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12:00 AM
Told you so.
 
 
2 hours later…
1:35 AM
@Thor Welcome back!
 
1:46 AM
hi @Ted
long time no talk!
please excuse the ad hoc entrance above
 
You’re up at weird hours yet again!
 
2:02 AM
Hi :-)
 
some habits die hard :P
that said, i probably should be going to sleep shortly
 
timezones...
It's 7:35 am here.
 
It's 10:07pm here
 
how do timezones affect economy?
if a country increases its number of timezones say from 2 to 5, then what happens to economy?
 
in terms of changes in spending?
 
2:11 AM
maybe depends on the size of the country?
 
@shintuku in terms of GDP
 
probably a slight drop in real GDP if people lose sleep or slight increase in real GDP if people gain sleep, wild guess tho
drop in rate of increase of real GDP and increase in rate of increase of real GDP***
every real way this would impact people's live would somehow influence real GDP growth rate
like, some trade procedures could be made more efficient because now the country can more adequately communicate with another country which was barely within common shared waking time
sending an email that would usually be barely missed wouldn't anymore, etc.
 
I think the optimal number of time zones $T$ to maximize GDP for a country of $X$ area is $X/T$
 
gotta factor in the expenses of implementing this change too
gotta factor in the increase in production due to the implementation of the new system too
 
well time zones are actually
bad for business
think about logistics/scheduling. It becomes more complex in the average case
having one single timezone would be best
 
2:28 AM
the first article I pull up on google suggest time zone differences have an impact on trade
so more time zones may slightly diminish that difference, etc.
 
2:51 AM
Professor @TedShifrin do you have any experience/opinion about the harkness table teaching method?
 
3:28 AM
@user4539917 No. seminar discussion works better in the humanities and perhaps at the graduate seminar level. I love discussions, but relying on students to run them and generate content effectively won’t get very far with content and developing intuition.
 
morning.
@TedShifrin :p
 
Thanks for responding sir.
 
teddy are you interested in publishing papers still (and would you be open for ideas, or cant be arsed?)
 
No.
 
: (
 
 
2 hours later…
5:50 AM
: )
 
: )
 
6:20 AM
What does P(A1, A2, ....., Ai) mean here?
Uh, never mind, I get it
 
Hey guys! I've been recently studying the properties of a coupler curve generated by a four-bar linkage. (Studying engineering like a good engineer :p) The book I have been following started considering the points at infinity/asymptotes of the coupler curve in a rather vague/loose fashion, so I went off and did some of my own research into projective geometry & homogeneous coordinates to get a better understanding.
Now, the question I want to ask relates to finding asymptotes of curves when the point(s) at infinity are complex, as in both x & y are complex numbers. The procedure I understand for fiding the equation of the asymptote of a curve $F(x,y) = 0$ when the point at infinity is real is as follows:
1. Homogenezie the equation $F(x, y) = 0$ by substiting $x = X/W$ and $y = Y/W$; simplify the expression
2. Find points at infinity by setting W = 0 and solving the equation for Y; let's say one of the points at infinity have the homogenous coordinates of [a, b, 0]
It seems like to me on the surface level this process still works when the point at infinity is a complex - for instance, finding the equation of the asymptote for a cirlce $x^2 + y^2 = R^2$ using the process described above gives me the same equation as my textbook
However, isn't the notion of a differential/total differential for a real function and a complex function entirely different? Is there some intuitive/heuristic reasoning why this process still works when the point at infinity is complex?
Furthermore, when points at infinty are normally introduced (at least on the web), people normally say the point at infinity is essentially the slope of the line or that it codifies this information. How does this notion translate to the complex numbers? I mean if both x and y are allowed to be complex, is there even a clean notion for the direction of a line?
 
 
3 hours later…
9:57 AM
is there another way to write $e^{\frac{d}{dx}}$ in operator theory?
And does this have any interpretation as an operator? $e^{\frac{d}{dx}A(x)}$
I see this written on the wiki page: $e^{t \beta(x) \frac d {dx}} f(x)$ but there's extra stuff here
In mathematics, and in particular functional analysis, the shift operator also known as translation operator is an operator that takes a function x ↦ f(x) to its translation x ↦ f(x + a). In time series analysis, the shift operator is called the lag operator. Shift operators are examples of linear operators, important for their simplicity and natural occurrence. The shift operator action on functions of a real variable plays an important role in harmonic analysis, for example, it appears in the definitions of almost periodic functions, positive definite functions, derivatives, and convolution....
 
