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How do we compute the $k$-th rational homology group of $#^infty S^infty$?
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How do we compute the $k$-th rational homology group of the connected sum $#^infty S^infty$, and hence the sequence of rational Betti numbers $b_kleft(#^infty S^infty;mathbbQright)=textrk H_kleft(#^infty S^infty;mathbbQright)$? Since finding the homology of a sphere is trivial, I am essentially asking how to find the homology of a union of topological spaces?
Context: The space under consideration is embedded in $mathbbR^infty$. In particular, if we consider an infinite-dimensional Hilbert manifold $M$, then we patch it via an atlas $bigcup_alphain AmathcalU_alpha$ for $A$ countable, but infinite and where $mathcalU_alpha$ is $S^infty$,$forallalphain A$.
Thanks in advance!
homology-cohomology
asked Aug 4 at 0:12
Multivariablecalculus
488313
 |Â
show 5 more comments
up vote
0
down vote
favorite
How do we compute the $k$-th rational homology group of the connected sum $#^infty S^infty$, and hence the sequence of rational Betti numbers $b_kleft(#^infty S^infty;mathbbQright)=textrk H_kleft(#^infty S^infty;mathbbQright)$? Since finding the homology of a sphere is trivial, I am essentially asking how to find the homology of a union of topological spaces?
Context: The space under consideration is embedded in $mathbbR^infty$. In particular, if we consider an infinite-dimensional Hilbert manifold $M$, then we patch it via an atlas $bigcup_alphain AmathcalU_alpha$ for $A$ countable, but infinite and where $mathcalU_alpha$ is $S^infty$,$forallalphain A$.
Thanks in advance!
homology-cohomology
asked Aug 4 at 0:12
Multivariablecalculus
488313
3
What do you mean by this union? In what ambient space?
– Eric Wofsey
Aug 4 at 0:15
Embedded in $mathbbR^infty$. In particular, if we consider an infinite-dimensional manifold $M$, then we patch it via an atlas $bigcup_alphain AmathcalU_alpha$ for $A$ countable, but infinite and where $mathcalU_alpha$ is $S^infty,forallalphain A$. @EricWofsey
– Multivariablecalculus
Aug 4 at 0:28
1
Every manifold is a union of contractible charts but most have interesting homology. You can say nothing if you don't know pairwise (triple, etc) intersections of your subsets.
– Mike Miller
Aug 4 at 0:53
@MikeMiller Couldn't we choose an "atlas" such that $mathcalU_alphacapmathcalU_beta=emptyset$ for all $alpha,beta$ and then the homology would simply be $H_kleft(sqcup_alphain AmathcalU_alpha;mathbbQright)=oplus_alphain AH_k(mathcalU_alpha;mathbbQ)$?
– Multivariablecalculus
Aug 4 at 1:38
1
For every $k$ the homology of $#^k S^infty$ is trivial. Note that $S^infty$ is contractible as well as $S^inftysetminusmathrmpt$. Use Mayer-Vietoris.
– Thomas Rot
16 hours ago
 |Â
show 5 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
How do we compute the $k$-th rational homology group of the connected sum $#^infty S^infty$, and hence the sequence of rational Betti numbers $b_kleft(#^infty S^infty;mathbbQright)=textrk H_kleft(#^infty S^infty;mathbbQright)$? Since finding the homology of a sphere is trivial, I am essentially asking how to find the homology of a union of topological spaces?
Context: The space under consideration is embedded in $mathbbR^infty$. In particular, if we consider an infinite-dimensional Hilbert manifold $M$, then we patch it via an atlas $bigcup_alphain AmathcalU_alpha$ for $A$ countable, but infinite and where $mathcalU_alpha$ is $S^infty$,$forallalphain A$.
Thanks in advance!
homology-cohomology
asked Aug 4 at 0:12
Multivariablecalculus
488313
How do we compute the $k$-th rational homology group of the connected sum $#^infty S^infty$, and hence the sequence of rational Betti numbers $b_kleft(#^infty S^infty;mathbbQright)=textrk H_kleft(#^infty S^infty;mathbbQright)$? Since finding the homology of a sphere is trivial, I am essentially asking how to find the homology of a union of topological spaces?
Context: The space under consideration is embedded in $mathbbR^infty$. In particular, if we consider an infinite-dimensional Hilbert manifold $M$, then we patch it via an atlas $bigcup_alphain AmathcalU_alpha$ for $A$ countable, but infinite and where $mathcalU_alpha$ is $S^infty$,$forallalphain A$.
Thanks in advance!
homology-cohomology
asked Aug 4 at 0:12
Multivariablecalculus
488313
edited 2 days ago
asked Aug 4 at 0:12
Multivariablecalculus
488313
asked Aug 4 at 0:12
Multivariablecalculus
488313
asked Aug 4 at 0:12
Multivariablecalculus
488313
488313
3
What do you mean by this union? In what ambient space?
– Eric Wofsey
Aug 4 at 0:15
Embedded in $mathbbR^infty$. In particular, if we consider an infinite-dimensional manifold $M$, then we patch it via an atlas $bigcup_alphain AmathcalU_alpha$ for $A$ countable, but infinite and where $mathcalU_alpha$ is $S^infty,forallalphain A$. @EricWofsey
– Multivariablecalculus
Aug 4 at 0:28
1
Every manifold is a union of contractible charts but most have interesting homology. You can say nothing if you don't know pairwise (triple, etc) intersections of your subsets.
– Mike Miller
Aug 4 at 0:53
@MikeMiller Couldn't we choose an "atlas" such that $mathcalU_alphacapmathcalU_beta=emptyset$ for all $alpha,beta$ and then the homology would simply be $H_kleft(sqcup_alphain AmathcalU_alpha;mathbbQright)=oplus_alphain AH_k(mathcalU_alpha;mathbbQ)$?
