Mixing
KO theory v.s. ko theory
It looks that there are different types of topological K-theories, with similar names but they are totally different outputs for the same input.
The first theory is called the KO theory. There are very limited information I can find on nlab.
The second theory is called the ko theory. Say the ko-Homology studied in Chapter 10 of this book.
My question is that is there a simple way to contrast the twos:
KO theory v.s. ko theory?
For example, we can compare:
$KO_d(BG)_p$
$ko_d(BG)_p$
with the spin bordism group:
- $Omega_d^{Spin}(BG)_p$
Here we use the subindex to denote the $p$-torsion part (mean $mathbb{Z}_{p^n}$ for some $n$). We can focus on $p=2$ and free part, for $dle 7$; since there is a theorem given here, saying that
$$ko_d(BG)_2=Omega_d^{Spin}(BG)_2.$$
algebraic-topology k-theory topological-k-theory cobordism
asked Nov 21 '18 at 23:24
wonderichwonderich
2,08631230
add a comment |
It looks that there are different types of topological K-theories, with similar names but they are totally different outputs for the same input.
The first theory is called the KO theory. There are very limited information I can find on nlab.
The second theory is called the ko theory. Say the ko-Homology studied in Chapter 10 of this book.
My question is that is there a simple way to contrast the twos:
KO theory v.s. ko theory?
For example, we can compare:
$KO_d(BG)_p$
$ko_d(BG)_p$
with the spin bordism group:
- $Omega_d^{Spin}(BG)_p$
Here we use the subindex to denote the $p$-torsion part (mean $mathbb{Z}_{p^n}$ for some $n$). We can focus on $p=2$ and free part, for $dle 7$; since there is a theorem given here, saying that
$$ko_d(BG)_2=Omega_d^{Spin}(BG)_2.$$
algebraic-topology k-theory topological-k-theory cobordism
asked Nov 21 '18 at 23:24
wonderichwonderich
2,08631230
1
Maybe this will help: $KO$ is a 8-periodic spectrum, and $ko$ is its connective cover.
– JHF
Nov 22 '18 at 22:01
add a comment |
It looks that there are different types of topological K-theories, with similar names but they are totally different outputs for the same input.
The first theory is called the KO theory. There are very limited information I can find on nlab.
The second theory is called the ko theory. Say the ko-Homology studied in Chapter 10 of this book.
My question is that is there a simple way to contrast the twos:
KO theory v.s. ko theory?
For example, we can compare:
$KO_d(BG)_p$
$ko_d(BG)_p$
with the spin bordism group:
- $Omega_d^{Spin}(BG)_p$
Here we use the subindex to denote the $p$-torsion part (mean $mathbb{Z}_{p^n}$ for some $n$). We can focus on $p=2$ and free part, for $dle 7$; since there is a theorem given here, saying that
$$ko_d(BG)_2=Omega_d^{Spin}(BG)_2.$$
algebraic-topology k-theory topological-k-theory cobordism
asked Nov 21 '18 at 23:24
wonderichwonderich
2,08631230
It looks that there are different types of topological K-theories, with similar names but they are totally different outputs for the same input.
The first theory is called the KO theory. There are very limited information I can find on nlab.
The second theory is called the ko theory. Say the ko-Homology studied in Chapter 10 of this book.
My question is that is there a simple way to contrast the twos:
KO theory v.s. ko theory?
For example, we can compare:
$KO_d(BG)_p$
$ko_d(BG)_p$
with the spin bordism group:
- $Omega_d^{Spin}(BG)_p$
Here we use the subindex to denote the $p$-torsion part (mean $mathbb{Z}_{p^n}$ for some $n$). We can focus on $p=2$ and free part, for $dle 7$; since there is a theorem given here, saying that
$$ko_d(BG)_2=Omega_d^{Spin}(BG)_2.$$
algebraic-topology k-theory topological-k-theory cobordism
algebraic-topology k-theory topological-k-theory cobordism
asked Nov 21 '18 at 23:24
wonderichwonderich
2,08631230
asked Nov 21 '18 at 23:24
wonderichwonderich
2,08631230
asked Nov 21 '18 at 23:24
wonderichwonderich
2,08631230
asked Nov 21 '18 at 23:24
wonderichwonderich
2,08631230
asked Nov 21 '18 at 23:24
wonderichwonderich
2,08631230
2,08631230
1
Maybe this will help: $KO$ is a 8-periodic spectrum, and $ko$ is its connective cover.
– JHF
Nov 22 '18 at 22:01
add a comment |
1
Maybe this will help: $KO$ is a 8-periodic spectrum, and $ko$ is its connective cover.
– JHF
Nov 22 '18 at 22:01
1
1
Maybe this will help: $KO$ is a 8-periodic spectrum, and $ko$ is its connective cover.
– JHF
Nov 22 '18 at 22:01
Maybe this will help: $KO$ is a 8-periodic spectrum, and $ko$ is its connective cover.
– JHF
Nov 22 '18 at 22:01
add a comment |
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Maybe this will help: $KO$ is a 8-periodic spectrum, and $ko$ is its connective cover.
– JHF
Nov 22 '18 at 22:01