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12:00 AM
$\mathbb{N}_0$ even though there was no $\mathbb{N}$ in the entire document.
 
12:25 AM
0
Q: Universal approximation theorem for multiset functions

user76284Let $\mathcal{M}, \mathcal{T}$ be subsets of a topological space. $\mathcal{M}$ approximates $\mathcal{T}$ iff $\mathcal{T} \subseteq \overline{\mathcal{M}}$. Let $\sigma$ be ReLU. Let $\mathcal{K} \subseteq \mathbb{R}^m$ be compact. The universal approximation theorem (UAT) says $$\{ g \circ \si...

Appreciate any input on this.
 
12:45 AM
Hello!! Does someone of you have an idea about my question :
2
Q: Is $f$ twice continuously partially differentiable?

Mary StarI have shown that a function is continuously partially differentiable. If it holds that $$\frac{\partial^2{f}}{\partial{y}\partial{x}}=\frac{\partial^2{f}}{\partial{x}\partial{y}}$$ forall $(x,y)\in \mathbb{R}^2$ does it follow that the function is twice continuously partially differentiable?

 
@Mary I have not vetted but math.stackexchange.com/questions/2095484/… addresses this issue
the construction seems plausible to me
 
So that means that $f$ is not necessarily twice partially differentiable, right? @leslietownes
 
yeah.
err, no.
it means it's twice differentiable but those things aren't continuous.
they're still equal everywhere, but they're both discontinuous at 0. i haven't written out the formulas but it seems likely to me that that's what happens.
it's kind of a two-variableization of one of the common examples used to show that derivatives don't have to be continuous.
 
1:06 AM
Twice differentiable in the Frechet/Gâteaux sense forces symmetry.
Just like differentiable is stronger than existence of partials.
 
Ok! And if we can find a point where the equality $\frac{\partial^2{f}}{\partial{x}\partial{y}}(x,y)=\frac{\partial^2{f}}{\partial{y}\partial{x}}(x,y)$ doesn't hold, then does it follow that $f$ is NOT twice partially differentiable?
 
Oh, you said partially differentiable. I don't think you meant that.
 
Oh I mean does it follow then that $f$ is not twice continuously partially differentiable? @TedShifrin
 
Leslie's link answers that.
 
That link was for my initial question(that the mixedpartial derivatives are equal), or not? If we can find a point where the mixed partial derivatives are not equal, what is then? @TedShifrin
So if the mixed partial derivatives are not equal everywhere, then f is not twice continuously partially differenatiable, right?
 
1:21 AM
Of course.
 
Thank you!! :-)
 
that's the usual theorem in contrapositive form, if i'm not mistaken.
i almost said converse, which shows how long it has been
 
Yes and grr.
But I made a more interesting point to you, @leslie, a bit earlier.
 
oh yeah, the frechet/gateaux.
the partial derivative is a funny concept. there's nothing like it in functional analysis. we do have that, though.
you apply a linear functional to some mess, it's the best you can hope for.
and often more than good enough
 
Well, it shows up for example with calculus of variations.
But one assumes differentiability in the strong sense, ordinarily.
 
 
2 hours later…
3:08 AM
my daughter can spend a surprising amount of time running around the house before she gets tired of it
 
Her energy/unit mass far outstrips yours
 
she is outmatched only by the cat, who has evolved not to be an endurance runner
our cat is conserving energy for tearing around the house at 3am and howling in the hallway for someone to play with her
 
my son got an offer from a uc. enormous relief here. repacking all my weapons again.
he had some community college+transfer plan that had me in the depths of despair
i had not realised just how stressed about it, i can hardly focus on my cup of tea now.
cats should be outdoors :-)
 
that's great news.
 
Let $f(x)=(x-3)^5(x+1)^4$ . Then we want minima and maxima, then we can use first derivative test or second derivative test
 
3:17 AM
I would've expected wine at this hour for you Copper
and good news for your son. So he is in the alma matar system then. ;)
 
@dc3rd @leslietownes THanks!
not quite, my alma mater is university college, cork, ireland :-)
 
are alma maters not counted for graduate work?
 
where george boole taught
 
CC+transfer is a viable route, most of my friends in late college were transfer students. but it's so much more work. if it isn't financial suicide, far better to take it easy.
 
well Boole has contributed a few things....
 
