Opposite Orientation of Boundary in BordismsWhat does having a bar on a manifold mean?Evaluation and Coevaluation maps of a TQFTAxiomatizing oriented cobordism(Topological quantum field theory) identifying objects of cobordism categoryDefinition of Bordism - Gluing Manifolds with StructureRelations between the homotopy class and the orientation of the connected sum of two manifoldsBordism between bordismsWhy are bordism groups of a point nontrivialHow are topological invariants obtained from TQFTs used in practice?Definition of connected sum and orientation problem
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Opposite Orientation of Boundary in Bordisms
What does having a bar on a manifold mean?Evaluation and Coevaluation maps of a TQFTAxiomatizing oriented cobordism(Topological quantum field theory) identifying objects of cobordism categoryDefinition of Bordism - Gluing Manifolds with StructureRelations between the homotopy class and the orientation of the connected sum of two manifoldsBordism between bordismsWhy are bordism groups of a point nontrivialHow are topological invariants obtained from TQFTs used in practice?Definition of connected sum and orientation problem
$begingroup$
In Lurie's "On the Classification of Topological Field Theories" (and certainly other places) he defines the category $mathbfCob(n)$ who objects are oriented $(n-1)$ manifolds. Given $M,NinmathrmCob(n)$ a morphism $Mto N$ is an $n$-dimensional manifold $B$ equipped with an orientation preserving diffeomorphism $partial Bsimeq overlineMcoprod N$ where $overlineM$ denotes the manifold $M$ equipped with the opposite orientation.
What is the necessity of having one part of the boundary have the reverse orientation? If we define it so that the above equation is simply $Mcoprod N$, do we run into problems?
Thanks!
algebraic-topology mathematical-physics quantum-field-theory cobordism topological-quantum-field-theory
edited 2 days ago
Andrews
1,2691421
asked Dec 31 '11 at 15:51
Jonathan BeardsleyJonathan Beardsley
2,4171238
$endgroup$
add a comment |
$begingroup$
In Lurie's "On the Classification of Topological Field Theories" (and certainly other places) he defines the category $mathbfCob(n)$ who objects are oriented $(n-1)$ manifolds. Given $M,NinmathrmCob(n)$ a morphism $Mto N$ is an $n$-dimensional manifold $B$ equipped with an orientation preserving diffeomorphism $partial Bsimeq overlineMcoprod N$ where $overlineM$ denotes the manifold $M$ equipped with the opposite orientation.
What is the necessity of having one part of the boundary have the reverse orientation? If we define it so that the above equation is simply $Mcoprod N$, do we run into problems?
Thanks!
algebraic-topology mathematical-physics quantum-field-theory cobordism topological-quantum-field-theory
edited 2 days ago
Andrews
1,2691421
asked Dec 31 '11 at 15:51
Jonathan BeardsleyJonathan Beardsley
2,4171238
$endgroup$
add a comment |
$begingroup$
In Lurie's "On the Classification of Topological Field Theories" (and certainly other places) he defines the category $mathbfCob(n)$ who objects are oriented $(n-1)$ manifolds. Given $M,NinmathrmCob(n)$ a morphism $Mto N$ is an $n$-dimensional manifold $B$ equipped with an orientation preserving diffeomorphism $partial Bsimeq overlineMcoprod N$ where $overlineM$ denotes the manifold $M$ equipped with the opposite orientation.
What is the necessity of having one part of the boundary have the reverse orientation? If we define it so that the above equation is simply $Mcoprod N$, do we run into problems?
Thanks!
algebraic-topology mathematical-physics quantum-field-theory cobordism topological-quantum-field-theory
edited 2 days ago
Andrews
1,2691421
asked Dec 31 '11 at 15:51
Jonathan BeardsleyJonathan Beardsley
2,4171238
$endgroup$
In Lurie's "On the Classification of Topological Field Theories" (and certainly other places) he defines the category $mathbfCob(n)$ who objects are oriented $(n-1)$ manifolds. Given $M,NinmathrmCob(n)$ a morphism $Mto N$ is an $n$-dimensional manifold $B$ equipped with an orientation preserving diffeomorphism $partial Bsimeq overlineMcoprod N$ where $overlineM$ denotes the manifold $M$ equipped with the opposite orientation.
What is the necessity of having one part of the boundary have the reverse orientation? If we define it so that the above equation is simply $Mcoprod N$, do we run into problems?
