I just finished an inductive proof which got a bit hairy and so I'm not sure if I did things correctly. I won't put it in here unless I can bother someone for the trouble of looking thru it with me, but no wrries if not...let me know :)
Well, put it this way, I think it's correct but it feels long for such a simple statement. Anyway, I will hold off and save any goodwill for future questions :)
I'm desperately trying to get an answer to this easy question about Sobolev spaces: math.stackexchange.com/questions/4847464/… Can anybody help me out? I don't think it is getting any views. :(
@BalarkaSen it's basically another parametrization of deformation space of quasifuchsian manifolds using bending laminations on two boundaries of its convex core instead of conformal structure at infinity (Bers' simultaneous uniformization in this case), originally conjectured by Thurston.
So essentially, $QF(S)\to\mathcal{ML}(S)\times\mathcal{ML}(S)$. Its image is already known by Bonahon. What's unknown is the injectivity (the main theorem of that paper).
It uses something I'm not quite familiar with, but I know some differential geometer I can ask so maybe I would read the paper with him in the near future.
@BalarkaSen Bending lamination is basically a measured lamination on a surface. If you see the boundary of the convex core, because it's basically a convex hull of the limit set, it's topologically a surface but "bent" along some geodesics (it's called pleated surface). Thurston proved in his note that it's actually a geodesic lamination (Thurston called pleated surface a uncrumpled surface)
btw, I recently learned that if $S$ is an incompressible surface in a closed hyperbolic 3-manifold, then either $S$ is quasi-fuchsian or virtual fiber(!!)
My guess is given a quasifuchsian rep $\rho : \pi_1(S) \to PSL_2(\Bbb C)$ there is a way to make a hyperbolic structure on $S \times \Bbb R$, namely take $\Bbb H^3/\mathrm{im}(\rho)$. Then this hyperbolic 3-manifold has a convex core
oh yeah I am thinking compact surface $S$ I suppose
But in this special case I think I understand: "flow backwards" along geodesics escaping to infinity in $S \times \Bbb R$ to deformation retract to a 2-skeleton in $S \times \Bbb R$. your claim is (or Thurston's theorem is) this is a topologically embedded subsurface homeomorphic to $S$, bent along some lamination
And this bending lamination contrasts the ending laminations on the two ends of $S \times \Bbb R$, which parametrizes the space of hyperbolic structures on $S \times \Bbb R$ as well (by Thurston's work)
@onepotatotwopotato Yeah that's the kind of example I have in mind
What is a non-geometrically finite hyperbolic structure on $S \times \Bbb R$ for $S$ closed?
@VivaanDaga Every element of $S_4$ induces an inner automorphism of $S_4$, call it $\phi$. Then we have an action $g\cdot x = \varphi(g) x \varphi(g)^{-1}$ of $S_4$ on the set $S_4$. This action has the property that $g\cdot x = x$ for all $g$ iff $x$ is in the center i.e. $x = e$. If two automorphisms $\varphi_1, \varphi_2$ determine the same action, then $\varphi_1(g)^{-1}\varphi_2(g) x = x\varphi_1(g)^{-1}\varphi_2(g)$ so $\varphi_1(g) = \varphi_2(g)$ i.e. $\varphi_1 = \varphi_2$.
So every element of $S_4$ determines a distinct action on the set $S_4$ with the property that the set of fixed points is its center. In particular we can pick one that is neither $g\cdot x = gxg^{-1}$ nor $g\cdot x = g^{-1}xg$
Do you guys know how can I translate a German paper into English? Google translate doesn't work that good, and I often have to switch back and forth to understand things.
But how is convex core of $S \times \Bbb R$ given by infinite cyclic covering of a mapping torus "the entire manifold"? The convex core has to be some topologically embedded copy of $S$ in $S \times \Bbb R$
It is not $S \times \{0\}$ because that is not quasi-convex (it's a virtual fiber!)
It has to be $S \times \{0\}$ bent along certain lamination
@Jakobian what if we add the condition that the action is not of the form \phi(g) x\phi(g^{-1})$ where $\phi$ is an automorphism. Cause such an example is not in the spirit of the Q.
Hyperbolic structures on $S \times \Bbb R$ come in three flavors. Where the representation in $PSL_2(\Bbb C)$ is fuchsian, where it's quasi-fuchsian, or where its a virtual fiber
In the fuchsian case $S \times 0$ is core
In the virtual fiber case, the entire $S \times \Bbb R$ is core. the information is captured by ending lamination
In the quasi-fuchsian case, there's a bent copy of $S$ inside which is the core. the bending lamination is the information
How useful to the asker do you think your answer is, considering they are not asking for a derivation but for guidance on what do assume to build a derivation? What if we ask the same question about the accepted and well upvoted answer?
