Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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".) –  Bruno Joyal Sep 15 '11 at 1:53
    
Should the denominator be $|1-z_iz_j|^2$? –  anon Sep 15 '11 at 2:18
1  
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. –  anon Sep 15 '11 at 3:06
    
Hi Guys, thank you very much for the help. I guess do you guys are right the denominator has to be modified. –  Leandro Sep 15 '11 at 3:09
2  
@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. –  t.b. Sep 15 '11 at 3:42

2 Answers 2

up vote 7 down vote accepted

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.

share|improve this answer
    
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. –  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.

share|improve this answer
    
Ah, thank you! That's much clearer. In particular, the Mobius transformation in question needn't preserve $\mathbb{D}$, which I hadn't realized. –  David Speyer Sep 15 '11 at 2:51
    
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? –  David Speyer Sep 15 '11 at 2:55
    
@David: Theo pointed out in the comments how to finish it. –  anon Sep 15 '11 at 4:21
    
Anon, now I agree that Theo's comment complete your argument. Thank you very much for post this. –  Leandro Sep 15 '11 at 5:03

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.