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00:00 - 19:0019:00 - 00:00

12:12 AM
Consider the set of all functions $\mathbb R^2\to\mathbb R^2$ defined by $f(\vec v)=\cos(\vec v\cdot\hat u+c)$ where $\hat u$ is a unit vector and $c$ is a constant
What is the topological shape of that set?
I've never heard of spreading out, Karim. What are you talking about?
DogAteMy, “topological shape”? And map to $\Bbb R^2$?
$\mathbb R^2\to\mathbb R$ I mean
Clearly $\cos(\vec v\cdot(-\hat u)-c)=\cos(\vec v\cdot\hat u+c)$
@TedShifrin so given projective variety we have defining equations
we can vary those equations and get X as a family I guess
What does this have to do with topology, DogAteMy?
but I don't like this way of getting a family.
12:15 AM
The set is a two-dimensional manifold
I don't think you capture all information. I wanted to ask you this question though.
because we have two degrees of freedom: $\hat u$ and $c$
There are plenty of families that are non-algebraic deformations.
there's an obvious map $S^1\times\mathbb{R}$ onto that set of functions, is it a quotient map?
12:16 AM
You should read basic Kodaira-Spencer deformation theory.
$S^1\times S^1$ even, since $c$ is only important mod $2\pi$
I assume yes
ah true
@TedShifrin where can I see it in Kodaira's book ?
so is the flip the only thing happening in the kernel
I guess it's $S^1\times S^1/((x,y)\sim(-x,-y))$
12:17 AM
which would boringly just be S^1xS^1
no wait, it wouldn't
It's discussed in Morrow-Kodaira.
Oh I wrote that wrong
$S^1\times S^1/((x,y)\sim(-x,\bar y))$
@TedShifrin Thanks
Lots of Kodaira's papers (which I no longer have).
because $-\hat u$ doesn't have a fixed point but $-c$ does
Is it a Klein bottle?
12:19 AM
Orientation preserving?
@TedShifrin btw my wife agreed to come to US with me.
After vaccines?
so once I am done with my PhD I will come to the U.S hopefully shipping all my books to the US will not be a problem
that is still 2 years away but yeah
Well, job first! :)
Yeah haha
12:22 AM
@AkivaWeinberger You saw my comment?
Oh that was for me?
What's orientation-preserving?
LOL ... Your map for equivalence relation.
Not the second one I wrote, no
It's not cx conj.
$x\mapsto-x$ preserves orientation (you're rotating around by 2pi) but $y\mapsto\bar y$ doesn't (you're reflecting the circle across the x-axis)
12:23 AM
Antipodal map is rotation.
thinking of $S^1$ as a subset of $\mathbb C$
Oh, I missed the bar.
Think back to the original relation
$\cos(\vec v\cdot(-\hat u)-c)=\cos(\vec v\cdot\hat u+c)$
Better notation: $\hat u$ lives in $S^1\subseteq\mathbb C$; $c$ lives in $\mathbb R/(2\pi)$
12:25 AM
Never mind. I'm on my iPad. I shouldn't chat on it.
We have $S^1\times(\mathbb R/(2\pi))/{\sim}$ where $(\hat u,c)\sim(-\hat u,-c)$
So it has to be the Klein bottle.
I believe so, yeah
These functions, visually, are monodirectional waves
(is that a term?)
Two-sheeted cover with orientation-reversing deck transformation.
The space of monodirectional waves is a Klein bottle
or unidirectional
(unidirectional is when they go to college)
That's a point: the difference between mono and uni is that you can get mono at uni but not the reverse
12:28 AM
"Monodirectional waves" does not make sense to me. If I saw that name I would be confused.
Anyway I agree that S^1 x S^1 quotiented by an orientation reversing free involution is necessarily a Klein bottle.
I don't know how else to describe the shape
It's the graph of $z=\cos x$ rotated and shifted
Uh... a corrugated sheet
The space of corrugated sheets is a Klein bottle
The reason I was thinking about this:
Grant Sanderson just tweeted:
> Treated as a topological space, ordered pairs of points on a loop are isomorphic to a torus, and unordered pairs are isomorphic to a Mobius strip.

Is there any analogous set that is isomorphic to a Klein bottle?
