Intersection form and geometric cycles
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Suppose you have a closed oriented smooth manifold $X$. And two orientable submanifolds $S_{1},S_{2}$ with $dim(S_{1}) + dim(S_{2}) = dim(X)$ intersecting transversally and with $[S_{1}] cdot [S_{2}] = 0$. Suppose that we are interested in finding a cycle $C in C_{dim(S_{1})}(X)$, such that $[C]=[S_{1}] in H_{dim(S_{1})}(X,mathbb{Z})$ and $C$ is disjoint from $S_{2}$.
I have heard cautionary tale that it is not always possible to find such a cycle. On the other hand, I have a vague recollection of an argument in geometric topology when we "bring together" (probably by some kind of homotopy of $S_{1}$) the +1 and -1 intersection points and they annihilate each other or something like that, although that was probably in some specific situation or is perhaps not applicable here.
I would be interested to know a counter-example, or some reference that addresses this in a reasonably general way. I wonder if this is a phenomina "linearises" in the sense algebraic topology for high dimensions i.e. is it possible to find such a $C$ if we assume that $dim(X)>>0$ or $dim(X)>> dim(S_{1})$?
reference-request algebraic-topology differential-topology smooth-manifolds
asked Nov 15 at 18:39
Nick L
1,151110
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Suppose you have a closed oriented smooth manifold $X$. And two orientable submanifolds $S_{1},S_{2}$ with $dim(S_{1}) + dim(S_{2}) = dim(X)$ intersecting transversally and with $[S_{1}] cdot [S_{2}] = 0$. Suppose that we are interested in finding a cycle $C in C_{dim(S_{1})}(X)$, such that $[C]=[S_{1}] in H_{dim(S_{1})}(X,mathbb{Z})$ and $C$ is disjoint from $S_{2}$.
I have heard cautionary tale that it is not always possible to find such a cycle. On the other hand, I have a vague recollection of an argument in geometric topology when we "bring together" (probably by some kind of homotopy of $S_{1}$) the +1 and -1 intersection points and they annihilate each other or something like that, although that was probably in some specific situation or is perhaps not applicable here.
I would be interested to know a counter-example, or some reference that addresses this in a reasonably general way. I wonder if this is a phenomina "linearises" in the sense algebraic topology for high dimensions i.e. is it possible to find such a $C$ if we assume that $dim(X)>>0$ or $dim(X)>> dim(S_{1})$?
reference-request algebraic-topology differential-topology smooth-manifolds
asked Nov 15 at 18:39
Nick L
1,151110
If the $S_i$ are connected and the manifold $X$ is simply connected of dim $geq 5$, the argument you have in mind is the Whitney trick: connect two intersection points of opposite sign by a path in $S_1$ and $S_2$, respectively, and then fill the union of those paths in with a disc; isotope the arc in $S_1$ across this disc until you can finally push it off of $S_2$. This is the so-called Whitney trick.
– Mike Miller
Nov 15 at 20:28
for curves on a surface this is very much impossible, though.
– Mike Miller
Nov 15 at 20:28
@ Mike Miller, thanks! could you sketch counter example?
– Nick L
Nov 15 at 21:00
I parsed your original question as asking for isotopies of $S_i$ as opposed to simply homologous cycles (through no fault of your own). The latter seems more plausible. I think it is true for oriented surfaces "by hand".
– Mike Miller
Nov 15 at 21:33
This question on MO seems to answer your question in the affirmative.
– Mike Miller
Nov 16 at 17:24
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Suppose you have a closed oriented smooth manifold $X$. And two orientable submanifolds $S_{1},S_{2}$ with $dim(S_{1}) + dim(S_{2}) = dim(X)$ intersecting transversally and with $[S_{1}] cdot [S_{2}] = 0$. Suppose that we are interested in finding a cycle $C in C_{dim(S_{1})}(X)$, such that $[C]=[S_{1}] in H_{dim(S_{1})}(X,mathbb{Z})$ and $C$ is disjoint from $S_{2}$.
I have heard cautionary tale that it is not always possible to find such a cycle. On the other hand, I have a vague recollection of an argument in geometric topology when we "bring together" (probably by some kind of homotopy of $S_{1}$) the +1 and -1 intersection points and they annihilate each other or something like that, although that was probably in some specific situation or is perhaps not applicable here.
I would be interested to know a counter-example, or some reference that addresses this in a reasonably general way. I wonder if this is a phenomina "linearises" in the sense algebraic topology for high dimensions i.e. is it possible to find such a $C$ if we assume that $dim(X)>>0$ or $dim(X)>> dim(S_{1})$?
reference-request algebraic-topology differential-topology smooth-manifolds
asked Nov 15 at 18:39
Nick L
1,151110
Suppose you have a closed oriented smooth manifold $X$. And two orientable submanifolds $S_{1},S_{2}$ with $dim(S_{1}) + dim(S_{2}) = dim(X)$ intersecting transversally and with $[S_{1}] cdot [S_{2}] = 0$. Suppose that we are interested in finding a cycle $C in C_{dim(S_{1})}(X)$, such that $[C]=[S_{1}] in H_{dim(S_{1})}(X,mathbb{Z})$ and $C$ is disjoint from $S_{2}$.
I have heard cautionary tale that it is not always possible to find such a cycle. On the other hand, I have a vague recollection of an argument in geometric topology when we "bring together" (probably by some kind of homotopy of $S_{1}$) the +1 and -1 intersection points and they annihilate each other or something like that, although that was probably in some specific situation or is perhaps not applicable here.