10:24 AM
last question, if you write $e^{s\frac{d}{dx}B(x)}$ does this mean $B(x)$ is being transformed by the operator?
 
 
2 hours later…
12:13 PM
If i draw a triangle in the real plane, with the same lenght of each side, lets say it is of lenght one. If i want to take the subset of points of one side, can i translate lenght into some attribute of the set. obviously i can not correlate to cardinality, since the cardinality will be infinity, but then i wont have a meassurement to compare lenghts of different sides, since the subsets will have the same cardinality
Infact, for any Line, side whatever, it does not have to be a triangle, any line draw anywhere, can i correspond the geometrical lenght of the line to an attribute of the set, containing those points.
 
the maximal distance between two points in the set?
 
that makes sense.
good comment.
How does this translate, if i pick the set as the whole triangle, and want to get the parameter?
Then i need to do some curve integrals. right, but from the set it self, i can not conclude anything?
 
I have no idea what you're asking
 
collect the points of the triangle and put them into a set
Can you read upon an attribute of the set of points, what is the parameter lenght of the triangle
I guess you can define something like to check if the set can be written as a sum of subsets where each subset consits only of the points spanned by the vector on the direction of the side, and build upon each subset the metric you mentioned and then sum
 
12:35 PM
@MadSpaces the set containing coordinates of end points of a side of the triangle
(if you are allowed to use the fact that given two distinct points, a unique line passes through them.)
 
 
2 hours later…
2:49 PM
Just found this YouTube channel
Seems really good!
 
3:25 PM
Guys could someone guide me with this problem pls, Let $A$ a matrix $n \times n$ with coefficients in a field $F$. show that there exist $a_{1}, a_{2}, \ldots, a_{n}$ de $F$ such that $A^{n}+a_{1} A^{n-1}+a_{2} A^{n-2}+\cdots+a_{n-1} A+a_{n} I_{n}=0_{n}$
 
Koro its not homework, its just me wondering, doesnt matter
@AlekMurt search for "Minimal Polynom"
 
4:11 PM
@AlekMurt Cayley-Hamilton comes to mind.
 
4:31 PM
@AlekMurt I think this paper deals with it but I forget how it goes: maa.org/sites/default/files/pdf/awards/Axler-Ford-1996.pdf
 
4:58 PM
Thank you for the answers guys
 
5:24 PM
@TedShifrin math.stackexchange.com/a/3291377/695930 To your answer here, i have calculated the form of the symplectic standard form , and it came to my attention, if n is equal to 1 , then you are multiply with the matrix representing the imaginary unit so you are rotating by 90 degrees. is there some meaning to this sorcerry
 
5:39 PM
0
Q: How can I draw $f(S)$ with $f(z)=z^{1+i}$ where $S=\{0<Arg(z)<\pi/6\}$?

Wave Let $f:\Bbb{C}\setminus(-\infty, 0]\rightarrow \Bbb{C}$ where $f(z)=z^{1+i}:=e^{(1+i)\log(z)}$. We consider $$S=\{0<Arg(z)<\pi/6\}$$ I want to draw $f(S)$. If I take $z\in S$ then $z=re^{it}$ where $r\in [0,\infty)$ and $t\in (0,\pi/6)$. Then $\log(z)=\log(r)+it$ where now $\log(r)\in (-\infty...

can maybe someone help me here
 
6:31 PM
Call a function $f:\Bbb Z\to\Bbb Z$ almost linear if $f(m+n)-f(m)-f(n)$ is bounded. Denote the set of almost linear functions by ${\rm AL}(\Bbb Z)$. Define an equivalence relation $\sim$ by $f\sim g$ iff $f-g$ is bounded.
Define $f<g$ if $f$ is eventually less than $g$.
Then $({\rm AL}(\Bbb Z)/\sim,+,\circ,<)$ is an ordered field
 
@MadSpaces For starters, one is a bilinear form and one is a linear map. But, yes, rotation by $\pi/2$ comes in there. Rotate $z$ and compute the real inner product of that with $w$. This is alternating.
 
