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Consider the trivial fibration $\mathbb{T}^2\to\mathbb{S}^1$, where $\mathbb{T}^2$ is the two-torus. Denote by $\mathbb{C}\to\mathbb{T}^2$ the trivial line bundle over $\mathbb{T}^2$, and equip it with the trivial connection. The cohomology bundle $H(\mathbb{S}^1, \mathbb{C}|_{\mathbb{S}^1})\to\mathbb{S}^1$ is then a trivial bundle (well every bundle over $\mathbb{S}^1$ is trivial). There is an induced connection, the Gauss-Manin connection $\nabla^{H(\mathbb{S}^1, \mathbb{C}|_{\mathbb{S}^1})}$, on $H(\mathbb{S}^1, \mathbb{C}|_{\mathbb{S}^1})\to\mathbb{S}^1$.

Does it make sense to ask the even (not odd) Chern character form $\textrm{ch}(\nabla^{H(\mathbb{S}^1, \mathbb{C}|_{\mathbb{S}^1})})$? Should $\textrm{ch}(\nabla^{H(\mathbb{S}^1, \mathbb{C}|_{\mathbb{S}^1})})$ be 2, the rank of $H(\mathbb{S}^1, \mathbb{C}|_{\mathbb{S}^1})\to\mathbb{S}^1$?

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  • $\begingroup$ Oh, I almost forgot that $H(\mathbb{S}^1, \mathbb{C}|_{\mathbb{S}^1})\to\mathbb{S}^1$ is a $\mathbb{Z}_2$-graded complex flat vector bundle.Thus its Chern character form should be zero. $\endgroup$
    – Ho Man-Ho
    Commented Jul 17, 2016 at 16:27

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