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Inspired by this question, I wonder if anyone can provide an example of a finite CW complex X for which the order of the torsion subgroup of $H^{even} (X; \mathbb{Z}) = \bigoplus_{k=0}^\infty H^{2k} (X; \mathbb{Z})$ differs from the order of the torsion subgroup of $K^0 (X)$, where $K^0$ is complex topological K-theory. This is the same as asking for a non-zero differential in the Atiyah-Hirzebruch spectral sequence for some finite CW complex X, since this spectral sequence always collapses rationally.

Even better, is there an example in which X is a manifold? An orientable manifold?

Tom Goodwillie's answer to the question referenced above gave examples (real projective spaces) where the torsion subgroups are not isomorphic, but do have the same order.

It's interesting to note that the exponent of the images of these differentials is bounded by a universal constant, depending only on the starting page of the differential! This is a theorem of D. Arrlettaz (K-theory, 6: 347-361, 1992). You can even change the underlying spectrum (complex K-theory) without affecting the constant.

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2 Answers 2

up vote 7 down vote accepted

This paper by Volker Braun shows that the orientable 8-manifold $X=\mathbb{RP}^3\times \mathbb{RP}^5$ gives an example. One has $$K^0(X) \cong \mathbb{Z}^2 \oplus \mathbb{Z}/4\oplus (\mathbb{Z}/2)^2$$ and $$H^{ev}(X) \cong \mathbb{Z}^2 \oplus (\mathbb{Z}/2)^5.$$ Braun does the calculations using the Künneth formulae for the two theories, with the discrepancy in size arising because the order of the tensor product of finite abelian groups is sensitive to their structure, not just their order.

One another remark is that Atiyah and Hirzebruch told us the $d_3$-differential in their spectral sequence. It's the operation $Sq^3 \colon H^i(X;\mathbb{Z})\to H^{i+3}(X;\mathbb{Z})$ given by $Sq^3 := \beta\circ Sq^2 \circ r$, where $r$ is reduction mod 2 and $\beta$ the Bockstein. As you say, Dan, if this is non-vanishing, K-theory has smaller torsion. This happens iff there's a mod 2 cohomology class $u$ such that $u$ admits an integral lift but $Sq^2 (u) $ does not. Can someone think of a nice example where this occurs?

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That's a surprisingly nice example! –  Dan Ramras Jul 7 '10 at 20:38

There are many spaces $X$ whose $K$ theory is trivial (isomorphic to that of a point), but whose ordinary cohomology is not. Famously, there is a 4-cell complex obtained by taking the homotopy cofiber of the "Adams map" $$f:S^{11}\cup_{3\iota} e^{12} \to S^7\cup_{3\iota} e^8.$$ Here, $3\iota$ represents a degree $3$ self-map of a sphere.

The Adams map induces an isomorphism in $K$-theory, so $K^*(X)\approx K^*(*)\approx Z$. But $H^*(X,Z)\approx Z\oplus Z/3\oplus Z/3$.

(Hopefully I have the dimensions correct now.)

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That was quick! –  Dan Ramras Jul 7 '10 at 0:38
The dimensions can't be right. $12-7$ should be $4=2p-2$ –  Tom Goodwillie Jul 7 '10 at 3:40
I think Tom is right. Letting $C_n (p)$ denote the cofiber of the degree p self map of $S^{n-1}$, the Adams map is a mapping $C_{n+2p-2} (p) \to C_n (p)$. Charles, under what conditions does the Adams map induce an isomorphism in K-theory? Is this in Adams' J(X) IV paper? –  Dan Ramras Jul 7 '10 at 20:37
Indeed, Tom is correct. I keep confusing the dimensions in the map, and the dimensions in the resulting cell complex (the cofiber). Argh. –  Charles Rezk Jul 8 '10 at 0:31

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