Is connectivity of a spectrum $E$ representing a cohomology theory equivalent to $E^{>1}(*)=0$?

Question

Let $E$ be a spectrum, $X$ a CW-complex and set $$ E^k(X)=[X_+,S^k\wedge E] $$ for an integer $k$ as the $k$-th cohomology group of $X$ associated to the spectrum where the brackets denote stable homotopy classes and where I suppress the symbol $\Sigma^\infty$ for the suspension spectrum in front of $X_+$.

A spectrum $E$ is called connective if it has no negative homotopy groups, i.e. if $$ [S^l,E]=0 $$ for $l<0$.

By adjunction $E^k(*)=[S^0,S^k\wedge E]=[S^{-k},E]$.

This reasoning implies that a spectrum $E$ is connective iff $E^k(*)=0$ for all $k\geq 1$. Is this true? I've read on Wikipedia that the complex $K$-theory spectrum $BU\times Z$ is not connective. Why isn't $K^k(*)=0$ always zero?

Answer 1

This is correct. Perhaps it's worth noting that the unreduced (co)homology of a point is precisely the reduced (resp., co)homology of $S^0$. Either of these graded groups is referred to as the coefficients of the spectrum $E$ (with the homological grading).

It's true that the spectrum $KU$ is not connective. This is an $\Omega$-spectrum with $KU_{2k} = BU \times \mathbb{Z}$ and $KU_{2k+1} = U$ for all $k \in \mathbb{Z}$. Maybe your confusion is in the difference between the spectrum $KU$ and the space $BU \times \mathbb{Z}$?