It is well kown that
If $\,\bar\partial_E$ defines a holomorphic structure on the complex vector bundle $E \to X$ and $K$ is a hermitian metric on (the fibres of) $E$, there is a unique connection $D$ on $E$ (the Chern connection) such that
- $D$ is metric, i.e. $K$ is parallel with respect to $D$
- $D^{0,1} = \bar\partial_E$
A Dolbeault-type operator (i.e. Leibniz' rule holds with respect to $\bar\partial$) $$ \bar\partial_E \colon A^0_X(E) \to A^{0,1}_X(E) $$ defines a holomorphic structure on $E$ if and only if satisfies the integrability condition $\,\bar\partial_E^2 = 0$. It seems to me that conditions 1. and 2. make perfectly sense even if the operator $\,\bar\partial_E$ is not integrable, so I conjectured that the statement above holds true in this more general setting.
I thought "maybe the proof itself does not make any use of the integrability of $\,\bar\partial$". But every proof I found begins: "Consider a holomorphic frame..." and this assumption can be made only if $\,\bar\partial$ actually defines a holomorphic structure, i.e. is integrable.
I thus slightly modified the proof in the following way. Does it work?
Conjecture:
If $\,\bar\partial_E$ is a Dolbeault-type operator on the complex vector bundle $E \to X$ and $K$ is a hermitian metric on (the fibres of) $E$, there is a unique connection $D$ on $E$ such that
- $D$ is metric, i.e. $K$ is parallel wth respect to $D$
- $D^{0,1} = \bar\partial_E$
Proof:
The hermitian metric $K$ allows us to reduce the structure group of the bundle to $U(r)$, i.e. we just consider unitary frames.
Suppose that such a connection $D$ exists. From the condition 1. we deduce that in any unitary frame we locally have
$$ D = \mathrm{d} + A $$
where $A$ is an antihermitian matrix of $1$-forms, i.e. $A + A^\dagger = 0$. Moreover, condition 2. says that $A^{0,1} = C$ where $C$ is the connection matrix of $\,\bar\partial_E$. These two conditions give the equation
$$ A = C - C^\dagger $$
So $D$ is uniquely determined by $K$ and $\,\bar\partial_E$. To show that $A = C - C^\dagger$ actually defines a connection on $E$ we must check how $A$ changes under gauge transformations $g$. But we know that
- $C \leadsto gCg^{-1} - \bar\partial g \, g^{-1}$
- we can take unitary gauge transformations, i.e. such that $g^{-1} = g^\dagger$
Under such $g$'s we easily see that
$$ A = C - C^\dagger \leadsto g A g^{-1} - \mathrm{d}g\,g^{-1} $$
as it should be.