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02:54
@SoumikMukherjee thanks, hope I get to the interview xD
03:20
@MaximeJaccon No idea, sent a mail to Desmos support. They reply pretty fast, so expecting a reply by tomorrow.
 
2 hours later…
05:22
cup product in de Rham cohomology is a wedge product. How about the cap product?
 
3 hours later…
08:33
@nickbros123 just say you know @Soumik and you'll pass the interview smoothly :D
 
2 hours later…
10:21
@SineoftheTime I didn't pass the interview myself ☠️
10:42
four spindles arranged in R^3 compete to be in quartic harmony
riddle^
11:00
what is this?
@SoumikMukherjee skull emoji
@HomesickIguana my bed sheet at 3 am
whatever it is I don't like it at all
11:26
0
Q: A quartic inquiry

Homesick IguanaIt is clear from this answer that a specific configuration of four spindles from spindle toruses, have at least three quartic planar slices. What is interesting is that, this configuration of spindles has quartic slices only for a subset of the entire quartic curve, suggesting a possible "extension"

Is it clear what I am asking?
Cone over the trefoil knot
For every knot we can construct the cone over the knot which is a disk in the 4-ball with the required property with the exception that it is not locally-flat or smooth at the singularity (it works for the trivial knot, though).
https://en.wikipedia.org/wiki/Slice_knot#Cone_construction
@SineoftheTime haha
12:02
I'm working this exercise in Rudin's PMA (exercise 6, chapter 1). It is about extending exponentiation to rationals (i.e. making sense of the expression $b^r$ for $b>1$ and $r\in\mathbb Q$). At one point, I feel like I desperately need to know the fact that $b^r$ is increasing for a fixed $r$. More specifically, I need to know $b^r>1^r$ for $r\in \mathbb Q$. Is this somehow "obvious" from the rules of exponentiation of integers?
Here's the exercise. After proving some basic identities for $b^r$ when $b>1$ and $r\in\mathbb Q$, Rudin asks one to show if $x$ is real, define $B(x)$ to be the set of all numbers $b^t$ where $t$ is rational and $t\leq x$. Prove that $b^r=\sup B(r)$.
So I need to show $b^r$ is an upper bound to $B(r) = \{b^t : t \in \Bbb{Q} \land t \leq r\}$.
I'm tempted to just argue $b^r=b^tb^{r-t}\geq b^t 1^{r-t}$ for any $t$, but you see, I'm using that $b^t$ is increasing for fixed rational $t$.
@onepotatotwopotato I don't understand, what does "is a wedge product" mean
12:45
@onepotatotwopotato We can transport the cap product from singular cohomology to de Rham cohomology via De Rham theorem
@hbghlyj this also is useful for multigraphs
sorry meant to reply to the previous statement above.
De Rham theorem says that the pairing of differential forms and chains, via integration, gives a homomorphism from de Rham cohomology $H_{\mathrm {dR} }^{k}(M)$ to singular cohomology groups $H^{k}(M;\mathbb {R} ).$
i.e. the boundary of the cone degerates to the graph structure.
obviously this would be a graph with smooth edges and vertices are self intersections.
What's even more interesting - attaching complexes to the boundary graph or knot depending on context of course.
Something I'm working on is when the boundary are essentially overlapping quartic functions of a specified type which extend to cones resp.
@Thorgott I mean defined using a wedge product.
but it seems the Poincare duality formulation gives me some idea. It should be something like $\alpha\frown\omega = \int_{\alpha}\omega$.
