I read the following statement, and I am trying to see why it is true.

Let $E\to M$ be a rank 2 complex vector bundle over a smooth manifold $M$, such that the determinant bundle $\det(E)=\Lambda^2(E)$ is the trivial line bundle over $M$, with trivialization $\phi:\det(E)\to M\times \mathbb{C}$. Then there exists a connection $\nabla^E$ on $E$ which induces the trivial connection on $\det(E)$, under $\phi$.

I wanted to construct $\nabla^E$ by defining local connections, which I then glue together by a partition of unity. Let $\{U_\alpha\}$ be a trivializing open cover of $M$. Over each $U_\alpha$, I have a local frame $\{e_1^\alpha,e_2^\alpha\}$ such that $e_1^\alpha\wedge e_2^\alpha$ corresponds to 1 under the diffeomorphism $\phi$. On each $U_\alpha$, I can define a flat connection $\nabla^\alpha$ (so by declaring the sections $e^\alpha_i$ to be flat). If $(\rho_\alpha)$ is a partition of unity subordinated to the trivializing open cover, I define $\nabla^E(s)=\sum\limits_\alpha \nabla^\alpha(\rho_\alpha s)$, which is indeed a connection. Next, I want to check that it induces the trivial connection on $\det(E)$ under $\phi$. Note that $\phi^{-1}(1)=\sum\limits_\alpha \rho_\alpha e_1^\alpha\wedge e_2^\alpha.$ Then \begin{align*} \nabla^{\det(E)}\big(\sum_\alpha \rho_\alpha e_1^\alpha\wedge e_2^\alpha\big)&=\sum_\alpha \nabla^{\det(E)}\big(\rho_\alpha e_1^\alpha\wedge e_2^\alpha)\\&=\sum_\alpha \nabla^E(\rho_\alpha e^\alpha_1)\wedge e_2^\alpha+\rho_\alpha e_1^\alpha\wedge \nabla^E(e_2^\alpha)\\&=\sum_\alpha \rho_\alpha\big(\nabla^E(e_1^\alpha)\wedge e_2^\alpha+e_1^\alpha\wedge \nabla^E(e_2^\alpha)\big). \end{align*} I want this to be zero, but the problem I encounter is that $\nabla^E(e_i^\alpha)\neq\nabla^\alpha(e_i^\alpha)=0$. Nevertheless, is this the correct approach to take and how can I fix it? Does summing over all $\alpha$ make sure it is correct again?

EDIT: This question seems to have been partially asked here: Connections on Bundles with Trivial Determinant But it seems that an answer is missing and I am very curious, it seems to take a different approach, by introducing a Hermitean metric on $E$.

  • 2
    $\begingroup$ Welcome to MSE. Nice first question! $\endgroup$ Nov 3, 2020 at 7:28
  • $\begingroup$ You haven't used that $\rho$ is subordinate to the given cover yet $\endgroup$
    – AHusain
    Nov 3, 2020 at 8:25
  • $\begingroup$ Could you please elaborate? $\endgroup$
    – user821819
    Nov 3, 2020 at 8:34

1 Answer 1


Maybe I'm missing something, but I think it goes like this. We have

$$ \nabla^E s = \sum_\alpha \nabla^\alpha(\rho_\alpha s) = \left( \sum_\alpha \rm{d} \rho_\alpha \right) \otimes s + \sum_\alpha \rho_\alpha \nabla^\alpha s = \sum_\alpha \rho_\alpha \nabla^\alpha s $$

since $\sum \rho_\alpha =1$ implies $\sum \rm{d} \rho_\alpha = 0$. Now I can continue where you left off. We have

$$ \nabla^E (e_1^\alpha) \wedge e_2 ^\alpha + e_1^\alpha \wedge \nabla^E (e_2^\alpha) = \sum_\beta \rho_\beta (\nabla^\beta (e_1^\alpha) \wedge e_2^\alpha + e_1^\alpha \wedge \nabla^\beta (e_2^\alpha) ) = \sum_\beta \rho_\beta \nabla^\beta (e_1^\alpha \wedge e_2^\alpha) =0 ,$$

where the last equality comes from the fact that $ \nabla^\beta (e_1^\alpha \wedge e_2 ^\alpha)=0 $ where defined, by construction. (I am abusing notation, writing $\nabla^\beta$ for the connection on the restriction of $\rm{det}(E)$ induced by $\nabla^\beta$.)


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