigma$ works for $N$. Further, $mathbb{H}P^k$ does not admit an orientation reversing diffeo for $k > 1$ (since $p_1$ is non-trivial), so this works for $M$.
– Jason DeVito
Nov 27 '18 at 20:46
2
And, as I'm sure you know, $overline{Sigma}cong underbrace{Sigma sharp ... sharp Sigma}_{6text times}$. Finally, this is the first time in my life that I've used an example whose dimension is in the thousands. Fun day!
– Jason DeVito
Nov 27 '18 at 20:49
1
@WarlockofFiretopMountain Once you know the definition of signature, it is immediate from knowing the cohomology ring of the two factors.
– Mike Miller
Nov 27 '18 at 21:42
1
@WarlockofFiretopMountain: Kosinski's book "Differential Manifolds" has a lot of this stuff about connect sums, as well as some of the facts I used about exotic spheres.
– Jason DeVito
Nov 27 '18 at 21:50
Do you know of an example of closed orientable (simply connected?) $M$ and $N$ which do not admit orientation reversing diffeos, but yet $Msharp N$ and $Msharp overline{N}$ are diffeomorphic?
– Jason DeVito
Nov 27 '18 at 20:20
Do you know of an example of closed orientable (simply connected?) $M$ and $N$ which do not admit orientation reversing diffeos, but yet $Msharp N$ and $Msharp overline{N}$ are diffeomorphic?
– Jason DeVito
Nov 27 '18 at 20:20
2
2
I wasn't aware of the result you just quoted. But here is an example. According to arxiv.org/pdf/1708.06582.pdf (remark 2.8), there is an $1240$-dimensional exotic sphere $Sigma^{1240}$ of order $7$ for which $mathbb{H}P^{310}sharp Sigma sharp ...sharp Sigma cong mathbb{H}P^{310}$ for any number of $Sigma$ summands. Now, an exotic sphere admits an orientation reversing diffeo iff it's order $2$, so $Sigma$ works for $N$. Further, $mathbb{H}P^k$ does not admit an orientation reversing diffeo for $k > 1$ (since $p_1$ is non-trivial), so this works for $M$.
– Jason DeVito
Nov 27 '18 at 20:46
I wasn't aware of the result you just quoted. But here is an example. According to arxiv.org/pdf/1708.06582.pdf (remark 2.8), there is an $1240$-dimensional exotic sphere $Sigma^{1240}$ of order $7$ for which $mathbb{H}P^{310}sharp Sigma sharp ...sharp Sigma cong mathbb{H}P^{310}$ for any number of $Sigma$ summands. Now, an exotic sphere admits an orientation reversing diffeo iff it's order $2$, so $Sigma$ works for $N$. Further, $mathbb{H}P^k$ does not admit an orientation reversing diffeo for $k > 1$ (since $p_1$ is non-trivial), so this works for $M$.
– Jason DeVito
Nov 27 '18 at 20:46
2
2
And, as I'm sure you know, $overline{Sigma}cong underbrace{Sigma sharp ... sharp Sigma}_{6text times}$. Finally, this is the first time in my life that I've used an example whose dimension is in the thousands. Fun day!
– Jason DeVito
Nov 27 '18 at 20:49
And, as I'm sure you know, $overline{Sigma}cong underbrace{Sigma sharp ... sharp Sigma}_{6text times}$. Finally, this is the first time in my life that I've used an example whose dimension is in the thousands. Fun day!
– Jason DeVito
Nov 27 '18 at 20:49
1
1
@WarlockofFiretopMountain Once you know the definition of signature, it is immediate from knowing the cohomology ring of the two factors.
– Mike Miller
Nov 27 '18 at 21:42
@WarlockofFiretopMountain Once you know the definition of signature, it is immediate from knowing the cohomology ring of the two factors.
– Mike Miller
Nov 27 '18 at 21:42
1
1
@WarlockofFiretopMountain: Kosinski's book "Differential Manifolds" has a lot of this stuff about connect sums, as well as some of the facts I used about exotic spheres.
– Jason DeVito
Nov 27 '18 at 21:50
@WarlockofFiretopMountain: Kosinski's book "Differential Manifolds" has a lot of this stuff about connect sums, as well as some of the facts I used about exotic spheres.
– Jason DeVito
Nov 27 '18 at 21:50
|
show 3 more comments
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