3:20 AM
the problem is that because of covid many have deferred a year which is why the crunch, so transfer is a big risk
and he does not have his sister's resume
 
i hadn't thought about that. the people i knew who went that route were really sharp on paper.
covid on top of everything else, yikes.
 
@dc3rd the alma mater is the 'mother' school, so i undergraduate would be how i interpret it
 
some people do use the term more widely. i think often you use the strict sense if you want to distance yourself from something.
as in, "i didn't go to harvard harvard, i only went to the trade school." ha, ha.
 
@copper.hat Congrats to him !
 
i knew a guy who always made that joke. it got old and it wasn't new to begin with.
 
3:22 AM
@TedShifrin Thanks. Quite a relief.
 
it sounded like you were working on a nice ulcer only 48 hours ago. happy news.
 
$f'(x)=5(x-3)^4(x+1)^4+4(x+1)^3(x-3)^5$
 
I was in a vile, dangerous & despondent mood.
Hope I am not stepping on toes. How do you know if someone attended Harvard?
They will tell you.
 
Is a yuge school a good fit for him?
 
that's a classic
rover, i agree.
 
3:23 AM
dear.....the pain my "mother" school has extolled on me has left me quite bitter, but it has also made me grow vastly.......
 
@TedShifrin I am not sure. But it is better than the community college+transfer plan he had earlier today (very risky in current context)
 
So, where f'(x)=0 gives x=-1,3,7/9
 
The UC size is much more of an issue than even in our days.
 
rover, we continue to agree.
 
@TedShifrin I think Santa Cruz is manageable. I did not want either to go to Berkeley for UG.
 
3:28 AM
@copper.hat so we should talk about alma pater?
 
@TedShifrin no doubt it should be almx xater
 
OK, UCSD and UCLA are also too yuge, perhaps.
 
what's wrong with Berkley for UG?...unecessary "competitive" stress?
 
The first derivative test for Maxima minima of Differential function States that : at a stationary point across with derivative changes sign from negative to positive moving from left to right is point of minimum and a stationary point across which derivative change is sign from positive to negative moving from right to left is point of Maxima
 
@dc3rd big yes.
 
3:29 AM
SC is as good as it gets, for that. i think.
rover, we agree once again.
 
@dc3rd berkeley is great for grad work, but easy to get lost as ug
 
going to U of T......and knowing Berkley's rep I'm not surprised.........
 
There's plenty of competitive stress at small places like MIT and CalTech … but Classes at Berkeley are 400+.
 
U of T suffers from the same thing.
 
they make it out as though there are a lot of tiny, smaller things you can plug in and not be just 1 person in a 400 person class. but you will be 1 person in 400 person class.
i was typing this as ted used the same figure. AGAIN.
not for your whole experience but the first year and perhaps even second depending on intended major.
it is a very bad fit for a lot of people
 
3:31 AM
400+ for only first year courses though right? then after that it "thins" out?
 
At UCSD upper division real analysis can be 140 students. Horrid.
 
dc it really depends, after that. when i was there the biology classes were that size (except for labs) until junior year.
in math, maybe 400 to 150 to 30 in fresh/soph/jr. you can probably multiply everything by 2 now.
 
i know a uta prof in cs. i had forgotten about him
 
class size got very low in the non-'hot' upper division subjects. my undergraduate second semester real analysis course had three people in it.
 
i was a ta for cs8 at some stage, for a guy called doug cooper. the lectures were 250_ if i recall
 
3:33 AM
Must have been a horrific teacher.
 
cooper?
he was an odd fellow, but a great teacher.
 
No, leslie's
 
profs you forget about tend not to be good, or really bad.
they're just in the middle
ted mostly it was just real analysis had this reputation of being a 'killer' subject and the required one semester of it was enough for just about everybody.
the second semester wasn't required for anything, it was just more analysis.
so, horrific teachers in the first semester, basically :)
 
there was a m104 teacher in berkeley who made a point of making it impossible
 
In 1978 when I taught differential topology as a grad student, I had 15 students.
 
3:36 AM
20-30 would be normal now i think. maybe more.
 