I would be interested to know a counter-example, or some reference that addresses this in a reasonably general way. I wonder if this is a phenomina "linearises" in the sense algebraic topology for high dimensions i.e. is it possible to find such a $C$ if we assume that $dim(X)>>0$ or $dim(X)>> dim(S_{1})$?
reference-request algebraic-topology differential-topology smooth-manifolds
reference-request algebraic-topology differential-topology smooth-manifolds
asked Nov 15 at 18:39
Nick L
1,151110
asked Nov 15 at 18:39
Nick L
1,151110
edited Nov 15 at 22:01
asked Nov 15 at 18:39
Nick L
1,151110
asked Nov 15 at 18:39
Nick L
1,151110
asked Nov 15 at 18:39
Nick L
1,151110
1,151110
If the $S_i$ are connected and the manifold $X$ is simply connected of dim $geq 5$, the argument you have in mind is the Whitney trick: connect two intersection points of opposite sign by a path in $S_1$ and $S_2$, respectively, and then fill the union of those paths in with a disc; isotope the arc in $S_1$ across this disc until you can finally push it off of $S_2$. This is the so-called Whitney trick.
– Mike Miller
Nov 15 at 20:28
for curves on a surface this is very much impossible, though.
– Mike Miller
Nov 15 at 20:28
@ Mike Miller, thanks! could you sketch counter example?
– Nick L
Nov 15 at 21:00
I parsed your original question as asking for isotopies of $S_i$ as opposed to simply homologous cycles (through no fault of your own). The latter seems more plausible. I think it is true for oriented surfaces "by hand".
– Mike Miller
Nov 15 at 21:33
This question on MO seems to answer your question in the affirmative.
– Mike Miller
Nov 16 at 17:24
add a comment |
If the $S_i$ are connected and the manifold $X$ is simply connected of dim $geq 5$, the argument you have in mind is the Whitney trick: connect two intersection points of opposite sign by a path in $S_1$ and $S_2$, respectively, and then fill the union of those paths in with a disc; isotope the arc in $S_1$ across this disc until you can finally push it off of $S_2$. This is the so-called Whitney trick.
– Mike Miller
Nov 15 at 20:28
for curves on a surface this is very much impossible, though.
– Mike Miller
Nov 15 at 20:28
@ Mike Miller, thanks! could you sketch counter example?
– Nick L
Nov 15 at 21:00
I parsed your original question as asking for isotopies of $S_i$ as opposed to simply homologous cycles (through no fault of your own). The latter seems more plausible. I think it is true for oriented surfaces "by hand".
– Mike Miller
Nov 15 at 21:33
This question on MO seems to answer your question in the affirmative.
– Mike Miller
Nov 16 at 17:24
If the $S_i$ are connected and the manifold $X$ is simply connected of dim $geq 5$, the argument you have in mind is the Whitney trick: connect two intersection points of opposite sign by a path in $S_1$ and $S_2$, respectively, and then fill the union of those paths in with a disc; isotope the arc in $S_1$ across this disc until you can finally push it off of $S_2$. This is the so-called Whitney trick.
– Mike Miller
Nov 15 at 20:28
If the $S_i$ are connected and the manifold $X$ is simply connected of dim $geq 5$, the argument you have in mind is the Whitney trick: connect two intersection points of opposite sign by a path in $S_1$ and $S_2$, respectively, and then fill the union of those paths in with a disc; isotope the arc in $S_1$ across this disc until you can finally push it off of $S_2$. This is the so-called Whitney trick.
– Mike Miller
Nov 15 at 20:28
for curves on a surface this is very much impossible, though.
– Mike Miller
Nov 15 at 20:28
for curves on a surface this is very much impossible, though.
– Mike Miller
Nov 15 at 20:28
@ Mike Miller, thanks! could you sketch counter example?
– Nick L
Nov 15 at 21:00
@ Mike Miller, thanks! could you sketch counter example?
– Nick L
Nov 15 at 21:00
I parsed your original question as asking for isotopies of $S_i$ as opposed to simply homologous cycles (through no fault of your own). The latter seems more plausible. I think it is true for oriented surfaces "by hand".
– Mike Miller
Nov 15 at 21:33
I parsed your original question as asking for isotopies of $S_i$ as opposed to simply homologous cycles (through no fault of your own). The latter seems more plausible. I think it is true for oriented surfaces "by hand".
– Mike Miller
Nov 15 at 21:33
This question on MO seems to answer your question in the affirmative.
– Mike Miller
Nov 16 at 17:24
This question on MO seems to answer your question in the affirmative.
– Mike Miller
Nov 16 at 17:24
add a comment |
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If the $S_i$ are connected and the manifold $X$ is simply connected of dim $geq 5$, the argument you have in mind is the Whitney trick: connect two intersection points of opposite sign by a path in $S_1$ and $S_2$, respectively, and then fill the union of those paths in with a disc; isotope the arc in $S_1$ across this disc until you can finally push it off of $S_2$. This is the so-called Whitney trick.
– Mike Miller
Nov 15 at 20:28
for curves on a surface this is very much impossible, though.
– Mike Miller
Nov 15 at 20:28
@ Mike Miller, thanks! could you sketch counter example?
– Nick L
Nov 15 at 21:00
I parsed your original question as asking for isotopies of $S_i$ as opposed to simply homologous cycles (through no fault of your own). The latter seems more plausible. I think it is true for oriented surfaces "by hand".
– Mike Miller
Nov 15 at 21:33
This question on MO seems to answer your question in the affirmative.
– Mike Miller
Nov 16 at 17:24