6:50 PM
The construction I gave is called the Eudoxus reals (compare, eg, the Dedekind reals or the Cauchy reals)
 
7:21 PM
Hi @TedShifrin, @AkivaWeinberger
 
Oh, hey
 
@Koro Congrats! The admission acceptance was well-deserved.
 
How goes it
 
Trying to understand what a loose chart is
Would you be interested to hear
 
I assume (from the name) it's like a chart but if you shake it it falls apart
Is this "chart" like "atlas"?
 
7:25 PM
@AkivaWeinberger Pretty much
Yes, chart as in atlas.
 
I'm guessing it defines a manifold-like object that somehow has less information
 
Actually, these are charts on manifolds with more structures.
Namely, certain submanifolds of so-called contact manifolds
Here's some background. A contact manifold is a $(2n+1)$-manifold where every point has a smoothly-varying hyperplane (rank $2n$) stuck at each point, such that locally it looks like $\Bbb R^{2n+1}$ with coordinates $(x_1, \cdots, x_n, y_1, \cdots, y_n, z)$ and hyperplane distribution $dz = \sum_i y_i dx_i$.
This is a "natural definition", in the sense that such hyperplane distributions are "maximally twisty"
In particular, there's no $2n$-submanifold which is tangent to all the hyperplanes on any neighborhood of any point, say.
 
So I'm 3D this is $dz/dx=y$?
 
Precisely.
So in 3D you have Legendrian knots, which are knots that satisfy that differential equation at every point (i.e., tangent to all the planes at every point). In general a Legendrian submanifold is an $n$-dimensional submanifold of a $(2n+1)$-dimensional contact manifold that is tangent to all the planes at every point. Turns out there's a lot of these.
 
@BalarkaSen A nonintegrability condition
@BalarkaSen Arright
I'm curious where this sort of idea comes from
 
7:32 PM
Correct. But it's stronger than that particular kind of nonintegrability; a contact distribution is in some sense "maximally nonintegrable". For example, $n$ is the maximal dimension where you can have submanifolds of tangency
 
(the idea behind contact geometry)
 
$n+1$ and everything fails
 
@BalarkaSen It is the worst infinitesimal jigsaw puzzle.
 
100%
 
I wonder what the discrete version of this would be
A tet mesh with specified subplanes of each tetrahedron, maybe? Dunno
(Fun fact: I'm pretty sure deciding whether a 5D simplicial mesh is manifold is undecidable)
 
7:34 PM
 
Either 5D or 4D, forget exactly
'cause you need each vertex figure to be a sphere, but telling whether something's a 4-sphere is also hard
 
This is the Cayley graph of the Heisenberg group, matrices of the form I + M where M is strictly upper triangular with integer entries.
 
Oh, interesting
Reminds me of nil geometry
(or is it Solv?)
 
If you let the mesh of this lattice go to 0, then this becomes the geometry of $(\Bbb R^3, \xi)$
 
One of the Thurston ones
 
7:36 PM
@AkivaWeinberger These are exactly the contact geometry, yeah
Not Solv, that can be terrible
@AkivaWeinberger This is related to Hauptvermutung I think
 
> German for main conjecture. It is an abbreviation for die Hauptvermutung der kombinatorischen Topologie, which translates as the main conjecture of combinatorial topology
 
i love Hauptvermutung. i have all of their records. they got a little too into synths in the 1980s.
 
- Wikipedia
You can't just go around calling a conjecture "main conjecture"
 
pages Munchkin and her fists
 
At least "decision problem" makes sense
I suppose in an alternate universe we'd talk about the Halteproblem
(solved by Türing)
So loose charts are charts with these contact structures? @BalarkaSen
In what sense are they "loose"?
 
7:41 PM
@AkivaWeinberger I think the origin story comes from wavefront and caustics. But I don't know much about that. I like the unicycle example; if you have a unicycle in 2D plane then it has three parameters $(x, y)$ for position and $\theta$ for angle of the axis of steering. These have to satisfy the ODE that $(x', y')$ is in the direction of $(\cos(\theta), \sin(\theta))$, so trajectories of the unicycle are paths in $S^1 \times \Bbb R^2$ s.t. $\cos(\theta) dx + \sin(\theta) dy = 0$
Or $dy = \tan(\theta) dx$.
 
