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In Quasi-cluster algebras from non-orientable surfaces by Dupont and Palesi, one can read the following on page 11:

Proposition

I don't understand why the 'following relations' in the image included hold. Applying $D$ to the point $(u,0)$ gives $(\mu u, 0)$. So surely $w = \mu u$.

What am I missing here? Am I being incredibly stupid?

MvG
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Ricky Blake
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  • There's a lot of missing context in this snippet. I don't know what $\lambda(U,V)$ is, I don't understand the notation $U=(u,h)$ for horocycles, ... – Lee Mosher Jul 04 '14 at 12:46
  • Sorry. Given a horocycle $U$ then we can denote $U = (u, h)$ where $u \in \mathbb{R}$ is the unique point on the real axis which the horocycle touches. And $h$ is the (euclidean) diameter of the horocycle.

    $\lambda(U,V)$ is the lambda length of the two horocycles $U$ and $V$.

     – Ricky Blake Jul 04 '14 at 12:50
  • I still don't know what "lambda length" means. – Lee Mosher Jul 04 '14 at 12:55
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    But I do know that $D(u) = \frac{\mu u + 0}{0 u - 1/\mu} = -\mu^2 u$. – Lee Mosher Jul 04 '14 at 13:02
  • What does $D(U) = V$ mean in this context then? Simply that $D(u) = v$ and $D(h) = k$? Originally I thought it meant that it took the horocyle $U$ to the horocycle $V$. But this clearly can't be the case now. – Ricky Blake Jul 04 '14 at 13:54
  • $D$ defines a function taking horocycles to horocycles, because it is an isometry of the hyperbolic plane. $D$ also defines a function taking $\mathbb{R} \cup \infty$ to itself, because $D$ is a fractional linear transformation with real coefficients. But $D$ does not define a function of horocycle diameter: $D$ may take two horocycles of the same diameter to two horocycles of different diameters. – Lee Mosher Jul 04 '14 at 14:58
  • In other words, $v$ depends only on $u$, but $k$ depends on both $u$ and $h$. ... @Lee Mosher: I think your comments cleared up the misunderstanding (the meaning of the matrix in the text). Would you like to post an answer? –  Jul 04 '14 at 15:16

2 Answers2

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Your key problem appears to be a misunderstanding of homogeneous coordinates. You homogenize by appending a one, not a zero:

$$ u\rightsquigarrow\begin{pmatrix}u\\1\end{pmatrix}\mapsto \begin{pmatrix}\mu&0\\0&-1/\mu\end{pmatrix} \begin{pmatrix}u\\1\end{pmatrix}= \begin{pmatrix}\mu u\\-1/\mu\end{pmatrix}\sim \begin{pmatrix}-\mu^2 u\\1\end{pmatrix}\rightsquigarrow -\mu^2u=w $$

The same holds for any other point in the plane (or complex line, depending on your point of view): the given matrix will scale every point by $-\mu^2$. Since $\mu>1$ it follows that $\mu\in\mathbb R$, so no special complex number magic applies. Therefore, the diameters of your circles will scale with the absolute value of that scale factor, i.e. with $\mu^2$.

MvG
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Here's a expanion of the discussion in the comments.

The quoted passage is missing some important information. First, an equation like $U=(u,h)$ means that $U$ is the horocycle in the upper half plane tangent to $u$ on the real axis and of diameter $h$. Second, I'm guessing that $\lambda(U,V)$ refers to some measurement of the "signed hyperbolic distance" between the horocycles $U$ and $V$, in any case there should be an isometry taking one pair $(U,V)$ to another pair $(U',V')$ if and only if $\lambda(U,V)=\lambda(U',V')$. Third, the question of how to apply a matrix like $D$ having negative determinant seems to be fishy.

To answer the explicit question, there's a miscalculation of the fractional linear transformation as applied to points on the real line: $$\begin{pmatrix} \mu & 0 \\ 0 & -1/\mu \end{pmatrix}(u) = \frac{\mu u + 0}{0u - 1/\mu} = -\mu^2 u $$

But there's also an implicit question regarding how the fractional linear transformation $D$ transforms diameters. In general, given a hyperbolic isometry acting on a horocycle in the upper half plane, the diameter of the image horocycle is not a well-defined function of the diameter of the given horocycle. On the other hand, this is true for the specific fractional linear transformation $D = \begin{pmatrix} \mu & 0 \\ 0 & -1/\mu \end{pmatrix}$: a calculation shows that given any complex number $x+iy$, the imaginary part of the image $D(x+iy)$ is a well-defined function of $y$. To put it another way, this $D$ takes horizontal lines to horizontal lines. The formula for how $D$ acts on the $y$-coordinate can be obtained by plugging in the pure imaginary number $yi$: $$\begin{pmatrix} \mu & 0 \\ 0 & -1/\mu \end{pmatrix}(iy) = \frac{\mu iy + 0}{0iy - 1/\mu} = - \mu^2 i y $$ BUT, how can the imaginary part by negative? I think maybe the formula for applying a negative determinant matrix such as $D$ to a point in the upper half plane should not be $z \mapsto D(z)$ but instead $z \mapsto \overline{D(z)}$, but I cannot tell from what is pasted in the question.

Assuming something like that is appropriate, the summary is that $D$ takes points with imaginary coordinate $y$ go to points with imaginary coordinate $\mu^2 y$, and it takes horocycles of diameter $y$ to horocycles of diameter $\mu^2 y$.

Lee Mosher
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