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In complex analysis, we are taught that instead of coordinates $x$, $y$ on the complex plane, one can use $z$, $\bar{z}$, then, for instance, the Cauchy-Riemann conditions become $\frac{\partial }{\partial \bar{z}}f(z, \bar{z})=0$, and $\Delta = \frac{1}{4}\frac{\partial^2}{\partial z \partial \bar{z}}$. This was explained to me in the following way: we simply perform a change of coordinates $z = x + y$, $\bar{z}=x-y$ on $\mathbb{R}^2$. However, if this was the case, functions with $\frac{\partial }{\partial \bar{z}}=0$ would be constant along lines $y=-x+C$, which is clearly not true (for instance, this would mean isolated zeros or isolated singularities are impossible). In general, this means that $\partial_{\bar{z}}$ can't be understood as a directional derivative in the complex plane, i.e. as a vector field on the complex plane with real coefficients, because holomorphic functions would then be constant along the integral lines of this vector field. So, I guess, either the derivative or the change of coordinates should be understood in some other sense. Reading old MSE answers to similar questions, I understood that

a) This confusion about the meaning of $\partial_{\bar{z}}$ is very common among people studying complex analysis, including, apparently, those who answer questions about Wirtinger derivatives (because often the answers boil down to treating $\partial_{z}$ and $\partial_{\bar{z}}$ as real vector fields).

b) The correct answer (as opposed to hand-waving the issue) has something to do with the structure of complex manifold on $\mathbb{C}$.

So my question is: where can I find an elementary introduction to the theory of complex manifolds that would carefully explain how $\partial_{z}$ and $\partial_{\bar{z}}$ fields work?

Edit: Perhaps, I should explain that my naive understanding of complex manifolds leads to further confusions. For instance, $\mathbb{C}$ is a one-dimensional complex manifold. The experience teaches me it should have one linearly independent vector at each tangent space, not two, which raises more questions about what $\partial_{z}$ and $\partial_{\bar{z}}$ really are.

