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I want to tag along on Johannes' answer, since I think it's helpful to see an explicit construction of the positive K-groups in terms of bundles of Clifford modules, due to Atiyah, Bott and Shapiro in Clifford Modules:

This construction works equally well for KU and KO, by taking complexified or real Clifford algebras, and I'll leave this out of the notation.

First some notation:

• Let $\mathbb{R}^{p,q}$ denote $\mathbb{R}^{p+q}$ equipped with a the non-degenerate quadratic form $-x_1^2-\ldots-x_q^2+x_{q+1}^2+\ldots+x_{q+p}^2$.
• Let $C^{p,q}$ denote the associated Clifford algebra, and $M^{p,q}(X)$ denote the Grothendieck group of graded bundles of modules for $X\times C^{p,q}$.
• Let $M^{p,q+1}(X)\rightarrow M^{p,q}(X)$ denote restriction of scalars associated to the embedding $\mathbb{R}^{p,q}\hookrightarrow \mathbb{R}^{p,q+1}$.

ABS construct a homomorphism $M^{p,q}(X)\rightarrow K(X\times B^{p+q}, X\times S^{p+q-1})$ which factors through $A^{p,q}(X):=M^{p,q}(X)/M^{p,q+1}(X)$. They use Bott periodicity to show that this is an isomorphism when $X$ is a point and $q=0$, and Karoubi essentially extends this for any $X$ showing under the identification $K^{q-p}(X)=K(X\times B^{p+q}, X\times S^{p+q-1})$ (disclaimer: Karoubi rephrases the construction in an algebraic formalism that I haven't completely translated into the ABS point of view. If anyone has, and can either confirm that it's the same or explain the difference, that would be really helpful).

The homomorphism is as follows:

Given a bundle of $X\times C^{p,q}$-modules $E=E^0\oplus E^1$, ABS observe that the Clifford action gives a bundle map $cl:E^0\rightarrow E^1$ on $X\times B^{p+q}$ which is an isomporphism on $X\times S^{p+q-1}$. This gives a class $[E^0,E^1,cl]\in K(X\times B^{p+q}, X\times S^{p+q-1})$.

If $E$ is in the image of $M^{p,q+1}(X)$, then the map $cl$ extends to an isomorphism on all of $X\times B^{p+q}$ (identify $B^{p+q}$ with the upper hemisphere of $S^{p+q}$, S^{p+q}$), and$E$is mapped to$0$, so the map indeed factors through$A^{p,q}(X)$. 1 I want to tag along on Johannes' answer, since I think it's helpful to see an explicit construction of the positive K-groups in terms of bundles of Clifford modules, due to Atiyah, Bott and Shapiro in Clifford Modules: This construction works equally well for KU and KO, by taking complexified or real Clifford algebras, and I'll leave this out of the notation. First some notation: • Let$\mathbb{R}^{p,q}$denote$\mathbb{R}^{p+q}$equipped with a the non-degenerate quadratic form$-x_1^2-\ldots-x_q^2+x_{q+1}^2+\ldots+x_{q+p}^2$. • Let$C^{p,q}$denote the associated Clifford algebra, and$M^{p,q}(X)$denote the Grothendieck group of graded bundles of modules for$X\times C^{p,q}$. • Let$M^{p,q+1}(X)\rightarrow M^{p,q}(X)$denote restriction of scalars associated to the embedding$\mathbb{R}^{p,q}\hookrightarrow \mathbb{R}^{p,q+1}$. ABS construct a homomorphism$M^{p,q}(X)\rightarrow K(X\times B^{p+q}, X\times S^{p+q-1})$which factors through$A^{p,q}(X):=M^{p,q}(X)/M^{p,q+1}(X)$. They use Bott periodicity to show that this is an isomorphism when$X$is a point and$q=0$, and Karoubi essentially extends this for any$X$showing under the identification$K^{q-p}(X)=K(X\times B^{p+q}, X\times S^{p+q-1})$(disclaimer: Karoubi rephrases the construction in an algebraic formalism that I haven't completely translated into the ABS point of view. If anyone has, and can either confirm that it's the same or explain the difference, that would be really helpful). The homomorphism is as follows: Given a bundle of$X\times C^{p,q}$-modules$E=E^0\oplus E^1$, ABS observe that the Clifford action gives a bundle map$cl:E^0\rightarrow E^1$on$X\times B^{p+q}$which is an isomporphism on$X\times S^{p+q-1}$. This gives a class$[E^0,E^1,cl]\in K(X\times B^{p+q}, X\times S^{p+q-1})$. If$E$is in the image of$M^{p,q+1}(X)$, then the map$cl$extends to an isomorphism on all of$X\times B^{p+q}$(identify$B^{p+q}$with the upper hemisphere of$S^{p+q}$, and$E$is mapped to$0$, so the map indeed factors through$A^{p,q}(X)\$.