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.

Is there a general description of the complex bundle $E \to M$ whose Chern character is the generator of $H^n(M)$?

  • $\begingroup$ Think about the grading. The question isn't too meaningful as written. The Chern character takes K(M) to the product of the even dimensional rational cohomology groups and K(\Sigma M) to the product of the odd dimensional groups. In particular, if n is odd the Chern character can't see the fundamental class in terms of bundles over M. $\endgroup$ – Peter May Aug 27 '13 at 3:00
  • $\begingroup$ Right, as written the question does only make sense for n even. But we can sure ask the analogous question for n odd, where we look for a bundle over SM. $\endgroup$ – AlexE Aug 27 '13 at 12:51

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.

  • $\begingroup$ Thanks! So we have reduced the problem (for n even) to finding a vector bundle which top Chern class is a generator of the top-dimensional cohomology. $\endgroup$ – AlexE Aug 27 '13 at 12:57

In algebraic geometry setting, this would be a structure sheaf of a point (or if you wish a linear combination of the vector bundles that provide projective resolution of such sheaf).


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