User grant lakeland - MathOverflow

Answer by Grant Lakeland for Should the formula for the inverse of a 2x2 matrix be obvious?

2012-02-22T15:33:04Z

I remember the inverse by looking at the corresponding linear fractional transformation. It sends $\frac{-d}{c}$ to $\infty$ and $\infty$ to $\frac{a}{c}$, so the inverse had better reverse this; it follows that the $c$ should stay put and the $a$ and $d$ should switch, and so the $b$ and $c$ get negated.

Answer by Grant Lakeland for Poincaré disk model: is this locus a known curve?

2010-08-17T17:59:38Z

I imagine the answer to this problem is known, but I don't have a reference for it; if someone has one, I would be interested to see it. Here is my calculation in the upper half-plane.

Following on from Will Jagy's comment, for a fixed angle $\theta$ at $P = x + iy$ and with the other two vertices in the upper half-plane placed at $i$ and $\infty$, the locus $S$ for the vertex $P$ is $$y^2 = 1 + x^2 - 2xy\cot{\theta}.$$ We can use this to extend to the case where the two fixed vertices are at $i$ and $e^h i$, where $h$ is the distance between the two points $A$ and $B$. Suppose $P$ makes the fixed angle $\theta$ with these two vertices, and denote by $\psi$ the angle at $P$ of the ideal triangle with vertices at $e^h i$ and $\infty$. We then have the two equations $$y^2 = e^{2h} + x^2 - 2xy\cot{\psi}$$ $$y^2 = 1 + x^2 - 2xy\cot{(\psi + \theta)}.$$ Using trig identities to remove $\psi$, from this we get $$y^4 - (e^{2h} + 1)y^2 + x^4 + (e^{2h} + 1)x^2 + 2xy(xy + \cot{\theta} - e^{2h}\cot{\theta}) = e^{2h}.$$ I am not personally aware of any particular significance of this locus, though I'd be interested to hear if there is one.

Answer by Grant Lakeland for Canonical fundamental domain for a discrete subgroup Γ of SL₂(R) acting on hyperbolic plane

2010-05-29T09:43:43Z

I found interesting the question of why we normally use "that" fundamental domain for the action of $PSL_2(\mathbb{Z})$ . I thought it worthy of its own answer, though I make no claims that the reasons I will list are exhaustive (and I'd be interested to see others!).

Once we have a fundamental domain, and as long as we keep track of the ways the sides get identified by the group, the Poincaré Polyhedron Theorem allows us to recover a presentation for the group, where side-pairings correspond to generators, and vertex cycles to relations. If, say, we were interested in finding minimal generating sets, that would correspond to finding fundamental domains with minimal numbers of sides. Since $PSL_2(\mathbb{Z})$ has rank 2, the "cleanest" fundamental domains we can find (in this sense) are those with 4 sides. (Here we regard the two halves of the bottom circle arc as distinct sides, since they are identified by the linear fractional transformation $z \rightarrow \frac{-1}{z}$, and we regard $i$ as a vertex.) Further, the two matrices which correspond to the side-pairings of the usual domain are $\left( \begin{smallmatrix} 1 & 1 \\ 0 & 1 \end{smallmatrix} \right)$ and $\left( \begin{smallmatrix} 0 & -1 \\ 1 & 0 \end{smallmatrix} \right)$ , which are frequently taken as generators for $PSL_2(\mathbb{Z})$. So, in this regard, the domain "fits in" with other existing conventions.
As Greg Kuperberg mentions, $PSL_2(\mathbb{Z})$ is a subgroup of index 2 in the $(2,3,\infty)$ triangle reflection group, and it's easy to see how our domain is just one copy of this triangle unioned with its reflection along a vertical side.
As coudy mentions, our domain is a Dirichlet domain, and the proof of this fact is fairly straightforward (and elegant, in my opinion). Certainly, it's easier to see with this particular domain than with other domains one might try to construct.
I suspect that one reason this domain is used so frequently is that short, elementary proofs exist that it is indeed a fundamental domain, which avoid any of the discussion listed above. In what little I have glanced at on automorphic forms, this type of proof is given at an introductory stage, so that one can move on to more content. I like to see the simplest arguments as being tied up with the previously mentioned Ford domains, though I'm not sure how useful others might find it. Essentially, the unit circle is the isometric circle of the map $z \rightarrow \frac{-1}{z}$, which (because this particular element has order 2, or is its own inverse) means the circle is sent to itself, and its exterior and interior are interchanged. This observation is part of showing that no two points of our domain are equivalent under the action of the group. Using other domains involves using more than one (isometric) circle, so our domain is simplest in this sense.