@AkivaWeinberger Well, far be it for me to use a vulgarity, but, shall we say, they have "been with" a large number of partners.
 
Which is pretty much $dz = y dx$
In some chart
 
@BalarkaSen Oh, interesting
 
Since you can travel from any point to any other point in a unicycle (well, most likely a bicycle, same difference) or park your car in tight spots, this tells you there's lots of Legendrians
Curves at least.
But also true in high dimensions
 
By the way, I searched your name in chat to remind myself where we were in our conversations
@BalarkaSen That's a very good intuition, I like that
 
7:45 PM
@AkivaWeinberger glomp?
 
@AkivaWeinberger Loose charts are certain charts on Legendrians, rather than on contact manifolds. Let's start with a prereq.
 
Well, we have to disallow $\tan\theta=\infty$, so $\theta$ is restricted to $(-\pi,\pi)$ (on the unicycle)
 
Yep. That was something I wrote on a chart
On the other chart you have $dx = \cot(\theta) dy$
 
Ah
and so your manifold is $\Bbb R^2\times S^1$
 
Indeed
Note that $(t^2/2, t, t^3/3)$ is a Legendrian on $(\Bbb R^3, \xi = \{dz - y dx = 0\})$.
Because $d(t^3/3) = t^2 dt = t \cdot d(t^2/2)$
Project to $xz$-plane (known as the front projection), you get a semicubical cusp $t^2/2, t^3/3$. This is like, you come in, park a car, and then unpark it out
 
7:50 PM
Sure. And as long as the projection is never vertical it's possible to pick *y
 
And if the only singularities, where both $z' = x' = 0$, are semicubical cusps.
Take a zig-zag:
This is the front projection of a Legendrian in $\Bbb R^3$. We call it a Legendrian zig-zag. Loose charts are certain higher-dimensional generalizations of these.
 
Is that isotopic to a straight line?
maintaining the contact
 
Nope!
 
@BalarkaSen Wait, how is this a chart?
 
This little piece is a chart on a Legendrian 1-submanifold of $\Bbb R^3$.
Or Legendrian 1-submanifolds of contact 3-manifolds in general
 
7:54 PM
What's the domain of the map, the whole circle or just the line?
('cause charts are maps, yeah?)
 
OK, let me say that very precisely. Let $C = [-1/2, 1/2]^3 \subset \Bbb R^3$, and $\Lambda_0 \subset C$ be the Legendrian segment which is a zig-zag in the middle, and matches with the $x$-axis in $C$ towards the boundary of $C$
$(C, \Lambda_0)$ is a zig-zag chart for a Legendrian curve in a contact 3-manifold.
That is, if you can produce a contactmorphism (a diffeomorphism preserving the tangent spaces/contact structure) from $C$ to some other closed neighborhood in a contact $3$-manifold $(M, \xi)$ at a point $p$ which takes $\Lambda_0$ to a a piece of the Legendrian curve passing through $p$, you say you have a zig-zag chart
 
Ah, and there are no contactmorphisms from this to a straight line.
 
Right.
Fixing the boundary.
 
Oh, OK.
What do you mean by "matches with the x-axis in C towards the boundary of C"?
Your picture doesn't look like it does that
 
The endpoints of the zig-zag as it exits $C$ should be $(-1/2, 0, 0)$ and $(0, 0, 1/2)$ and the tangent vector should agree with $\partial/\partial x$ as it does that.
@AkivaWeinberger Yeah that's because I pilfered it off the internet, but it's easy to arrange that.
 
8:04 PM
Like this?
 
Yup
 
Gotcha
So does that mean a closed Legendrian loop has a fixed number of zigzags?
 
Ah no, but what is Legendrian isotopy invariant is #(total number of signed crossings) - 1/2 #(number of cusps)
This is the so-called Thurston-Bennequin number. You can cancel cusps at the cost of crossings, or vice versa.
Think of the Legendrian version of the 1st Reidemeister move
 
Makes sense
Does that still make sense in an arbitrary manifold?
 
