14
$\begingroup$

As known, any non-trivial finite spectrum $X$ can not have non-zero homotopy groups $\pi_i(X)$ only for finite number of $i$. As I understand, the same is true for any spectrum $X$ with finitely generated homology groups $H\mathbb{Z}_i(X)$ which are non-zero only for finite number of $i$.

There is a "dual" statement (?) that a spectrum $Y$ with finitely generated homotopy groups $\pi_i(Y)$ which are non-zero only for finite number of $i$ can not has the same property for homology groups $H\mathbb{Z}_i(Y)$.

If all this true, is there a conceptual reason for such a "duality" in complexity? Can these statements being proof uniformly?

$\endgroup$
15
$\begingroup$

Here are two ways of thinking about it. The first comes from the way one proves the final statement you cited: if $X$ has finitely many nonzero homotopy groups which are all finitely generated, then it is an extension of Eilenberg-MacLane spectra, which necessarily have infinitely many cells to kill most of their homotopy groups. This necessitates $X$ having huge (i.e., not finitely generated) homology, etc. Using this observation, one can also prove your first statement that any nonzero finite spectrum must have infinitely many nonzero homotopy groups: every finite spectrum is harmonic (i.e., $\bigvee_{n\geq 0} K(n)$-local, where $K(n)$ is the $n$th Morava $K$-theory), but torsion Eilenberg-MacLane spectra are all dissonant. One now concludes using the observation that $\pi_\ast X\otimes\mathbf{Q} = \mathrm{H}_i(X;\mathbf{Q})$ must be bounded. One can similarly prove the second statement you made. This shows where the duality between Postnikov and cellular decompositions of spectra comes from.

The second way of thinking about this comes from Brown-Comenetz duality. Let us $p$-localize everywhere. Let $I_{\mathbf{Q/Z}}$ denote the Brown-Comenetz dualizing spectrum, defined so that $\pi_\ast F(X, I_{\mathbf{Q/Z}}) = \mathrm{Hom}(\pi_{-\ast} X, \mathbf{Q/Z})$. In his answer to my question at https://mathoverflow.net/a/301928/102390, Strickland argues that if $X$ is a spectrum, then $X$ is $I_{\mathbf{Q/Z}}$-acyclic if and only if $F(X, S^0)$ is contractible. If $X$ is finite, then this happens if and only if $X$ is itself contractible. In other words, nontrivial finite spectra are never $I_{\mathbf{Q/Z}}$-acyclic. However, almost every non-finite spectrum you can concoct (e.g., any spectrum with unbounded cohomology) will be $I_{\mathbf{Q/Z}}$-acyclic; see Strickland's answer for more. (One interesting non-example is the infinite stunted projective space $P^\infty_{-\infty}$, which doesn't look like a finite spectrum, but is in fact equivalent to the $2$-complete sphere by Lin's theorem.) In fact, Drew's answer to my question discusses the dichotomy conjecture, which states that if $E$ is any spectrum which is not $I_{\mathbf{Q/Z}}$-acyclic, then there is a finite spectrum $X$ such that $\langle E \rangle \geq \langle X \rangle$. (One interesting example of a non-finite spectrum $E$ which satisfies this is $\mathbf{C}P^\infty$; an unpublished theorem of Hopkins says that it satisfies $\langle \mathbf{C}P^\infty\rangle = \langle S^0 \rangle$.)

By the way, note that most of your claims are also true unstably. For example, the claim that any nonzero finite complex must have infinitely many nonzero homotopy groups is also true unstably: this is the McGibbon-Neisendorfer theorem. Also note that the distinction you talk about disappears rationally: in the stable world, the rational sphere is just the Eilenberg-MacLane spectrum $\mathbf{Q}$, and unstably, I'm pretty sure that the only spaces whose Hurewicz map is an isomorphism are given by the $K(\mathbf{Q},n) \simeq S^n_\mathbf{Q}$ for odd $n$. Finally, you should also look up Eckmann-Hilton duality (and, in particular, Fuks duality).

$\endgroup$
  • $\begingroup$ Thanks a lot for this answer, but i don't understand some things. "Strickland argues that if X is a spectrum, then X is IQ/Z-acyclic if and only if F(X,S0) is contractible." I understand his arguments for prime p, but I don't understand how to deduce the full statement. In his argument is critical that F(I,I) = I, which is not true for genuine Brown-Comenetz. $\endgroup$ – Q. Q. Nov 5 '18 at 19:08
  • 1
    $\begingroup$ I'd implicitly p-localized. Note that since the difference you are after disappears after rationalization, it suffices to even work in the p-complete stable category to observe this distinction. Also, F(I,I) is not equivalent to I. $\endgroup$ – skd Nov 5 '18 at 19:12
  • $\begingroup$ Yes, I mean F(I,I) = S^0 but can not edit now. I am also interested in this statement for all spectra and now wondering if it is true in such generality? It seems to agree on HZ: HZ^I=0 and F(HZ,S^0)=0. $\endgroup$ – Q. Q. Nov 6 '18 at 7:08
  • $\begingroup$ What do you mean by "such generality"? You don't need X to be finite to conclude that X smash I = 0 if and only if F(X, S^0) = 0. $\endgroup$ – skd Nov 6 '18 at 17:15
  • $\begingroup$ @skd, just a small remark, unless I missed something, I think you need simply connected for McGibbon-Neisendorfer, otherwise $S^1$ is a counter example. $\endgroup$ – Shay Ben Moshe Nov 6 '18 at 20:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.