I can't find this result anywhere, but it seems pretty straightforward. I want to avoid silly mistakes, but I can't see any fault. I'd love to receive some feedback
Let $X$ be a $\Omega$-spectrum (of compactly generated topological spaces). By definition for $q<0$, $$\pi_q(X)=\lim_k \pi_{k+q <k}X_k$$
But since $\pi_i(X_k)\cong \pi_{i}(\Omega X_{k+1})\cong \pi_{i+1}(X_{k+1})$ the spaces are more and connected, in particular, $X_k$ is $k-1$-connected and therefore $\pi_q(X)=0$.
Is that correct?
$\endgroup$2 Answers
$\begingroup$A $\Omega$-spectrum is not necessarily a connective one. In your argument, there is no reason "the spaces are more and [more] connected". For a concrete example, consider the $K$-theory spectrum $KU$. In its usual incarnation (with $BU \times \mathbb{Z}$ even and $\Omega(BU \times \mathbb{Z})$ odd), it is already a $\Omega$-spectrum by Bott periodicity. But it's 2-periodic, so its homotopy groups extend infinitely in negative degrees.
In general, remember then every spectrum can be converted into a $\Omega$-spectrum by a telescope construction, without changing its homotopy groups. So if every $\Omega$-spectrum was connective, this would mean then every spectrum is connective, which should strike you as patently false. This gives another bunch of counterexamples: take any spectrum, desuspend it a bunch of times so that it has negative homotopy groups, and then "$\Omega$-spectrify".
$\endgroup$ 6 $\begingroup$Direct answers would be helpful. I believe you are saying for connective $\Omega$-spectrum $X$, $\pi_i(X)=0$ for all $i\leq -1$. For non connective $\Omega$-spectrum, there are nontrivial negative homotopy groups.
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