@AkivaWeinberger There is a higher dimensional Thurston-Bennequin number, yeah. This is in the Appendix of the paper I am reading
So I'd have to read and report back to you how it works
 
8:09 PM
@BalarkaSen In an arbitrary 3-manifold even
 
Oh yeah, it makes sense in arbitrary 3-manifolds.
#(total number of signed crossings) is the writhe of the knot
Which makes sense anywhere. Take some pushoff and count the linking number.
 
Ah, sure. Instead of picking a plane to project onto, you pick a ribbon
 
Yeah
That works too I think
 
I mean, you have to pick a ribbon for "pushoff" to make sense
It's a choice of pushoff directions
though, with contact geometry being as it is, I think that actually chooses it for you effectively
I bet that's what this is
 
Good point. Yes, there's a Reeb-direction; the orthocomplement of each plane.
 
8:13 PM
@BalarkaSen Wait, what does "linking number" mean if our loops aren't nullhomotopic
 
Er, too many hard questions.
Not sure, would have to think
Yeah don't think this makes sense except for nullhomologous knots
 
I feel like morally it should still make sense (for linking numbers of ribbons, at least rather than arbitrary pairs of loops) but I don't know how
 
Try $S^1 \times \Bbb R^2$. $S^1 \times \{0\}$ is a Legendrian, but if you perturb I feel like you can introduce arbitrarily many squiggles
 
Perhaps
 
$S^1 \times \{0\}$ is your car standing still. Nearby you have a car which does little short parks in a circularly arranged parking lot, circle of radius $\varepsilon > 0$
These feel Legendrian isotopic?
Or maybe I'm just dreaming
Ah OK. It's not a smooth isotopy in $S^1 \times \Bbb R^2$.
Maybe
Heh, I guess $S^1 \times \{0\}$ is like one of those 360 rotation car stunts
Yeah I'm convinced there's a smooth isotopy like this.
@AkivaWeinberger So, want to hear what a loose Legendrian is, finally?
 
8:30 PM
Oh, sure
I will say, though -
I'm on an airplane and we're gonna take off soonish
(heading to Seattle)
 
I can write it down, you can check it out later
 
so I'm gonna lose Internet soon
@BalarkaSen Sure
 
Let me explain what a loose Legendrian looks like in $\Bbb R^5$. Consider $\Bbb R^5 = \Bbb R^3 \times \Bbb R^2$ with coordinates $(x, y, z, p, q)$, and contact distribution $dz = ydx + p dq$. A Legendrian here will be 2-dimensional
 
Say hi to Seattle from me!
 
Let $B \subset \Bbb R^2$ be the box $B = \{|p| < \rho, |q| < \rho\}$, and $Q \subset B$ be the subset $\{p = 0\}$. Then $\Lambda_0 \times Q$ is a Legendrian contained in $C \times B$, because we have $dz = ydx$ and $pdq = 0$.
This is what a loose Legendrian is
@AkivaWeinberger Looks like this x interval
Two parallel cuspidal ridges
That seems fine and dandy. There's a very important caveat, which is that one also puts the hypothesis that $\rho > 1$ in the definition of a loose Legendrian, and this is somehow crucial
 
8:36 PM
A loose Lagrangian is one where one of the coordinates is 0?
and its pair is everything?
 
Yeah
 
So why "loose"?
 
zig-zag x {one axis of the left out pair}
But the length of the other factor matters, apparently
 
Oh, specifically a zig-zag
 
Yes
 
8:38 PM
I imagine you can generalize this to higher dimensions by having $p_1,\dots,p_n,q_1,\dots,q_n$
 
@BalarkaSen In the sense that, I guess, if the length $\rho < 1$ it's not going to be contactomorphic to the one with length $\rho > 1$. Pretty bizarre.
@AkivaWeinberger Yeah, just set $Q = \{p_1 = \cdots = p_n = 0\}$
 
So it's a generalization of the zig-zag
('cause that's $n=0$)
 
Right.
 
@BalarkaSen Wait, heh?
How
 
Yeah this boggles my mind.
I am not even sure if this is the correct interpretation of the constraint $\rho > 1$.
 
8:41 PM
Arright, I have to go now
Bye!
 