At UGA I never got that big.
 
the analysis classes were always the smallest when i was in school. except PDE. probably hasn't changed much since then
 
i took 104 as a refresher when i started in berkeley, the teacher was visiting from Venezuela and he was very helpful to me.
 
my 104 instructor was not helpful. she was captious.
that may have been where i picked that word up. trying to find a word for it.
 
wonderful word.
 
3:40 AM
going through the dictionary word by word until i found it.
aha!
 
i have found the captious nature to be the worst enemy of brain storming
i told my daughter that if you need to solve a problem, you must separate the idea generation from the evaluation part otherwise you will not move
 
I got the point while commenting
I was getting different answers from two methods
 
she liked asking gotcha questions where she was clearly expecting someone to give the wrong answer and when they got the right one it visibly annoyed her and she'd switch tacks in some way to make the answer sound nonresponsive. she would have been good at keeping certain people out of universities in the USSR.
rover, never a good sign. but hopefully it was just a sign error or something.
 
she might be the teacher i was thinking of. she announced to one class that no one would be getting an a
i mean really.
 
@Leslie surely not Ratner
 
3:45 AM
it wasn't, this was someone on the first year of a postdoc.
i loved marina.
 
I had an occasional class with no As
 
looking back on it, i sorta get it. you're beginning a postdoc and they ask you to teach something that has a reputation for being a 'killer,' and it's not your subject. but that's only an excuse to teach poorly, not to also be captious.
 
i can understand a class with no a's, but declaring it at the start...
 
that seems like malpractice to me.
 
Not your subject? Doubtful. Analysis is half of math.
 
3:47 AM
that's like that the evil prof in the paper chase.
 
paper chase the legal show?
 
i gathered that she had taken the minimum amount of analysis needed to get a BA, and had focused elsewhere in graduate scohol.
 
i loved that
 
Kingsfield was fabulous.
 
i was referring to the film on which the show was based. which was based on a book which i haven't read.
but the same character i think.
 
3:49 AM
I am skeptical, leslie. That's not how course assignments go.
 
saw both & read the book, all good as far as i am concerned
 
it could be she just had no experience in front of a classroom, or was just phenomenally bad at teaching. i asked about some of the harder sequences and series problems that she didn't assign and was completely blown off.
 
i need to be up at 7:30am for a ride, and have work to finish, but all i can think of now is polishing off my bottle of new zealand something or other
 
didn't go to those office hours again
copper, enjoy your ride
 
Surely no experience, esp if foreigner.
 
3:53 AM
@leslietownes thanks! good night folks.
 
Night!
 
my wife had a rocky adjustment. got a tenure track job with essentially no teaching experience. the students noticed less than they might have had the subject not been more of a free-for-all than math tends to be
my twisted syntax betraying a desire to phrase everything as indirectly as possible
 
 
2 hours later…
6:15 AM
How to integrate 1/(1+x^5) w.r.t x? It's getting horribly long.
Wolfram gives a scary answer.
 
they do get long. exact expression for an antiderivative of 1/(1 + x^n) is not fun for large n
nothing's wrong, that's just what it is. if you're only working on (-1,1) you can use a series expansion, 1/(1+x^n) has the form 1/(1 - u) with u = -x^n. geometric series formula
2
if i'm wrong, please pipe up, anybody.
i used to give that as an example of how you can write down very simple formulas where the exact, non-series expressions become very hard to come up with. i don't think there's a nice recursion or anything like that. as there sometimes is in the exercises you see in calculus books
i gave that example in office hours, of course. i wouldn't dream of taking that in a classroom setting
i used to use that to break wolfram alpha. i forget how high i could go before it stopped answering.
 
6:35 AM
I was solving this question. It took at least 30 minutes alone to partially decompose this.
1/(x+1)(x^4-x^3+x^2-x+1)
 
in general, it will be a lot of work by hand.
 
i love maxima so i tried maxima
 
if i were writing a textbook i wouldn't try to give the impression that you can integrate rational functions by hand. in theory, yes, and with a good computer, maybe. but anything that involves factoring polynomials is not really suitable for work by hand
except in the very low degree things you see in calculus books
 
EM4
wolfram is scary.
 
