To my understanding, in a 2D CFT with complex coordinates, the coordinates $z$ and $\bar{z}$ are to be treated as independent, and only at the end of the calculation should one take $\bar{z}=z^*$. But I'm not sure what this means for a contour integral. What exactly is the meaning of integrating a function of both coordinates over just one? For example, in

$$\oint \phi(z,\bar{z}) dz,$$

do I integrate only over $z$ so that the result of the integral is a function of $\bar{z}$, or do I get a number (i.e. the result is independent of $\bar{z}$)? And in the latter case, how is the integral calculated?

Also, how is this integral different from the integral

$$\oint_\gamma dz \oint_\delta d\bar{z}\phi(z,\bar{z}),$$

where $\gamma$ is the contour in $z$ and $\delta$ is the contour in $\bar{z}$? And in particular, does $\delta$ need to be the conjugate of $\gamma$, i.e. $\delta=\gamma%*$? (For an example of an integral of this kind, see Di Francesco Eq. 6.7)


2 Answers 2


This is a pretty interesting question, since there are two different ways of looking at this. Let me try to give an idea of both viewpoints:

(for whoever can't be bothered to read the whole thing, just read the ``Note'' in both cases as a representative example)

  1. The mathematician's point of view (which is also the viewpoint that I'm familiar with). Then $z$ and $\bar z$ are not in any way independent, as one would have expected on first sight. We just work with the complex plane, originally defined in the two real coordinates $x$ and $y$, and any point can equivalently be described by $z := x+iy$. The complex conjugate is then simply a function of $z$, defined by $\Re (z) - i \Im(z) = x-iy$ and denoted $\bar z$. Moreover, one defines the derivative $\frac{\partial}{\partial z} := \frac{\partial}{\partial x} - i\frac{\partial}{\partial y}$ and its complex conjugate $\frac{\partial}{\partial \bar z} := \frac{\partial}{\partial x} + i\frac{\partial}{\partial y}$. These definitions are not chosen arbitrarily, indeed one can derive that e.g. $\frac{\partial}{\partial z} z = 1$ and $\frac{\partial}{\partial \bar z} z = 0$. If we then have a function on the complex plane, then in principle we could just write $f(z)$ (since it is only a function of $z$), but sometimes it is helpful to write $f(z,\bar z)$ to denote the fact that we don't presume that $\frac{\partial}{\partial \bar z} f = 0$. So in this case writing $f$ as a function of both $z$ and $\bar z$ is not literal, and it's not implying we treat them as independent variables. It is just to denote that it is a very general function. If our function $f$ is special such that $\frac{\partial}{\partial \bar z} f = 0$, we call $f$ holomorphic and denote it as $f(z)$. If we want to integrate over the complex plane, we can do this by the most obvious way $\int f dxdy$. Note that this is a two-dimensional integral. Equivalently, since $z = x+iy$ we have that $dz = dx + idy$ and $d\bar z = dx - idy$, such that $dz d\bar z = -2i dxdy$ (where we use $(dx)^2 = 0$), we can also write $\frac{i}{2} \; \int f \; dz d\bar z$. We might also want to look at contour integrals, in which case we integrate over a 1-dimensional subset of the complex plane. Let $\gamma(t)$ be such a curve. Then writing $\int_\gamma f(z,\bar z) dz$ is defined to mean $\int f(\gamma(t),\bar \gamma(t)) \; \gamma'(t) dt$. Moreover, if we consider contour integrals of holomorphic functions, then there is the whole branch of complex analysis with some beautiful results like the Cauchy integral formula $f(z_0) = \int_\gamma \frac{f(z)}{z-z_0} dz$ where $\gamma$ is any contour encircling $z_0$. [Note: so in this way of looking at things, the expression $\oint \phi(z,\bar z) dz$ would just be a number, and not a function of $\bar z$, since by definition it's $\int \phi(\gamma(t),\bar \gamma(t)) \gamma'(t) dt$ where $\gamma(t)$ is the curve integrated over.]
  2. And apparently the CFT point of view. There it seems $z$ and $\bar z$ are treated ``independently''. The main question is then: how exactly, and how does it compare to the previous viewpoint? After all if you flip open a math book you will most likely see the treatment following (1), so it's good to know how one should compare things. The main point seems to be the following: while $z$ and $\bar z$ are strictly speaking dependent on each other (as discussed in viewpoint 1), they are not algebraically dependent! That is, you cannot write $\bar z$ as some polynomial in $z$. This has an important consequence: if one decomposes a function $f(z,\bar z)$ algebraically in terms of $z$ and $\bar z$ (e.g. through a power series), then this is not ambiguous. E.g. once I write $f(z) = z + \bar z ^2 + 2 z \bar z$, I know there is no other way of writing it in terms of $z$ and $\bar z$. (This would be entirely different if $\bar z$ could be expressed as $g(z)$ where $g$ is algebraic.). This means that it is entirely consistent of me to consider $f(z,\bar z)$ as a function of two independent variables, as long as I only consider algebraic operations. This way of looking at things is really nice: whereas in viewpoint (1) a general function $f(z,\bar z)$ does not have nice methods for it (unlike holomorphic functions), in viewpoint (2) we can consider $f(z,\bar z)$ as being decomposed into holomorphic functions of our two variables. Example: the function $f(z,\bar z) = z + 2 z \bar z$ uniquely defines the function $g(u,v) = u + 2uv$ which is holomorphic in the two complex coordinates $u$ and $v$, and $f(z,\bar z) = g(z, \bar z)$. The point is that we can now use our holomorphic tricks (Cauchy etc) on this function $g$. That is now what is implicitly happening in expressions such as (6.7) of Di Francesco. I.e. there we really act as if $z$ and $\bar z$ are two complex numbers that have nothing to do with each other, such that we can decompose general functions $\phi(z,\bar z)$ into a power series where the coefficients are determined by the Cauchy integral formula, just as we are used to for holomorphic functions of one variable. [Note: so in this way of looking at things, the expression $\oint \phi(z,\bar z) dz$ is indeed still a function of $\bar z$, with the expression being defined by first expanding $\phi(z,\bar z)$ into a power series of $z$ and $\bar z$, then temporarily replacing $z \to u$ and $\bar z \to v$ and then employing two-variable complex analysis (with a contour integral for the first variable) and then plugging back in $u \to z$ and $v\to \bar z$. Of course once one gets the point, this is not done explicitly and is merely implied in the expressions.]
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    $\begingroup$ Fantastic! Clarified many lingering doubts that I've had since I learnt CFTs many years ago. Thanks! $\endgroup$
    – Prahar
    Aug 23, 2016 at 23:24