And this was my attempt at an answer
I guess I could also say it's the space of rows of unit-spaced parallel lines
(like lined paper)
Keeping in mind that the Möbius band is the space of lines in the plane, this makes it clear how two Möbius bands make up the plane
(divide it in two based on if the origin has distance <1/4 or >1/4 to the nearest line)
2 hours later…
2:43 AM
What's the word for when a 2-form is a wedge of 1-forms
like sometimes doesn't happen in n>3
or simple
or decomposable
1 hour later…
3:58 AM
happiness is like peeing in pants, everyone can see it but only you can feel its warmth
let's see how many people will get butthurt by my beautiful quotes
4:30 AM
So many stars... Even more than the number of stars in the sky...
then you must be living in polluted city
4:52 AM
the unity of rational numbers is 1 right?
@EM4 yes
1 hour later…
6:05 AM
Hi All..
$\int \frac{dx}{\sqrt{4x - x^2}} = arcsin(\frac{x-2}{2})$
Hi @TedShifrin Pls see my integral. I found that integral of L.H.S does satisfy arcsin(x-2)/2
$\bf{slope}$ of L.H.S function at x-coordinate is not equal to R.H.S function $\bf{y-coordinate}$ at every point.
6:35 AM
@123 isn't the slope of RHS supposed to be equal to the integrand on LHS instead?
It's correct.
As Leaky said, you're upside-down.
1 hour later…
7:51 AM
The question I uploaded yesterday, i.e. The complement of a closed discrete subspace (S) of $\Bbb R^n$ is simply connected if $n\geq 3$. Using the set $S$ is discrete, for each $x_i\in S$, we can find small n-1 sphere $S_i$ centered at $x_i$ such that $S_i$'s are disjoint.
How can I fill the rest of the space in $\Bbb R^n$? i.e $\Bbb R^n-\bigcup_i S_i$
What do you mean by fill?
I wonder how to attach $n$-cell
$\Bbb R^n-S$ deformation retract to $\Bbb R^n-\bigcup_i S_i$
The whole thing deformation retracts to $S_i$ union some line segments.
@love_sodam This is nonsense. You mean it deformation retracts to $\Bbb R^n \setminus \bigcup_i B_i$, where $B_i$ is the ball bounded by $S_i$.
Ah yes
Question ask to use proposition 1.26 so I tried to use 1.26(b) by attaching $n$-cell to $S_i$'s to make $\Bbb R^n-\bigcup_i B_i$
Using the fact that the space deformation retract to $S_i$ union some line segments is homotopy equivalent to wedge sum of $S_i$'s and $\pi_1(S_i)=0$ so using van Kampen, the whole space is simply connected. This is what you're saying right?
8:05 AM
Oh, I can't use proposition 1.26(b) since $X$ is assumed to be path connected which is not in my case
By the way why the closed condition is included in the problem?
limit point?
It is still true if you drop the closed condition but harder to argue.
If I drop the closed condition then what should be changed?
What should be changed where?
I don't understand the question
The sketch proof I wrote above
What sketch proof?
8:10 AM
line segment and $S_i$'s?
It is not true that it deformation retracts to a wedge of spheres.
By thinking for more than 2 seconds
Such subset would be like infinite set deleting limit point right?
I mean open discrete subspace of $\Bbb R^n$
8:32 AM
open discrete?

I am looking at the question :
Check the convergence of the sequence $a_n=\left (\frac{n+2000}{n-2000}\right)^n$, $n>1$. If it converges calculate the limit.

We have $$a_n=\left (1+\frac{4000}{n-2000}\right) ^{n-2000}\left (1+\frac{4000}{n-2000}\right) ^{2000}\to e^{4000}$$ Having found the limit means that the sequence is also convergent, right? But could we have shown the convergence also in an other way?
Hmm not closed I would say
9:13 AM
@MaryStar You can also look at the numerator and denominator separately after cancelling $n$: $\lim\limits_{n\to\infty}\left(1+\frac{2000}n\right)^n=e^{2000}$ and $\lim\limits_{n\to\infty}\left(1-\frac{2000}n\right)^n=e^{-2000}$
@robjohn Ah and since both, numerator and denominator converge then also the fraction?
