Given a holomorphic function $f:\Omega\subset\mathbb{C}^n\rightarrow\mathbb{C}$,with $\Omega$ open and connected, when is it possible to find a holomorphic $m-$primitive, i.e. a holomorphic function (in every variable) $g:\Omega\rightarrow\mathbb{C}$ such that \begin{equation} f=\dfrac{\partial g}{\partial z_m}? \end{equation} Maybe, it is simply a consequence of the 1-variable version of the problem of finding a primitive of a holomorphic function, where it is enough to assume $\Omega$ simply connected, so in the several variables version, if one define \begin{equation} \Omega_m:=\{w\in\mathbb{C}\mid (z_1,...,z_{m-1},w,z_{m+1},...,z_n)\in\Omega\} \end{equation} then if $\Omega_m$ is simply connected, such an $m-$primitive $g$ exists? Can someone give me a rather formal proof? Thank you in advance.
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1$\begingroup$ Can you say what exactly is $\Omega$: an open set, a connected open set, a simply-connected open set...? (You mention simply-connected, but it's not entirely clear to me if that's an additional condition, or if you're assuming the slice in the $z_m$-axis is simply-connected, or what.) $\endgroup$Andrew D. Hwang– Andrew D. Hwang2022-01-12 15:27:42 +00:00Commented Jan 12, 2022 at 15:27
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1$\begingroup$ Interestingly, it's enough to assume $\Omega$ is simply-connected by your argument, where we allow deformations of path in $\Omega$, not merely in the slice $\Omega_{m}$. (Incidentally, this observation is related to Hartogs' phenomenon: If $\Delta$ is a polydisk, $K\subset \Delta$ is compact, and $f$ is holomorphic in $\Delta \setminus K$, then $f$ has a holomorphic extension to $\Delta$.) $\endgroup$Andrew D. Hwang– Andrew D. Hwang2022-01-12 17:27:30 +00:00Commented Jan 12, 2022 at 17:27
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1$\begingroup$ The geometric idea I had in mind is, there is a smaller polydisk $\Delta'$ containing $K$ and compactly contained in $\Delta$. The difference $\Delta \setminus \overline{\Delta'}$ is topologically a punctured $2n$ ball, so is simply-connected for $n > 1$. For Hartogs, we're using a $1$-form integrand from the Cauchy integral formula rather than $f(\dots, w, \dots)\, dw$, but either way we have a closed $1$-form, which is exact because the domain is simply-connected. $\endgroup$Andrew D. Hwang– Andrew D. Hwang2022-01-12 18:15:02 +00:00Commented Jan 12, 2022 at 18:15
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1$\begingroup$ I have a question for what concerns allowing deformations of path out of the slice $\Omega_m$, in which the other variables are not freezed anymore: does it holds a Cauchy theorem (namely, the integral of a holomorphic function over a closed path in a simply connected domain is always zero) in several variables? If it's not so, how could we let deformations out of $\Omega$? $\endgroup$Jules Binet– Jules Binet2022-01-13 09:41:35 +00:00Commented Jan 13, 2022 at 9:41
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1$\begingroup$ A holomorphic $1$-form $\eta$ in a domain is closed (locally, there exists a holomorphic function $f$ such that $\eta = \partial f = df$), so Stokes' theorem guarantees the integral of $\eta$ over a path is invariant under homotopy. :) $\endgroup$Andrew D. Hwang– Andrew D. Hwang2022-01-13 13:53:18 +00:00Commented Jan 13, 2022 at 13:53
1 Answer
My attempt.
Define $g:\Omega\rightarrow\mathbb{C}$ by
\begin{equation}
g(z_1,...,z_n):=\int_{z^0_m}^{z_m}f(z_1,...,z_{m-1},w,z_{m+1},...,z_n)dw,
\end{equation}
where $z^0=(z^0_1,...,z^0_n)$ is any point in $\Omega$.
Note that $g$ is well defined since it does not depend on the choice of the path between $z^0_m$ and $z_m$. Indeed, let $\gamma_1$ and $\gamma_2$ any two paths between $z^0_m$ and $z_m$, then $\gamma_1\cup -\gamma_2$ is closed and since $\Omega_m$ is simply connected, Cauchy's Theorem applied to the holomorphic function $f$ guarantees that $$\int_{\gamma_1}f-\int_{\gamma_2}f=0.$$
Now, by the fundamental theorem of calculus it holds $\dfrac{\partial g}{\partial x_m}=f$, moreover $g$ is holomorphic, since
$$\dfrac{\partial g}{\partial{\bar{z}_j}}=\int_{z^0_m}^{z_m}\dfrac{\partial f}{\partial{\bar{z}_j}}dw=0.$$