Of course, like everyone else, the Danish government wants a Corona-app. And like elsewhere, the government and experts are arguing over how best to implement it.
Mathematicians like to ask covering various sets with open intervals
and the answers to these riddles have strange tendencies to become
strange lemma's or theorems Heine-Borel Theorem, lebesgue's Number
Lemma, Vitali Covering Lemma, Besicovitch Covering Theorem, (etc)
-3blue1brown
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In the mathematical area of topology, a train track is a family of curves embedded on a surface, meeting the following conditions: The curves meet at a finite set of vertices called switches. Away from the switches, the curves are smooth and do not touch each other. At each switch, three curves meet with the same tangent line, with two curves entering from one direction and one from the other.
I mean branched manifold. If $X_1$ is a manifold, $X_2$ is isometric to $X_1$ and they are transversal, with $N^2$ double points, that are associated to lattice with intrinsic positive curvature, then consider diff(B), the diffeomorphism group of the branched manifold B consisting of $N^2$ double points. Diff(B) is a discrete group correct?
so, the mapping class group could be used to represent the symmetries of the branched manifold B?
$MCG(B)=\pi_0(Diff(B)).$ not sure if this holds tho
What is the direct sum of unitary representations? How is it defined? From what I understand, it doesn't correspond to the direct sum of the underlying Hilbert spaces.
Also the usual notion of dimension like the Lebesgue covering dimension pick the maximum of the "local dimensions"
Like if you have a ball with a plane emanating from its equator the covering dimension is $3$, because the ball is $3$ dimensional, but (with the standard metric inherited from $\Bbb R^3$) the asymptotic dimension is $2$, because it looks $2$-dimensional at infinity (outside of a compact set)
I see. Because the ball is compact, at the large scale, its as good as not being there at all. Like I could have a 7d-ball and a ray sticking out of it, it will be 1-d.
@Alessandro ah of course Gromov! :) do you have any idea how you'd compute the asymptotic dimension of a mapping class group of an orientable finite type surface? It must have finite dimension of course. But is it hard to compute?
in case anyone wants to jump in...A surface S has finite type if there exists a compact surface F and a finite subset A⊂F such that S is homeomorphic to F−A
$f$ is a continuous function from a compact metric space $X$ to metric space $Y$, then I think $f$ is uniformly continuous and hence $f(x)$ is also continuous?
@geocalc33 Its from Bott&Tu btw. They use it in all the place: proof of finite-dimensionality of de Rham cohomology, Poincaré duality, Kunneth formula etc...
@feynhat ah cool. Consider a surface $U=(0,1)^2$ in the real plane. Decompose $U$ into an infinite set of real analytic functions which form a family, $F_s=\{f_{s}(x):s\in \Bbb R_{>0}\},$ with real parameter $s$, s.t. $f_{s_n}(x)f_{s_k}(x)=f_{s_n+s_k}(x)$ for some $n,k\in \Bbb N.$ Require $f(0)=1$ and $f(1)=0.$ Decompose $U$ again into $F^*_s=\{f_s(1-x):s \in \Bbb R_{>0}\}$ requiring $f(0)=0$ and $f(1)=1,$ and $f_{s_n}(1-x)f_{s_k}(1-x)=f_{s_n+s_k}(1-x).$ Take $K=F_s \cup F^*_s$ to be topologically equivalent to the surface $U.$
The above should still be true but I don't know the equivalence. I can sketch one direction but it will be sketchy.
Do you believe me that a compact manifold with boundary admits a finite "good-with-boundary" cover, in which every intersection of open sets is empty, diffeomorphic to Euclidean space, or diffeomorphic to Euclidean half-space?
I can find a finite good cover for the interior, and then for the boundary. I will extend these boundary neighborhoods to interior... and the take the union of this collection.
what. no. how can I find finite good cover for interior? this is not true (my argument, i mean).
@feynhat You can find a finite good cover near the boundary (do a sort of "halfspace exponentiation", where you have chosen the metric so that the boundary is totally geodesic, aka the exponential actually preserves the boundary). This covers an open set containing the boundary, whose complement is compact.
