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Let $X$ be a $k$-connected spectrum for $k \in \Bbb{Z}$.

I want to deduce how connected the counit of $(\Sigma^\infty, \Omega^\infty)$- adjunction is, that is, how connected is the map $$ \Sigma^\infty\Omega^\infty X \to X. $$

Any help would be appreciated.

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  • $\begingroup$ For $k\geq 0$, it is $2k$-connected by Freudenthal's suspension theorem. $\endgroup$ Sep 5 '18 at 11:07
  • $\begingroup$ @FernandoMuro I think you are slightly off. $\endgroup$
    – John Klein
    Sep 5 '18 at 11:31
  • $\begingroup$ @JohnKlein could be, but I don't think so. Freudenthat's suspension theorem shows that the homotopy groups of $\Sigma^\infty\Omega^\infty X$ coincide up to dimension $2k$ with those of $\Omega^\infty X$, which in turn are those of $X$. Hence my comment. $\endgroup$ Sep 5 '18 at 11:41
  • $\begingroup$ If the spectrum $X$ is $r$-connected, then the map $\Sigma^\infty\Omega^\infty X \to X$ is $(2r+2)$-connected. $\endgroup$
    – John Klein
    Sep 5 '18 at 11:51
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    $\begingroup$ @JohnKlein sure, I didn't claim my bound was optimal, it's obviously not optimal for $k=0$ since $\Omega^\infty X$ has abelian fundamental group. Actually, my argument above shows that the map is $(2k+1)$-connected. Let me provide an alternative answer below with the same bound as you get. $\endgroup$ Sep 5 '18 at 13:51
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If the spectrum $X$ is $r$-connected, then the map $\Sigma^\infty\Omega^\infty X \to X$ is $(2r+2)$-connected.

Here's a sketch: apply the functor $\Omega^\infty$ to get the map of spaces $$ Q(\Omega^\infty X) \to \Omega^\infty X $$ where $Q = \Omega^\infty\Sigma^\infty$. It will be enough to identify the connectivity of the latter.

This map has a section so it is surjective on homotopy in all degrees. In particular your map of spectra is surjective on homotopy in all degrees.

It's enough to compute the degree of injectivity. Let $S^j \to Q(\Omega^\infty X)$ be a map, $j \le 2r+1$. Suppose that its image in $\pi_j(\Omega^\infty X)$ is trivial.

By Freudenthal, the map $$ \pi_j(\Omega^\infty X) \to \pi_j(Q(\Omega^\infty X)) $$ is surjective in this degree ($j \le 2r+1$). So our map desuspends to give a map $S^j \to \Omega^\infty X$. But the composite $$ \pi_j(\Omega^\infty X) \to \pi_j(Q(\Omega^\infty X))\to \pi_j(\Omega^\infty X) $$ is the identity. This means that our desuspension is trivial, hence its iterated suspensions are as well.

Here is the essence of the above argument: suppose $A\to B\to C$ are maps of based spaces in which $A\to B$ is $k$-connected and $A\to C$ is $(k+1)$-connected. Then $B\to C$ is $(k+1)$-connected as well. Apply this to $$ \Omega^\infty X \to Q(\Omega^\infty X) \to \Omega^\infty X . $$

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    $\begingroup$ Thanks for the answer John, it was very illuminating, and nice to be able to avoid using Snaith Splitting/ Taylor tower where possible. $\endgroup$ Sep 5 '18 at 15:03
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Indeed, as John Klein shows, the map is $(2k+2)$-connected. Let me offer an alternative proof of the fact that, for $X$ a $k$-connective spectrum, $k\geq 0$, the homomorphism $\pi_i\Sigma^\infty\Omega^\infty X\rightarrow\pi_i X$ is an isomorphism for $i\leq 2k+1$. As indicated by John, this homomorphism is surjective for all $i$ and it suffices to show that the canonical homomorphism $\pi_i\Omega^\infty X\rightarrow \pi_i\Sigma^\infty\Omega^\infty X$ is an iso for $i\leq 2k+1$. We will actually prove that this is more generally true replacing $\Omega^\infty X$ with a $k$-connected space $Y$ with abelian fundamental group and vanishing Whitehead products, e.g. a single loop space. I though that someone might eventually find this useful, even if it is not what you're asking about.

The Freudenthal suspension theorem shows that, for an $k$-connected space $Y$, the suspension operator \[ \Sigma_*\colon \pi_i Y \longrightarrow \pi_{i+1}\Sigma Y \] is an isomorphism for $i\leq 2k$ and an epimorphism for $i=2k+1$. This and the fact that $\Sigma^j Y$ is $(j+k)$-connected clearly shows that the natural map from the homotopy groups of $Y$ to those of its suspension spectrum $\Sigma^\infty Y$ \[ \pi_i Y \longrightarrow \pi_{i}\Sigma^\infty Y, \] which is a transfinite composition of suspension operators, is an iso for $i\leq 2k$ and epi for $i=2k+1$.

Moreover, by the theorems of Blakers and Massey on the homotopy groups of triads, we have an exact sequence in the critical dimension \[ \pi_{k+1}Y\otimes \pi_{k+1}Y\stackrel{[-,-]}\longrightarrow \pi_{2k+1}Y\stackrel{\Sigma_{*}}\twoheadrightarrow \pi_{2k+2}\Sigma Y \] where the first arrow is the Whitehead product. This still makes sense for $k=0$ using the non-abelian tensor product of groups and the Brown-Loday non-abelian Van Kampen theorems. In this case $[-,-]$ is the commutator product.

If Whitehad products in $\pi_*Y$ vanish and the fundamental group is abelian, then the suspension operator is also an isomorphism for $i=2k+1$, and we get an isomorphism between the homotopy groups of $Y$ and $\Sigma^\infty Y$ in this dimension too.

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