homology of a subspaces of product of circles.fundamental groups of complement of lines in $mathbbC^2$Wedge Sum of Two Spheres Homotopy Equivalent to a Compact Manifold?Homology of 5-manifoldShow $H_2(M, mathbbZ) = mathbbZ^r$ if $M$ is orientable, $mathbbZ^r-1 oplus mathbbZ_2$ if nonorientableIntuition of non-free homology groups?Compute homology groups of space $Bbb RP^2$ attached with Mobius band using Mayer VietoriesHomology and Neighborhood$3$-manifold has same homology groups as a $3$-sphere.Homology groups with coefficientsComputing the homology of genus $g$ surface, using Mayer-Vietoris
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homology of a subspaces of product of circles.
fundamental groups of complement of lines in $mathbbC^2$Wedge Sum of Two Spheres Homotopy Equivalent to a Compact Manifold?Homology of 5-manifoldShow $H_2(M, mathbbZ) = mathbbZ^r$ if $M$ is orientable, $mathbbZ^r-1 oplus mathbbZ_2$ if nonorientableIntuition of non-free homology groups?Compute homology groups of space $Bbb RP^2$ attached with Mobius band using Mayer VietoriesHomology and Neighborhood$3$-manifold has same homology groups as a $3$-sphere.Homology groups with coefficientsComputing the homology of genus $g$ surface, using Mayer-Vietoris
$begingroup$
I do not how to approach the computation of the homology of the following space
$X = (z_1, z_2, z_3) in (S^1)^2 times D_2 : z_1 + z_2 + z_3 = 0 $
$S^1$ denotes the unit circle and $D_2$ the unit ball in $mathbbR^2$
From the regular value theorem, $X$ is a compact orientable connected manifold of dimension $2$, so $H_0(X) = H_2(X) = mathbbZ$.
But what can I say about $H_1$ (or the number of $1$-dimensional holes)?
algebraic-topology manifolds homology-cohomology
asked Mar 11 at 20:52
SamSam
161
New contributor
Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I do not how to approach the computation of the homology of the following space
$X = (z_1, z_2, z_3) in (S^1)^2 times D_2 : z_1 + z_2 + z_3 = 0 $
$S^1$ denotes the unit circle and $D_2$ the unit ball in $mathbbR^2$
From the regular value theorem, $X$ is a compact orientable connected manifold of dimension $2$, so $H_0(X) = H_2(X) = mathbbZ$.
But what can I say about $H_1$ (or the number of $1$-dimensional holes)?
algebraic-topology manifolds homology-cohomology
asked Mar 11 at 20:52
SamSam
161
New contributor
Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
I just wanted to double check, you're sure the disc has unit radius? If it had radius 2, then $X$ would be a torus.
$endgroup$
– Michael Albanese
Mar 11 at 21:15
add a comment |
$begingroup$
I do not how to approach the computation of the homology of the following space
$X = (z_1, z_2, z_3) in (S^1)^2 times D_2 : z_1 + z_2 + z_3 = 0 $
$S^1$ denotes the unit circle and $D_2$ the unit ball in $mathbbR^2$
From the regular value theorem, $X$ is a compact orientable connected manifold of dimension $2$, so $H_0(X) = H_2(X) = mathbbZ$.
But what can I say about $H_1$ (or the number of $1$-dimensional holes)?
algebraic-topology manifolds homology-cohomology
asked Mar 11 at 20:52
SamSam
161
New contributor
Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I do not how to approach the computation of the homology of the following space
$X = (z_1, z_2, z_3) in (S^1)^2 times D_2 : z_1 + z_2 + z_3 = 0 $
$S^1$ denotes the unit circle and $D_2$ the unit ball in $mathbbR^2$
From the regular value theorem, $X$ is a compact orientable connected manifold of dimension $2$, so $H_0(X) = H_2(X) = mathbbZ$.
But what can I say about $H_1$ (or the number of $1$-dimensional holes)?
algebraic-topology manifolds homology-cohomology
algebraic-topology manifolds homology-cohomology
asked Mar 11 at 20:52
SamSam
161
New contributor
Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 11 at 20:52
SamSam
161
New contributor
Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 11 at 20:52
SamSam
161
New contributor
Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 11 at 20:52
SamSam
161
asked Mar 11 at 20:52
SamSam
161
161
New contributor
Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
I just wanted to double check, you're sure the disc has unit radius? If it had radius 2, then $X$ would be a torus.
$endgroup$
– Michael Albanese
Mar 11 at 21:15
add a comment |
$begingroup$
I just wanted to double check, you're sure the disc has unit radius? If it had radius 2, then $X$ would be a torus.
$endgroup$
– Michael Albanese
Mar 11 at 21:15
$begingroup$
I just wanted to double check, you're sure the disc has unit radius? If it had radius 2, then $X$ would be a torus.
$endgroup$
– Michael Albanese
Mar 11 at 21:15
$begingroup$
I just wanted to double check, you're sure the disc has unit radius? If it had radius 2, then $X$ would be a torus.
$endgroup$
– Michael Albanese
Mar 11 at 21:15
add a comment |
1 Answer
1
active
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$begingroup$
Correct me if I am wrong (my smooth manifold knowledge is not great), but won't the preimage have boundary, so you cannot use Poincare duality?
I mention this because I believe the manifold is a cylinder, so my calculation of the homology will disagree with you. I believe the following is a deformation retraction to a circle (so its 2 dimensional homology would be 0).