General formula for integration of 1/(x^n+1) would be helpful.
 
6:44 AM
the power series is the only thing i know at that level of generality
 
EM4
Sometimes I wonder how wolfram gets this answers ;)
 
I don't know how wolfram came up with that hypergeometric function. Anyways power series is also not going to be easy here
 
EM4
how you guys doing by the way?
 
algooooooooooooooooooooooooooooooooooooorithms
and daaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaata
 
6:51 AM
here's one from the vault. the integral of heaviside(sin(x)-(pi/2)*cos(2*x)+1/8) from 0 to 2pi. it's about 3, and this is also clear from the graph of the function without the heaviside( ) around it. but if you're not careful how you input it, a lot of CAS's will mess it up, giving you nonsense like 0 or stuff near 5. sometimes depending on how many digits you tell it to work to.
i never figured out why but i think with some ways of asking a CAS to do a definite integral, it first tries to find a symbolic antiderivative and gets stuck on that.
those parameters aren't magic numbers, they're just the ones i had when i discovered this in something i was trying to do.
even fairly dumb numerical packages get it right or close to right.
i think it has something to do with symbolic integration
 
7:13 AM
can complex analysis be used to integrate 1/(1+x^5) w.r.t. x?
 
over $\mathbb{R}$?
you mean you want $\lim_{R \rightarrow \infty} \int_{-R}^{R} \frac{1}{1 + x^5} dx$/
 
7:35 AM
I suppose that there is no program to which if I feed $n$ and it would give me how many transitive relations can be formed on the set
 
probably not, but it seems like something you could count
 
Thank you. I thought so. Of course if the number $n$ is small then we can count
 
transitivity is hard.
 
For large n, it’s possible but it will take its own sweet time
Yeah , right @leslietownes.
 
8:05 AM
couldnt you use inclusion exclusion to count the number of non transitive relations
i dont think the formula would be useful, but it seems doable
 
 
4 hours later…
12:08 PM
Hello, just need some advice. Would it be wise to study Algebraic Topology without a background in Topology?
If yes, where do I start? If no, would it be wise to learn Topology from Munkres first?
and is Kosnioskwi's Algebraic Topology self-contained?
 
Greetings, I have a fairly easy problem that I'm wanting to check the solution to and I haven't been able to online, could I ask here?
Basic calculus, just don't have the answer in my worksheet to confirm if I'm right or wrong.
 
1:05 PM
If you take a lattice in the plane and circularly rotate the lattice points about the origin, then you'll get predictable periodic orbits right?
 
@TedShifrin what I was thinking of was one could collapse the part of the one skeleton consisting of the three edges emanating from the apex of the three simplex towards the vertices of the base two simplex, and then the resulting space would be the one skeleton of the two simplex with vertices identified, which would be homotopy equivalent to $S^1 \lor S^1 \lor S^1$ if i'm not mistaken
 
1:32 PM
If A is a d by d matrix, how can I get this inequality $$|AA^{T}A-A|\leq C(1+|A|^3)$$ with the norm being any norm on $$\mathbb{R}^{d\times d}$$ ?
 
 
1 hour later…
2:33 PM
@epsilon-emperor You need to be familiar with the basic notions of general topology to study algebraic topology, but it's not an extensive background requirement. I haven't read Kosniowski's book, but looking at the table of contents, it introduces all the general topology one would need to know before starting with the algebraic topology, so it's probably fine to read.
 
i think a more important prereq than point set topology is being comfortable with things like squares/regular n-gons/'most' disk like shapes are topologically all disks
does anyone have any ideas about this? What is $$\mathbb{RP}^{3} = S^3 / \sim$$ where $$\sim$$ identifies antipodal points a covering space of, besides itself of course
 
2:51 PM
hmm, maybe it covers $S^3 / \sim$ where $\sim$ is the equivalence relation generated by identifying $(x_1,x_2,x_3,x_4)$ with all points $(+/- x_1,+/- x_2, +/- x_3, +/- x_4)$
 
