Mathematics

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May 30 22:50
@XanderHenderson John Gabriel would probably say that's because you're one of them.
May 30 22:48
Actually the fad these days is John Gabriel
May 30 22:46
Coming very close to N J A Wildberger territory here, folks.
May 30 20:28
HELP as in Homotopy Extension and Lifting Properties, of course.
May 30 20:28
@BenSteffan Always remember, Peter May is there to HELP you
May 30 20:19
@VladimirLysikov Biblically-accurate $\Bbb Z/p\Bbb Z$
May 28 14:46
I honestly do not know what the point is
May 28 14:37
@Thorgott welcome to homotopically stratified sets
May 5 18:40
@BenSteffan average comma 2-comonads comrade
May 3 14:45
@onepotatotwopotato Having to think about them, unfortunately, yeah.
May 2 18:17
tall talk from someone who studied fractals
May 2 18:17
lol
May 2 17:50
I think the answer is "no" but I have been struggling to prove this.
May 2 17:50
Question: Is there a homotopy $f_t : S \to \mathbf{R}^2$ between $f_0$ and $f_1$ such that (a) $f_t$ moves points by a uniformly bounded distance, i.e. there is a $C > 0$ such that for all $x \in S$, the path $\{f_t(x) : t \in I\}$ has diameter at most $C$ in $\mathbf{R}^2$, and (b) for every $t \in I$, $f_t(S) \subset \mathbf{R}^2$ is a $1$-dimensional piecewise smooth CW complex
May 2 17:48
There is a continuous, nearest point projection $p_0 : S \to K_0$ and a projection $p_1 : S \to K_1$. Let $\pi : \mathbf{R}^3 \to \mathbf{R}^2$ be the projection to the $xy$-plane. Let $f_i : \pi \circ p_i$, $i = 0, 1$. Then $f_0, f_1 : S \to \mathbf{R}^2$ are two maps whose images are the grids given by joining the nearest neighbour vertices of the lattices $\mathbf{Z}^2$ and $\mathbf{Z}^2 + 1/2(1, 1)$, respectively.
May 2 17:46
May 2 17:46
I have a very strange question. Let $K_0, K_1 \subset \mathbf{R}^3$ be the graphs given by joining the nearest neighbour vertices by edges of the lattices $\mathbf{Z}^3$ and $\mathbf{Z}^3 + 1/2(1, 1, 1)$, respectively. Let $S$ be the infinite genus surface ("infinite jungle gym") which is equidistant from both $K_0$ and $K_1$, here is an image:
May 2 17:22
@BenSteffan Yes, first choose a function $f : M \to \mathbf{R}$ such that $f^{-1}(c) = \partial M$ for some $c \in \mathbf{R}$. Then approximate $f$ rel $\partial M$ to be Morse. Then the CW structure on $M$ rel $\partial M$ is given by union of all the stable cells corresponding to all the Morse critical points. That is, this tells you how to obtain $M$ from $\partial M \times [0, \varepsilon]$ by attaching cells to $\partial M \times \{\varepsilon\}$.
Apr 23 21:51
I think for me $3\Bbb{RP}^2 = T^2#\Bbb{RP}^2$ is better. You want to go from $aba^{-1}b^{-1}cc$ to $aabbcc$. Slide $b^{-1}$ over $c$ to get $aba^{-1}cbc$. Then slide $a^{-1}$ over $c$ to get $abcabc$. This is disk with three Mobius strips attached, everything interlocking with everything.
Apr 23 21:46
@Thorgott Yes. If I remember correctly you need a couple more handleslides, because just doing the one you mention will tangle all the handles together.
Apr 23 12:54
it was just a string of arrows on a line, spanning 10 lines lol
Apr 23 12:54
they expect me to cut and paste polygons but i refuse
Apr 23 12:53
@leslietownes np. i remember writing a proof of (Klein bottle) # RP^2 = T^2 # RP^2 using "one dimensional Kirby diagrams" like this in my quals. pretty sure the guy who checked it was like wtf
Apr 22 19:00
dont feed the Ben Steffan
Apr 22 18:59
i had to call a quantum mechanic a few days ago to fix my vents
Apr 22 18:58
@Semiclassical aka "good enough"
Apr 22 18:53
lol
Apr 22 18:52
Correction: when I wrote red circle above I really meant red segments/arcs.