9:40 AM
Sowe have that $$\lim\limits_{n\to\infty}a_n=\lim\limits_{n\to\infty}\frac{\left (1+\frac{2000}{n}\right )^n}{\left (1-\frac{2000}{n}\right )^n}=\frac{\lim\limits_{n\to\infty}\left (1+\frac{2000}{n}\right )^n}{\lim\limits_{n\to\infty}\left (1-\frac{2000}{n}\right )^n}=\frac{e^{2000}}{e^{-2000}}=e^{2000+2000}=e^{4000}$$ right? @robjohn
9:51 AM
@MaryStar yes
being undergrad it is shame I got stucked at Chinese elementary school problem lol
a=2.6^10-2.6^9,b=2.6^{11}-2.6^{10}, and c=(2.6^{11}-2.6^9)/2
and tells me to sort in ascending order
and solve it without calculator
how will you guys solve it
asking it here is like using thermonuclear weapon to kill a cockroach
10:17 AM
ok it is now i get it
@robjohn Great!! Thank you!! :-)
10:49 AM
@eryceriousmatherfunker b=2.6a so a<b, c=(a+b)/2 so c is between a and b
F, handed in an empty sheet in the exam. really caught me off guard
2008 arxiv cringe:
that's not how you count nodes in an infinite tree
@LeakyNun mine is bit dumber a=2.6^9×(2.6-1)=2.6^9×1.6 b with kinda same method and c too except divide by 2 so b>c>a
using calculator makes my arithmetic skills dumb nowadays
11:07 AM
@Secret I'm more interested in which university and which people approved this paper in the first place
Exercise one: a) All homotopy, homology, cohomology groups of klein bottle times 2-torus. Hell idk how to prove the fundamental group of the klein bottle on the spot, higher homotopies are easy though. b) show reduced zeroth homology of S^n minus smooth closed embedded smaller-dimensional submanifold is equal to n-1th reduced cohomology of the manifold, so the k=0 case of alexander duality c) asks whether K^2 times T^2 embeds into R^5

Exercise 2: a) show that klein bottle is an S^1-bundle over S^1. This is easy and was on the exercises but I wasted too much time doing this carefully until
@Secret looks like they're from a certain business school, which might explain why
@user2103480 a literal empty sheet?
@LeakyNun well there were 2 words and my name on it to signify that I took part
11:16 AM
but it was harsh
last week there were still lectures, I have 3 other exams in the coming weeks and all this intersection form stuff we only covered in the week before last week
@Secret The problem in Proposition 2.1 is that there is no integer for the "node" representing $1/3$.
so no bijection
he has counted the finite decimals.
but well everyone who said anything said that the time was absurdly short lol
11:36 AM
Hi, why is this true? Let $r_i \geq 0 $ such that $\sum_{i=1}^k r_i = 1$. Then if $\{z_i \}_{i=1}^k \subset S^1 \subset \mathbb{C}$, and $\sum_{i=1}^k z_i r_i = 1$, all $z_i = 1$ necessarily
Or rather, for all $i$ such that $r_i > 0$, $z_i = 1$
@porridgemathematics try to do it for k=2
11:59 AM
am I supposed to use a trig identity somewhere?
or is it even more obvious than that
oh I thought about using that
yes I do
not really seeing how that will be helpful though
12:16 PM
Cauchy-Schwartz allows you to find when the triangle inequality is an equality
but cauchy schwartz is going to give me a bunch of square terms
use probability brain
Gotta work this out still haha lemme chick if that actually works
Compare $||u+v||^2$ and $(||u||+||v||)^2$
I mean that's how you derive the triangle inequality from CS
And that also gives you the condition for the equality in the triangle inequality
Which is what you want
Also, do you see the picture ?
Hi, I have a question. Let $G$ be a group acting on a topologiacl space $X$ and $H_1,H_2$ be two infinite cyclic subgroups of $G$ generated by $\varphi_1,\varphi_2$, respectively. Define $f:X/H_1\to X/H_1$ as $[x]_1\longmapsto[x]_2$. Is $f$ a well-defined homeomorphism?
