# Construction of a connection on a vector bundle $E\to M$ with trivial determinant bundle

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 $$E$$.

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

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