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Let $G$ be a compact, connected Lie group. There is an Atiyah–Hirzebruch spectral sequence

$$H^*(BG;K^*) \implies K^*(BG)$$

connecting $H^*BG$, which generally contains torsion, with $K^*BG \cong \widehat R(G)$, which does not.

Is it known whether the torsion situation "only improves"? More precisely, let $P_r \subsetneq \mathbb N$ be the set of torsion primes of the $r$th page:

  1. Is it always the case that $P_{r+1} \subseteq P_r$?

  2. Is it at least true that $P_r \subseteq P_2$?

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    $\begingroup$ I haven't thought about this very much, but a priori it is possible that the $E_∞$-page contains torsion while the $K^*BG$ does not (remember: the $E_∞$-page is just the associated graded of a filtration on $K^*BG$). Moreover it is not even clear that the AHSS converges in this case... $\endgroup$
    – Denis Nardin
    Commented Apr 6, 2017 at 22:23
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    $\begingroup$ Convergence: Pick a CW-structure on $BG$ and look at $n$-skeleta; the AHSS unambiguously converges for these, and the $\varprojlim^1$ term in the Milnor exact sequence vanishes, so it is enough to look at these finite approximations. Their torsion primes will be amongst those of $BG$. I thought about putting this in the question but decided it would just take up space. $\endgroup$
    – jdc
    Commented Apr 6, 2017 at 22:45
  • $\begingroup$ Filtration: Could you give me an example of the type of situation you might expect torsion in the associated graded? I mean, $\widehat R(G)$ is a power series ring over $\mathbb Z$ in finitely many indeterminates; I do not have intuition about this, but would you expect it to be possible, say, that $2x$ lies in a lower filtrand than $x$ for some $x$? $\endgroup$
    – jdc
    Commented Apr 6, 2017 at 22:49
  • $\begingroup$ @jdk the classical example is the Bockstein spectral sequence, that is build for this. Another one is the Adams spectral sequence, where 2∊π_0S is represented by an element h_0 in Adams filtration 1. $\endgroup$
    – Denis Nardin
    Commented Apr 6, 2017 at 23:11
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    $\begingroup$ The representation ring $R(C_2)$ is $\mathbb{Z}[T]/(T^2-1)$ with $T$ the sign representation. Taking additive basis $1$ and $H= T-1$, so the augmentation ideal $I$ consists of the multiples of $H$, we have $H^2 = -2H$. Thus the associated graded for the $I$-adic filtration is $\mathbb{Z} \oplus (\mathbb{Z}/2)^\infty$. This has torsion. $\endgroup$
    – Oscar Randal-Williams
    Commented Apr 7, 2017 at 17:27

1 Answer 1

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Of course, in any spectral sequence $E_{r+1}$ is a subquotient of $E_r$ (the kernel of $d_r$ divided by the image of $d_r$). And in general new torsion can appear in the sense of torsion elements in $ker/im$ that are not represented by torsion elements in $ker$.

But this cannot happen when the spectral sequence is rationally trivial, that is, when the image of $d_r$ is in the torsion subgroup. So in that case the torsion subgroup of $E_{r+1}$ is a subquotient of that of $E_r$.

This holds for the Atiyah-Hirzebruch spectral sequence of any simply connected space and any cohomology theory $k^\ast$. The reason is that $k^\ast\otimes \mathbb Q$ is a product of ordinary cohomology theories.

So, yes the torsion situation only improves.

Edit: I overstated things a bit. Tyler Lawson makes a point that I had not appreciated: at the level of generality of my answer, tensoring the AHSS with $\mathbb Q$ does not yield the AHSS for another cohomology theory. The argument sketched above applies to the homology AHSS of any simply connected space, or spectrum, but in the case of cohomology the conclusion is false in general. Think of the case when $k$ has only two nontrivial coefficient groups $k^0(\ast)=A$ and $k^n(\ast)=B$, $n>0$. The one nontrivial differential $d_{n+1}$ is a stable cohomology operation $H^\bullet(X;B)\to H^{\bullet +n+1}(X;A)$, corresponding to an element of $H^{n+1}(HB;A)\cong Hom(H_{n+1}(HB),A)\oplus Ext(H_{n}(HB),A)$. This group need not be a torsion group, even though $H_n(HB)$ and $H_{n+1}(HB)$ are torsion groups for any abelian group $B$. But it's OK if $B$ is finitely generated.

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    $\begingroup$ Is there an easy way to get past the need to relate $k^*(X) \otimes \Bbb Q$ and $(k \otimes \Bbb Q)^* (X)$ in the case where $X$ isn't of finite type? (Though not relevant to the question at hand.) $\endgroup$
    – Tyler Lawson
    Commented Apr 7, 2017 at 17:52
  • $\begingroup$ I'll have to think about that. $\endgroup$
    – Tom Goodwillie
    Commented Apr 7, 2017 at 20:14

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