While I was reading a paper about random analytic function I found a statement that I was not able to prove and after try brute force and search for some references I decided to ask for a help here. The statement is the following: Denote by $\mathbb{D}$ the unit dic on the complex plane, and consider the function $p:\mathbb{D}^n\to\mathbb{C}$ given by

$$ p(z_1,\dots,z_n)\, =\, 4^{-n}\, \det\left(\frac{(1-|z_i|^2)(1-|z_j|^2)}{|1-z_i\overline{z}_j|^2}\right)_{i,j=1}^{n}\, $$ then for any fixed $u\in\mathbb{D}$ the function $p$ is invariant for the Möbius transformation
$$ \phi(u,z)=\frac{u+z}{1+\overline{u}z}, $$ in the sense that
$$ p(z_1,\dots,z_n)\, = p(\phi(u,z_1),\ldots,\phi(u,z_n)). $$

I appreciate any reference or hint to prove this fact.

Edition: The denominator was modified as point out for David and Anon.

  • $\begingroup$ Have you tried small values of $n$ to see what happens? It's what I'd try first! (On a side note, you seem to be misusing the term "conformally invariant" - it doesn't mean "invariant under a Möbius transformation".) $\endgroup$ – Bruno Joyal Sep 15 '11 at 1:53
  • $\begingroup$ Should the denominator be $|1-z_iz_j|^2$? $\endgroup$ – anon Sep 15 '11 at 2:18
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    $\begingroup$ Hmm, I noticed each matrix entry is $(z^{-1},w^{-1};\bar{z},\bar{w})$, but Speyer pointed out this doesn't show it's a Möbius invariant because $\phi^{-1}\ne\bar{\phi}$ generally. $\endgroup$ – anon Sep 15 '11 at 3:06
  • $\begingroup$ Hi Guys, thank you very much for the help. I guess do you guys are right the denominator has to be modified. $\endgroup$ – Leandro Sep 15 '11 at 3:09
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    $\begingroup$ @anon: I think that this expression is invariant under the transformations preserving the unit disk: you have in Leandro's notation $\phi(u,z^{-1}) = \phi(\bar{u},z)^{-1}$ and $\phi(u,\bar{z}) = \overline{\phi(\bar{u},z)}$. But then $[z^{-1},w^{-1};\bar{z},\bar{w}] = [\phi(u,z^{-1}),\phi(u,w^{-1});\phi(u,\bar{z}),\phi(u,\bar{w}] = [\phi(\bar{u},z)^{-1},\phi(\bar{u},w)^{-1};\overline{\phi(\bar{u},z)},\overline{\phi(\bar{u},w)}]$ shows that it is invariant under applying $\phi(\bar{u},\cdot)$. Replace $\bar{u}$ by $u$ and get what you want. $\endgroup$ – t.b. Sep 15 '11 at 3:42

I suspect you have a typo, and the denominator should be $|1-z_i \overline{z_j}|^2$. At least, I will show that the quantity is Möbius invariant with this modification. It is possible, of course, that both are invariant.

The determinant is a red herring; each individual entry in the matrix is invariant under Möbius transformations. I begin by quoting a fact from hyperbolic geometry: Define $$\delta(u,v) = \frac{|u-v|^2}{(1-|u|^2)(1-|v|^2)}$$ for $u$ and $v$ in $\mathbb{D}$. Then this quantity is invariant under Möbius transformations. I can't give an intuition for this fact, but its easy to give a reference.

So $$1+\delta(u,v) = 1+\frac{(u-v)(\overline{u} - \overline{v})}{(1-u \overline{u}) (1-v \overline{v})} = \frac{\left( 1-u \overline{u} - v \overline{v} + u \overline{u} v \overline{v} \right) + \left( u \overline{u} - u \overline{v} - \overline{u} v + v \overline{v} \right)}{(1-u \overline{u}) (1-v \overline{v})}$$ $$=\frac{1- u \overline{v} - \overline{u} v + u \overline{u} v \overline{v}}{(1-u \overline{u}) (1-v \overline{v})} = \frac{(1-u \overline{v})(1-\overline{u} v)}{(1-|u|^2)(1-|v|^2)} = \frac{|1-u \overline{v}|^2}{(1-|u|^2)(1-|v|^2)}$$ is invariant under Möbius tranformations.

Replacing $u$ and $v$ by $z_i$ and $z_j$, this is the reciprocal of the $(i,j)$ entry in your matrix. So every element of your matrix is Möbius invariant, as claimed.

UPDATE: I just realized a nice way to express anon's solution below. Let $\sigma$ denote Schwarz reflection in the boundary of $\mathbb{D}$. Since Schwarz reflection is a conformally invariant operation, if $\phi$ is a Möbius transformation of $\mathbb{P}^1$ preserving $\mathbb{D}$, then $\phi(\sigma(z)) = \sigma(\phi(z))$. Explicitly, $\sigma(z) = \overline{z}^{-1}$.

Then the $(i,j)$ entry in your matrix is the cross ratio $(z_i, z_j; \sigma(z_i), \sigma(z_j))$, by a computation very similar to anon's.

  • $\begingroup$ Hi David, thanks for you help. Your feeling about the modulus seems to be right, I spend long time this afternoon trying to prove without modulus with no success. $\endgroup$ – Leandro Sep 15 '11 at 3:11

More pointedly, each entry in the matrix is a cross-ratio: $$\frac{(1-|z|^2)(1-|w|^2)}{|1-z\bar{w}|^2}=\frac{(1-z\bar{z})(1-w\bar{w})}{(1-\bar{z}w)(1-z\bar{w})}=(z^{-1},w^{-1};\bar{z},\bar{w}).$$ This can be seen by dividing both numerator and denominator by $zw$ appropriately. Cross-ratios are invariant under Möbius transformations, in that sense that if $f$ is one, then $$(a,b;c,d)=(f(a),f(b);f(c),f(d)).$$

EDIT: Ah, the last ingredient is provided by Theo Buehler. Since $\phi(u,z)^{-1}=\phi(\bar{u},z)$ and similarly we have $\overline{\phi(u,z)}=\phi(\bar{u},z)$, we can say that $\phi(u,\cdot)$ acting on both $z$ and $w$ is equivalent to the Möbius transformation $\phi(\bar{u},\cdot)$ acting on each individual component the $4$-tuple $(z^{-1},w^{-1},\bar{z},\bar{w})$, hence the cross-ratio is preserved under it.

  • $\begingroup$ Ah, thank you! That's much clearer. In particular, the Mobius transformation in question needn't preserve $\mathbb{D}$, which I hadn't realized. $\endgroup$ – David E Speyer Sep 15 '11 at 2:51
  • $\begingroup$ Wait, something is slightly off. It is not usually true that $f(z^{-1})=f(z)^{-1}$ or that $f(\overline{z}) = \overline{f(z)}$ for $f$ a Mobius function. So your nice observation that this is $(z^{-1}, w^{-1} ; \overline{z}, \overline{w})$ doesn't actually show that this is Mobius invariant. You're definitely onto something though; can you add some more detail and explain how to fix this? $\endgroup$ – David E Speyer Sep 15 '11 at 2:55
  • $\begingroup$ @David: Theo pointed out in the comments how to finish it. $\endgroup$ – anon Sep 15 '11 at 4:21
  • $\begingroup$ Anon, now I agree that Theo's comment complete your argument. Thank you very much for post this. $\endgroup$ – Leandro Sep 15 '11 at 5:03

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