You can view X as a subspace of $T^2$ by forgetting the third coordinate since it is fully determined by the first two. Let's call it $X' subset T^2$ and it is given by pairs $(x,y) in T^2$ such that the magnitude of their sum is at most 1. Let $F: X' times I rightarrow X'$ be given by $(x,y,t) rightarrow (frac(1-t)x - ty(1-t)x-ty,y)$. The denominator is never zero because $y neq x$, and it isn't hard to convince yourself that it lies inside $X'$ (though I'm sure that there is an argument with inequalities).
Thus, there is a deformation retraction to the anti-diagonal of $T^2$ (pairs of coordinates such that the first is the negative of the second) which is homeomorphic to $S^1$, so $H_1(X)=mathbbZ$.
answered Mar 11 at 22:27
Connor MalinConnor Malin
489110
$endgroup$
add a comment |
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$begingroup$
Correct me if I am wrong (my smooth manifold knowledge is not great), but won't the preimage have boundary, so you cannot use Poincare duality?
I mention this because I believe the manifold is a cylinder, so my calculation of the homology will disagree with you. I believe the following is a deformation retraction to a circle (so its 2 dimensional homology would be 0).
You can view X as a subspace of $T^2$ by forgetting the third coordinate since it is fully determined by the first two. Let's call it $X' subset T^2$ and it is given by pairs $(x,y) in T^2$ such that the magnitude of their sum is at most 1. Let $F: X' times I rightarrow X'$ be given by $(x,y,t) rightarrow (frac(1-t)x - ty(1-t)x-ty,y)$. The denominator is never zero because $y neq x$, and it isn't hard to convince yourself that it lies inside $X'$ (though I'm sure that there is an argument with inequalities).
Thus, there is a deformation retraction to the anti-diagonal of $T^2$ (pairs of coordinates such that the first is the negative of the second) which is homeomorphic to $S^1$, so $H_1(X)=mathbbZ$.
answered Mar 11 at 22:27
Connor MalinConnor Malin
489110
$endgroup$
add a comment |
$begingroup$
Correct me if I am wrong (my smooth manifold knowledge is not great), but won't the preimage have boundary, so you cannot use Poincare duality?
I mention this because I believe the manifold is a cylinder, so my calculation of the homology will disagree with you. I believe the following is a deformation retraction to a circle (so its 2 dimensional homology would be 0).
You can view X as a subspace of $T^2$ by forgetting the third coordinate since it is fully determined by the first two. Let's call it $X' subset T^2$ and it is given by pairs $(x,y) in T^2$ such that the magnitude of their sum is at most 1. Let $F: X' times I rightarrow X'$ be given by $(x,y,t) rightarrow (frac(1-t)x - ty(1-t)x-ty,y)$. The denominator is never zero because $y neq x$, and it isn't hard to convince yourself that it lies inside $X'$ (though I'm sure that there is an argument with inequalities).
Thus, there is a deformation retraction to the anti-diagonal of $T^2$ (pairs of coordinates such that the first is the negative of the second) which is homeomorphic to $S^1$, so $H_1(X)=mathbbZ$.
answered Mar 11 at 22:27
Connor MalinConnor Malin
489110
$endgroup$
add a comment |
$begingroup$
Correct me if I am wrong (my smooth manifold knowledge is not great), but won't the preimage have boundary, so you cannot use Poincare duality?
I mention this because I believe the manifold is a cylinder, so my calculation of the homology will disagree with you. I believe the following is a deformation retraction to a circle (so its 2 dimensional homology would be 0).
You can view X as a subspace of $T^2$ by forgetting the third coordinate since it is fully determined by the first two. Let's call it $X' subset T^2$ and it is given by pairs $(x,y) in T^2$ such that the magnitude of their sum is at most 1. Let $F: X' times I rightarrow X'$ be given by $(x,y,t) rightarrow (frac(1-t)x - ty(1-t)x-ty,y)$. The denominator is never zero because $y neq x$, and it isn't hard to convince yourself that it lies inside $X'$ (though I'm sure that there is an argument with inequalities).
Thus, there is a deformation retraction to the anti-diagonal of $T^2$ (pairs of coordinates such that the first is the negative of the second) which is homeomorphic to $S^1$, so $H_1(X)=mathbbZ$.
answered Mar 11 at 22:27
Connor MalinConnor Malin
489110
$endgroup$
Correct me if I am wrong (my smooth manifold knowledge is not great), but won't the preimage have boundary, so you cannot use Poincare duality?
I mention this because I believe the manifold is a cylinder, so my calculation of the homology will disagree with you. I believe the following is a deformation retraction to a circle (so its 2 dimensional homology would be 0).
You can view X as a subspace of $T^2$ by forgetting the third coordinate since it is fully determined by the first two. Let's call it $X' subset T^2$ and it is given by pairs $(x,y) in T^2$ such that the magnitude of their sum is at most 1. Let $F: X' times I rightarrow X'$ be given by $(x,y,t) rightarrow (frac(1-t)x - ty(1-t)x-ty,y)$. The denominator is never zero because $y neq x$, and it isn't hard to convince yourself that it lies inside $X'$ (though I'm sure that there is an argument with inequalities).
Thus, there is a deformation retraction to the anti-diagonal of $T^2$ (pairs of coordinates such that the first is the negative of the second) which is homeomorphic to $S^1$, so $H_1(X)=mathbbZ$.
answered Mar 11 at 22:27
Connor MalinConnor Malin
489110
edited Mar 11 at 22:39
answered Mar 11 at 22:27
Connor MalinConnor Malin
489110
answered Mar 11 at 22:27
Connor MalinConnor Malin
489110
answered Mar 11 at 22:27
Connor MalinConnor Malin
489110
489110
add a comment |
add a comment |
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$begingroup$
I just wanted to double check, you're sure the disc has unit radius? If it had radius 2, then $X$ would be a torus.
$endgroup$
– Michael Albanese
Mar 11 at 21:15