3:12 PM
Does this sound familiar to anyone? $\gcd(2^a-1,3^b-1)=1,638$ for coprime naturals $a,b$
 
uh
not sure that gcd is gonna be even
 
683 sorry
 
(22,31) is a solution
there are many others
 
yeah I have it coded in c++, but no idea why that's happening
I expected the gcd to always be 1 for some reason
 
so you want a general understanding of the circumstances under which that happens? or something less than that but more than a family of solutions
 
3:24 PM
I want to understand what's happening
 
so the former
you would like a characterization of the a and b for which that happens
 
why 683? What if we change the bases of the exponentials? Are there other primes the gcd can be a multiple of?
I get that we have this because ord_{683}(2) and ord_{683}(3) are coprime
but then what?
 
this is too broad of a question for me but maybe a number theorist can step in
 
I mean I'm willing to take any useful ideas
 
3:37 PM
there are certainly other primes that that gcd can be a multiple of (or in fact be). gcd(2^4 - 1, 3^4 - 1) = 5, gcd(2^3 - 1, 3^6 - 1) = 7.
with n and m between 1 and 500, these are the prime gcds you can get: [5,7,11,13,23,31,43,47,59,71,83,103,107,127,151,157,167,179,191,223,227,229,233,239,263,283,311,347,359,367,383,431,439,443,467,479,491,503,601,643,647,683,719,733,743,839,863,887,911,971,983,1021,1093,1399,1471,1609,2281,2441,4057,4513,5113,6553,6563,7753,11113]
i'll stop spamming the chat. i would focus on 2 or 3 and maybe just the question of what that gcd can be. no changing bases of the exponentials.
 
4:01 PM
No other solutions up to a million
 
4:15 PM
@Thorgott Got it, thanks!
 
4:52 PM
hello good afternoon
 
this seems intuitively obvious but im not sure how to show it: Let $U \subset \mathbb{R}^2$ be an open set not containing $(0,0)$, and suppose $U$ is simply connected. Show that there is at least one ray $\{(r, \theta) : r \geq 0, \theta = \theta_{0} \}$ for some fixed $\theta_{0}$ that does not intersect $U$
 
what about a really swirly open set
 
this is false, no?
 
Yeah, I had the wrong counterexample in my head, but it's false.
 
draw a spiral around the 0 (stay away from 0), go a few times around, and then thicken it up a bit
 
5:02 PM
take a noodle and thicken it up
 
Damn, Leslie has transmogrified.
 
(what leslie said)
 
this isn't leslie it's the bugs who ate his brain
 
yeah ok, that is an easy counterexample.. i actually asked this because I read that on any simply connected open subset of $\mathbb{C}$ we can define a branch of the logarithm that is holomorphic on the open set
but it seems like the spiral provides a counterexample
 
you can do it on the spiral
 
5:06 PM
ok, so i guess the branch cut would be a curve
er by that i mean not just a straight line
 
take pizza dough with flour covering it and drop it over (0,0)
 
if the spiral hits the negative axis in two nearby places, the fact that it's a spiral means you don't need to worry about the values of the log being related to one another the way you'd expect if you could go on a straight line from one intersection point to the other
 
we assume the pizza dough is open and the flower makes it not stick to itself
or alternatively take an hdmi and wrap it around a nail (which we assume is (0,0) )
 
@porridgemathematics Think integrals.
 
we assume the cable inside the hdmi chord is open for some reason
and the casing of the cable separates the interior of the chord from itself
or basically what leslie said
 
5:08 PM
i only have cat 5 cable, will that work?
 
the point is that going around the origin leads to monodromy
i.e. if you go around the origin once, the log increases by 2pi i
but as long as you don't end up at the same point after going around once, this not an issue
 
ah yes, that makes sense to me intuitively
 
does cat5 have power on lan?
Oh I meant power over ethernet
power on lan is possible on all the cats
also, power on lan is commonly called wake on lan it seems
 
then cat5 is suitable
 
5:14 PM
since you know covering theory, the exponential map is a holomorphic covering map $\exp\colon\mathbb{C}\rightarrow\mathbb{C}\setminus\{0\}$ and picking a branch of the logarithm for a holomorphic function $f\colon U\rightarrow\mathbb{C}\setminus\{0\}$ is equivalent to lifting $f$ through $\exp$, which is possible iff $\pi_1(U)$ is trivial in $\pi_1(\mathbb{C}\setminus\{0\})$ i.e. iff $U$ doesn't contain non-trivial loops about $0$
 
oh wow, that is a really nice perspective
i only know pretty basic complex analysis, but logarithms being a lift w.r.t exp makes total sense
do you know a resource for learning about this sort of complex analysis
like complex analysis which assumes knowledge of algebraic topology and deals with monodromy/analytic continuation
 
