# What object exactly is $\frac{d \bar{z}}{dz}$?

I know very basic differential geometry, that is i am aware of definitions of tangent spaces and differential forms, and I believe I know what is meant by $$dz$$ and $$d \bar{z}$$.

In a book I'm reading on complex dynamics i found the notation $$\frac{d \bar{z}}{dz}$$, what kind of object is that? Differential forms are linear functionals, as far as i understand division of one by another does not make sense.

EDIT: After getting a few replies, which were helpful in their own right, I realized i should have included some more context.

The context for this question is the document "Complex dynamics and renormalization" by Curtis McMullen. On page 47 there is the sentence:

"A line field is the same as a Beltrami differential $$\mu = \mu (z) d \bar{z} / dz$$".

I was curious about the notation i.e. what is $$d \bar{z}/dz$$.

• There is a dependence between $z$ and $\bar{z}$, and this is the differential measuring that dependence. It's the same procedure as the one you use to calculate the new form $dy$ when you have a change of coordinates from $x$ to $y$ on your manifold. Mar 30 '21 at 11:48

Long answer. Let me start by saying what $$d\bar z/dz$$ is not.

It's not a quotient of differential forms -- as you noted, such a quotient doesn't make sense.

It's not an ordinary derivative -- the only way it makes sense to take an ordinary derivative with respect to a complex independent variable like $$z$$ is if the function you're differentiating is holomorphic, which $$\bar z$$ emphatically is not.

It's not a partial derivative either -- if you try to think of $$z$$ and $$\bar z$$ as independent variables, you might think that $$d\bar z/dz$$ would mean taking the derivative of the function $$\bar z$$ with respect to $$z$$ while holding $$\bar z$$ (and any other variables that happen to be around) fixed. But if you hold $$\bar z$$ fixed, then $$z$$ remains fixed too, so the "partial derivative" of every function of $$z$$ and $$\bar z$$ would be zero.

But there is an important and rigorous sense in which the operator $$d/dz$$ (more commonly and more properly denoted by $$\partial /\partial z$$) can be applied to functions that are not necessarily holomorphic in $$z$$. Here's how that works. For this discussion, I'll assume that we're talking about functions defined on an open subset of $$\mathbb C$$, although similar considerations apply on Riemann surfaces and more general complex manifolds.

When considering smooth complex-valued functions on an open subset $$\Omega\subseteq\mathbb C$$, it is natural to consider differentials of such functions. These are sections of the complexified cotangent bundle $$T^*\Omega \otimes \mathbb C$$, which is the bundle whose fiber at a point $$p\in \Omega$$ is the complex vector space $$T_p^*\Omega \otimes \mathbb C$$, where the tensor product is taken over the reals. More concretely, we can view $$T_p^*\Omega\otimes\mathbb C$$ as the space of real-linear functionals from $$T_p\Omega$$ to $$\mathbb C$$, or as the space of linear combinations of ordinary covectors with complex coefficients. (It's a useful exercise to show that these various characterizations are naturally isomorphic.) So, for example, $$dz$$ is a smooth section of $$T^*\Omega\otimes\mathbb C$$, as is $$d\bar z$$.

Now things get interesting. Now that we've introduced the complexified cotangent bundle, it turns out also to be useful to look at its dual bundle over $$\mathbb C$$ -- this is the complexified tangent bundle $$T\Omega\otimes \mathbb C$$, whose fiber at $$p\in\Omega$$ is the complex vector space $$T_p\Omega\otimes \mathbb C$$. It doesn't have quite such an intuitive interpretation as the complexified cotangent bundle, but we can just think of a section of $$T\Omega\otimes \mathbb C$$ as a linear combination of ordinary vector fields with complex-valued coefficient functions -- a thing like $$z^2\, \partial/\partial x + i\bar z \, \partial/\partial y$$.