Thanks!
algebraic-topology mathematical-physics quantum-field-theory cobordism topological-quantum-field-theory
algebraic-topology mathematical-physics quantum-field-theory cobordism topological-quantum-field-theory
edited 2 days ago
Andrews
1,2691421
asked Dec 31 '11 at 15:51
Jonathan BeardsleyJonathan Beardsley
2,4171238
edited 2 days ago
Andrews
1,2691421
asked Dec 31 '11 at 15:51
Jonathan BeardsleyJonathan Beardsley
2,4171238
edited 2 days ago
Andrews
1,2691421
edited 2 days ago
Andrews
1,2691421
edited 2 days ago
Andrews
1,2691421
1,2691421
asked Dec 31 '11 at 15:51
Jonathan BeardsleyJonathan Beardsley
2,4171238
asked Dec 31 '11 at 15:51
Jonathan BeardsleyJonathan Beardsley
2,4171238
asked Dec 31 '11 at 15:51
Jonathan BeardsleyJonathan Beardsley
2,4171238
2,4171238
add a comment |
add a comment |
1 Answer
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$begingroup$
Yes, it's essential. The boundary of an oriented manifolds inherits an orientation, the determination of which involves (depending on the definition of orientation), e.g. the inner normal at the boundary.
Consider two copies of $S^1$ making up the boundary of an annulus. (In general consider $Mtimes [0,1]$) The induced orientation (in this case representable by a nonzero one form, hence by an vector field on which this form is $1$) on the first $S^1$ will be pointing into a 'different direction' than the other one, because in both cases you are using the inner normal to determine it.
In order to have $S^1$ -- $M$ in general -- being bordant to itself (what you always want, it's supposed to be an equivalence relation) you have to revert the orientation of one of the copies.
$endgroup$
$begingroup$
Ah yes that makes sense. Basically if your arrows are going to be pointing outwards, the boundary circles have to be oriented oppositely. Haha, that's a really infantile way of describing it, but I'm pretty sure I get what you're saying.
$endgroup$
– Jonathan Beardsley
Dec 31 '11 at 16:25
3
$begingroup$
@JBeardz Why infantile? Of course a rigouros proof would need additional reasioning, but the geometric picture is what is really behind it, and in case of $S^1$ it's possible to easily visualize it. I'd claim it's absolutely to the point.
$endgroup$
– user20266
Dec 31 '11 at 17:15
add a comment |
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$begingroup$
Yes, it's essential. The boundary of an oriented manifolds inherits an orientation, the determination of which involves (depending on the definition of orientation), e.g. the inner normal at the boundary.
Consider two copies of $S^1$ making up the boundary of an annulus. (In general consider $Mtimes [0,1]$) The induced orientation (in this case representable by a nonzero one form, hence by an vector field on which this form is $1$) on the first $S^1$ will be pointing into a 'different direction' than the other one, because in both cases you are using the inner normal to determine it.
In order to have $S^1$ -- $M$ in general -- being bordant to itself (what you always want, it's supposed to be an equivalence relation) you have to revert the orientation of one of the copies.
$endgroup$
$begingroup$
Ah yes that makes sense. Basically if your arrows are going to be pointing outwards, the boundary circles have to be oriented oppositely. Haha, that's a really infantile way of describing it, but I'm pretty sure I get what you're saying.
$endgroup$
– Jonathan Beardsley
Dec 31 '11 at 16:25
3
$begingroup$
@JBeardz Why infantile? Of course a rigouros proof would need additional reasioning, but the geometric picture is what is really behind it, and in case of $S^1$ it's possible to easily visualize it. I'd claim it's absolutely to the point.
$endgroup$
– user20266
Dec 31 '11 at 17:15
add a comment |
$begingroup$
Yes, it's essential. The boundary of an oriented manifolds inherits an orientation, the determination of which involves (depending on the definition of orientation), e.g. the inner normal at the boundary.
Consider two copies of $S^1$ making up the boundary of an annulus. (In general consider $Mtimes [0,1]$) The induced orientation (in this case representable by a nonzero one form, hence by an vector field on which this form is $1$) on the first $S^1$ will be pointing into a 'different direction' than the other one, because in both cases you are using the inner normal to determine it.
In order to have $S^1$ -- $M$ in general -- being bordant to itself (what you always want, it's supposed to be an equivalence relation) you have to revert the orientation of one of the copies.
