Relation between the homological groups of a set and the ones of its boundary (for pseudo-manifolds) Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)If $Z$ is an oriented manifold with boundary such that $partial Z=M$ where $M$ is a compact and oriented manifold then $chi(M)=0mod(2)$..How to use Mayer-Vietoris to show $chi(X)=2chi(M)-chi(partial M)$ where $X$ is the double of $M$?Removing the boundary from a manifold with boundaryA question about the boundary points of manifolds.What geometric information can be recovered from $L^2(X)$ for a manifold $X$?Homotopy type of manifolds homeomorphic to the interior of a compact manifold with boundaryCompact manifold without boundary is boundary of another manifoldHonology, Cohomology, Euler Number for Non Orientable ManifoldsWhy is the boundary of an oriented manifold with its (opposite oriented) copy the empty set?Non-vanishing vector field on a manifold..
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Relation between the homological groups of a set and the ones of its boundary (for pseudo-manifolds)
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)If $Z$ is an oriented manifold with boundary such that $partial Z=M$ where $M$ is a compact and oriented manifold then $chi(M)=0mod(2)$..How to use Mayer-Vietoris to show $chi(X)=2chi(M)-chi(partial M)$ where $X$ is the double of $M$?Removing the boundary from a manifold with boundaryA question about the boundary points of manifolds.What geometric information can be recovered from $L^2(X)$ for a manifold $X$?Homotopy type of manifolds homeomorphic to the interior of a compact manifold with boundaryCompact manifold without boundary is boundary of another manifoldHonology, Cohomology, Euler Number for Non Orientable ManifoldsWhy is the boundary of an oriented manifold with its (opposite oriented) copy the empty set?Non-vanishing vector field on a manifold..
$begingroup$
I heard that when $X$ is a connected compact topological $n$-manifold in $mathbbR^n$ whose boundary $partial X$ is a connected compact (n-1)-manifold, then we can deduce the Betti numbers of $X$ from the Betti numbers of $partial X$. Also, we can deduce the Euler characteristic of $X$ from the one of $partial X$.
However, I do not know if some similar results exist when we work with connected compact topological sets whose boundary is a combinatorial pseudo-manifold (that means that the boundary can own "pinches").
I would be very grateful if someone can help me on this.
Thanks for your help.
manifolds homology-cohomology
$endgroup$
add a comment |
$begingroup$
I heard that when $X$ is a connected compact topological $n$-manifold in $mathbbR^n$ whose boundary $partial X$ is a connected compact (n-1)-manifold, then we can deduce the Betti numbers of $X$ from the Betti numbers of $partial X$. Also, we can deduce the Euler characteristic of $X$ from the one of $partial X$.
However, I do not know if some similar results exist when we work with connected compact topological sets whose boundary is a combinatorial pseudo-manifold (that means that the boundary can own "pinches").
I would be very grateful if someone can help me on this.
Thanks for your help.
manifolds homology-cohomology
$endgroup$
2
$begingroup$
Is the fact that $X$ has the same dimension as the ambient space $mathbbR^n$ on purpose or is it an oversight?
$endgroup$
– Najib Idrissi
Apr 1 at 14:52
$begingroup$
It is on purpose: I want my boundary to own separation properties.
$endgroup$
– Nicolas Boutry
Apr 1 at 17:29
add a comment |
$begingroup$
I heard that when $X$ is a connected compact topological $n$-manifold in $mathbbR^n$ whose boundary $partial X$ is a connected compact (n-1)-manifold, then we can deduce the Betti numbers of $X$ from the Betti numbers of $partial X$. Also, we can deduce the Euler characteristic of $X$ from the one of $partial X$.
However, I do not know if some similar results exist when we work with connected compact topological sets whose boundary is a combinatorial pseudo-manifold (that means that the boundary can own "pinches").
I would be very grateful if someone can help me on this.
Thanks for your help.
manifolds homology-cohomology
$endgroup$
I heard that when $X$ is a connected compact topological $n$-manifold in $mathbbR^n$ whose boundary $partial X$ is a connected compact (n-1)-manifold, then we can deduce the Betti numbers of $X$ from the Betti numbers of $partial X$. Also, we can deduce the Euler characteristic of $X$ from the one of $partial X$.
However, I do not know if some similar results exist when we work with connected compact topological sets whose boundary is a combinatorial pseudo-manifold (that means that the boundary can own "pinches").
I would be very grateful if someone can help me on this.
Thanks for your help.
manifolds homology-cohomology
manifolds homology-cohomology
asked Apr 1 at 13:06
Nicolas BoutryNicolas Boutry
287
287
2
$begingroup$
Is the fact that $X$ has the same dimension as the ambient space $mathbbR^n$ on purpose or is it an oversight?
$endgroup$
– Najib Idrissi
Apr 1 at 14:52
$begingroup$
It is on purpose: I want my boundary to own separation properties.
$endgroup$
– Nicolas Boutry
Apr 1 at 17:29
add a comment |
2
$begingroup$
Is the fact that $X$ has the same dimension as the ambient space $mathbbR^n$ on purpose or is it an oversight?
$endgroup$
– Najib Idrissi
Apr 1 at 14:52
$begingroup$
It is on purpose: I want my boundary to own separation properties.
$endgroup$
– Nicolas Boutry
Apr 1 at 17:29
2
2
$begingroup$
Is the fact that $X$ has the same dimension as the ambient space $mathbbR^n$ on purpose or is it an oversight?
$endgroup$
– Najib Idrissi
Apr 1 at 14:52
$begingroup$
Is the fact that $X$ has the same dimension as the ambient space $mathbbR^n$ on purpose or is it an oversight?
$endgroup$
– Najib Idrissi
Apr 1 at 14:52
$begingroup$
It is on purpose: I want my boundary to own separation properties.
$endgroup$
– Nicolas Boutry
Apr 1 at 17:29
$begingroup$
It is on purpose: I want my boundary to own separation properties.
$endgroup$
– Nicolas Boutry
Apr 1 at 17:29
add a comment |
0
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2
$begingroup$
Is the fact that $X$ has the same dimension as the ambient space $mathbbR^n$ on purpose or is it an oversight?
$endgroup$
– Najib Idrissi
Apr 1 at 14:52
$begingroup$
It is on purpose: I want my boundary to own separation properties.
$endgroup$
– Nicolas Boutry
Apr 1 at 17:29