Now do the usual kind of Riemannian exponential for your charts for points near that compact set.
(You had the right idea, you just did them in the wrong order!)
Now deleting the boundary gives you a finite good cover (in the usual sense) of a noncompact manifold.
It's not hard to see that a surface with a finite good (or good-with-boundary) cover has H^1(S;Z/2) finite dimensional. And it follows without too much difficulty from the fact that noncompact surfaces have compact exhaustions, and the classification of compact surfaces, that such a surface is obtained by a compact surface from deleting finitely many points in the interior
(does anyone have any idea how to compute the asymptotic dimension of a mapping class group of an orientable finite type surface?) If the surface is $S=[0,1]^2$ then the asymptotic dimension is probably $<\infty$
I don't know what you can say in higher dimensions, there's a theorem of Stallings characterizing manifolds which you can compactify to one with boundary
Does anyone here have any experience with Google Classroom? If it says, "Due Tomorrow, 9:00AM" and its 3AM right now, does it mean its due this morning? Or the next one?
I'm confused why we have one $\lambda$ for all $f$. I only see that they argued that for all non-zero scalar multiples of (our chosen) $f$ we have the same $\lambda$, but what about an $f'$ which is not a scalar multiple of $f$?
Here $\hat\gamma$ sends an operator $U$ to the operator defined by $\hat\gamma U(\phi)=\overline{Uf}$, where $\phi=\overline f$.
My argument now is: let $f$ be as given. Consider $\mathbb C f$. We know its orthogonal complement exists. So we can write any $g\in\mathbb H$ as $g=f^\perp+\mu f$. Using linearity of $U$, we obtain $U(g)=U(f^\perp+\mu f)=\lambda' f^\perp +\lambda\mu f$. For $U$ to be truly linear, we need $\lambda'=\lambda$.
@ShaVuklia Something to do with the fact that if every subspace is invariant under a linear map, then the map is a scalar multiple of identity. (just guessing though. I don't have much idea what going on there)
In general the way they've defined this (the image of unitaries in $\text{Aut}(\Bbb P)$) is probably the least insightful way you could possibly define this object
Lifting those to the universal cover seems the way to go, and from there on I have to determine irr. unitary representations of the special linear group
Oh, interesting. Well, I've been doing all my regular exercises for my neck, so I need the actual massages and manipulations, so I haven't tried tele-stuff for this. It's bad enough for teaching mathematics :(
@ShaVuklia OK. I need to think for a while to see if I have anything to say. I don't know the literature on this subject, but it's related to things I know. Are you sure your question is answered in that book?
I'll catch you both up later, I'm basically checking this for a moment in the middle of work with a deadline --- I should focus instead of getting into a long chat :)
@MikeMiller I believe so yes. I went through a bunch of possible literature, and I seemed most pleased with the treatment in this one. My plan is to go roughly through the relevant theory for me tomorrow. If I feel unsatisfied about the treatment, I might ask you tomorrow or the day after if something came up to you.
And I sure believe they make simplifying assumptions on the spaces they're working with, as their goal is physics-oriented, which probably makes the treatment a bit less abstract (and I'm fine with that for the moment).
What am I missing here? I need to show that $\int_D \frac{\dot{\beta}(t)\dot{\beta}(t)}{\vert\vert\dot{\beta}(t)\vert\vert} dt = \vert\vert\dot{\beta}(t)\vert\vert$.
Let $f$ be monotonically increasing on $(a, b)$. Then for any point $x$ of $(a, b)$, we have Both $f(x^+)$ and $f(x-)$ exists ? I think it should be $f(x^+) > f(x^-)$ ?
Mathematicians like to ask covering various sets with open intervals
and the answers to these riddles have strange tendencies to become
strange lemma's or theorems Heine-Borel Theorem, lebesgue's Number
Lemma, Vitali Covering Lemma, Besicovitch Covering Theorem, (etc)
-3blue1brown
By...