well-defined and makes sense but didn't check if it's really the cap product (but seems correct to me)
13:01
...these quartics intersect - objectives split here (again depending on context) - if you like algebraic topology you want to study these overlapping algebraic curves in terms of line bundles, divisors etc. - graph theory folks want to look at combinatorial graphs (edge lengths normalized to 1) from which you can study a plethora of situations, even spectral properties/laplacian matrix etc...Ihara zeta appears here naturally. Ihara zeta is by no means exclusive in this representation
de Rham theorem is not helping here. I'm intentionally avoiding singular stuff. Understanding using form is more intuitive to me now.
I want to know Stiefel-Whitney class at some point. Seems (very) important
I am slightly interested in Hasse-weil zeta function over the corresponding varitie(s) but I don't know exactly how it arises i.e. if line bundles over said varieties indeed support divisors capable of creating a "well-defined" weil zeta function w/ appropriate meromorphic continuation blah blah blah
had to get that out.
@onepotatotwopotato so you're talking about de Rham cohomology? context is everything
@onepotatotwopotato this is only the Kronecker pairing
@Thorgott Yes I'm talking de Rham cohomology. The degree of $\alpha$ and $\omega$ does not have to be the same.
13:27
then this is not well-defined as written
you probably get the right thing if you restrict $\alpha$ to an appropriate front or back face
14:13
you can undelete a question you deleted, right?
@onepotatotwopotato then why are you avoiding singular stuff?
SW-classes live in singular theory with $\mathbb{F}_2$-coefficients, not in dR theory
how do u feel about statements like "reasoning is not the only way to approach truth"?
this idea is used to argue in favor of god
if you want me to take part in your psychological study you'll have to pay me
2
hmmm
@BenSteffan u can also do it as charity :P
I'm not feeling charitable :)
you need god :P
14:28
heaven forbid :)
@RyderRude mu
@XanderHenderson what does this mean
In the Sinosphere, the word 無, realized in Japanese and Korean as mu and in Standard Chinese as wu, meaning 'to lack' or 'without', is a key term in the vocabulary of various East Asian philosophical and religious traditions, such as Buddhism and Taoism. == Etymology == The Old Chinese *ma (無) is cognate with the Proto-Tibeto-Burman *ma, meaning 'not'. This reconstructed root is widely represented in Tibeto-Burman languages; for instance, ma means 'not' in both Tibetan and Burmese. == Pronunciations == The Standard Chinese pronunciation of wú (無; 'not', 'nothing') historically derives from the...
what is the buddha...
14:31
i thought you were trying to dismiss reasoning in ur reply :P @XanderHenderson
No, only you.
$\mu$ :)
无 :)
That works, too, according to Wikipedia.
I don't speak or read any of the indicated languages.
14:36
it does in China but not in Japan
24 mins ago, by Ben Steffan
you can undelete a question you deleted, right?
@BenSteffan Yes.
ok cool, thanks
15:34
Is there any community policy about non-English answers like this one on MSE?
As far as I understand it the site is English-only.
That answer also violates the guideline of "don't rely on images to rely critical information as much as possible"
16:03
@SohamSaha There is a FAQ on meta (math.meta.stackexchange.com/questions/tagged/faq).
18
Q: What is Math SE's policy on posts in languages other than English?

