Let $L\to V$ be complex line bundle and $F_{t}:V\to V$, $t\in [0,1]$, be a loop of diffeomorphisms, $F_0=F_1=$ identity.

For every $x\in V$, we get a loop $\gamma_x(t)=\{F_t(x)\}$ whose class in $\pi_1(V)$ is independent of $x\in V$.

I call a Hermitian connection $\nabla$ on $L$ semi-flat (with respect to $\{F_t\}$) if $Hol^\nabla_{\gamma_x}=Id$ for every $x\in V$.

Can you describe a purely topological condition (i.e. it only depends on the class of $\{F_t\}\in \pi_1(Diff_0(V)$) which insures the existence of such connection.

One trivial such condition is the triviality of $L$.

  • $\begingroup$ Let $\over{V}$ be the quotient of $V$ with respect to equivalence relation $x\sim F_t(x)$. Then we most probably want $L$ to be pull-back of some line bundle $\over{L}$ from $\over{V}$! How bad $\over{V}$ could be? $\endgroup$ Nov 3, 2014 at 0:26

1 Answer 1


As pointed out to me by Guangbo Xu, if the $S^1$-action gives $V$ the structure of an $S^1$-bundle $\pi:V\to B$, then such connection exists if and only if $\pi_*(c_1(L))=0 \in H_1(B)$, where $\pi_*$ is the integration map along the fibers of the Gysin sequence. However, one still has to think what happens if we choose another loop of diffeomoerphisms homotopic to this one.

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    $\begingroup$ I don't think it is as easy as this. You did not specify that $F_{t+s}=F_t\circ F_s$, so the family of diffeomorphisms may not even define an $S^1$ action. In fact, without restricting to a sufficiently nice subspace of the space of maps from $S^1$ to the diffeomorphism group, the individual loops $\gamma_x$ could be really wild. $\endgroup$ May 30, 2016 at 7:53

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