  • I believe your change should be $z=x+iy$ and $\bar z=x-iy$. – GReyes Jan 26 '24 at 14:49
  • This coordinate change is not valid in $\mathbb{R}^2$, so it becomes unclear, what is, exactly, the geometric structure we are working with. – Daigaku no Baku Jan 26 '24 at 14:53
  • Regarding your edit, when thinking of $\mathbb C$ as a one-dimensional complex manifold, its tangent space at a particular point is a one-dimensional vector space *over the complex numbers $\mathbb C$*. So yes, that tangent space cannot have two linearly independent vectors over $\mathbb C$. – Lee Mosher Jan 26 '24 at 15:11
  • Then what are $\partial_{z}$ and $\partial_{\bar{z}}$ geometrically? – Daigaku no Baku Jan 26 '24 at 15:18
  • @DaigakunoBaku I agree with you that my posting doesn't completely answer the posted question. Actually, I am not qualified to answer this question. So, I have deleted my answer. – user2661923 Jan 26 '24 at 15:36
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    I am quite sure this was discussed before on MSE. A brief answer is that these are really sections of the complexified tangent bundle and not of the tangent bundle itself. – Moishe Kohan Jan 26 '24 at 15:45
  • The question itself was discussed multiple times, as I noted in the post, but all the answers were "duh, this is just a change of coordinates", and then the usual computation which shows that, given these derivatives work as expected, they work as expected. I was able to find one answer that mentioned that one really needs to consider the structure of the complex tangent bundle, without further details. What I am asking about is not "how can I perform the informal calculation that is used to loosely show that this works", but "how do I strictly define the geometric structure... – Daigaku no Baku Jan 26 '24 at 15:53
  • ... we are working with, and how do I derive from its definition the properties of $\partial_{z}$ and $\partial_{\bar{z}}$" – Daigaku no Baku Jan 26 '24 at 15:54
  • The $\partial_{z_i}$ locally span the holomorphic tangent bundle $TM$, and $\partial_{\bar{z}_i}$ locally span the antiholomorphic tangent bundle $\bar{TM}$. Both of these can be viewed as the "eigen-bundles of the complex structure $J$ acting on the complexification of the tangent bundle viewed as a real $2n$ vector bundle. – Chris Jan 26 '24 at 16:44
  • The coordinate change in question is perfectly valid on $\mathbb{R}^2$. When identifying $\mathbb{C}$ with $\mathbb{R}^2$, we take $1$ to be $(1,0)$ and $i$ to be $(0,1)$. So @GReyes is correct. – Quaere Verum Jan 26 '24 at 19:46
  • My answer here might be helpful to resolve the issue appearing in the last paragraph of your question. As for a reference, I suggest Griffiths and Harris "Foundations of algebraic geometry." – Moishe Kohan Jan 26 '24 at 20:04
  • Quare Verum, again, if this was the coordinate change in $\mathbb{R}^2$ you describe, isolated zeros of holomorphic functions would not exist. – Daigaku no Baku Jan 27 '24 at 15:04
  • Chris, thank you, where can I learn more? – Daigaku no Baku Jan 27 '24 at 15:05
  • It's true by definition that $z=x+iy$ and $\overline{z}=x-iy$. What you wrote is $z=x+y$ and $\overline{z}=x-y$. It should be rather obvious that this is incorrect. – Quaere Verum Jan 27 '24 at 15:14
  • The coordinate change $z=x+iy$, $\bar{z}=x-iy$ is undefined in $\mathbb{R}^2$. If you identify $i$ with unit vector along $y$, and $1$ with unit vector along $x$, then the coordinate change you describe is precisely $z=x+y$, $\bar{z}=x-y$, with real coefficients. If you use complex coefficients, this means you are not working in $\mathbb{R}^2$. – Daigaku no Baku Jan 27 '24 at 15:17
  • You seem to have some fundamental misunderstanding of what $x$ and $y$ are. They are the coordinate functions. Coordinates on $\mathbb{R}^2$ are given by $(x,y)$, i.e. $x\cdot(1,0)+y\cdot(0,1)$. The coordinate change $z=x+y$ and $\overline{z}=x-y$ would be a linear coordinate change on $\mathbb{R}^2$ where $z$ and $\bar{z}$ are now the coordinate functions. That's clearly not what they are; $z$ and its conjugate are complex valued, not real valued. So no, $z=x+iy$ and this is the same as stating $(x,y)=x\cdot(1,0)+y\cdot(0,1)$. It's really a change of notation, not coordinates. – Quaere Verum Jan 27 '24 at 15:32
  • If $z=x+i y$ was the same as stating $(x, y)=x\cdot (1, 0) + y \cdot (0, 1)$, then the corresponding coordinate change would be $z = x + y$ and $\bar{z}=x - y$, with new basis vectors $\frac{1}{\sqrt{2}}(1+i)$ and $\frac{1}{\sqrt{2}}(1-i)$. And this would mean $\partial_{\bar{z}}=0$ corresponds to the rate of change of a function being zero in the direction of the basis vector $\frac{1}{\sqrt{2}}(1-i)$, which would mean no zero of a holomorphic function is isolated. This is not the case exactly because the change of coordinates in question is not really equivalent to that in $\mathbb{R}^2$. – Daigaku no Baku Jan 27 '24 at 15:38
  • Formulate the Cauchy-Riemann conditions in $(x, y)$ coordinates and then trace what change of coordinates $(x,y)$ is needed to transform them into $\partial_{\bar{z}}=0$. – Daigaku no Baku Jan 27 '24 at 15:41
  • I'll just repeat one last time: we have $z:\mathbb{R}^2\to\mathbb{C}$. It is a complex-valued function, if you will. The pair $(z,\overline{z})$ isn't a pair of different real coordinates for the complex plane. The coordinate $z$ already parameterises the entire complex plane. By your logic, $z$ would parameterise a $1$-dimensional real subspace. As such, it should be self-evident that the equations you wrote down are incorrect. – Quaere Verum Jan 27 '24 at 17:23

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$\newcommand{\dd}{\partial}\newcommand{\Cpx}{\mathbf{C}}$tl; dr: Holomorphic vector fields don't literally have a geometric interpretation the way real vector fields do. Instead, they're complex-valued differential operators, or complex linear combinations of real vector fields (their real and imaginary parts).

That said, we can view a holomorphic vector field geometrically as twice its real part.


