Let $M$ be a closed, oriented manifold of dimension $n$. We know that the Chern character induces an isomorphism $K^\ast(M) \otimes \mathbb{Q} \cong H^\ast(M; \mathbb{Q})$ and now I was wondering how the preimage of the generator of $H^n(M)$ looks like.



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> Is there a general description of the complex bundle $E \to M$ whose Chern character is the generator of $H^n(M)$?
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Preimage of $1 \in H^n(M^n)$ under Chern character
Suppose $M$ is a 2n dimensional manifold and $F$ is a rank $n$ complex vector bundle on $M$ with $c\_n(F)=k \in H^{2n}(M)$. Then 


$$ \sum\_{i=0}^n (-1)^i[\Lambda^iF^\*] $$


is an element in $K(M)$ which is the preimage of $k$ under the chern character map. Assuming that $k\neq 0$, this will be the preimage of a generator of $H^{2n}(M,\mathbb{Q})$.


Edit: To address Alex's comment, I think that one can construct a bundle $F$ with $c\_n(F)=1$ as follows. Take an open cover of $M$ consisting of an open ball around a point and its complement. The overlap of these two sets is homotopic to $S^{2n-1}$ and and so a bundle $F$ which is trivialized on this open cover is determined by the homotopy class of the transition function $S^{2n-1}\to U(n)$. Since $U(n)$ has the rational homotopy type of $S^1\times S^3\times \cdots\times S^{2n-1}$, there is an element which generates the factor of $\mathbb{Z}$ in $\pi\_{2n-1}(U(n))$ corresponding to the $S^{2n-1}$ factor in $S^1\times S^3\times \cdots\times S^{2n-1}$. The bundle with this transition function will have Euler class equal to 1. 


