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3
votes
1
answer
744
views
What does the trace of a loxodromic Mobius transformation tell us about how it rotates?
A matrix $X=\begin{pmatrix}a&b\\c&d\end{pmatrix}\in\mathrm{PSL}_2(\mathbb{C})$
acts isometrically on the upper half-space model $\mathbb{H}^3$
via isometric extension of the Mobius transformation on $ …
2
votes
1
answer
179
views
Real slices of Minkowski space, using a complex quadratic form
Ordinary Minkowski space is $\mathbb{R}^{3,1}:=(\mathbb{R}^4,\phi)$
where $\phi:\mathbb{R}^4\rightarrow\mathbb{R}$
is a quadratic form of signature $(3,1)$.
Lying within this is a hyperboloid model fo …
0
votes
Accepted
Real slices of Minkowski space, using a complex quadratic form
$S$ cannot be a 3-dimensional real hyperboloid unless $\psi$ is of the same form as $\phi$, though it can be a 2-dimensional real hyperboloid*.
Excluding the possibilities achievable when $\psi$
is a …
0
votes
Accepted
How can we explicitly verify a canonical Dirichlet domain for this hyperbolic punctured torus
As discussed in the comments, the contradiction is that $D$ isn't really the Dirichlet domain for $G$ centered at $i$. The red edge is in fact an edge of the domain.
4
votes
1
answer
234
views
Trace field of a hyperbolic $3$-manifold with a totally geodesic subsurface
Let $X$
be a complete finite-volume orientable hyperbolic $3$-manifold,
and let $\Gamma$
be a Kleinian representation of $\pi_1(X)$.
Let $K\Gamma:=\mathbb{Q}\big(\{\mathrm{tr}\mid\gamma\in\Gamma\}\big …
0
votes
Accepted
Trace field of a hyperbolic $3$-manifold with a totally geodesic subsurface
The answer to this follows easily from section 9.5 of Machlachlan and Reid -- which somehow I managed not to see every other time I checked the book! The thing stated in my question is not true, but t …
3
votes
0
answers
231
views
Pairs of non-isometric subsurfaces of a hyperbolic 3-manifold, with the same genus
When I say manifold below, I mean a complete orientable finite-volume hyperbolic $3$-manifold and when I say subsurface, I mean immersed closed totally geodesic subsurface.
Whenever a manifold has a …
2
votes
1
answer
107
views
Find the fixed geodesic of an orientation-preserving isometry of the $3D$ hyperboloid model
Let $\mathcal{I}^3\subset\mathbb{R}^4$
be the standard hyperboloid model for hyperbolic $3$-space and consider the usual $\mathrm{SO}(3,1)$
action of $\mathrm{PSL}(2,\mathbb{C})$
on $\mathcal{I}^3$.
G …
5
votes
1
answer
527
views
How can we explicitly verify a canonical Dirichlet domain for this hyperbolic punctured torus
This is a follow-up question to a question from math.stackexchange: https://math.stackexchange.com/q/1436253/67563
Had it not been for the exchange there between myself and @Lee_Mosher in the comments …
3
votes
1
answer
286
views
How many non-commensurable non-arithmetic manifolds have a quaternion algebra like this?
I am interested in realizing commensurability classes of hyperbolic $3$-manifolds whose quaternion algebra (note: not invariant quaternion algebra) is isomorphic to one of the form $\Big(\frac{a,b}{F( …
4
votes
1
answer
274
views
How many quadratic fields occur as trace fields of hyperbolic knot complements?
I am interested in when the trace field of a knot complement has the form $F(\sqrt{-d})$ for $F\subset\mathbb{R}$ and $d\in F^+$ (squarefree). Does this occur for infinitely many choices of pairs $(F, …
1
vote
Accepted
How many non-commensurable non-arithmetic manifolds have a quaternion algebra like this?
The comments from @IanAgol above lead to an affirmative answer in the compact case. This paper gives an affirmative answer in the non-compact case: http://www.math.umt.edu/chesebro/AIMCLC.pdf
2
votes
1
answer
240
views
Why do noncocompact arithmetic Kleinian groups have quadratic trace fields?
I realize there are a few different ways of going about proving this, depending on one's background, but there's a particular number theoretic aspect that I am just blanking on, and can't seem to find …
0
votes
1
answer
158
views
Does this count as a canonical decomposition for non-elementary hyperbolic 3-orbifolds?
Let $\Gamma$ be a Kleinian group and let $\mathbb{H}^3$ be the upper half-space model for hyperbolic 3-space.
Then $\mathbb{H}^3/\Gamma$ is an orientable hyperbolic 3-orbifold (with the group action d …
0
votes
Accepted
Does this count as a canonical decomposition for non-elementary hyperbolic 3-orbifolds?
No. (See Ian Agol's comment.)