I still say you should understand how to define log by an integral. And simple connectivity gives path independence .
 
im guessing we would be integrating $\frac{1}{z}$ along contours from some fixed based points in the spiral/simply connected open set to whatever point we are taking log at?
and the choice of contour doesn't matter as you say because the open set is simply connected
 
Right.
Do you know about closed and exact forms?
 
i dont (yet)
 
5:25 PM
OK. You can get path independence from Green's Theorem.
 
yeah, our perspectives are somewhat dual. I'm making a point about the nature of $\pi_1(U)$ in $\pi_1(\mathbb{C}\setminus\{0\})$ and Ted is making a point about $H_{dR}^1(U)$ in $H_{dr}^1(\mathbb{C}\setminus\{0\})$ (and these are related by the canonical maps $\pi_1(U)\rightarrow H_1(U)\rightarrow H_1(U;\mathbb{R})$ and the de Rham pairing)
sorry, second "in" should be reversed of course
 
Mine is actually more naive … just multivariable calc and Cauchy's theorem.
 
yeah, I'm just giving the more "modern" perspective
@porridgemathematics I think most standard complex analysis texts are somewhat outdated in these matters, but a great one that does speak in this perspective is Narasimhan and Nievergelts Complex Analysis in One Variable
 
Hi all
 
thanks for the recommendation, and help @TedShifrin @Thorgott
 
5:32 PM
A Balarka!
 
oh and for the counterexampe @leslietownes
 
I saw your message but forgot to respond. Thanks!
 
Of course!
 
this perspective leads fairly quickly to considering Riemann surfaces, in which case Forsters book is a good reference too
he has a very clear one or two chapters on analytic continuation and the Riemann surface defined by a germ
 
Only natural to do sheaf cohomology and the Cousin problems, Thor.
 
5:34 PM
@leslietownes do you remember yesterday's question ?
 
it is natural, I have no excuse for not having properly done it yet
you probably want to explain branches of the logarithm in terms of the exponential sheaf sequence
 
 
@TedShifrin Here are the questions that were asked, all discussions removed. 1. Are $S^2$ with $n$ points removed and $T^2$ with $m$ points removed ever homeomorphic? 2. What's a simple example of a manifold with $H_1 = 0$ but $\pi_1 \neq 0$? 3. Equip $D^2 \setminus \{0, 1/2\}$ with the "Poincare metric", what's the isometry group? 4. Prove that a finite subset of $M_n(\Bbb R)$ which is closed under multiplication has a matrix with integral trace.
 
oh, that. yes.
 
We can say it has maxima at x=-1 ?
 
5:37 PM
a local maximum, yes. it does not appear to have other local maxima
 
@BalarkaSen so 3. is a doubly punctured hyperbolic space or what is meant?
 
Yeah you can take that.
 
would $T^2$ with a point removed be homeomorphic to $S^2$ with three points removed ?
 
No never
 
so it's asking which Möbius transformations are fixing 0 and 1/2?
 
5:39 PM
oh whoops, homotopy equivalent
 
@porridgemathematics It's a good guess
 
i misinterpretedf
 
Yes, homotopy equivalent.
 
@leslietownes But, at this point if we find f"(x), it comes 0
 
@Thorgott Yup.
Not fixing, preserving {0, 1/2}.
 
5:39 PM
right
and I guess you need to account for orientation-reversing ones too
 
They meant orientation-preserving isometries, so you can take that.
 
rover: that's fine. note e.g. g(x) = -x^4 has g''(0) = 0 but 0 is still a local maximum
 
ok, so 3. just comes down to some calculations, yes?
 
sometimes the 'first derivative test' is conclusive when the 'second derivative test' is inconclusive.
 
feel like I'm missing something
 
5:41 PM
No calculation. It's clearly Z_2.
 