12:41 PM
Does anyone have examples of questions where I have to find the equation of a tangent to a curve, GIVEN AN EXTERNAL point? I am having trouble finding worksheets or papers of this kind. many thanks
12:52 PM
probability brain doesn't work out nicely, one can in principle view this as a real random variable with finitely many values and $\varphi_X(1) = \Bbb E[e^{iX}] = 1$, and then try to prove it must have values in $2\pi \Bbb Z$. Wouldn't know how to do that though
@User what's your opinion?
I think if $G$ acts properly discontinuously on $X$ i.e. $X\to X\G$ is a covering, then it is true.
Am I right? @AlessandroCodenotti
It's somewhere between false and obviously false
Your map is clearly not usually well-defined which you'd see if you wrote out what it means for it to be well-defined
If the actions are conjugate (so the generators are conjugate elements in $G$) then this is true but that's the best you're going to get
Otherwise the actions are basically unrelated...
There is an action of $\Bbb Z$ on the cylinder $S^1 \times \Bbb R$ whose quotient is the Klein bottle, but so that $2\Bbb Z$ just acts by a translation
1 hour later…
2:08 PM
@BalarkaSen I lovingly selected a small memedump for you as a celebration of a successful exam and then I got rekt. Now it's not appropriate anymore smh
Sorry pal
yeh happens. Consolation is that even those who had a 5 page LaTeX written solution ready for every exercise session had massive problems, and I did not invest that level of effort
Shit is that I now have 3 other exams in the next 17 days and invested lots of time into the topology thing. I could probably have passed but that's not worth it for a possibly bad grade
*possibly very bad
Yeah that really sucks
I hate grades
2:26 PM
@user2103480 Back to back exams suck
They drain me out
Just get done with it
It's not worth worrying about grades so much
@BalarkaSen I still try to get back into some scholarship
Ironically, scholarship & job applications stole a nonnegligible time where I should have been studying
It's a vicious cycle in pursuit of something that never happens, I know
Study for exams/prepare for entrances/fill application forms -> Takes away time from actually studying -> Fail, repeat
Also, the time wasted by worrying about/procrastinating those things
I would much prefer just teaching and not evaluating
Then I can use tests as an educational tool for feedback
Instead of a bludgeon
2:34 PM
It's not possible to learn in an academic setup
What you do is not teaching, and what I do is not learning. It's a formalism.
@porridgemathematics look at the real part
@BalarkaSen noo dont say that, me cramming 200 pages into my head to spurt the rough details out in about one hour is totally perfect
... because I retained soo much information from the last x times
my memory is good and lecturers were kind so orals were not a problem for me yet, but this is so tiring
back to maths: are there simple rules (for most examples) for calculating tensor products of polynomial rings modulo some relations?
Dunno but probably the answer you want is universal property
k[x]/(f) tensor k[y]/(g) enjoys a map from k[x] o k[y] = k[x,y] with (f) and (g) in the kernel, so it enjoys a map from k[x,y]/(f,g)
So the question is whether that map has extra kernel you didn't notice
I'm gonna go with no, that's an Isomorphism
I vaguely remember from the AG course that it should be an iso
But maybe there were some assumptions I'm forgetting about now
2:51 PM
Probably, so the cohomology ring of $K^2 \times K^2$ with $\Bbb Z_2$ coefficients is $\Bbb Z_2[x_1,x_2,y_1,y_2]/(x_1x_2,x_1^2-x_2^2,y_1y_2,y_1^2-y_2^2)$?
Hm but then these would not reach the fourth cohomology with any combination I see
3:31 PM
@MikeMiller $k[x_1,\dots,x_n]/(f_1,\dots,f_r) \otimes_k k[y_1,\dots, y_m]/(g_1,\dots,g_q)\cong k[x_1,\dots,x_n,y_1,\dots,y_m]/(f_1,\dots,f_r,g_1,\dots,g_q)$
Ok cool so this holds for fields, and the above should be true. I missed that $x_1^2y_1^2$ is not zero in this ring
you don't need $k$ to be a field
Hm ok. Still messed up something there I think
$$f(x,y)= \begin {cases} \frac {x cos (y) +y cos (x)}{\sqrt {x^2+y^2}} & (x,y)\ne (0,0) \\ 0 & (x,y) = (0,0) \end {cases}$$
f is differentiable at (0,0)

f is continuous at (0,0)


f is not differentiable at (0,0)
i am getting d..
setting y=0, and letting x=0+h
and letting h --->0
f(0+h, 0)-f(0,0)/h
we will find that the limit isnt finite
4:06 PM
anyone? thoughts?
