ag.algebraic-geometrycomplex-geometrydg.differential-geometry
More precisely formulated: We are given a hermitian holomorphic vector bundle $(E,\langle\cdot,\cdot\rangle,\bar{\partial})$ on a $\mathbb{C}$-manifold M such that as a topological bundle, it is flat (i.e. there is some connection, not necessarily compatible with $\langle\cdot,\cdot\rangle$ or $\bar{\partial}$, whose curvature vanishes).
Is it true that the curvature of the Chern connection necessarily vanishes?
In the affine algebraic case there is always an algebraic connection: Let $A$ be a commutative unital ring and let $E$ be a finite rank projective $A$-module. There is the Atiyah sequence
$$ 0 \rightarrow \Omega^1_A \otimes E \rightarrow J^1(E) \rightarrow^{\pi} E \rightarrow 0$$
and since $E$ is projective and $\pi$ surjective, it follows $\pi$ has an $A$-linear split $s: E \rightarrow J^1(E)$.The splitting $s$ corresponds to a connection
$$\nabla: E \rightarrow \Omega^1_A \otimes E$$
and $\nabla$ is "seldom" a flat algebraic connection. There is an explicit formula for the curvature $R_{\nabla}$ of a connection/covariant derivative $\nabla$: Let
$$p: A^n \rightarrow E$$
be a surjection and let $u_1,..,u_n$ be a basis for $F:=A^n$ as $A$-module. Let $s$ be an $A$-linear splitting of $p$
Let $\phi:=s \circ p: F \rightarrow F$ be the idempotent of $E$ corresponding to $p,s$ and let $\phi:=(a_{ij})$ with $a_{ij}\in A$ be the matrix of $\phi$ in the basis $u_i$. Define for any pair $z,x\in \operatorname{Der}(A)$ the matrix
$$L1.\text{ }M:= [(z(a_{ij})), (x(a_{ij}))] \in A^{n \times n}\cong \operatorname{End}_A(F).$$
You may prove that $M $ induce a map $M^*\in \operatorname{End}_A(E)$ and the following formula holds:
Theorem 1: $R_{\nabla}(z,x):= M^* \in \operatorname{End}_A(E)$.
From formula L1 it follows $\nabla$ is seldom a flat connection. Hence in the affine situation it is easy to give explicit examples of non-flat algebraic connections: You must calculate a splitting $s$ of $p$ and the idempotent element $\phi$.
In the projective case it follows $E$ may not have a connection - there is always a cohomology class $a(E)$ which is zero iff $E$ has a connection. If $X$ is a complex projective algebraic manifold, there is a correspondence between flat connections on finite rank vector bundles on $X$ and finite dimensional complex representations of the topological fundamental group $\pi(X)$ of $X$. Given a flat algebraic connection $(E,\nabla)$ on $X$ it follows $E^{\nabla}$ is a local system of finite dimensional complex vector spaces on $X$, and to $E^{\nabla}$ you get a finite dimensional complex representation
$$\rho: \pi(X) \rightarrow GL(V).$$
This correspondence (the "Riemann-Hilbert correspondence") is an "equivalence of categories" in an appropriate sense (this is "vague"). Hence you consider the category of pairs $(E,\nabla)$ where $E$ is a finite rank vector bundle on $X$ with a flat connection $\nabla$, and "maps of connections". You also consider the category of finite dimensional complex representations of $\pi(X)$ and maps of representations. Hence there is "as many" flat connections as representations of the topological fundamental group. This is a well developed theory (originating in some papers of Weil I believe, many people have contributed to this study) from the 40s and 50s. In the below mentioned book you will find this further developed in the framework of "holonomic D-modules" - this is a well developed theory. The book also gives many references.
In the affine situation there is always a space
$$C:=\operatorname{Hom}_A(\operatorname{Der}(A), \operatorname{End}_A(E))$$
of connections, and if $\Omega^1_A$ is a finite rank projective $A$-module you get
$$ C\cong \Omega^1_A \otimes \operatorname{End}_A(E),$$
hence the "parameter-space of connections" is a finite rank vector bundle on $A$. Hence given a connection $\nabla$ with curvature given by Theorem 1, you may add a potential $\phi \in C$ to get a new connection $\overline{\nabla}:=\nabla + \phi$. Hence if you want to study the problem if $E$ has a flat algebraic connection you must study the "moduli space" $C$. Similar for $X$ - this again is a well devleoped theory - "moduli spaces of connections". If you consider the "set of potentials" $\phi \in C$ with the property that the curvature is zero
$$L2.\text{ }R_{\overline{\nabla}}=0$$
you get a subvariety $M^{fl}(E) \subseteq \mathbb{V}(C^*)$ parametrizing flat connections on $E$, and in high dimension and low rank, the system of equations defining $M^{fl}(E)$ will be "overdetermined". Hence it is not clear if $E$ has a flat algebraic connection in general. Hence in the affine algebraic situation there is always a connection which is non-flat in general by formula L1, and it is an open problem to determine if $E$ has a flat algebraic connection. You must study the subvariety you get from equation L2.