Locus of equal area hyperbolic triangles

2010-05-11T23:12:20Z

Henry Segerman and I recently considered the following question:

Given a fixed area $A < \pi$ and two fixed points in the upper half-plane model for hyperbolic $2$-space, what is the locus of points which give rise to a hyperbolic triangle of the given area?

We found it a fun exercise in hyperbolic geometry to show that the answer is a Euclidean straight line, or an arc of a Euclidean circle. As this requires only elementary properties of hyperbolic geometry, we strongly suspect it should be known, but have thus far been unable to find a reference for it. Does anyone know whether it's known, and if so, where one can find it?

Comment by Grant Lakeland

2012-06-12T20:18:12Z

@Lee: OK, great! I guess this shows any two convex, finite-sided fundamental domains are comparable.

Comment by Grant Lakeland

2012-06-12T18:06:51Z

Here's an approach to proving it. It's true if the group is cocompact, so suppose it is not. Take a maximal $\Gamma$-invariant set of horoballs, call its union $H$. Then $D_2 \setminus H$ has compact closure, so is covered by finitely many translates of $D_1$. We should also be able to cover each of the finitely many horoball components of $D_2$ with finitely many translates of $D_1$, but I don't yet see how to show this cleanly.

Comment by Grant Lakeland

2012-02-22T21:27:36Z

@Martin: Yes, thank you. @François: Yes, true; I should mention that I was assuming the "switch one pair and negate the other" comment from the original question!

Comment by Grant Lakeland

2012-01-25T01:08:38Z

This question could be related: mathoverflow.net/questions/33977/…

Comment by Grant Lakeland

2010-05-28T22:28:08Z

@Sam: OK, thanks, I see what you're saying in your edit now. Do you have a reference for that? I've always taken a Ford domain to be just a region of hyperbolic space (exterior to all isometric spheres etc.), whereas you seem to be taking it to be my region with some identifications already built-in.

Comment by Grant Lakeland

2010-05-28T17:49:05Z

(sorry, I had two comments and they were too long to fit together) Secondly, if we use coudy's definition of Dirichlet domain, I believe it's usually required that the basepoint for a Dirichlet domain is not fixed by any non-trivial element of the group, because the half-spaces we intersect must be well-defined. However, taking any point on the imaginary axis above $i$ will lead to the standard fundamental domain for $PSL_2(\mathbb{Z})$.

Comment by Grant Lakeland

2010-05-28T17:47:58Z

@Sam: Firstly, in this case the Ford domain is not quite "canonical" in your sense, because we need to choose a fundamental region for the action of the stabilizer of $\infty$. Obviously, each finitely generated group will indeed have finitely many nice choices.

Comment by Grant Lakeland

2010-05-13T00:11:09Z

Yes, and in fact the same is true for any area: once you have one vertex $v$ giving the right area, with $0$ and $\infty$, the locus is the straight line through $0$ and $v$. The proof we constructed actually used this as a warm-up case, and the proof when all vertices are non-ideal is built from this.

Comment by Grant Lakeland

2010-05-12T20:43:44Z

@Will: The endpoints of the locus are actually distinct from the geodesic through the two fixed points, and hence define a separate geodesic (which necessarily does not intersect the first, even on the boundary). If, say, the two fixed points lie on a vertical line, we get one "banana" curve to the right of the line, and its reflection on the other side, but these do not share endpoints.

Comment by Grant Lakeland

2010-05-12T17:29:37Z

Thanks! The KP paper seems to be what we were after.

Comment by Grant Lakeland

2010-05-12T17:27:59Z

@Will: They seem to be equidistant curves only. A related question would be: Does the geodesic with the same endpoints have any significance?