ok, I just redid them and came to the conclusion the absolute value sign is not an issue
so I again stand behind my calculations and my claim
the hyperbolic distance between $iz,iy$ is $\ln\max(y,z)/\min(y,z)$, so the point lying $r$ hyperbolic units above $iy$ on the vertical ray is $ie^ry$ and the one lying $r$ hyperbolic units below $iy$ is $ie^{-r}y$ and the Euclidean distance between $ie^ry$ and $iy$ is greater than that between $iy$ and $ie^{-r}y$
@Balarka isnt the issue with your argument that while the reflection maps the disks and the hyperbolic centers to one another, it neednt map the euclidean center of one disk to the euclidean center of the other
but i guess the translation between the two induces the asymmetry, cause half of a diameter becomes a short interval, whereas the other half becomes a very stretched interval
it should definitely not be a CW complex, because I think I've once seen an example of a non-Hausdorff space covered by a nice space, but homotopy equivalent to a CW complex?
"25. Let ϕ:R2→R2 be the linear transformation ϕ(x,y) = (2x,y/2). This generates an action of Z on X = R2 − {0}. Show this action is a covering space action and compute π1(X/Z). Show the orbit space X/Z is non-Hausdorff, and describe how it is a union of four subspaces homeomorphic to S1×R, coming from the complementary components of the x axis and the y axis."
I've though about whether one can cover the line with two origins by a Hausdorff space and have not been able to come up with anything (though of course that's far from saying it's not possible)
ok, I think I'm starting to see where the cylinders are coming from
the orbits are discrete subspaces of hyperbolas and each orbit in a hyperbola is represented by an element of some half-open interval along that hyperbola, so each hyperbola gives rise to some S^1 and that should give the S^1XR for each open quadrant and then these are glued together some way
@MikeMiller Let's do geometrization theorem for nonHausdorff 3-manifolds
Should be wide open
at the end of the joint paper by us we'd write
GOTCHA THURSTON WHOS THE HYPERBOLIC GEOMETER NOW
There's a vector field tangent to this foliation (the boundary circles are included)
The flow is defined throughout $\Bbb R$. This is an action of $\Bbb R$ on the annulus, restrict to $\Bbb Z$
That's definitely a covering map, yeah?
The quotient is a nonHausdorff manifold-with-boundary $Q$ given by collapsing the leaves to points. Set-theoretically, $Q = S^1 \cup \{b_1\} \cup \{b_2\}$ where $b_i$ corresponds to the two boundary leaves.
Given any point $p \in S^1$, any neighborhood of $p$ intersects any neighborhood of either $b_1$ or $b_2$
Imagine neighborhoods of $b_1$ as spiral arcs coming out of $S^1$ but only an infinitesimal part of it escapes the circle
Same for $b_2$
Any neighborhood of $b_i$ contains any given point on $S^1$
What's the Hausdorffification? It's a point, right?
But it has fundamental group $\Bbb Z$ (simply wind around the $S^1$ part)
A homotopy equivalence to a CW complex has to be a constant map but domain has $\pi_1 = \Bbb Z$ so cannot be isomorphism
@Thorgott @MikeMiller
What was even the definition of a covering space action?
I think the Hausdorffification is the thing you get when you identify the four axes all to a common point. Then your Hausdorffification of the R action is just an infinite X.
Then upstairs I imagine Hausdorffification is S^1 x R sqcup_{S^1 x 0} S^1 x R
I reject this approach! I will Carl Gustav Jacob Jacobi it. Take $[0, 1)$ and $(0, 1]$. Identify the two along $(0, 1)$. What you get I am pretty sure is the leafspace of the Reeb foliation on the tube, $\Bbb R \times I$, on which the flow action of $\Bbb Z$ is in fact completely properly discontinuous
Show the orbit space X/Z is non-Hausdorff, and describe how it is a union of four subspaces homeomorphic to S 1×R, coming from the complementary components of the x axis and the y axis.
It's S^1 x R sqcup_{S^1 x (0,infty)} ... done four times cyclically. The Hausdorffification is going to be S^1 x R sqcup_{S^1 x [0,infty)} ... done cyclically. This is clearly S^1 x (R sqcup_{[0, infty)} R) done cyclically, which is quite clearly S^1 x an X shape.
The R quotient is like if you took the set |xy| = 1 except forced the hyperbolae together and then pushed the seam down to the origin except for an infinitesimal bit so that the seams themselves don't join
The transitions between the seams are each points
But then you push slight further to get a Hausdorff space and the seams are identified, X
Let $D^n$ be the $n$-dimensional unit disk in euclidean $\mathbb{R}^n$.
Larry Guth's Sponge Problem asks: Does there exist a constant $\epsilon=\epsilon_n$ such that every open subset $U\subset \mathbb{R}^n$ satisfying $vol(U)< \epsilon_n$ admits an expanding embedding $f: U\hookrightarrow D^n$?
...
what's the best stack to ask "what algorithms are available to solve this problem in a computationally optimal way?" here? or one of the coding stacks?
I was looking and I was undecided between here and SO. I can't find another coding stack where they seem to take this kind of questions. And I'm scared of SO
That question seems to be either not knowing the statement of Dehn-Nielsen-Bahr or not understanding the definition of proper map and proper homotopy. I suppose it could also be mainly interested in the case of infinite topological type but there are some comments which suggest it's really one of the former options ...
Here the short proof of the fact that every f.g. group of subexponential growth is amenable.
Let $S$ be a symmetric generating subset with 1. If $G$ has subexponential growth, then clearly $\liminf |S^{n+1}|/|S^{n}|=1$. So we can extract from $(S^n)$ a Følner sequence.
The converse fails: many...
I've been reading Löh's book and I like it. Very clear exposition of proof strategies, nice illustrations, lots of interesting references/applications to related areas of maths, instructive exercises.
I think I just had a small brain fart. I got why each of the antecedent and the consequent were true
yeah, also saw why it was irreducible, but remembering that quotienting by the ideal generated by an irreducible poly gave you an extension field was this basic fact I tripped over
Somebody can help explain this: "If the unit group of Z[\sqrt(3)] were finite then all units would be roots of unity (+-1)"? I don't understand why "all units would be roots of unity" :(
A train is moving towards east and a car is along north, both with same speed. The observed direction of car to the passenger in the train is.
I drew the diagram like this (blue one)and got answer as north east or east north direction.
My questions are
It is normal right if relative velocity and...
mapping $y$ to $1-y$ induces an automorphism of $\mathbb{Z}/2\mathbb{Z}[y]$ that switches the ideals $(y)$ and $(1-y)$, so this shouldn't be too surprising
Whoa it's as if the three berlin universities take turns with going wild. Last semester TU had quite a few advanced lectures, this semester HU is going crazy and FU will offer a lot
I don't have a clue what to choose. Apparently there will be Random Dynamical Systems, Probability IV: Stochastic Homogenisation, Nonlinear Functional Analysis, Bifurcation Theory (Dyn. Systems III), Computational Dynamics, "Gaussian measures in infinite dimensions and invariant measures for PDEs" and "Functional Analysis Applied to Modeling of Molecular Systems"
Here the short proof of the fact that every f.g. group of subexponential growth is amenable.
Let $S$ be a symmetric generating subset with 1. If $G$ has subexponential growth, then clearly $\liminf |S^{n+1}|/|S^{n}|=1$. So we can extract from $(S^n)$ a Følner sequence.
The converse fails: many...
