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 $d \bar{z}/dz$.

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    $\begingroup$ 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. $\endgroup$
    – S.Farr
    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.

  • $\begingroup$ 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. $\endgroup$ Mar 31 '21 at 19:49
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    $\begingroup$ Ah, I see. OK, I'll add another comment. $\endgroup$
    – Jack Lee
    Mar 31 '21 at 20:01
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    $\begingroup$ 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. $\endgroup$
    – 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:


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:



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