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Claim: Suppose that $f$ is a conformal map from the unit disk to the unit disk, $f(z_1) = w_1$ , $f(z_2) = w_2$, then we it can be derived that $|\frac{z_1-z_2}{1-\bar z_1 z_2}| = |\frac{w_1-w_2}{1-\bar w_1 w_2}|$. Let $z_1$ approaches $z_2$, we have $\frac{|dz|}{1-|z|^2}= \frac{|dw|}{1-|w|^2}$, so can conclude that the Riemannian metric $ds= \frac{2|dz|}{1-|z|^2}$ is invariant under the conformal mapping.

Questions:

  1. What does $dz$ actually mean here? Is it some sort of complex form? How should we conclude that $\frac{|dz|}{1-|z|^2}= \frac{|dw|}{1-|w|^2}$ by letting $z_1$ tends to $z_2$.

  2. Is $|dz| = (dx)^2+(dy)^2$ which is the usual Riemannian metric. I feel that to understand the claim, one needs some rigorous treatment of the complex form maybe? Is there any reference on this matter?

More details of the claim can be found in Ahlfors, 'Conformal Invariants, Topics in Geometric Function Theory' page 2.

Questions (1) and (2) have been addressed in the comments, but I am still confused about the procedure of letting '$z_1$ approach $z_2$', since it seems to be a rather vague idea, how do you really formalize this idea to arrive the conclusion such that the given metric is invariant under the conformal mapping?

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    $\begingroup$ Although $dz$ can indeed be formalized as a complex form, for example as $dz=dx+ i \, dy$ where $dx$ and $dy$ are the usual real forms, and although it can very often aid understanding, it is not necessary to do this in order to do calculations. Also, $$|dz|=\sqrt{(dx)^2 + (dy)^2}$$ $\endgroup$ – Lee Mosher Mar 5 '17 at 17:23
  • $\begingroup$ Then how would $ds$ be a Riemannian metric? Can it be expressed as a linear combination of bases? $\endgroup$ – Keith Mar 5 '17 at 18:03
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    $\begingroup$ This uses a common notation for special 2-dimensional Riemannian metrics, namely $$ds^2 = f(x,y) dx^2 + g(x,y) dy^2$$ where $f(x,y)$ and $g(x,y)$ are smooth, positive functions (what's special here is that there is no mixed term, i.e. no $dx\,dy$ term). The equation $ds = \frac{2 |dz|}{1-|z|^2}$ translates into this notation as $$ds^2 = \frac{2}{1 - x^2 - y^2} dx^2 + \frac{2}{1-x^2-y^2} dy^2$$ $\endgroup$ – Lee Mosher Mar 5 '17 at 19:26
  • $\begingroup$ I see, thank you very much. Do you know why it is true that $\frac{|dz|}{1+|z|^2} = \frac{|dw|}{1+|w|^2}$? $\endgroup$ – Keith Mar 5 '17 at 19:47
  • $\begingroup$ That's harder to guess at, because I don't know the details of the formalities used in your textbook (or whatever it is you are getting this from). Clearly they are letting $\Delta z = z_1 - z_2$ go to zero and replacing that by $dz$, and similarly for $\Delta w$. But how your textbook justifies this in any formal fashion is impossible for me to guess at. $\endgroup$ – Lee Mosher Mar 5 '17 at 20:01
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The $dz$ is the canonical one-form of the complex plane: $dz=dx+idy$, if complex numbers are written as $x+iy$ for $x,y\in\mathbb R$.

This is the one-form used in complex integration. If you multiply it by a complex function $f(z)$, you get a new one-form $f(z)dz$. You can integrate this one-form over any $C^1$ curve $\gamma$ to get $\int_\gamma f(z)dz$. This will precisely match the definition of the complex integral given in any complex analysis course.

Treating $dz$ as a complex one-form is by no means necessary for complex integration. The same holds for your problem: you can formalize it using complex one-forms, but the object $dz$ is typically left undefined one relies on more informal geometric reasoning. The symmetrized tensor product of $dz=dx+idy$ and its complex conjugate $\overline{dz}=dx-idy$ is indeed $dx\otimes dx+dy\otimes dy$, the usual Riemannian metric.

More details on the invariance of the metric

Here is some elaboration on what the hyperbolic metric is in terms of Riemannian geometry and why it is preserved. You can also read from here why $|dz|$ is related to $|dw|$ as claimed. If you want more details on how to relate that claim to the one about preserved hyperbolic metric, I suggest you ask a follow-up question.

Let me denote $c(z)=\frac12(1-|z|^2)$ and let $e$ be the usual Euclidean metric (as a Riemannian metric). Then the claim is that the conformally Euclidean metric $g(z)=c^{-2}(z)e(z)$ is invariant under conformal mappings.

That is, if $f\colon D\to D$ is conformal, then $f^*g=g$. It is enough to show that for any $z\in D$ and any $v\in\mathbb R^2(=T_zD)$ we have $$ g_z(v,v) = g_{f(z)}(df_zv,df_zv). $$ This can be rephrased in terms of the Euclidean metric for which $e_z(v,v)=|v|^2$ and the conformal factor $c$: $$ c^{-2}(z)|v|^2 = c^{-2}(f(z))|df_zv|^2. $$

Now take any $C^1$ curve $\gamma$ with $\gamma(0)=x$ and $\dot\gamma(0)=v$. Then $$ |v|=\lim_{t\to0}\left|\frac{\gamma(t)-z}{t}\right| $$ and $$ |df_zv|=\lim_{t\to0}\left|\frac{f(\gamma(t))-f(z)}{t}\right|. $$ Using $z$ and $\gamma(t)$ for $z_1$ and $z_2$ in your lemma (namely $|\frac{z_1-z_2}{1-\bar z_1 z_2}| = |\frac{w_1-w_2}{1-\bar w_1 w_2}|$) gives $$ |df_zv|=\lim_{t\to0} \left|\frac{1-f(\gamma(t))\overline{f(z)}}{1-\gamma(t)\overline{z}}\right| \cdot \left|\frac{\gamma(t)-z}{t}\right|, $$ whence $$ |df_zv| = \frac{c(f(z))}{c(z)} |v|. $$ This is what we were after.

If the notation is not clear, please let me know.

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  • $\begingroup$ Then how do you conclude that the hyperbolic metric is preserved by conformal maps? $\endgroup$ – Keith Apr 30 '17 at 20:22
  • $\begingroup$ @Keith You can denote $|dz|=\sqrt{(dx)^2+(dy)^2}$ without defining what $dz$ is. Can you see how to do the calculation with this starting point? I can add some details, but it'll take some time. $\endgroup$ – Joonas Ilmavirta Apr 30 '17 at 20:26
  • $\begingroup$ I will think about it, but can you add some more details? $\endgroup$ – Keith Apr 30 '17 at 20:39
  • $\begingroup$ @Keith I expanded the answer. The way I did it is certainly not the only way (and perhaps not the easiest one either), but it is a way of thinking that's useful in Riemannian geometry. $\endgroup$ – Joonas Ilmavirta May 1 '17 at 1:23

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