@tigre Yes, I was writing a special case since it didn't seem like one gained anything from additional notation
Anyway, I agree that the canonical map from one side to the other is presumably an isomorphism I just haven't bothered to justify it
4:18 PM
@satan29 your options are not mutually exclusive lol
@MikeMiller Oh, I thought you would know the general case, I probably shouldn't have even tagged it, I was just writing it down for user2103480
Ah, I just got confused that you were pinging me. I thought you were trying to tell me "Your fact can be generalized..." which I knew.
I got you now.
@robjohn Could we use also the ratio test to check the convergence?
@LeakyNun hmm
but for sure, d is correct, no?
explicit what you computed to get that the limit isn't finite
4:49 PM
1 hour ago, by satan 29
f(0+h, 0)-f(0,0)/h
@MaryStar The ratio test is for series. That was a product. How would you apply the ratio test?
we will get 1/|h|
as h-->0
Sorry I misread you, yes that is true
What's f(x,0)?
4:52 PM
Is it continuous at 0 ?
So does f have a chance of being continuous at 0?
b is wrong
So does f have a chance of being differentiable at 0?
4:55 PM
@robjohn Ah yes, so there is no criteria to check the convergence,only to find the limit, right?
@MaryStar You can check convergence. You can turn products into sums of logarithms.
thanks again, @Astyx
5:27 PM
@LeakyNun Does a commutative ring suffice?
ok nice thank you
howdy, @Leaky, @robjohn, @user2103480
Salut @Astyx
Hi Ted
Salut !
5:30 PM
salve @TedShifrin
Nothing too exciting at the moment, I guess.
Definitely not
Let me tell you the notion of Benjamini-Schramm convergence of graphs
It can be generalized to manifolds but let me stick to graphs for now
Fix $d \geq 0$. Denote $G_d$ to be the space of connected finite graphs of degree at most $d$
what's the degree of a graph?
number of neighbors
5:38 PM
So every vertex has degree at most $d$?
Ok (are your graphs finite?)
Edited yeah
$G_d$ is a space in the following sense. I know how to define distance between rooted finite graphs of degree at most $d$
If $\Gamma_1, \Gamma_2$ are two graphs with roots $o_1, o_2$, I define $d(\Gamma_1, \Gamma_2) = 1/k$ if $k$ is the maximum number for which the $k$-balls in $\Gamma_1$ and $\Gamma_2$ centered at $o_1$ and $o_2$ resp are isomorphic
This is the rooted distance
But how do you turn an arbitrary graph into a rooted graph? The root need not be god-given
Can you not take the maximum over all pairs of nodes ?
5:46 PM
@TedShifrin 'tag
Yeah but then it's a boring notion because small parts of the graph can have massive degree
You're only looking at that for a long time
I'll give you an example soon
Solution: Choose a root uniformly, cuz your graph is finite. That makes any graph in $G_d$ a random rooted graph, a distribution on the space $RG_d$ of rooted connected graphs of degree at most $d$.
$RG_d$ is the metric space, mind you, with the rooted distance. It's a compact metric space.
So finite graphs are just probability distributions on $RG_d$.
@TedShifrin is guessing an exciting part of your day?
You say a sequence of graphs in $G_d$ is Benjamini-Schramm convergent if the corresponding probability measures on $RG_d$ converge in the weak*-sense
Of course.
The winds have finally died down here. We've had pretty high winds for about a week. If we're lucky, we might have some rain on Wednesday.
5:49 PM
What is this thing you call rain?
@Astyx I will now tell you why this is a wrong notion. Ok, consider the path graph $P_n$. Does it converge in the Benjamini-Schramm sense?
As $n \to \infty$?
@TedShifrin well, it is a 25% chance, and that has a tendency to get less as the time gets closer
Remind me what weak convergence is?
Say measures $\mu_n \to \mu$ converge if $\int f d\mu_n \to \int f d\mu$
Or convergence in distribution of random variables
@Astyx it doesn't work out as much as strong convergence. You can only really notice it when compared to other functions.
