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Given two cohomology theories $h^{\bullet}$ and $k^{\bullet}$, we define their direct sum $(h \oplus k)^{\bullet}$ as the cohomology theory $(h \oplus k)^{n}(X) := h^{n}(X) \oplus k^{n}(X)$. If I'm not wrong, this is another cohomology theory, whose spectrum is the wedge product of the corresponding spectra. For example, the singular cohomology with coefficients in $\mathbb{Z} \oplus \mathbb{Z}$ is the direct sum of two copies of the one with coefficients in $\mathbb{Z}$.

Let us call a theory $h^{\bullet}$ irreducible if it is not decomposable as a non-trivial direct sum. Is it true that, if $h^{\bullet}$ is irreducible, then the cohomology groups of the point are also irreducible? In other words, is it true that $h^{n}(pt) = \mathbb{Z}, \mathbb{Z}/p\mathbb{Z}$ or $0$, but never a non-trivial direct sum of them?

If the answer is no, is it true for multiplicative theories?

Added later: The answer is no, as shown by Eric Wofsey. So I modify the question: is it true that the free part of $h^{n}(pt)$ is $\mathbb{Z}$ or $0$, but never $\mathbb{Z}^{n}$ with $n > 1$?

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Are you aware that there are non-cyclic indecomposable abelían groups? –  Fernando Muro Jan 4 '13 at 15:27
    
Yes, it's true, but I'm supposing that $h^{n}(X)$ is finitely generated. –  Fabio Jan 4 '13 at 15:50
    
I hope $X$ does not dessignate an arbitrary space. In that case the only example is the trivial cohomology theory! –  Fernando Muro Jan 4 '13 at 17:38
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2 Answers 2

up vote 8 down vote accepted

The sphere spectrum, representing stable cohomotopy, is irreducible (you can see this, for instance, from the fact that its homology is irreducible and any summand would be connective and thus would have to have nontrivial homology). But the associated cohomology of a point is the stable homotopy groups of spheres, which are certainly not irreducible in general.

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Thank you, I could think by myself of this example! In this case, even if the group is not irreducible, the free part is $\mathbb{Z}$ or $0$, never $\mathbb{Z}^{n}$ with $n > 0$. Is this true in general for irreducible theories? –  Fabio Jan 4 '13 at 16:59
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No. Just take a 3-cell complex $X$ with two 0-cells and one $n$-cell which is attached to the 0-cells by (say) two $mathbb{F}_p$-linearly independent $p$-torsion classes in $\pi_{n-1}(S)$. By looking at homology, if $X$ split it would have to have a copy of the sphere spectrum as a summand, but by construction no class in $\pi_0(X)=\mathbb{Z}^2$ which generates a summand of $\mathbb{Z}^2$ can generate a free module over $\pi_*(S)$. So $X$ is irreducible, but $\pi_0(X)=\mathbb{Z}^2$. –  Eric Wofsey Jan 4 '13 at 17:20
    
Thank you, now I have nothing more to ask! –  Fabio Jan 4 '13 at 17:26
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A generalized cohomology theory is indecomposable iff its classifying spectrum is. This follows from Brown representability. Baues and Drozd have a paper in Topology about finite-dimensional indecomposable stable homotopy types withney fg free homology. In diension 4, you have 3 indecomposable spectra with rank 1 homology in dimensions 0 and 4, rank 2 homology in dimension 2, and 0 elsewhere, see definition 1.7. Whitehead's long exact sequence for the Hurewicz homomorphism looks as follows towards the end:

$$\pi_2\rightarrow H_2=\mathbb Z^2 \rightarrow \Gamma_1=H_0\otimes \mathbb Z/2=\mathbb Z/2$$

This implies that the second generalized homology of the point wrt this spectrum, which is $\pi_2$, contains a subgroup isomorphic to $\mathbb Z^2$.

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