One of the lemmas at the foundation of Teichmuller theory is as follows. Let $Q(x,y)$ be a positive definite quadratic form. Then there exists unique $\lambda \in \mathbb{R}$ and $\mu \in \mathbb{C}$ with $\lambda > 0$ and $\|\mu\| < 1$ such that the following hold. Let $Q_{\mu}(x,y)$ be the quadratic form $$Q_{\mu}(x,y) = \|z + \mu \overline{z}\|^2 \quad \quad (z=x+i y)$$ Then $$Q(x,y) = \lambda Q_{\mu}(x,y).$$ I can prove this using brute force (it comes down to solving a system of 3 equations in 3 unknowns). However, this shed no light on why it should be true. Does anyone know a conceptual reason for it?
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
This is a special case of a general fact, that the space of positive definite quadratic forms on $\mathbb{R}^n$ with determinant $=1$ is a symmetric space for the Lie group $PSL(n,\mathbb{R})$. A matrix acts on a quadratic form by change of basis, and a stabilizer is a conjugate of $SO(n,\mathbb{R})$. When $n=2$, the symmetric space for $PSL(2,\mathbb{R})$ is $\mathbb{H}^2$, hyperbolic 2-space, and you are considering the identification with the unit disk model. From this perspective, your question reduces to: why is upper half-space the symmetric space for $PSL(2,\mathbb{R})$? Of course, this is a special fact related to the existence of $\mathbb{C}$, a commutative field containing $\mathbb{R}$. The action here is the restriction of the action of $PSL(2,\mathbb{C})$ on $\mathbb{CP}^1$ to the subgroup $PSL(2,\mathbb{R})$, so that the point stabilizers of $\mathbb{CP}^1-\mathbb{RP}^1$ are compact. I don't know a good explanation for this, other than the fact that the eigenvalues of the two complex conjugate eigenvectors must be unit complex conjugates, which forms a compact group $U(1)\cong PSO(2)$.