Xander HendersonWhat is Math SE's policy on questions, answers, and comments which are posted in languages other than English? What should I do if I see something written in another language?

The more problematic part of that answer is that it is an image of text, and not text.
ben is right, though. that answer would not be OK even if the text on the image were in english
as xander said as i was typing that :)
63
A: Why image cannot be used for explaining my maths problem?

Xander HendersonThis is largely an accessibility issue. Images are not searchable. Whatever you think of the quality of the built-in search engine, or external tools such as approach0, and their ability to find good matches for search terms, they do exponentially worse when the only information is given in ima...

It is kind of crazy the speed at the potus is operating - scary
even if you don't have visual impairment, images are often very hard to describe. you're basically at the mercy of the graphic design skills of whoever designed the image (which, when someone is pasting in an image, is almost guaranteed to be somebody other the OP). and the ability of the OP to understand text references to the image (which they probably did not create). its an extra layer of potential confusion over everything.
images are also not searchable, at that
16:09
@leslietownes welp I messed up facepalm.
or editable. nobody can just pop in using site tools and fix the mathjax, or whatever.
or rather their content cannot be found through a search engine
I tend to avoid images - but when I do I mostly use them properly and within the sites guidelines
...and if I do slip up you'll bet I'll croak to the mods.
there's also a whole lot of equation spew in that answer. even if it were typed in, i skip right by answers like that, as i am not going to read through like 20 related equations to see if maybe there is a mistake. which means even if its upvote worthy i'll never upvote it.
Yeah images in answers I think are generally a peg below that of images in questions
16:14
thats not a site policy thing, just my own thing. a lot of people put in stuff like that because they think it is indicative of effort (in a question) or makes it easier to follow because it "shows all the steps" (in an answer). in my head, it is indicative of low effort and harder to follow than something that someone took the time to make short
@leslietownes If I am going to write an answer with a lot of derivation, I tend to try to give a summary at the top. :/
this is making me feel a little uncomfartable
because i put 4 images in my last question :/
the figures in that answer are of higher quality than one usually sees in image-only posts, though. i'll say that. it's also typeset in something instead of someone just randomly popping text labels over something in ms paint.
homesick nobody is talking about "putting images in a post," we are talking about "making a post that is entirely an image"
oh - i see leslie.
@HomesickIguana Images are not, in and of themselves, a problem. It is images of text that are a real problem.
16:16
and i'm veering off into my own rant about equation spew, which is only tangentially related to the issues raised before my rant :)
no images allowed >:(
@BenSteffan what about diagrams?
well there's amscd :)))))
you never need anything that isn't square do you :))
true ig :)
i only just started using overleaf 3 months ago - late to the party perhaps? i might use one of those commutative diagrams just to see how to code them up
16:27
I've never really used Overleaf. I've been using Plain TeX since the early 2000s, and switched over to LaTeX around the end of my masters program in 2012ish.
I have created a couple of very small Overleaf documents to show my students some basic LaTeX.