@leslietownes ok πŸ‘πŸ»
 
If some automorphism of the disk fixes a pair of points it's identity. If it reverses it's an involution.
 
oh, yeah, what I was missing is that we have to consider Möbius transformations that also fix the disk lol
preserve, not fix
 
Interesting @Balarka.
I have no clue on 4 yet.
 
It's a trick hah.
 
5:44 PM
That you thought of?
 
Yup, I was able to answer all. I got confused in 3 for a legitimate reason, but only for 2 minutes.
 
I'm also thinking about 4., seems weird
 
Kudos to you, a Balarka!
 
Thanks!! I am happy to have done well. They seemed reluctant to bother me too much but also I think mine was the hardest interview of them all, just because they know me.
 
why is it that a sphere with some points removed can never be homeomorphic to a torus with some points removed?
 
5:46 PM
some end bs, I bet
 
There are a few arguments. My argument was, first do it for diffeomorphism. For that, look at meridian and longitude and their intersection number, which is 1 in the torus.
 
good job for answering them all btw
 
There are no pair of circles on the sphere with intersection number 1.
Upgrade to TOP by cup product in compactly supported cohomology.
I can tell you less technology arguments if you want
(Following @Thorgott's suggestion)
 
ahh, im afraid this is currently beyond what i know (i haven't met diffeomorphism yet)
 
@BalarkaSen yikes
 
5:48 PM
OK, so I will tell you an easier argument
 
So we have to think of trace of matrices of finite order.
 
@TedShifrin Hah, you are almost there
 
there's gonna be a projection in there
so not just an integer but something in {0, ..., n}
 
@leslietownes Right.
 
Why a projection? Why not rotations?
 
5:50 PM
every finite semigroup contains an idempotent
 
that
there could be rotations in there too
 
Which follows easily from Ellis's theorem that every compact semitopological semigroup has an idempotent, after putting the discrete topology on it.
(just kidding it's an easy calculation)
 
horrible
 
hahaha
 
loool
@Ted Well, if $I$ is there you are done, since $I$ has integer trace.
So you should think about $A^n = A^m$
 
5:53 PM
Right.
 
@porridgemathematics For any manifold $M$, consider the space of proper rays $[0, \infty) \to M$ upto unbased proper homotopy. This gives a number (number of equivalence classes of such), which is a homeomorphism invariant of $M$. If $M$ is sphere with $n$ punctures, this number is $n$ (rays going into any of the punctures cannot be properly homotoped to a ray which goes into another puncture).
Similarly, for $T^2$ minus $m$ points, this number is $m$
So $n = m$
But now compute fundamental group.
I could have written a textbook about 1 and 2 but they seemed to be aware and moved on very quickly
 
ah, I got to n=m, but wasn't sure what comes after
 
wow, i can't say i've heard of the space of proper rays, but I can visualize your argument and it does make sense
 
Now after some thought I realize I can also write a textbook about 3
 
I think complements of compacta is easier than proper rays here
 
5:56 PM
And Alessandro will write one for 4
@Thorgott Yeah, anything works, but you're right.
 
Like a half-hour interview?
 
It was very quick, yeah.
 
naively i was thinking, we can draw a loop that avoids removed points of a torus, so that the complement of this slice is still path connected, but it does not seem like this can ever work on the sphere with some number of points removed
complement of this *loop
 
That's actually a working argument isn't it?
 
like in the torus you can always wrap around some $S^1$ which fails in $S^2$
 
5:58 PM
My topology/geometry oral qualifying at Berkeley they asked all sorts of things not on the syllabus because they sorta knew me. Kirby asked Thom-Pontryagin, Wu asked by a manifold with connection had to be paracompact. Kirby led me through thecThom-Pontryagin, which at that point I'd never seen. That was fun.
 
well idk how to formalize it
 
if I remove n points from a compact manifold, then any compact subspace of the resulting space is contained in a compact subspace whose complement has n connected components (by removing some small neighborhoods of the punctures too)
 
@porridgemathematics You want Jordan curve theorem. It should work.
 
yeah, that this doesn't work on S^2 is precisely what Jordan curve says
 
Very nice. It's actually the heart of my intersection number argument.
 
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