Which spaces have trivial K-theory?
What is known about spaces $X$ with the property that $K^*(text{point})to K^*(X)$ is an isomorphism?
The same question for $K$-homology $K_*(X)to K_*(text{point})$; I don't even know whether these conditions are equivalent.
Note that replacing $K$-theory with integral homology one gets very interesting (I think) class of spaces, studied in "Acyclic spaces" by E. Dror.
at.algebraic-topology kt.k-theory-and-homology
asked 4 hours ago
მამუკა ჯიბლაძე
8,066244112
add a comment |
What is known about spaces $X$ with the property that $K^*(text{point})to K^*(X)$ is an isomorphism?
The same question for $K$-homology $K_*(X)to K_*(text{point})$; I don't even know whether these conditions are equivalent.
Note that replacing $K$-theory with integral homology one gets very interesting (I think) class of spaces, studied in "Acyclic spaces" by E. Dror.
at.algebraic-topology kt.k-theory-and-homology
asked 4 hours ago
მამუკა ჯიბლაძე
8,066244112
3
All acyclic spaces have trivial K-theory (it's a simple argument with the Atiyah-Hirzebruch spectral sequence), but I think you get more things
– Denis Nardin
4 hours ago
@DenisNardin I would rather think that all K-trivial spaces would be acyclic, while K-theory could distinguish more spaces than "ordinary" homology. I just don't know enough, but I think the Adams $e$-invariant detects K-theory classes killed by the Chern character, cannot one find something along these lines?
– მამუკა ჯიბლაძე
2 hours ago
I don’t know a characterization, but it’s bigger than you might expect: K(Z/p, 2) is such a space.
– Eric Peterson
6 mins ago
add a comment |
What is known about spaces $X$ with the property that $K^*(text{point})to K^*(X)$ is an isomorphism?
The same question for $K$-homology $K_*(X)to K_*(text{point})$; I don't even know whether these conditions are equivalent.
Note that replacing $K$-theory with integral homology one gets very interesting (I think) class of spaces, studied in "Acyclic spaces" by E. Dror.
at.algebraic-topology kt.k-theory-and-homology
asked 4 hours ago
მამუკა ჯიბლაძე
8,066244112
What is known about spaces $X$ with the property that $K^*(text{point})to K^*(X)$ is an isomorphism?
The same question for $K$-homology $K_*(X)to K_*(text{point})$; I don't even know whether these conditions are equivalent.
Note that replacing $K$-theory with integral homology one gets very interesting (I think) class of spaces, studied in "Acyclic spaces" by E. Dror.
at.algebraic-topology kt.k-theory-and-homology
at.algebraic-topology kt.k-theory-and-homology
asked 4 hours ago
მამუკა ჯიბლაძე
8,066244112
asked 4 hours ago
მამუკა ჯიბლაძე
8,066244112
asked 4 hours ago
მამუკა ჯიბლაძე
8,066244112
asked 4 hours ago
მამუკა ჯიბლაძე
8,066244112
asked 4 hours ago
მამუკა ჯიბლაძე
8,066244112
8,066244112
3
All acyclic spaces have trivial K-theory (it's a simple argument with the Atiyah-Hirzebruch spectral sequence), but I think you get more things
– Denis Nardin
4 hours ago
@DenisNardin I would rather think that all K-trivial spaces would be acyclic, while K-theory could distinguish more spaces than "ordinary" homology. I just don't know enough, but I think the Adams $e$-invariant detects K-theory classes killed by the Chern character, cannot one find something along these lines?
– მამუკა ჯიბლაძე
2 hours ago
I don’t know a characterization, but it’s bigger than you might expect: K(Z/p, 2) is such a space.
– Eric Peterson
6 mins ago
add a comment |
3
All acyclic spaces have trivial K-theory (it's a simple argument with the Atiyah-Hirzebruch spectral sequence), but I think you get more things
– Denis Nardin
4 hours ago
@DenisNardin I would rather think that all K-trivial spaces would be acyclic, while K-theory could distinguish more spaces than "ordinary" homology. I just don't know enough, but I think the Adams $e$-invariant detects K-theory classes killed by the Chern character, cannot one find something along these lines?
– მამუკა ჯიბლაძე
2 hours ago
I don’t know a characterization, but it’s bigger than you might expect: K(Z/p, 2) is such a space.
– Eric Peterson
6 mins ago
3
3
All acyclic spaces have trivial K-theory (it's a simple argument with the Atiyah-Hirzebruch spectral sequence), but I think you get more things
– Denis Nardin
4 hours ago
All acyclic spaces have trivial K-theory (it's a simple argument with the Atiyah-Hirzebruch spectral sequence), but I think you get more things
– Denis Nardin
4 hours ago
@DenisNardin I would rather think that all K-trivial spaces would be acyclic, while K-theory could distinguish more spaces than "ordinary" homology. I just don't know enough, but I think the Adams $e$-invariant detects K-theory classes killed by the Chern character, cannot one find something along these lines?
– მამუკა ჯიბლაძე
2 hours ago
@DenisNardin I would rather think that all K-trivial spaces would be acyclic, while K-theory could distinguish more spaces than "ordinary" homology. I just don't know enough, but I think the Adams $e$-invariant detects K-theory classes killed by the Chern character, cannot one find something along these lines?
– მამუკა ჯიბლაძე
2 hours ago
I don’t know a characterization, but it’s bigger than you might expect: K(Z/p, 2) is such a space.
– Eric Peterson
6 mins ago
I don’t know a characterization, but it’s bigger than you might expect: K(Z/p, 2) is such a space.
