What is the geometric intuition for the $\bar \partial$-Poincare lemma, or for $\bar \partial$ more generally?

The one variable $\bar \partial$-Poincare lemma is proven in Huybrechts and Forster and so on in essentially the same way: one shows that for a local form $f d \bar z$, with $$g(z) := \frac{1}{2\pi i} \int_{B_\varepsilon} \frac{f(w)}{w-z} dw \wedge d\bar w$$ we have $\bar \partial g = fd\bar z$. I can follow the proof on a formal, line by line level, but the manipulations used to obtain the result really don't have any geometric meaning to me, partially because I really don't understand how I should think of $\bar \partial$ except as "some operator with kernel the holomorphic functions, and which we really really want to be locally exact so that we can use sheaf cohomology." My questions here are: is there any intuitive reason the lemma would be true at all, and why should this integral, which suspiciously resembles that of the Cauchy integral formula, be the correct $g$, and is there more geometric intuition for the $\bar \partial$ operator generally than 'it kills holomorphic functions'?

$\def\dbar{\overline{\partial}}\def\zbar{\overline{z}}$This isn't geometric, but here is a non-rigorous argument suggesting this formula that I gave when I taught this lemma.

Consider the case where $g$ is the $\delta$ function $\delta_w$ at $w$. Can we find a smooth function $f$ such that, in some sense, $\tfrac{\partial f}{\partial \zbar} = \delta_w$? In other words, for any disc $D$, we want $$\int_D \frac{\partial f}{\partial \zbar} d(\mathrm{Area}) = \begin{cases} 1 & w \in D \\ 0 & w \not\in D \end{cases}.$$ The wedge product $d z \wedge d \zbar$ is $2i$ times the area form, so we want $$\int_D \frac{\partial f}{\partial \zbar} dz \wedge d \zbar = \begin{cases} 2 i & w \in D \\ 0 & w \not\in D \end{cases}.$$ Since $df = \tfrac{\partial f}{\partial z} dz + \tfrac{\partial f}{\partial \zbar} d\zbar$, we can rewrite the left hand side as $$- \int_D df \wedge dz = - \int_D d \left(f dz \right) =- \int_{\partial D} f dz.$$ So we want a smooth function $f$ such that $$-\int_{\partial D} f dz = \begin{cases} 1 & w \in D \\ 0 & w \not\in D \end{cases}.$$ The obvious choice is $f(z) = \tfrac{1}{\pi (w-z)}$.

Now, consider a general function $g(z)$. We have $$g(z) = \int_w g(w) \delta_w(z) \ d(\mathrm{Area}).$$ Since $\dbar$ is linear and we "have" $\tfrac{\partial}{\partial \zbar} \tfrac{1}{\pi(w-z)} = \delta_w(z)$, we should have $$f(z) = \frac{\partial}{\partial \zbar} \int_w g(w) \frac{1}{\pi(w-z)} \ d(\mathrm{Area}).$$

The last formula is correct for any compactly supported smooth $g$ and is the formula you ask for, up to the relation $dw \wedge d \overline{w} = (2 i) d (\mathrm{Area})$.

PS: You might wonder what happens if we replace $\tfrac{1}{\pi (w-z)}$ by some other meromorphic $h(w,z)$ with a pole of residue $\tfrac{-1}{\pi}$ at $w$ and no other poles. If $h$ blows up to higher order at $w$, then $\int_w g(w) h(w,z) d(\mathrm{Area})$ won't converge absolutely even for compactly supported $g$, so we will have to specify what the integral means more carefully, which is a pain. So we want $h(w,z)$ to be $\tfrac{1}{\pi (w-z)}$ plus an entire function of $z$. If that entire function is nonzero, the result is still right for any compactly supported $g$. Making the simplest choice $\tfrac{1}{\pi (w-z)}$ rather than, say $\tfrac{1}{\pi (w-z)} + e^{wz}$, has the advantage of making the integral still converge if $|g(z)| = O(1/|z|^{1+\epsilon})$. I don't think that is a big advantage though; we still eventually want to prove the Dolbeault lemma for all $g$ and, as far as I know, that still needs a partition of unity argument, which I did in my next lecture.

This isn't geometric, but the $\bar{\partial}$-Poincare lemma $$ g({\rm Re}(z),{\rm Im}(z)) ~:=~ \frac{1}{\pi} \iint_{B_\varepsilon} \frac{f({\rm Re}(w),{\rm Im}(w))}{z-w} \mathrm{d}{\rm Re}(w) \wedge \mathrm{d}{\rm Im}(w)$$ $$\qquad\Rightarrow\qquad \frac{\partial g({\rm Re}(z),{\rm Im}(z))}{\partial \bar{z}}~=~f({\rm Re}(z),{\rm Im}(z)) \tag{1}$$ is tied to the distributional identity $$ \frac{\partial}{\partial \bar{z}}\frac{1}{z-w}~=~\pi~\delta({\rm Re} (z\!-\!w))~\delta({\rm Im} (z\!-\!w))~=~\lim_{\varepsilon\searrow 0^+} \frac{\varepsilon}{(|z-w|^2+\varepsilon)^2} \tag{2} $$ for the 2D Dirac delta distribution. The second equality in eq. (2) is a well-known representation of the 2D Dirac delta as a generalized function. (Its proof is very similar to the 1D case.) The first equality in eq. (2) can be understood intuitively via regularization $$\frac{1}{z-w}~=~\lim_{\varepsilon\searrow 0^+}\frac{\bar{z}-\bar{w}}{|z-w|^2+\varepsilon},\tag{3}$$ of the singular term by a smooth function. Differentiation $$ \frac{\partial}{\partial \bar{z}}\frac{\bar{z}-\bar{w}}{|z-w|^2+\varepsilon}~=~\frac{\varepsilon}{(|z-w|^2+\varepsilon)^2} \tag{4}$$ yields the rhs. of eq. (2). $\Box$