So the useful part of the answer is the link, and link only answers are discouraged on MSE. An answer summarizing the contents of the link (which the accepted answer essentially is) would have been better
I'm not really knowledgeable about the Riemann zeta and its relation to differential topology so I'm not sure what the video is about. But this seemed to be the gist of it
maybe @BalarkaSen does
but yeah the feedback would probably be that you don't explain what you're talking about clearly enough. For example that its about the Riemann zeta function
If you keep the notebook on a table and hold the camera from above then it will be easier for you to explain things as then you can see which part you are pointing at.
@Jakobian on "R^2" in the video, we have $$\Delta_t(x)=\lim_{r\to \infty} \frac{1}{r} \sum_{n=1}^\infty e^{\frac{t\log n}{\log r \log x}}$$ which is a controlled deformation of the riemann zeta function composed with -1/lnx
then direct analytic continuation we can obtain the meromorphic zeta function on the entire complex plane aside from a simple pole and a branch from the logarithm
$\Delta_t(x)$ uniformly converges to $\Delta_t(x)=e^{\frac{t}{\log x}} $ so that one can use that as the base graph and construct a section over it with a prescribed topology. Then you have to deform the real analytic fibers on the lifted "surface" into holomorphic ones. since the $\Delta_t(x)$ is a solution to the well posed diffusion equation running forward in time $\frac{\partial^2}{\partial t^2}\Delta_t(x)=-x\frac{\partial}{\partial x}\Delta_t(x) $ it's possible to construct an anolougs pde
trying to think of a pithy way to say: "the only reason i spotted your mistake as fast as I did is because I've made that exact mistake many times myself"
@oneofvalts Its very simple, for a pair of functions $(\sigma, \tau)\in S_k\times S_m$ consider the map $f(i) = \sigma(i)$ for $1\leq i\leq k$ and $f(k+i) = k+\tau(i)$ for $1\leq i\leq m$
Ah, it's @Semiclassical. Maybe I am telling you this the second time, but anyways: Here I asked my first question about seven years ago. I was in high school. I think it was a question about the definition of derivative. You were very helpful and patient. Now I am in my third year in university. Thank you :)
there is many copies of $S_k\times S_m$ in $S_n$ as you saw (first choice we made was splitting of the set $\{1, ..., n\}$ and second is in the choice of our bijections), but the point is that one such copy can be obtained
and then once you get one copy, you get that $\frac{n!}{k!m!}$ is an integer
I suppose, bonus exercise: Calculate the amount of copies of $S_k\times S_m$ in $S_n$ using this method
Hi, I'm reading about distribution theory and I'm not understanding this: the notion of the (standard) convergence in $\mathcal D$ (space of test functions) is not compatible with any norm, but we can define seminorms.
note that there is some variation in how people define 'test functions' and the relevant seminorms, but those questions/answers are wrestling with exactly that issue
@SineoftheTime $\mathcal{D}$ is a complete vector space, if it were metrizable then by exhausting your base set by compact sets $K_n$, we'd have that $\mathcal{D}$ is a union of closed sets $\mathcal{D}_{K_n}$ of functions with support contained in $K_n$
But $\mathcal{D}_{K_n}$ have empty interior so that would contradict Baire category theorem
i would mentally file this under "there is an explanation for why any analysis of convergence issues is slightly more complicated than norm convergence"
but the explanation itself is maybe not all that illuminating until you are steeped in the subject matter
the meta-meta-mathematical explanation for why there isn't a norm that gives you the notion of convergence that you want is that if there were, every textbook would use it, instead of doing what they actually do
sort of an economists argument. it must be efficient because it's what everybody does, textbook writers have every incentive to arbitrage away any inefficiencies arising from the use of seminorms
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I'm stuck at something basic. Suppose $f$ is continuous, strictly increasing and say, surjective (these are assumptions made in a video I was watching about image measures). I'm trying to show $f^{-1}([a,b))=[f^{-1}(a),f^{-1}(b))$. Here's my attempt.
If $x\in f^{-1}([a,b))$, then $f(x)\in[a,b)$. Since $f$ is strictly increasing, so is the inverse, and from this it follows that $a\leq f(x)< b\iff f^{-1}(a)\leq x< f^{-1}(b)$, which shows $\subset$.
