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Aloff--Wallach spaces were discussed in this question. They are quotients of SU(3) by U(1) indexed by a lattice of rank 2. Am I correct in guessing that the $K$-theory group $K_0$ of these spaces is the same for all elements in the lattice? (I am also assuming here, perhaps falsely, that the algebraic, topological, and smooth $K_0$ groups coincide.)

What is the case in the more general Eschenburg space setting?

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Since it was not quite specified what sort of K-theory is relevant, I'll try to argue that topological complex K-theory is not the same for all Aloff–Wallach spaces.

The Eschenburg spaces are parametrized by $k=(k_1,k_2,k_3)$ and $l=(l_1,l_2,l_3)$ such that $k_1+k_2+k_3=l_1+l_2+l_3$ plus a gcd condition. The case $l_i=0$ should be the Aloff–Wallach spaces, and in this case the gcd condition means only that the $k_i$ are pairwise coprime.

The integral cohomology of Eschenburg spaces has been computed by Eschenburg, cf.

  • J.H. Eschenburg. New examples of manifolds with strictly positive curvature. Invent. Math. 66 (1982), 469–480.

  • J.-H. Eschenburg. Cohomology of biquotients. Manuscripta Math. 75 (1992), no. 2, 151–166.

The formulas for the additive structure can also be found in the thesis of Pongdate Montagantirud; it is given as follows, where $r=\sigma_2(k_1,k_2,k_3)=k_1k_2+k_1k_3+k_2k_3$ is an odd integer: $$ {\rm H}^i(X,\mathbb{Z})=\left\{\begin{array}{ll} \mathbb{Z} & i=0,2,5,7\\ \mathbb{Z}/r\mathbb{Z} & i=4\\ 0 & \textrm{otherwise}\end{array}\right. $$

Now we can compute topological K-theory using the Atiyah–Hirzebruch spectral sequence $$ E^{p,q}_2=H^p(X,KU^q(\ast))\Rightarrow KU^{p+q}(X). $$ with the differentials going $d_r:E^{p,q}_r\to E^{p+r,q-r+1}_r$. Note that KU is concentrated in even degrees and therefore the differentials can only be non-trivial for $q$ even and $r$ odd. Moreover, since the Atiyah–Hirzebruch sequence collapses rationally, the only nontrivial differentials can be the ones with $H^4(X,\mathbb{Z})\cong\mathbb{Z}/r\mathbb{Z}$ as target. Therefore, all differentials are trivial and the spectral sequence collapses at $E_2$. There is no extension problem involving the odd cohomology, so we find that $$ KU^n(X)=H^{\rm odd}(X,\mathbb{Z})\cong\mathbb{Z}^2 $$ for $n$ odd. There is one extension problem for even cohomology: however, the filtration would have smallest step $H^4$ and the two subsequent subquotients $\mathbb{Z}$. This is the direction in which extensions must necessarily split, so we find $$ KU^n(X)=H^{\rm ev}(X,\mathbb{Z})\cong \mathbb{Z}^2\oplus\mathbb{Z}/r\mathbb{Z} $$ for $n$ even.

Since there are Aloff–Wallach spaces with different $r$, there are Aloff–Wallach spaces whose topological K-theories differ (but only on the torsion subgroup in $KU^{2i}$). I don't know what would happen for $KO$. Probably the fact that the only torsion in the cohomology is odd could be exploited, but I think the calculations for $KO$ would be a lot messier.

Two parenthetical remarks:

  • I don't know much about smooth K-theory and what it would look like for Aloff–Wallach spaces, so I can't say anything about that.

  • I don't think that algebraic K-theory applies here. (I suppose that there are no natural lifts of Aloff–Wallach spaces to algebraic varieties; and I suppose that the Waldhausen K-theory of finite spaces over Aloff–Wallach manifolds has not been computed. The Waldhausen-type algebraic K-theory should split off a copy of the stable homotopy type of the underlying space, so these could also be distinguished by the torsion part $r$.)

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