Now that we have these two complex vector bundles, we should think about some natural local frames for them. Of course, because $$\{dx,dy\}$$ is a local frame (actually global in this case) for $$T^*\Omega$$ (over $$\mathbb R$$), it follows that the same two $$1$$-forms constitute a frame for $$T^*\Omega\otimes\mathbb C$$ over $$\mathbb C$$. Similarly, $$\{\partial/\partial x,\ \partial/\partial y\}$$ forms a frame for $$T\Omega\otimes\mathbb C$$ over $$\mathbb C$$. But we can also look for complex-valued frames. Note that we can write $$dz = dx + i\, dy$$ and $$d\bar z = dx - i\, dy$$, and conversely $$dx = \frac 12 (dz + d\bar z)$$, $$dy = \frac{1}{2i} (dz - d\bar z)$$. Thus $$\{dz,d\bar z\}$$ also forms a frame for $$T^*\Omega\otimes \mathbb C$$.

What is the dual frame for $$T\Omega\otimes \mathbb C$$? It will be a pair of complexified vector fields $$A,B$$ with the property that $$dz(A) = d\bar z(B) = 1$$, $$dz(B) = d\bar z(A) = 0$$. A little linear algebra shows that the unique solutions are $$A = \frac 1 2 \left(\frac{\partial}{\partial x} - i \frac{\partial}{\partial y}\right), \qquad B = \frac 1 {2} \left(\frac{\partial}{\partial x} + i \frac{\partial}{\partial y}\right).$$ By analogy with the notation $$\{\partial/\partial x, \partial/\partial y\}$$ for the frame for $$T\Omega\otimes \mathbb C$$ dual to $$\{dx,dy\}$$, it is standard to denote these two vector fields by $$\frac{\partial}{\partial z} = \frac 1 2 \left(\frac{\partial}{\partial x} - i \frac{\partial}{\partial y}\right), \qquad \frac{\partial}{\partial \bar z} = \frac 1 {2} \left(\frac{\partial}{\partial x} + i \frac{\partial}{\partial y}\right). \tag{1}$$

At this point, this is just a definition, and these complexified vector fields cannot be interpreted as partial derivatives. However, they are closely related to partial derivatives in the following way. Suppose $$f\colon \mathbb C\to \mathbb C$$ is any polynomial function of $$(x,y)$$. By making the substitutions $$x = \frac12(z+\bar z)$$ and $$y = \frac{1}{2i}(z-\bar z)$$, we can rewrite it as a polynomial in $$z$$ and $$\bar z$$: $$F(z,\bar z) = f\left( \tfrac12(z+\bar z),\ \tfrac{1}{2i}(z-\bar z)\right).$$ The original polynomial $$f$$ defines a holomorphic function if and only if the expression for $$F$$ involves only powers of $$z$$ with no occurrences of $$\bar z$$.

Here's where the operators $$\partial/\partial z$$ and $$\partial/\partial \bar z$$ come in. The complex equation $$\partial F/\partial \bar z \equiv 0$$ is equivalent to the Cauchy-Riemann equations for $$f$$, as you can check. Thus in a certain sense, $$\partial/\partial \bar z$$ seems to be taking a derivative of $$F$$ with respect to $$\bar z$$ while "holding $$z$$ fixed." Here's how to make rigorous sense of that. In the same way that we defined the polynomial function $$F$$ above, we can define a polynomial function $$G$$ of two complex variables by formally substituting $$w$$ in place of $$\bar z$$: $$G(z,w) = f\left( \tfrac12(z+w),\ \tfrac{1}{2i}(z-w)\right).$$ Then our original polynomial function $$f$$ is equal to $$G(z,\bar z)$$, and $$f$$ is holomorphic if and only if $$G$$ is independent of $$w$$. Similar considerations apply to any real-analytic function $$f$$, with the same formulas.

If $$f$$ is merely smooth and not real-analytic, this construction doesn't work, because it doesn't make sense to plug complex values like $$\tfrac12(z+w)$$ into $$f$$. However, $$\partial f/\partial \bar z \equiv 0$$ still characterizes holomorphic functions, so it's useful to think of this equation as meaning intuitively that "$$f$$ depends only on $$z$$ and not on $$\bar z$$," and indeed from the theory of holomorphic functions we know that this implies that $$f$$ can be expressed (locally, at least) as a convergent power series in $$z$$. Similarly, $$\partial f/\partial z \equiv 0$$ means that $$f$$ is anti-holomorphic, and thus can be written locally as a convergent power series in $$\bar z$$.