there really are levels to this game
overleaf is... not great
but for a start it's okay I guess
im not super satisfied with my latex game
Suppose $V$ is a vector space over $L$ as well as over $K$, and $K \subset L$ is a field extension. I am trying to prove $[V:K]=[V:L][L:K]$. I write an arbitrary vector $v \in V$ as $v=\sum_{i=1}^{[V:L]}l_i v_i $ for a basis $\{v_i\}$ over $L$. But each $l_i$ can be written as $\sum_{j=1}^{[L:K]} k_{ij}\beta_{j}$ for $\{\beta_j\}$ a basis for $L$ over $K$. Together this yields $v=\sum_{i=1}^{[V:L]}\sum_{j=1}^{[L:K]}k_{ij}\beta_j v_i$.

This equation can take care of span, and also linear independence. But is this proof fine? It seems kind of suspect that those vectors are basis for $V$ over
@XanderHenderson plain TeX?? you scare me
16:30
@XanderHenderson I use texlive on VS code
@BenSteffan could you elaborate? I just started using it to make a clean write-up for something, and I still have chance to switch over to something else
@BenSteffan I am actually shocked to my core. I thought that overleaf was the best 0:
I gotta get plain tex?
overleaf is fine for documents less than like 20 pages
what happens after 20 pages?
idk, for me rendering takes time
16:31
compiling gets slow
@nickbros123 My phd thesis takes about 30 seconds to compile.
From the command line.
how big was ur thesis
100+?
About 120 pages (i.e. too long).
did you get 120% on it @XanderHenderson?
@HomesickIguana That isn't how PhD theses work.
16:33
sorry - bad joke
@SohamSaha for a start it's limited by the fact that it's an editor in a browser window. limited customizability, having to navigate around browser key bindings, etc. the way suggestions etc. are implemented is... okay, but not great.
@XanderHenderson Searching for plain tex on google is a pain. It always gets autocorrected to plain text
Am I the only one who uses TeXstudio?
also overleaf is showing signs of going down the drain. they recently limited how many people can work together on a project on their free plan to just two.
they also have some new AI gadget which you cannot properly turn off which is just... eh
16:35
@BenSteffan time for me to jump ship then
@BenSteffan what's the alternative?
Any better alternative? Plain tex?
im jumping over...leaf?
overboard.
@BenSteffan texlive on VS code with common git repository open is a good way to collaborate i guess
^ sounds pretty good
I swap out the vs code component for neovim
but that's how I go about things
16:36
ah, i can smell a techbro
:)
old habits die hard
bro is two steps ahead
essentially (favourite text editor) + texlive + latexmk does just fine
there are also "IDEs" such as TexStudio
Since we are talking about tex rendering and stuff, does anyone know any good blogging platform that supports mathjax+markdown?
Plain TeX by the way is orthogonal to this (pretty much): it refers to a TeX format
since there seems to be some confusion
16:39
@BenSteffan Yeah, Plain TeX is to LaTeX as C is to C++.
it's the original, vanilla tex. Knuth's tex. It's very low-level
that's an apt comparison
I think so.
@SohamSaha define "blogging platform." You can set up a static site generator that eats some form of markdown pretty easily
Something like Hugo would do the trick
getting mathjax to work shouldn't be very hard: all you have to do is load a script, and I'm certain people have come up with templates to do just that
16:40
Online website where I can login and just start writing, like inbuilt mathjax+md
Currently I am trying to edit my Blogger html code, but md doesn’t work on the index page, i don’t know why
Let $(Y,y)$ be a pointed space. The *loop space* $\Omega Y$ of $Y$ is the subspace of the path space $Y^I$ (with compact-open topology) consisting of the loops in $Y$ with base point $y$, i.e.,

