# (Anti-) Holomorphic significance?

What are holomorphic and anti-holomorphic components? Why don't we call them complex components and their conjugates? What is holomorphic coordinate transformation?

## 2 Answers

$\newcommand{\dd}{\partial}\newcommand{\Reals}{\mathbf{R}}\newcommand{\Cpx}{\mathbf{C}}$The $z^{\beta}$ are emphatically not holomorphic coordinates in general. "Holomorphic" refers to overlap maps being holomorphic, i.e., satisfying the Cauchy-Riemann equations, not merely to the coordinates being complex-valued.

In more detail, fix an identification of $\Cpx^{n}$ and $\Reals^{2n}$, say $z^{\beta} = x^{\beta} + iy^{\beta}$, and identify $$(z^{1}, \dots, z^{n}) \leftrightarrow (x^{1}, \dots, x^{n}, y^{1}, \dots, y^{n}).$$ Define the complex-valued differential operators $$\frac{\dd}{\dd z^{\beta}} = \frac{1}{2}\left(\frac{\dd}{\dd x^{\beta}} - i\frac{\dd}{\dd y^{\beta}}\right),\qquad \frac{\dd}{\dd \bar{z}^{\beta}} = \frac{1}{2}\left(\frac{\dd}{\dd x^{\beta}} + i\frac{\dd}{\dd y^{\beta}}\right),$$ and their dual one-forms $dz^{\beta} = dx^{\beta} + i\, dy^{\beta}$, $d\bar{z}^{\beta} = dx^{\beta} - i\, dy^{\beta}$.

A $C^{1}$ function $f:\Cpx^{n} \to \Cpx$ is holomorphic if $\dfrac{\dd f}{\dd \bar{z}^{\beta}} = 0$ for all $\beta$, and is anti-holomorphic if $\dfrac{\dd f}{\dd z^{\beta}} = 0$ for all $\beta$, i.e., if $\bar{f}$ is holomorphic. A smooth mapping $f:\Cpx^{n} \to \Cpx^{n}$ is holomorphic if each component function is holomorphic.

Holomorphic maps constitute a (rich but) thin subset of all smooth mappings. Among their many nice properties, a holomorphic mapping $F$ is complex analytic (representable by a convergent power series in a neighborhood of each point), and is locally invertible near a point $p$ (i.e., is a local biholomorphism at $p$) if and only if its derivative $dF(p)$ (which is a complex-linear mapping) is invertible (non-singular).

The chain rule guarantees a composition of holomorphic maps is holomorphic, so the set of local biholomorphisms forms a pseudo-group. (Loosely, forms a group, modulo fine print about composability.) It therefore makes sense to speak of a "holomorphic structure" or a "holomorphic coordinate system" by declaring two $\Cpx^{n}$-valued coordinate systems to be compatible if the change of coordinates (a.k.a. the overlap map) from one to the other is holomorphic. This notion of compatibility is an equivalence relation; particularly, compatibility is transitive.

Anti-holomorphic maps are conjugate-analytic in an obvious sense, and are locally invertible at $p$ if and only if their Jacobian is non-singular. The derivative of an anti-holomorphic map is not complex-linear, however, and a composition of anti-holomorphic maps is not anti-holomorphic. Consequently, there is no notion of an "anti-holomorphic structure" or "anti-holomorphic coordinate system": The corresponding notion of "compatibility" is not transitive.

Loosely, "holomorphic" and "anti-holomorphic" are not "dual" or "perfectly symmetric" concepts; the distinction between them is not arbitrary. That said, the definition ultimately rests on a choice of complex square root $i$ of $-1$, and arguably such a choice is arbitrary. But this choice is fixed by the time one even asks about complex differentiability.

• thanks Andrew for the answer. Very elaborate. I still have a question about the transformation. Yesterday, as I was looking deeper into my notes, I saw this: The ‘splitting’ of an index b into $\beta$ and $\bar{\beta}$ is not preserved by general transformation of complex coordinates $z'^{\beta} = f ^{\beta}(z, \bar{z})$, but it is preserved under the special class of ‘holomorphic’ cordinate transformations $z'^{\beta}=f^{\beta}(z)$. So, accordingly, I am wondering about two things. (I will write them in a new comment) – user291321 Nov 23 '15 at 13:58
• 1)I you have written in the beginning of your answer that this has nothing to do with being complex-values, why is there a repeated tendency to compare between transformation of complex coordinates and holomorphic coordinate transformation? 2) I still haven't really understood what do we mean by holomorphic coordinate transformation. – user291321 Nov 23 '15 at 14:00
• For 2 (and possibly for 1), my answer to a question about $z$ and $\bar{z}$ versus $x$ and $y$ may be of interest. Formally, $z^{\beta}$ and $\bar{z}^{\beta} = z^{\bar{\beta}}$ may be treated as independent variables, alternative to the real variables $x^{\beta}$ and $y^{\beta}$. A holomorphic map depends only on the $z^{\beta}$, not on the "conjugate" variables. – Andrew D. Hwang Nov 24 '15 at 1:19
• Oh right.. i see.. so, is there any reason behind the fact that a holomorphic map depends only on $z^{\beta}$ and not on the conjugate variables? – user291321 Nov 26 '15 at 13:34
• That's the definition of "holomorphic": $\dd f/\dd \bar{z}^{\beta} = 0$, i.e., "$f$ is constant with respect to $\bar{z}^{\beta}$". – Andrew D. Hwang Nov 26 '15 at 14:13

You can talk about homomorphic and antiholomorphic functions, but a general complex-valued smooth function will not be the sum of a holomorphic function with an antiholomorphic one (for example, $|z|^2$ on $\mathbb{C}$). Also, the constant functions are both holomorphic and antiholomorphic.

On the other hand, complex-valued smooth $1$-forms do decompose into holomorphic and antiholomorphic components. In local coordinates $(z_i)$, the holomorphic $1$-forms are spanned by the $dz_i$, with smooth (not necessarily holomorphic) coefficient functions, and the antiholomorphic forms by their conjugates. Of course, you should check that this definition is independent of the choice of (holomorphic) coordinates.

I do not know why you would call the holomorphic components "complex," because they are no more complex than the antiholomorphic components. Usually, they would be called forms of type $(1, 0)$ or $(0, 1)$, because this generalizes to higher differential forms, and a "holomorphic $1$-form" is typically required to have holomorphic coefficients.

It is an instructive exercise to check if a holomorphic coordinate transformation is just a holomorphic map with a holomorphic inverse or not. Does the domain and/or codomain have to be an open subset of $\mathbb{C}$?