– Eric Peterson
6 mins ago
add a comment |
1 Answer
1
active
oldest
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I'll just give an answer for finite complexes $X$. The condition $widetilde{K}^*(X)=0$ only depends on the suspension spectrum of $X$ so this is naturally regarded as a question in stable homotopy theory. The condition also implies that $widetilde{H}^*(X;mathbb{Q})=0$ and thus that $n.1_X=0$ as a stable map for some $n>0$. From this it follows that $X$ splits stably as a wedge of finitely many $p$-torsion finite spectra for different primes $p$, so we are really looking at a question in $p$-local stable homotopy theory. The condition $widetilde{K}^*(X)=0$ then says that $X$ has chromatic type at least two. For the general theory of chromatic type of finite spectra you can read Ravenel's book "Nilpotence in stable homotopy theory". For type two, it is possible to be a little more explicit than for higher types. For example, Adams constructed a certain self-map of the mod $p$ Moore spectrum which induces an isomorphism in $K$-theory, so the cofibre of that map has type two.
answered 1 hour ago
Neil Strickland
36.2k694186
add a comment |
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I'll just give an answer for finite complexes $X$. The condition $widetilde{K}^*(X)=0$ only depends on the suspension spectrum of $X$ so this is naturally regarded as a question in stable homotopy theory. The condition also implies that $widetilde{H}^*(X;mathbb{Q})=0$ and thus that $n.1_X=0$ as a stable map for some $n>0$. From this it follows that $X$ splits stably as a wedge of finitely many $p$-torsion finite spectra for different primes $p$, so we are really looking at a question in $p$-local stable homotopy theory. The condition $widetilde{K}^*(X)=0$ then says that $X$ has chromatic type at least two. For the general theory of chromatic type of finite spectra you can read Ravenel's book "Nilpotence in stable homotopy theory". For type two, it is possible to be a little more explicit than for higher types. For example, Adams constructed a certain self-map of the mod $p$ Moore spectrum which induces an isomorphism in $K$-theory, so the cofibre of that map has type two.
answered 1 hour ago
Neil Strickland
36.2k694186
add a comment |
I'll just give an answer for finite complexes $X$. The condition $widetilde{K}^*(X)=0$ only depends on the suspension spectrum of $X$ so this is naturally regarded as a question in stable homotopy theory. The condition also implies that $widetilde{H}^*(X;mathbb{Q})=0$ and thus that $n.1_X=0$ as a stable map for some $n>0$. From this it follows that $X$ splits stably as a wedge of finitely many $p$-torsion finite spectra for different primes $p$, so we are really looking at a question in $p$-local stable homotopy theory. The condition $widetilde{K}^*(X)=0$ then says that $X$ has chromatic type at least two. For the general theory of chromatic type of finite spectra you can read Ravenel's book "Nilpotence in stable homotopy theory". For type two, it is possible to be a little more explicit than for higher types. For example, Adams constructed a certain self-map of the mod $p$ Moore spectrum which induces an isomorphism in $K$-theory, so the cofibre of that map has type two.
answered 1 hour ago
Neil Strickland
36.2k694186
add a comment |
I'll just give an answer for finite complexes $X$. The condition $widetilde{K}^*(X)=0$ only depends on the suspension spectrum of $X$ so this is naturally regarded as a question in stable homotopy theory. The condition also implies that $widetilde{H}^*(X;mathbb{Q})=0$ and thus that $n.1_X=0$ as a stable map for some $n>0$. From this it follows that $X$ splits stably as a wedge of finitely many $p$-torsion finite spectra for different primes $p$, so we are really looking at a question in $p$-local stable homotopy theory. The condition $widetilde{K}^*(X)=0$ then says that $X$ has chromatic type at least two. For the general theory of chromatic type of finite spectra you can read Ravenel's book "Nilpotence in stable homotopy theory". For type two, it is possible to be a little more explicit than for higher types. For example, Adams constructed a certain self-map of the mod $p$ Moore spectrum which induces an isomorphism in $K$-theory, so the cofibre of that map has type two.
answered 1 hour ago
Neil Strickland
36.2k694186
I'll just give an answer for finite complexes $X$. The condition $widetilde{K}^*(X)=0$ only depends on the suspension spectrum of $X$ so this is naturally regarded as a question in stable homotopy theory. The condition also implies that $widetilde{H}^*(X;mathbb{Q})=0$ and thus that $n.1_X=0$ as a stable map for some $n>0$. From this it follows that $X$ splits stably as a wedge of finitely many $p$-torsion finite spectra for different primes $p$, so we are really looking at a question in $p$-local stable homotopy theory. The condition $widetilde{K}^*(X)=0$ then says that $X$ has chromatic type at least two. For the general theory of chromatic type of finite spectra you can read Ravenel's book "Nilpotence in stable homotopy theory". For type two, it is possible to be a little more explicit than for higher types. For example, Adams constructed a certain self-map of the mod $p$ Moore spectrum which induces an isomorphism in $K$-theory, so the cofibre of that map has type two.
answered 1 hour ago
Neil Strickland
36.2k694186
answered 1 hour ago
Neil Strickland
36.2k694186
answered 1 hour ago
Neil Strickland
36.2k694186
answered 1 hour ago
Neil Strickland
36.2k694186
36.2k694186
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3
All acyclic spaces have trivial K-theory (it's a simple argument with the Atiyah-Hirzebruch spectral sequence), but I think you get more things
– Denis Nardin
4 hours ago
@DenisNardin I would rather think that all K-trivial spaces would be acyclic, while K-theory could distinguish more spaces than "ordinary" homology. I just don't know enough, but I think the Adams $e$-invariant detects K-theory classes killed by the Chern character, cannot one find something along these lines?
– მამუკა ჯიბლაძე
2 hours ago
I don’t know a characterization, but it’s bigger than you might expect: K(Z/p, 2) is such a space.
– Eric Peterson
6 mins ago