Safe landing
 
@BalarkaSen I'll be sure to tell the pilots
 
Hope the wings don't go loose
 
out of all the silly terminology in mathematics, "contactomorphic" has to be a top contender
 
 
1 hour later…
9:48 PM
@AkivaWeinberger: Note that if you scale $(x, y, z, p, q) \mapsto (cx, cy, c^2 z, c p, cq)$ (this is a contactomorphism as if $dz - y dx - p dq = 0$ before scaling, so is it afterwards), then both the "height", let's call it $a$, of the zig-zag and the width $2\rho$ shrinks.
$a$ shrinks to $c^2 a$, and $2\rho$ shrinks to $c 2\rho$
Because the height is the $z$-direction in the front projection of the zig-zag.
So the quantity $\rho^2/a$ seems to be scale-invariant
When the author speaks about zig-zags, they have a fixed $a$ in mind, which is fine. Fix the zig-zag from the get-go.
And it seems the claim is that if you have a long loose chart, i.e., $\rho > 1$ large then it'll have arbitrarily long loose charts $\bar{\rho} \gg \rho > 1$ contactomorphically embedded inside.
As opposed to short loose charts
 
10:20 PM
Here is a sort of picture for why I think this is true
Take a loose chart in $\Bbb R^5$, the project it to the $xzq$-plane. It'll look like the surface in the top picture.
$a$ is the height of the zig-zag, $\rho$ is the width of the chart.
If it's long enough, we should be able to apply a compactly supported contactomorphism to squeeze in the middle so that the height goes down arbitrarily small $a \delta$ for a substantial $\rho' > 0$ width in the middle
Scale everything by $1/\sqrt{\delta}$ (in the above sense, so $(x/\delta, y/\delta, z^2/\delta, p/\delta, q/\delta)$ to make height $a$ and width $\rho'/\sqrt{\delta}$, which is as large as you want as $\delta \to 0$
So it's an arbitrarily long loose chart sitting inside the given loose chart
Why is it important that the chart was long enough to start with? Because else, to do the squeeze, we would need a large derivative $dz/dq$. Since $dz = y dx + p dq$, $p = dz/dq - y dx/dq$. Here, $dz/dq$ is extremely large, but we need $|p| < \rho$.
I guess $dx/dq = 0$? Yeah I think so, we can always parametrize it that way.
 
11:01 PM
Looks like you're playing with paper airplanes, @Balarka.
 
@TedShifrin: Hopefully Akiva has landed by now because after what I'm doing to the airplanes, they will definitely never feel the gentle breeze rush over them.
 
If $X$ is topological space (CW complex?) that is $n$ connected, is its reduced suspension $\Sigma X$ then $n+1$ connected?
they seem to use this when arguing about the eventual stability of $\pi_{n+k}(\Sigma^kX)$ where $X$ is a (pointed) CW complex, using the Freudenthal suspension thm
 
@ShaVuklia In general $\pi_k(X) \to \pi_{k+1}(\Sigma X)$ given by suspending a map $S^k \to X$ is an isomorphism on $k \leq 2n - 1$ if $X$ is $(n-1)$-connected.
Freudenthal suspension theorem is a corollary of this more quantitative fact
 
my notes say for $k\leq 2n-2$
$k=2n-1$ only yields surjectivity
 
Yeah, maybe that's right. A surjection on $k = 2n - 1$ or something
 
11:09 PM
@BalarkaSen my notes call this the Freudenthal suspension thm, so I'm alright with that
if I can argue that applying $\Sigma$ increases connectivity by 1, then I understand why we have eventual stability
 
But this is enough, yes? If $X$ is $(n-1)$-connected, $\Sigma X$ will be $n$-connected since $n-1 \leq 2n - 2$ for all $n \geq 1$.
 
yes, it is enough
I just need to prove that connectivity increases
 
I am not sure what you mean, this proves that.
 