Let's carefully examine the complex line equipped with its standard holomorphic coordinate $z$, its complex structure $J$ given by "multiplication by $i$ at each point," and its holomorphic coordinate vector vield $\dd_{z}$.

One might imagine a world where these are primitive notions, where we do calculus without decomposing a complex number into real and imaginary parts, where we're blissfully unaware of complex conjugation (!), where differentiable means holomorphic. In this world, the "tangent bundle" of $\Cpx$ may be identified with $\Cpx \times \Cpx$. A point $(z, a)$ in the Cartesian product is viewed as the vector $a\, \dd_{z}$ at the point $z$. A holomorphic function $f$ may be identified with a holomorphic vector field $\bigl(z, f(z)\bigr)$. A flow line of this field is a holomorphic mapping $\Phi$ satisfying $\Phi' = f \circ \Phi$. (Compare the situation for vector fields on the real line.)

In our world, by contrast, we're irresistably drawn to write a complex number $z$ in the form $x + iy$ with $x$ and $y$ real; to form a new independent complex variable $\bar{z} = x - iy$; to consider real vector fields such as $\dd_{x}$, $\dd_{y}$, and $$ X = X_{1}(x, y)\, \dd_{x} + X_{2}(x, y)\, \dd_{y}, $$ and interpret their flows as real paths; etc., etc. How are we supposed to make sense of holomorphic coordinates and vector fields in this real setting?

The standard answers boil down to:

  1. We can pass from a holomorphic manifold of complex dimension $m$ to a real manifold of real dimension $n = 2m$ equipped with an extra complex structure $J$ that effects multiplication by $i$ at each point.
  2. Conversely, given complex manifold—a smooth manifold $N$ of real dimension $n = 2m$ and a complex structure $J$—the real tangent bundle $TN$ acquires the structure of a smooth complex vector bundle of complex rank $m$. We extend $J$ to the complexified tangent bundle $TN \otimes \Cpx$ by $J(v \otimes \alpha) = J(v) \otimes \alpha$ for all tangent vectors $v$ and all complex $\alpha$. The complexified bundle then decomposes into complementary real subbundles, often denoted $T^{1,0}N$ and $T^{0,1}N$, on which the extension of $J$ acts as multiplication by $i$ and by $-i$. These are the well-known formulas \begin{align*} v^{1,0} &= \tfrac{1}{2}(v \otimes 1 - J(v) \otimes i), \\ v^{0,1} &= \tfrac{1}{2}(v \otimes 1 + J(v) \otimes i). \end{align*} (If, further, $J$ satisfies a differential equation—vanishing of the Nijenhuis tensor, we can assemble the $2m$ real coordinates into $m$ complex coordinates so that changes of coordinates are holomorphic with holomorphic inverse, and our complex manifold becomes a holomorphic manifold.)

Particularly, the real vector bundles $(T^{1,0}N, i)$ and $(TN, J)$ are isomorphic as complex vector bundles. This, at last, tells us "what a holomorphic vector (field) is geometrically" and explains the (technically sloppy) way people speak of "flow lines" of holomorphic vector fields: A "holomorphic vector" $v^{1,0}$, which from the real perspective is not a vector at all, corresponds to (or "is" if we're being sloppy) the honest real vector $v = v^{1,0} + v^{0,1}$.

Specifically, a holomorphic vector field $X$ "is" (ahem) twice its real part, $X + \bar{X}$. For example, \begin{align*} z\, \dd_{z} &= (x + iy)\, \tfrac{1}{2}(\dd_{x} - i\, \dd_{y}) \\ &= \tfrac{1}{2}\bigl[(x\, \dd_{x} + y\, \dd_{y}) - i(-y\, \dd_{x} + x\, \dd_{y})\bigr] \end{align*} is the Euler field $x\, \dd_{x} + y\, \dd_{y}$, while the rotation field $-y\, \dd_{x} + x\, \dd_{y}$ is $iz\, \dd_{z}$.

  • Is the isomorphism of complex vector bundles $(T^{1,0}N,i)$ and $(TN,J)$ natural in the sense of https://en.wikipedia.org/wiki/Natural_transformation ? – user505117 Jan 16 '25 at 15:51