I'm finding myself a little confused about Koszul-Malgrange holomorphic structures in a certain context.

Suppose $M$ is a complex manifold, $N$ is a smooth manifold with a smooth complex vector bundle $V$ and bundle connection $\nabla$, and $f:M\to N$ is a smooth map. We form the pullback bundle $f^*V$ over $M$ and give it the pullback connection $f^*\nabla$. Let $P: f^*V \to V$ be the associated smooth projection map. Assuming $f^*\nabla^{0,1}\cdot f^*\nabla^{0,1}=0$, a theorem of Koszul-Malgrange says there is a holomorphic structure on $f^*V$ that turns it into a holomorphic vector bundle over $M$.

My question is about the following. Suppose we have a (possibly locally defined) holomorphic section $s$ of $f^*V$. In a suitable trivialization over some open set $\Omega$, so that $f^*V$ splits as $\Omega\times \mathbb{C}^n$, we can write $$s(z) = (z, h(z))$$ where $h:\Omega\to \mathbb{C}^n$ is holomorphic.

Now, consider the map $$P\circ s: \Omega \to V$$ I'm basically wondering if it makes sense to write $P\circ s =(f(z), h(z))$ for some choice of trivialization of $V$. The sort-of roadblock that I'm hitting is that the trivialization of the pullback we used should come from the Koszul-Malgrange theorem, rather than be pullback of a trivialization of $V$. So, I'm finding it hard to fiddle the definition of the pullback bundle.

I'm mainly interested in the case where $\dim M \leq \dim N$ and $f$ is regular everywhere. Then in my trivizalization over $\Omega$, the map $f_*$ should be an isomorphism of $f^*V|_{\Omega}\to V|_{f(\Omega)}$ that acts by $(z,v)\mapsto (f(z),v)$. Of course, this induces a trivialization over $f(\Omega)$. Then we can find an open set $U$ containing $f(\Omega)$ in which we have a smooth trivialization that extends our trivialization over $f(\Omega)$. I think that in this trivialization, $P\circ s$ should assume the desired form.


1 Answer 1


Okay I thought about it a bit more, and here is what I can say in general. We can find a neighbourhood $U\subset N$ such that $V|_U$ splits diffeomorphically as $U\times \mathbb{C}^n$ and the section $s$ takes the form $s(z) = (f(z), A(z)\circ h(z))$, where $A(z)$ is a smoothly varying family of invertible $n\times n$ complex matrices.

Let $(U,\varphi)$ be a trivialization of $V$ over some open set $U\subset N$ and $\Omega\subset M$ such that $f(\Omega) \subset U$. In $\Omega$ we have the pullback trivialization $\psi$, and in these trivializations the map $P$ takes $\Omega\times \mathbb{C}^n\to U\times \mathbb{C}^n$ via $(z,v)\mapsto (f(z),v)$. Meanwhile, according to Koszul-Malgrange (by the way, the easiest exposition I've found is in here http://gtnmk.droppages.com/2019/250B-2019-set.pdf), then for $p\in \Omega$ we can find $\Omega_p\subset \Omega$ and a holomorphic trivialization $k:f^*V|_{\Omega_p}\to \Omega_p \times \mathbb{C}^n$. Let us just restrict $\Omega$ so that $\Omega_p=\Omega$. This should make the notation easier.

If $s(z) = (z,b(z))$ with respect to the trivialization $\psi$, then $P\circ s= (f(z),b(z))$ with respect to the trivializations $\psi,\varphi$. Under $k$, $s$ assumes the form $s(z) = (z,h(z))$, with $h$ holomorphic. The relation between $b$ and $h$ is as follows. $\varphi$ and $k$ are trivializations, so in coordinates, we can write $$\varphi\circ k^{-1}(z,v) = (z, A(z)v)$$ for some smoothly varying family of invertible matrices $A(z)$. $A(z)$ then satisfies $$b(z) = A(z) \circ h(z)$$ which gives the result.


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