Example: If you let $n:=dim(A)\geq 10$ and $rk(E)=2$ you get an overdetermined system of equations defining $M^{fl}(E)$, hence in this case you may get an empty moduli space
$$M^{fl}(E)=\emptyset.$$
For such $E$ you always have a non-flat connection from equation L1 and Theorem 1. The system defining $M^{fl}(E)$ has $\binom{n}{2}$ equations and $rk(\Omega^1\otimes \operatorname{End}(E))=4n$. For $n>>0$ it follows this system does not "have a solution".
Question: "What about (1) ⟹ (2) and (3) ⟹ (4)?"
Answer: I believe it is an old conjecture that if $E$ has an algebraic/holomorphic connection, then $E$ has a flat algebraic /holomorphic connection, but I do not have a precise reference (maybe the paper of Atiyah from 1956 in TrAMS).
You'll find this statement
Citation: "Non-flat algebraic connections for bundles on complex projective manifolds are virtually non-existent (we know of none)"
in the introduction of
Note 1: In Borel's book page 226 you find the following construction: If $X \subseteq \mathbb{P}^n_{\mathbb{C}}$ is a smooth quasi projective algebraic variety and if $\omega^1_X$ is the canonical bundle of $X$, it follows the Lie derivative induce a right $D_X$-module structure on $\omega^1_X$. This does not imply there is a flat algebraic connection
$$\nabla: \omega^1_X \rightarrow \Omega^1_{X}\otimes \omega^1_X.$$
Since $D_X$ is a sheaf of non-commutative rings, there is no obvious relation between left $D_X$-modules and right $D_X$-modules. From a right $D_X$-module $E$ we get a left $D_X$-module $\omega^{-1}_X \otimes E$ and to the canonical bundle $\omega^1_X$ we get the trivial bundle $\mathcal{O}_X$. To a left $D_X$-module we get canonically a flat connection.
Note 2: If $X$ is a complex projective manifold and $E$ is an indecomposable finite rank vector bundle on $X$, it follows $\Gamma(X,\operatorname{End}(E))$ is a finite dimensional algebra over $\mathbb{C}$.
Borel, A. Algebraic D-modules. Perspectives in Mathematics, Vol. 2, Boston etc.: Academic Press, Inc., Harcourt Brace Jovanovich, Publishers. xii, 355 p.; $ 29.95; £25.00 (1987).
Note 3: There is the following general result: If $(L,a)$ is a Lie-Rinehart algebra and $(E,\nabla)$ is a connection, there is a non-abelian extension (the non-abelian Atiyah extension)
$$N1.\text{ } 0 \rightarrow \operatorname{End}_A(E) \rightarrow L(E,\nabla) \rightarrow L \rightarrow 0 $$
of Lie-Rinehart algebras which splits iff $E$ has a flat connection. There is a "cohomology set" $\operatorname{Ext}^1(L,(E,\nabla))$ classifying such non-abelian extensions. By definition
$$L(E,\nabla):=\operatorname{End}_A(E)\oplus L$$
with the following Lie product:
$$[(\phi,x),(\psi,y)]:=([\nabla(x),\psi]-[\nabla(y), \phi]+R_{\nabla}(x,y),[x,y])$$
for all $\phi, \psi\in \operatorname{End}_A(E)$ and $x,y\in L$. You must check that the sequence $N1$ splits iff there is an $A$-linear map $P: L \rightarrow \operatorname{End}_A(E)$ such that $\overline{\nabla}:=\nabla+P$ is a flat connection. A similar construction exists globally.
Hence given a holomorphic vector bundle $E$ on a complex projective manifold $X$, there are two obstructions: The Atiyah class $a(E)$ which is zero iff $E$ has a holomorphic (or algebraic) connection. The extension class $n(E)$ given by sequence $N1$ is zero iff $E$ has a flat holomorphic connection. The sequence $N1$ exists whenever $E$ has a holomorphic (or algebraic) connection.