$endgroup$
$begingroup$
Ah yes that makes sense. Basically if your arrows are going to be pointing outwards, the boundary circles have to be oriented oppositely. Haha, that's a really infantile way of describing it, but I'm pretty sure I get what you're saying.
$endgroup$
– Jonathan Beardsley
Dec 31 '11 at 16:25
3
$begingroup$
@JBeardz Why infantile? Of course a rigouros proof would need additional reasioning, but the geometric picture is what is really behind it, and in case of $S^1$ it's possible to easily visualize it. I'd claim it's absolutely to the point.
$endgroup$
– user20266
Dec 31 '11 at 17:15
add a comment |
$begingroup$
Yes, it's essential. The boundary of an oriented manifolds inherits an orientation, the determination of which involves (depending on the definition of orientation), e.g. the inner normal at the boundary.
Consider two copies of $S^1$ making up the boundary of an annulus. (In general consider $Mtimes [0,1]$) The induced orientation (in this case representable by a nonzero one form, hence by an vector field on which this form is $1$) on the first $S^1$ will be pointing into a 'different direction' than the other one, because in both cases you are using the inner normal to determine it.
In order to have $S^1$ -- $M$ in general -- being bordant to itself (what you always want, it's supposed to be an equivalence relation) you have to revert the orientation of one of the copies.
$endgroup$
Yes, it's essential. The boundary of an oriented manifolds inherits an orientation, the determination of which involves (depending on the definition of orientation), e.g. the inner normal at the boundary.
Consider two copies of $S^1$ making up the boundary of an annulus. (In general consider $Mtimes [0,1]$) The induced orientation (in this case representable by a nonzero one form, hence by an vector field on which this form is $1$) on the first $S^1$ will be pointing into a 'different direction' than the other one, because in both cases you are using the inner normal to determine it.
In order to have $S^1$ -- $M$ in general -- being bordant to itself (what you always want, it's supposed to be an equivalence relation) you have to revert the orientation of one of the copies.
answered Dec 31 '11 at 16:15
user20266
$begingroup$
Ah yes that makes sense. Basically if your arrows are going to be pointing outwards, the boundary circles have to be oriented oppositely. Haha, that's a really infantile way of describing it, but I'm pretty sure I get what you're saying.
$endgroup$
– Jonathan Beardsley
Dec 31 '11 at 16:25
3
$begingroup$
@JBeardz Why infantile? Of course a rigouros proof would need additional reasioning, but the geometric picture is what is really behind it, and in case of $S^1$ it's possible to easily visualize it. I'd claim it's absolutely to the point.
$endgroup$
– user20266
Dec 31 '11 at 17:15
add a comment |
$begingroup$
Ah yes that makes sense. Basically if your arrows are going to be pointing outwards, the boundary circles have to be oriented oppositely. Haha, that's a really infantile way of describing it, but I'm pretty sure I get what you're saying.
$endgroup$
– Jonathan Beardsley
Dec 31 '11 at 16:25
3
$begingroup$
@JBeardz Why infantile? Of course a rigouros proof would need additional reasioning, but the geometric picture is what is really behind it, and in case of $S^1$ it's possible to easily visualize it. I'd claim it's absolutely to the point.
$endgroup$
– user20266
Dec 31 '11 at 17:15
$begingroup$
Ah yes that makes sense. Basically if your arrows are going to be pointing outwards, the boundary circles have to be oriented oppositely. Haha, that's a really infantile way of describing it, but I'm pretty sure I get what you're saying.
$endgroup$
– Jonathan Beardsley
Dec 31 '11 at 16:25
$begingroup$
Ah yes that makes sense. Basically if your arrows are going to be pointing outwards, the boundary circles have to be oriented oppositely. Haha, that's a really infantile way of describing it, but I'm pretty sure I get what you're saying.
$endgroup$
– Jonathan Beardsley
Dec 31 '11 at 16:25
3
3
$begingroup$
@JBeardz Why infantile? Of course a rigouros proof would need additional reasioning, but the geometric picture is what is really behind it, and in case of $S^1$ it's possible to easily visualize it. I'd claim it's absolutely to the point.
$endgroup$
– user20266
Dec 31 '11 at 17:15
$begingroup$
@JBeardz Why infantile? Of course a rigouros proof would need additional reasioning, but the geometric picture is what is really behind it, and in case of $S^1$ it's possible to easily visualize it. I'd claim it's absolutely to the point.
$endgroup$
– user20266
Dec 31 '11 at 17:15
add a comment |
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