Possibly, you are not completely convinced by my answer on PO. But I will try to make that answer a bit clearer here. I hope it will help.

To my understanding, the complex coordinates are introduced by complexification of spacetime. It is just a generalization of continuation. Specifically, we first take $\sigma_1$, $\sigma_2$ as complex numbers and take the transformation; $$ z_1=\sigma_1+i\sigma_2,\\ z_2=\sigma_1−i\sigma_2. $$ Here we should take $z_1$ , $z_2$ as completely independent complex variables since now $\sigma_1$ and $\sigma_2$ are also complex variables too. This understanding is justified by that the above transformation makes no sense if we insist that $\sigma_1$, $\sigma_2$ are real numbers because the transformation has complex coefficients and which means that we are now working $\mathbb{C}^2$ instead of $\mathbb{R}^2$. Now all original fields such as $ϕ(\sigma_1,\sigma_2)$ become fields on $\mathbb{C}^2$ (4 real dimensional). All these are simply the very same thing as we continue a real function to a complex function. If we want to come back to the real function, we just limit us at the real axis. So similarly, if we want to come back to the physical case, we just constrain $\sigma_1$, $\sigma_2$ as real variables, equivalently, constrain $z_2=\bar z_1$. But before we look at the final result, all the intermediate process should be treated that $z_1$, $z_2$ are independent complex variables. So to avoid confusion, just replace $z$, $\bar z$ with $z_1$ and $z_2$ in the intermediate calculations and only set $z_1=z$, $z_2=\bar z$ at last step (in order to go back to the real surface). And indeed, $∮dzϕ(z_1,z_2)$ is generally a function of $z_2$. Also, the integral coutours of $z_1$, $z_2$ are independent too. One your confusion is that, how can a final result like $\phi(z_2)$ which is only antiholomorphic can go back to the physical case by setting $z_2=\bar z_1$. Actually, at last, we just need to set $z_1=z$, $z_2=\bar z$. For instance, when we get a final expression $\phi(z_2)$ after integrating $z_1$, then you the physical result is simply $\phi(\bar z)$ and it means that we got an operator decouples from the left-moving degrees of freedom. See here for the more authoritative argument you guys would like to believe. (The last paragraph below Eq.(5.11), "The first question that comes into mind....").

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    $\begingroup$ I disagree with this answer. Complexification of spacetime is wholly unrelated to using the complex notation $z = x+iy$. I'm pretty sure the correct answer is that $z$ and $\bar z$ are in fact not independent variables. One might be tempted to think they are independent because people consider both derivatives $\frac{\partial}{\partial z}$ and $\frac{\partial}{\partial \bar z}$, but they are merely defined via $\frac{\partial}{\partial z} = \frac{1}{2} \left( \frac{\partial}{\partial x} - i \frac{\partial}{\partial y} \right)$ and analogously for $\bar z$. [cont] $\endgroup$ Aug 23, 2016 at 10:28
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    $\begingroup$ As for $\mathrm dz \mathrm d\bar z$, if you write them out in terms of their definition $ z=x+iy$ and $\bar z = x-iy$, you will see this just gives $dx dy$. I have to run now but if this hasn't been properly answered later, I can give a crack it. $\endgroup$ Aug 23, 2016 at 10:30
  • $\begingroup$ @RubenVerresen, Please do not say words like "Complexification of spacetime is wholly unrelated to using the complex notation..." but without any explanations. See, e.g. right below Eq.(5.11) of CFT by diFrancesscon (see the link in the answer). And $dzd\bar z=dxdy$ does not violate the idea of complexification, on the contrary, it supports that point of view. $\endgroup$
    – Wein Eld
    Aug 23, 2016 at 10:58
  • $\begingroup$ Thanks for that link! Pretty interesting. I don't understand why Di Francesco et al claim that would be the proper approach. $\endgroup$ Aug 23, 2016 at 19:47

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