It might be possible to gain some intuition for the $\partial/\partial \bar{z}$-derivative by considering the physics of light. Light is a transverse wave and monochromatic light then comes with a polarisation. A general monochromatic wave is an elliptic wave. Such a light wave can be thought to be composed of a horisontal polarisation part and a vertical polarisation part. Each of these polarisation parts has an amplitude and a phase so we encode the light wave with two complex numbers, say $a$ and $b$ for the horisontal and vertical polarisation respectively. When the two polarisation parts are added together we get the general elliptic wave.

We can also split the light wave in other polarisation parts. The linear polarisation planes may not necessarily be in the horisontal/vertical directions. The polarisation planes are also commonly taken to be tilted by $\pi/4$, and in fact any tilting of the polarisation planes for some angle between $0$ and $\pi$ makes a valid polarisation decomposition. But there are more ways to decompose the polarisation. The light wave can also be split into circular polarisation parts. The light wave is then thought to be composed of a right-handed circular wave and left-handed circular wave. Each of these circular waves has an amplitude and a phase so we can also encode the light wave with two other complex numbers, say $c_+$ and $c_-$ for the right-handed and left-handed parts respectively.

Now consider an $\mathbf{R}$-differentiable function $f:\mathbf{C}\rightarrow\mathbf{C}$, which also can be the thought of as a differentiable function $f:\mathbf{R}^2\rightarrow\mathbf{R}^2$. The differential $df: \mathbf{R}^2\rightarrow\mathbf{R}^2$ is a linear function and for example maps a circle to a general ellipse. When we express the differential as $df=f_xdx+f_ydy$ this gives us a decomposition of the differential into horisontal/vertical polarisation parts. The partial derivatives are $\mathbf{R}^2$-valued which we also consider as complex numbers. So we get at each point two complex numbers $f_x$ and $f_y$ that encodes the horisontal/vertical polarisation parts of $df$.

What about circular polarisation decomposition of $df$? To compute the circular polarisation parts of $df$ we can calculate the average change of $f$ over a circle when compared to a same-oriented respectively anti-oriented circle with small radius and let this radius tend to zero. The partial derivatives $\partial/\partial z$ and $\partial/\partial\bar{z}$ as circular polarisation computation operators are then defined as $$ \frac{\partial f}{\partial z}=\lim_{r\rightarrow 0}\frac{1}{2\pi}\int_0^{2\pi}\frac{f(z+re^{i\theta})-f(z)}{re^{i\theta}}d\theta $$ $$ \frac{\partial f}{\partial\bar{z}}=\lim_{r\rightarrow 0}\frac{1}{2\pi}\int_0^{2\pi}\frac{f(z+re^{i\theta})-f(z)}{re^{-i\theta}}d\theta $$ A quick calculation gives that $\partial z/\partial z=1$, $\partial \bar{z}/\partial z=0$, $\partial z/\partial\bar{z}=0$ and $\partial\bar{z}/\partial\bar{z}=1$ so these definitions agree with the usual definitions of $\partial/\partial z$ and $\partial/\partial\bar{z}$ given by \begin{align*} \frac{\partial}{\partial z}&=\frac{1}{2}\left(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y}\right)\\ \frac{\partial}{\partial\bar{z}}&=\frac{1}{2}\left(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\right) \end{align*} The usual definitions can be recognised as the Jones vectors of circular polarisation (with an extra factor $\frac{1}{\sqrt{2}}$). The Jones vectors of different polarisation states are unit $\mathbf{C}^2$-vectors and form together $S^3$. If we consider the polarisation states differing only by an overall phase factor as equal we get a projection from $S^3$ to $S^2$. This $S^2$ is the Poincaré sphere.

The circular polarisation decomposition of $df$ is then given by $$ df=f_zdz+f_{\bar{z}}d\bar{z} $$ For a holomorphic function $f_{\bar{z}}=0$ and the differential is $df=f_zdz$, meaning that a tangent space circle is mapped to a circle magnified a factor $|f_z|$ and turned an angle $\arg f_z$. This is in some text books called the "amplitwist" property of a holomorphic funtions. For a general function, not necessarily holomorphic, an amplitwisted anti-clockwise oriented circle is added with an amplitwisted clockwise oriented circle and this gives an ellipse.