Apr 22 18:49
The point, then, is one cannot switch the order of two points on a line by an isotopy (ie without the points colliding with each other), but one can switch the order of two disks on a plane (just swoosh it around by a 180 rotation).
Apr 22 18:48
Same for (b). When taking product with $I = [0, 1]$ of either picture one just fattens the picture laterally, and every attachment arc becomes an attachment square (or, as I drew, a disk).
Apr 22 18:47
Which is a torus with a hole.
Apr 22 18:47
@leslietownes What one gets is the second picture that leslie drew here
Apr 22 18:47
The key to interpret it is to think about (a) as follows: take a disk, and identify the black line with an arc on the boundary circle. Attach a strip by pasting opposite ends of it to the red circle (the vector tells you how to identify the strip -- whether it will be a standard strip or a twisted strip). Do the same with the blue arcs.
Apr 22 18:45
Yeah these are called Kirby diagrams. Any manifold has one.
Apr 22 18:44
@hbghlyj It's an obnoxiously cryptic proof of this.
Apr 22 18:40
@leslietownes @Thorgott
Apr 22 18:40
Feb 23 18:41
@Thorgott Looks right to me.
Feb 23 18:40
@Thorgott Take a circle $S^1 \subset \Bbb R^2$ centered at $0$ and of radius $1/2$ and consider $f : S^1 \to [-1, 1]$ as the height function. Then $f$ is submersive on $f^{-1}[\pm 1-\varepsilon, \pm 1 + \varepsilon]$, as this is the empty set. Any map from the empty set is vacuously submersive(?)
Feb 22 21:53
And this does not seem to me to have much to do with $(\pi, f) : W \times [a_0, a_1] \to X \times [a_0, a_1]$ being a fiber bundle. As the example above shows, that can be false.
Feb 22 21:52
So they glue to give a fiber bundle $\pi : W \to X$.
Feb 22 21:51
Essentially following the proof of Ehresmann's theorem. This suggests to me that the trivializations are compatible over $X$. That is, in fact, $\pi : (\pi, f)^{-1}(X \times (a_i- \epsilon, a_i+\epsilon)) \to X$ is a fiber bundle and the trivializations are compatible with that of $\pi : W[a_0, a_1] \to X$.
Feb 22 21:49
@Thorgott The trivializations of $(\pi, f) : (\pi, f)^{-1}(X \times (a_i-\epsilon, a_i+\epsilon)) \to X \times (a_i-\epsilon, a_i+\epsilon)$ (for $i = 0, 1$) and $\pi : W[a_0, a_1] \to X$ are both obtained from the following fashion:

In the first case, choose local vector fields $\partial/\partial x_1, \cdots, \partial/\partial x_n$ and $\partial/\partial t$, lift them up by the submersivity property of $(\pi, f)$ and then flow along them to generate a trivializing chart. In the second case, use only $\partial/\partial x_1, \cdots, \partial/\partial x_n$, lift them up by the submersivity p
Feb 22 21:33
It satisfies all the conditions you are asking for, yet $(\pi, f) : W[-1, 1] \to X \times [-1, 1]$ is not a fiber bundle. It's the standard height function on a circle.
Feb 22 21:30
This situation seems possible to me. Take $X = \{pt\}$ and $W \subset X \times (-1-\epsilon, 1+\epsilon) \times \Bbb R$ be a circle embedded in a square.
Feb 22 21:21
So the only way that $(\pi, f)$, restricted to the interior of $W[a_0, a_1]$, can fail to be submersive is if $f$ has a critical point in the interior of $W[a_0, a_1]$?
Feb 22 21:10
Am I right in saying that $(\pi, f)$ maps the interior of $W[a_0, a_1]$ to the interior of $X \times [a_0, a_1]$?
Feb 22 21:05
Don't you want to know whether $(\pi, f) : W[a_0, a_1] \to X \times [a_0, a_1]$ is a fiber bundle or not?
Feb 22 21:01
I think if you can show that you are in good shape. If $q : (M, \partial M) \to N$ is a proper map and $q|M^\circ : M^\circ \to N$ and $q|\partial M : \partial M \to N$ are both surjective submersions, then $q$ is a fiber bundle.