TITLE: To what extent can we characterise the image of the topological Chern character?
QUESTION [21 upvotes]: For a finite CW complex $X$, the Chern character gives an isomorphism
of finite-dimensional vector spaces:
$$
  ch : K^*(X)\otimes \mathbb{Q} \to H^*(X, \mathbb{Q}).
$$
The vector space $V = H^*(X, \mathbb{Q})$ thus comes equipped with two natural maximal-rank lattices:

$L_H = H^*(X, \mathbb{Z}) / T$ (where $T$ is the torsion),
$L_K = ch(K^*(X) / T')$ (where $T'$ is the torsion).


What we can say about the relationship between $L_H$ and $L_K$?

Here is simple necessary relationship. Let $n=[\dim X/2]$; using
denominators of size at most $n!$, the Chern character can be
expressed as a linear combination of Chern classes. Chern classes
are integral, so we must have:
$$
  n! \cdot L_K \subseteq L_H.
$$

What other necessary conditions are known? Is there a known set of
necessary and sufficient conditions? I.e., any $(V, L_H, L_K)$
satisfying them can be realised by some $X$.

For what it's worth, my motivation comes from spheres. Suppose we define a topological invariant $k$ to be the smallest natural number such that $k\cdot L_K \subseteq L_H$ (which presumably captures only a tiny bit of the relationship). Then the fact that $k=1$ for spheres allows a slick proof that the spheres admit no almost complex structure above dimension six. (In fact $L_K = L_H$ for spheres.)
Incidentally, the above ignores the $\mathbb{Z}/(2)$-grading. I haven't thought about this but I'd even be interested in the case when $X$ is even-dimensional with no odd-dimensional cohomology (e.g., a non-singular complex quadric).

REPLY [2 votes]: I remember, dimly, work of J.Frank Adams,published in "Topology" (journal), in the early 1960's, on this kind of problem,
Bruno Harris