– Multivariablecalculus
Aug 4 at 1:38
1
For every $k$ the homology of $#^k S^infty$ is trivial. Note that $S^infty$ is contractible as well as $S^inftysetminusmathrmpt$. Use Mayer-Vietoris.
– Thomas Rot
16 hours ago
 |Â
show 5 more comments
3
What do you mean by this union? In what ambient space?
– Eric Wofsey
Aug 4 at 0:15
Embedded in $mathbbR^infty$. In particular, if we consider an infinite-dimensional manifold $M$, then we patch it via an atlas $bigcup_alphain AmathcalU_alpha$ for $A$ countable, but infinite and where $mathcalU_alpha$ is $S^infty,forallalphain A$. @EricWofsey
– Multivariablecalculus
Aug 4 at 0:28
1
Every manifold is a union of contractible charts but most have interesting homology. You can say nothing if you don't know pairwise (triple, etc) intersections of your subsets.
– Mike Miller
Aug 4 at 0:53
@MikeMiller Couldn't we choose an "atlas" such that $mathcalU_alphacapmathcalU_beta=emptyset$ for all $alpha,beta$ and then the homology would simply be $H_kleft(sqcup_alphain AmathcalU_alpha;mathbbQright)=oplus_alphain AH_k(mathcalU_alpha;mathbbQ)$?
– Multivariablecalculus
Aug 4 at 1:38
1
For every $k$ the homology of $#^k S^infty$ is trivial. Note that $S^infty$ is contractible as well as $S^inftysetminusmathrmpt$. Use Mayer-Vietoris.
– Thomas Rot
16 hours ago
3
3
What do you mean by this union? In what ambient space?
– Eric Wofsey
Aug 4 at 0:15
What do you mean by this union? In what ambient space?
– Eric Wofsey
Aug 4 at 0:15
Embedded in $mathbbR^infty$. In particular, if we consider an infinite-dimensional manifold $M$, then we patch it via an atlas $bigcup_alphain AmathcalU_alpha$ for $A$ countable, but infinite and where $mathcalU_alpha$ is $S^infty,forallalphain A$. @EricWofsey
– Multivariablecalculus
Aug 4 at 0:28
Embedded in $mathbbR^infty$. In particular, if we consider an infinite-dimensional manifold $M$, then we patch it via an atlas $bigcup_alphain AmathcalU_alpha$ for $A$ countable, but infinite and where $mathcalU_alpha$ is $S^infty,forallalphain A$. @EricWofsey
– Multivariablecalculus
Aug 4 at 0:28
1
1
Every manifold is a union of contractible charts but most have interesting homology. You can say nothing if you don't know pairwise (triple, etc) intersections of your subsets.
– Mike Miller
Aug 4 at 0:53
Every manifold is a union of contractible charts but most have interesting homology. You can say nothing if you don't know pairwise (triple, etc) intersections of your subsets.
– Mike Miller
Aug 4 at 0:53
@MikeMiller Couldn't we choose an "atlas" such that $mathcalU_alphacapmathcalU_beta=emptyset$ for all $alpha,beta$ and then the homology would simply be $H_kleft(sqcup_alphain AmathcalU_alpha;mathbbQright)=oplus_alphain AH_k(mathcalU_alpha;mathbbQ)$?
– Multivariablecalculus
Aug 4 at 1:38
@MikeMiller Couldn't we choose an "atlas" such that $mathcalU_alphacapmathcalU_beta=emptyset$ for all $alpha,beta$ and then the homology would simply be $H_kleft(sqcup_alphain AmathcalU_alpha;mathbbQright)=oplus_alphain AH_k(mathcalU_alpha;mathbbQ)$?
– Multivariablecalculus
Aug 4 at 1:38
1
1
For every $k$ the homology of $#^k S^infty$ is trivial. Note that $S^infty$ is contractible as well as $S^inftysetminusmathrmpt$. Use Mayer-Vietoris.
– Thomas Rot
16 hours ago
For every $k$ the homology of $#^k S^infty$ is trivial. Note that $S^infty$ is contractible as well as $S^inftysetminusmathrmpt$. Use Mayer-Vietoris.
– Thomas Rot
16 hours ago
 |Â
show 5 more comments
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3
What do you mean by this union? In what ambient space?
– Eric Wofsey
Aug 4 at 0:15
Embedded in $mathbbR^infty$. In particular, if we consider an infinite-dimensional manifold $M$, then we patch it via an atlas $bigcup_alphain AmathcalU_alpha$ for $A$ countable, but infinite and where $mathcalU_alpha$ is $S^infty,forallalphain A$. @EricWofsey
– Multivariablecalculus
Aug 4 at 0:28
1
Every manifold is a union of contractible charts but most have interesting homology. You can say nothing if you don't know pairwise (triple, etc) intersections of your subsets.
– Mike Miller
Aug 4 at 0:53
@MikeMiller Couldn't we choose an "atlas" such that $mathcalU_alphacapmathcalU_beta=emptyset$ for all $alpha,beta$ and then the homology would simply be $H_kleft(sqcup_alphain AmathcalU_alpha;mathbbQright)=oplus_alphain AH_k(mathcalU_alpha;mathbbQ)$?
– Multivariablecalculus
Aug 4 at 1:38
1
For every $k$ the homology of $#^k S^infty$ is trivial. Note that $S^infty$ is contractible as well as $S^inftysetminusmathrmpt$. Use Mayer-Vietoris.
– Thomas Rot
16 hours ago