Example: If $\mathbb{P}^n_k$ is complex projective space and if $\mathcal{O}(d)$ is the invertible sheaf with $d \geq 1$, it follows the Atiyah sequence
$$0 \rightarrow \Omega^1 \otimes \mathcal{O}(d) \rightarrow J^1(\mathcal{O}(d)) \rightarrow \mathcal{O}(d) \rightarrow 0$$
does not split, hence $\mathcal{O}(d)$ does not have a holomorphic (or algebraic) connection.
Any finite rank projective $A$-module $E$ with a non-zero class in $\operatorname{H}^2_{DR}(A)$ will have an algebraic connection and no flat algebraic connection. Hence the Atiyah sequence splits, but the sequence $N1$ does not split. I belive there is an explicit example in Loday's book "Cyclic Homology".
I will write in terms of the holomorphic frame bundle, i.e. the bundle of choices of complex linear bases of tangent spaces, a holomorphic principal $\operatorname{GL}_n$-bundle. Any sum $\gamma=\sum h_a \gamma_a$ with $\sum h_a=1$ of $(1,0)$-connection forms on the holomorphic frame bundle is a $(1,0)$-connection. If all $\gamma_a$ are torsion-free, so is $\gamma$. To see this, take the soldering forms $\omega=(\omega^{\mu})$ on the frame bundle, and then the structure equations of a connection are $d\omega=-\gamma\wedge\omega+\frac{1}{2}a\omega\wedge\omega+b\omega\wedge\bar\omega+c\bar\omega\wedge\bar\omega$, with $a,b,c$ the components of the torsion. (Note that $c$ is the Nijenhuis tensor, so $c=0$ on a complex manifold.) If these vanish for all of the $\gamma_a$, they still vanish for $\gamma$. Therefore if the $h_a$ form a partition of unity, for local choices of torsion-free $(1,0)$-connections, then $\gamma$ is a torsion-free $(1,0)$-connection.
Best Answer
The answer is yes if $M$ is compact Kähler (EDIT: and you allow a conformal change in the metric) and $L$ is a line bundle and no in general, already in the case of line bundles.
Take for example $M$ to be the standard Hopf surface $(\mathbb{C}^2\backslash \{(0,0)\})/[(z_1,z_2)\sim (2z_1,2z_2)]$. Then $M$ is diffeomorphic to $S^1\times S^3$ so every line bundle on $M$ has first Chern class zero, therefore it is topologically trivial and it admits a flat connection.
Let us now see that the canonical bundle $K_M$ is not Chern flat (for any Hermitian connection). First of all, write down the explicit Hermitian metric on $M$
$$g_{i\overline{j}}=\frac{\delta_{ij}}{r^2}, r^2=|z_1|^2+|z_2|^2.$$
It induces a Hermitian metric on $K_M$ whose curvature is the first Chern form which can be easily calculated in local coordinates
$$\alpha=-i\partial\overline{\partial}\log\det (g_{i\overline{j}})=\frac{2}{r^2}\left(\delta_{kl}-\frac{\overline{z}_k z_l}{r^2}\right)i dz_k\wedge d\overline{z}_\ell,$$
and $\alpha$ is a nonnegative-definite Hermitian form, which is not identically zero. If you had another Hermitian connection on $K_M$ with zero first Chern form, by the transgression formula it would imply that $\alpha=i\partial\overline{\partial}u$ so $u$ would be a global plurisubharmonic function on $M$, which must be constant by the maximum principle, and you'd get that $\alpha=0$ which is false.
However, if $M$ is compact Kähler then the $\partial\overline{\partial}$-lemma shows that topologically trivial holomorphic line bundles admit flat Hermitian connections. Just start with any Hermitian connection, its Chern curvature form will be $\alpha$, say, which is cohomologous to zero, so $\alpha=i\partial\overline{\partial}u$ by the $\partial\overline{\partial}$-lemma. Then conformally scaling the Hermitian metric by $e^u$ gives you a new Hermitian connection with vanishing Chern curvature.
For a higher rank holomorphic bundle $E$ over a compact Kähler manifold $(M,\omega)$ you have that $E$ admits a flat Hermitian metric iff it has a Hermitian-Yang-Mills connection $H$ with $\omega^{n-1}\wedge F_H=0$ and its second Chern class vanishes, iff it is $[\omega]$-polystable and its first and second Chern classes vanish. This is essentially the Donaldson-Uhlenbeck-Yau Theorem.
So if $E$ admits a flat connection its Chern classes indeed vanish, so it will admit a flat Hermitian metric iff it is $[\omega]$-polystable. I think there should be plenty of examples of such bundles which are not polystable.