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For cell complexes${}^1$ $X$ we have an isomorphism

$$ K^*(X)\otimes \mathbb{Q}\cong H^{*}(X;\mathbb{Q}), $$

which is induced by the Chern character.

What is the analogous statement for $KO(X)$?

${}^1$:Hatcher states finite, but I've seen arbitrary CW-complexes stated as well.

edit: The footnote seems wrong, as by the comments.

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    $\begingroup$ Shouldn't it be $K^0(X)\otimes\mathbb{Q}\cong H^{2\ast}(X;\mathbb{Q})$ instead? $\endgroup$
    – archipelago
    Commented Feb 9, 2016 at 12:30
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    $\begingroup$ If $X$ is an arbitrary cell complex, then $\operatorname{ch}(E)$ could be an infinite sum of non-zero terms, and hence not belong to $H^*(X; \mathbb{Q})$ which is the direct sum of the groups $H^k(X; \mathbb{Q})$. Instead the $\operatorname{ch}(E)$ would belong to the direct product of the groups. Note, if one assumes $X$ is a finite cell-complex, then there are only finitely many non-zero cohomology groups, there is no problem (the direct sum and direct product coincide). $\endgroup$
    – Michael Albanese
    Commented Feb 9, 2016 at 12:58
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    $\begingroup$ It can't work for infinite complexes: consider X = BG for a finite group G. $\endgroup$
    – Gijs Heuts
    Commented Feb 9, 2016 at 13:05
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    $\begingroup$ An example where the Chern character of a vector bundle doesn't belong to the direct sum is $X = \mathbb{CP}^{\infty}$ and $E$ the tautological line bundle. Then $\operatorname{ch}_k(E) = c_1(E)^k/k! \neq 0$, so $\operatorname{ch}(E) \not\in H^*(\mathbb{CP}^{\infty}; \mathbb{Q})$, but rather $\operatorname{ch}(E) \in \prod H^k(\mathbb{CP}^{\infty};\mathbb{Q})$. $\endgroup$
    – Michael Albanese
    Commented Feb 9, 2016 at 13:06
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    $\begingroup$ @MichaelAlbanese I think this is a matter of convention. If you want to have a universal Chern character (or a universal total Chern class or whatever) on $BU$, then you have to interpret $H^\bullet$ as an infinite product. This should not come as a surprise as cohomology tends to have infinite products where homology has infinite sums. It is clear that this convention will raise problems in other places, but for characteristic classes, it seems to be the correct choice. $\endgroup$
    – Sebastian Goette
    Commented Feb 9, 2016 at 14:30

1 Answer 1

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In the following, $X$ is a finite complex. The Adams operator $\psi^{-1}$ (complex conjugation) acts on $K^0(X)$. After inverting $2$, the group $KO^0(X) \otimes \Bbb Z[1/2]$ maps isomorphically to the subset of $K^0(X) \otimes \Bbb Z[1/2]$ fixed by $\psi^{-1}$.

We can then tensor this with $\Bbb Q$. Then $K^0(X) \otimes \Bbb Q \cong \prod_{n \geq 0} H^{2n}(X;\Bbb Q)$. The operator $\psi^{-1}$ acts on this by fixing the factors with $n$ even and negating the factors with $n$ odd. The fixed set is $KO^0(X) \otimes \Bbb Q \cong \prod_{m \geq 0} H^{4m}(X; \Bbb Q)$.

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  • $\begingroup$ Thank you for your answer. I will have to read up on the Adams operations, but this seems nice. I think I will have to read more about this stuff. The point that puzzles me is the fact that $KO$ is 8 periodic. Does this mean that after tensoring with $\mathbb{Q}$ this periodicity reduces to 4 periodicity? $\endgroup$
    – Thomas Rot
    Commented Feb 9, 2016 at 14:33
  • $\begingroup$ The original question has $H^*$ and $K^*$, not $K^0$ and $H^{4*}$. In the complex setting, this works if one regards $K^*$ as a $\mathbb Z/2$-graded theory. Real $K$-theory is $\mathbb Z/8$-graded. @Thomas Rot: I just read your comment. The solution is that you get a direct sum of two copies of $H^*$ if you take the full 8-periodic $KO$-theory. $\endgroup$
    – Sebastian Goette
    Commented Feb 9, 2016 at 14:34
  • $\begingroup$ Before asking this question I thought that one might need the Pontryagin or Stiefel-Whitney character. To me this did not seem to map into the right dimensions of the cohomology, which prompted me to ask this question. Is this map realized by the Pontryagin character? $\endgroup$
    – Thomas Rot
    Commented Feb 9, 2016 at 14:38
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    $\begingroup$ @ThomasRot Yes on both counts. There is a class $w \in KO^4(pt)$ such that $w^2$ is twice the generator of $KO^8(pt)$ inducing Bott periodicity. After inverting $2$, the result is a 4-periodic theory. The isomorphism is indeed implemented by the Pontryagin character. $\endgroup$
    – Tyler Lawson
    Commented Feb 9, 2016 at 14:55
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    $\begingroup$ @SebastianGoette Fair enough. I was aiming for a clean statement. In reduced $KO$-theory then there are isomorphisms $\widetilde{KO}^0(\Sigma^r X) \cong \widetilde{KO}^{-r}(X) \cong \widetilde{KO}^{8m-r}(X)$, and so one can deduce the other $KO$-groups of $X$ from this. $\endgroup$
    – Tyler Lawson
    Commented Feb 9, 2016 at 14:58

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