5:52 PM
Right, I remember
Went for dinner
It should converge right?
Yeah, why?
By some Cesaro sum argument
$\int fdP_n = {1\over n}\sum _{k=1}^n f(P_n^k)$ where $P_n^k$ is the path graph rooted at the k-th node
6:08 PM
@Astyx The weights should be different. $P^k_n$ gets weight $2/n$ for leaf roots
The rest of the weights is all where the root is in the interior
That's the distribution of the path graph $P_n$ as a random rooted graph
$d(P_n^k, P_n^{k'})=\max(1/(n-k), 1/k, 1/n-k', 1/k')$ so for every $\epsilon = {1\over K}$, we can separate the sum as $\sum_{k=1}^K f(P^k_n)+f(P^{n-k}_n) +\sum_{k=K}^{N-K}f(P_n^k)$
The left sum is bounded and the right sum is close to n times some limit
Yeah that's right
The distribution it converges to deserves to be called $\Bbb Z$
@BalarkaSen Do you mean we need to make the distinction between $P^k_n$ and $P^{n-k}_n$ ?
Since the corresponding graphs are iso
No, I got what you were doing. I was saying they're exactly the same graphs coz isomorphic as rooted graphs
I was writing the sum without multiplicities
Oh $RG_d$ is not restricted to finite graphs
6:16 PM
Hi @TedShifrin
So the limit I mentioned is precisely $f(P)$ where $P$ is the path graph
@Astyx yeah
I got it
bcoz if two rooted graphs are isomorphic upto vision R for all R
then they're isomorphic
I thought my argument for compact only worked for finite graphs but I was wrong
right right
6:19 PM
6:31 PM
@Astyx Oh I forgot to say why your notion is wrong
Consider $T_3$, $3$-regular tree. Take large balls inside $T_3$. These will not Benjamini-Schramm converge to $T_3$
Because half of the vertices are leaves. Subgraphs of $T_3$ approximates $T_3$ very badly
If you chose a random root with probability 1/2 it'll be a leaf root
So you're getting screwed and the amount of time you're getting screwed is not probabilistically vanishing
Oh right
But if you took max degree you won't see this
The Cesaro argument doesn't work any more because the left sum is dominant or something
fun stuff
6:46 PM
Hi, I'm happy with the definition of a sheaf of $\mathcal{O}_X$-modules, but it occurred to me that I don't know how these are defined as functors. In functorial language, perhaps atleast dealing with the presheaf part, it's a pair of functors $\mathcal{O}_X\times \mathcal{M}:\text{Open}(X)^{\text{op}}\to \text{CRing}\times \Bbb Z\text{-mod}$ and $\mathcal{M}:\text{Open}(X)^{\text{op}}\to \text{CRing}\times \Bbb Z\text{-mod}$, along with a natural transformation
$\alpha:\mathcal{O}_X\times \mathcal{M}\implies \mathcal{M}$
Maybe make them coproducts of categories
Actually, nah, this doesn't work
If O is a presheaf of rings, then a presheaf F of O-modules ought to be a presheaf of abelian groups equipped with a map of presheaves O x F -> F satisfying the usual relations, yes?
I think you're looking for the notion of ring object in an abelian category, and then module object over a ring object in that category.
@MikeMiller I guess I was confused about the targets of those functors O x F -> F, and the natural transformation is a natural transformation between functors on which categories, that sort of thing
Btw, the only reason I was trying to do something like this is I want to take stalks on a complex of sheaves of O_X-modules, but wanted to state something about functoriality
Like you want to apply $\varinjlim_{U\ni x}: \mathcal{C}\to\mathcal{D}$ for whatever categories (in theory)
You could take them to land in Set, but ask that they factor through Ring, AbGrp etc I guess
Like $O:Open(X)^{op}\to CRing, F:Open(X)^{op}\to \Bbb Z-mod$, so the natural transformation doesn't immediately make sense to me functorially
say $M$ is a smooth manifold (with boundary, if that matters) and $f\colon M\rightarrow[0,1]$ is a submersion, can I always connect two points in different fibers of $f$ by a path tranverse to the fibers of $f$?
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