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




share|cite|improve this question

















  • 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














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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




share|cite|improve this question

















  • 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












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




share|cite|improve this question














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





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share|cite|improve this question




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edited 2 days ago
























asked Aug 4 at 0:12









Multivariablecalculus

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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












  • 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















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