Now back to your original question. The only reasonable interpretation of $$d\bar z/dz$$ is to rewrite it as $$\partial \bar z/\partial z$$, where $$\partial/\partial z$$ it the operator defined by ($$1$$). Using that definition, you can compute easily that $$\partial \bar z /\partial z \equiv 0$$.

EDIT: Now that the OP has clarified the context of the question, I realize that my answer is not quite to the point (although not completely irrelevant, either). Here's a better answer.

The paper being referenced is describing what it means for a map between plane domains to be quasiconformal. The basic definition is this: if $$f\colon \Omega \to \Omega'$$ is an orientation-preserving $$C^1$$ diffeomorphism, it is said to be quasiconformal if it satisfies $$\frac{\partial f}{\partial \bar z} = \mu \frac{\partial f}{\partial z}$$ for some complex-valued function $$\mu$$ that satisfies $$\sup_\Omega|\mu|<1$$, where $$\partial/\partial z$$ and $$\partial/\partial \bar z$$ are the operators defined by $$(1)$$. (The definition can be extended to maps with less regularity, but this will suffice for our purposes.)

The function $$\mu$$ depends on the choice of coordinates. But if you change coordinates, it transforms as a well-defined section of a certain complex line bundle. Let $$\Lambda^{1,0}\Omega$$ and $$\Lambda^{0,1}\Omega$$ denote the subbundles of $$T^*\Omega\otimes\mathbb C$$ spanned by $$dz$$ and $$d\bar z$$, respectively; and let $$T^{1,0}\Omega$$ and $$T^{0,1}\Omega$$ be the subbundles of $$T\Omega\otimes \mathbb C$$ spanned by $$\partial/\partial z$$ and $$\partial/\partial \bar z$$. (It's an easy exercise to prove that these subbundles are independent of the choice of local holomorphic coordinates.) Then $$\Lambda^{0,1}\Omega \otimes T^{1,0}\Omega$$ is a complex line bundle, locally spanned by $$d\bar z \otimes \partial/\partial z$$, and the section $$\mu(z)d\bar z\otimes \partial/\partial z$$ is well-defined, independent of coordinates. It's called the Beltrami differential of $$f$$.

Now the set of isomorphism classes of complex line bundles on $$\Omega$$ forms a group under tensor product, and with that group structure, the inverse of a bundle $$L$$ is (the isomorphism class of) its dual bundle $$L^*$$. Since $$T^{1,0}\Omega$$ is naturally dual to $$\Lambda^{1,0}\Omega$$, it's reasonable to think of the Beltrami differential as a section of $$\Lambda^{0,1}\Omega \otimes (\Lambda^{1,0}\Omega)^{-1}$$, and to write a local frame for this tensor product bundle as $$d\bar z \otimes dz^{-1}$$. That's the meaning of the expression $$d\bar z/dz$$ in McMullen's paper.

• This was a very helpful answer overall. At the same time the original context for my question was the sentence "A line field is the same as a Beltrami differential μ=μ(z)dz¯/dz" from McMullen's book "Complex dynamics and renormalization", and judging from this context the right hand side probably isnt supposed to be zero. I realized I should have included more information, my bad and i edited the question now. Mar 31 '21 at 19:49
• Ah, I see. OK, I'll add another comment. Mar 31 '21 at 20:01
• Why does the complexified tangent bundle "doesn't have quite such an intuitive interpretation as the complexified cotangent bundle"? I think it can be realised as $\text{Der}_\mathbb{C}(C^\infty_{X,p,\mathbb{C}},\mathbb{C})$, see math.stackexchange.com/questions/577501/… for example.
– Zero
Apr 3 '21 at 1:23

I think you are looking at a curve specified by:

$$z=z\left(t\right) \quad\&\quad \bar{z}=\bar{z}\left(t\right) \quad \mbox{etc.}$$

Where $$t$$ is the parameter you choose to parametrize your curve with respect to. The $$\bar{z}$$ component of the tangent to the curve is:

$$\frac{d\bar{z}}{dt}$$

And then there was a choice to look at the curves that correspond to varying $$z$$ and keeping other variables fixed, so you parametrize that curve by $$z$$. Now $$\bar{z}$$-component of the tangent to that curve will be:

$$\frac{d\bar{z}}{dz}$$