# To what extent can we characterise the image of the topological Chern character?

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).