Connectivity increased from $n - 1$ to $n$
 
11:11 PM
ur rightt...
I considered that for a second, but somehow my mind convinced me not to try that route, because we already used Freudenthal suspension in the argument
(the argument for eventual stability)
thx x)
 
Sure!
 
freudenthal sussypension
 
Freud dental sus pension
 
btw Thorgott, I finished reading the geometric proof of representability of cohomology, and it was quite doable in hindsight
 
11:16 PM
i quite like the geometric proof, its just a simple corollary of the yoneda embedding theorem
 
I was surprised Hatcher didn't do it this way, because he hinges on the geometric side
@BalarkaSen o, that sounds neat:D
 
i find hatcher to do a lot of ugly things in the homotopy theory chapter
 
why is his proof ugly
tbh i never read most of Ch 4
 
though part of that is just homotopy theory being inherently ugly
 
i only read the Blakers-Massey theorem which i like
 
11:19 PM
lol thats the worst
 
its just PL topology
cope harder
 
which is why its the worst
just like his cellular approximation proof
 
I was first a bit confused about the explicit correspondence between cubical and spherical descriptions of the homotopy groups, but Tammo wrote out the iso's, and now I'm happy with homotopy thy xD
 
thats actually genuinely confusing
too many dimensions. once you get used to it its fine
@Thorgott i never read a proof of that because its obvious
make the maps cellwise smooth and then find a regular value
project
 
it isn't obvious
it is a bunch of technical nonsense that is arguably obvious + $\pi_k(S^n)=0$ for $k<n$
 
11:21 PM
cope harder bro
 
i agree $\pi_k(S^n)=0$ for $k<n$ is nicely argued from a differential topology perspective
hatcher gives a stupid PL proof that makes me zone out before even finishing reading the claims
 
i just make it cellwise smooth
a CW complex is a stratified space. no big deal
 
sounds like a more complicated way of saying the same thing
 
i agree smooth is better for that one
its just technically easier to say things in PL
+ hatcher doesnt want to assume you know difftop
 
PL is too technical for my algebra brain
 
11:25 PM
homotopy theory is best learnt through hearsay unless you want to be a homotopy theorist
 
he could just prove Hurewicz in a more efficient manner than he does and deduce the relevant fact from there lol
 
in which case you probably need to read Ravanel, "Complex Cobordism and Stable Homotopy Groups of Spheres" or something
@Thorgott i never remember the proof of Hurewicz tbh
but its alright
 
homotopy theory is learned by drawing stupid pictures of squares and pretending they communicate deep insight
 
theres a simpler proof using geometric homology theory
 
and i mean literal squares, not commutative squares
@BalarkaSen i know like three except i actually do not really understand any of them
 
11:28 PM
obvious using $H^n(X; \Bbb Z) = [X, K(\Bbb Z, n)]$.
 
you can deduce it easily from the serre SS, but that doesnt tell you the iso is actually the right map
you can deduce it with a cell-by-cell argument if you have the result comparing relative homotopy with quotient homotopy, but that rests upon the awful homotopy excision stuff
and you can prove it using the eilenberg complex, which is the most intuitive as the proof is basically the same as in the $n=1$ case except for a technicality ive been failing this entire afternoon to work out
 
or rather, the dual $H_n(X; \Bbb Z) = \pi_{\infty + n} (X \wedge K(\Bbb Z, \infty))$
 
$\infty+k$ lol
 
ze ztable ztem
@Thorgott deducing it from serre SS is pathetic
 
id like it if i knew why that gives me the right map
but i cant figure that out
that type of stuff seems more convenient in the algebraic settings
 
11:34 PM
too much work man
i believe most of homotopy theory
the details dont pay off for me
 
im allergic to believing
which makes me a subpar mathemtician
 
not really
i mean yeah it does
but if thats your ethical framework you should do it
 
my ethical framework is beating myself up over technical details
 
eg i am incapable of understanding most of modern geometric topology because i have ethical issues with learning to think like an analyst
2
 
thinking is cool, i should do it more
 
11:40 PM
i spent last 12 hours on why lengths of certain charts should matter at all and half the time i was raging because it seemed like an analytic nonissue
it wasnt
but at least this was the only inequality in the whole paper
 
grats
at least your last 12 hours yielded results
 
nothing is result until you publish
 
i meant, like, a literal result, not an academic result
 
what is a literal result
i dont think i have heard of these before
will it get me a job
 
a what
 
11:51 PM
money
you know
to eat
 
i think it's a machine for eating money
 
have you heard of the earthly activity of the homo sapiens called eating, dear sir?
 
not sure, it's what I gather from the conversation
 
never eaten money tbh
you eat the coin or the paper ones?
 
i prefer them crunchy
 

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