$$ \Omega Y = \{ w\in Y^I | w(0)=w(1)=y \}. $$
To finish that train of thought: don't use plain tex. almost nobody does anymore.
Is it possible to use multiple basepoints at the same time?
@HomesickIguana elaborate?
@BenSteffan would try out texlive then
16:43
you want your paths to go between any number of points, rather than just one fixed one?
@BenSteffan yeah - except I haven't figure out much in this direction
maybe I need to keep reading
@SohamSaha TexLive is the name of the most common LaTeX distribution. It includes plain tex, latex, etc. You will need it whatever you do with TeX anyways (pretty much)
@HomesickIguana well that's not very difficult
@BenSteffan oh, ok
Define the space of paths starting at some point $x$ and ending at some point $y$, and then take the union of those for all the points your interested in
They're all subspaces of $\operatorname{Map}(I, Y)$ after all :)
Oh okay @BenSteffan I get that thanks
16:50
@BenSteffan of course, this will be a disjoint union in most reasonable circumstances, so not particularly interesting
it is what it says on the tin :)
@BenSteffan Yeah, but the tin is written in Arabic, or Sanskrit, or something.
it is written in the algebraic topologist's vernacular
Yeah, like I said, "or something".
A sphere with two points identified is homotopy equivalent to $S^1 v S^2.$ I took an $S^2$ with two points, defined an attaching map onto antipodal vertices of a cubical cell complex (which is not contractible). Is that well-defined from the standpoint of Hatcher's alg. top.? (I'm on p. 13 if that helps).
First sentence I get. Second is stretching my brain a little
17:05
wdym by "cubical cell complex"
also what are you attaching
actually let me say cube for now (which is contractible). I am attaching the "cone points" of the $S^2$ (I think that is precise terminology) to $v_1$ and $v_2$ resp. for this question let $v_1=(0,0,0)$ and $v_2=(1,1,1)$.
okay yes
why would there be an issue of well-definedness with that?
I guess I should say this is in R^3 too.
I'd guess you shouldn't. That seems irrelevant.
okay
Then it seems well-defined to take $v_3,v_4$ and $v_5,v_6$ and $v_7,v_8$ and define $S^2$ copies for each vertex pair using the same construction as initially. Is there anything stopping me from taking the union of these four $S^2$ copies (with their cone points having already been attached to the vertices specified)?
17:19
take the union in which space? you can just attach each of these copies in sequence, or all at the same time
I'll have to think about which space. But your last sentence makes sense
Don't try to work in an ambient space; this will only make your life more complicated.
17:44
Speaking of overleaf and the like, has anyone done beamers before and enjoyed it? Or are you currently practicing it? My past experience with beamers was...challenging, to say the least (mainly because I am very inexperienced with beamers). I don't know if I should get better at it or just give up and well....Windows powerpoint! :D There are so many templates for beamer. It's hard to know where to start...
there is a tendency for people to put too much on slides for a math talk, even people who are well aware of this tendency and try to avoid it end up doing it
so to any extent that a tool for creating math slides is annoying to use and causes people to do less, it might be a feature and not a bug
yes, the beamer I was doing once got so big with so many images, I reached the compile timeout on overleaf
i guess it might depend on what kind of environment you use for texing stuff but i always found it really easy to make batch changes to beamer slides (its just latex underneath, any text editor can do it) vs. PPT (there's "search and replace," which sucks, and other than that i guess you can write custom vb script)
its the classic effort curve that seems to accompany all things latex. if you're just doing something once, or even once every six months, some MS product is probably just as good, at least if you're starting from a premade template that you like. latex is only clearly more convenient if you are going to be using it a lot
yes
although there's also the trap of "i don't plan on doing this a lot so i'll just do it in ms office this one time" or "i don't have time to learn this week so i'll use MS this one time" and you blink and a few years later "this one time" was 20 times, you've wasted a nonzero percentage of your life on fiddling with word/PPT quirks, and you still don't know latex
Joe
Joe
18:11
I have a somewhat silly question. Is there a reason why short exact sequences are often written as $0\to M'\to M\to M''\to 0$? That is, why do we use the letters $M'$, and $M$, and $M''$, in that order? Is this some kind of mnenomic that I am missing?
the middle part is 'made up' of the two outer parts, I think that's the entire reason
@psie I have no problems with beamer. Though I have never tried to use it with Overleaf.
These are the black-and-white handout version of my slides from a talk at the JMMs in 2020: yozh.org/wp/wp-content/uploads/2020/01/notes.pdf .
And the template I created for my work at UCR: yozh.org/wp/wp-content/uploads/2017/05/slides.zip (it isn't too hard to modify it with other colors / logos).
18:31
@XanderHenderson nice! during my education, I felt like they put a lot of effort into teaching the students how to write LaTex, but nothing in how to present mathematics (for a thesis presentations, if I remeber correctly, slides were mandatory). I remember being very lost with how to do anything in beamer before a very important presentation.
@psie That was my experience, too. Except that there was no focus on teaching LaTeX, either. Everything I know, I figured out on my own.
it is hard to teach, outside of extreme generalities that people probably already know (e.g. "don't put too much on any one slide, and 'too much' is probably less than you think it is"). so much depends on specifics of the audience and what the talk is actually doing. it is more helpful to hear specific feedback on specific works in progress, but often, nobody has the time for that.
its hard to evaluate slides in a vacuum. the best advice might be "you need to put way more in about X," or "you don't need to say most of that stuff about X," for the exact same deck. depending on who the audience is. and it is really easy to infer the wrong things from examples that are recommended to you as 'good,' e.g. maybe it's good because of specifics of that audience that won't hold true for your audience.
at my last academic job there were two kinds of thesis defense talks. ones that spent way too much time setting the stage, and ones that spent no time at all setting the stage. i often got the impression people were re-using large portions of decks on their material that they had already used somewhere else, and hadn't adjusted for the audience.
«Il semble que la perfection soit atteinte non quand il n'y a plus rien à ajouter, mais quand il n'y a plus rien à retrancher.» — Antoine de Saint Exupéry
"Perfection is achieved, not when there is nothing more to add, but when there is nothing left to take away." - Antoine de Saint-Exupery
18:46
weird advice coming from a guy with a long ass name
Yeah, he should just be called Tony St Exp.
Or maybe just "Tony".
yeah, I imagine it being hard to teach. I don't know how I'd organize it myself, but I remember at one math course I took, one mandatory assignment in the course was to present solutions to exercises in front of everyone at the blackboard/whiteboard. Maybe one could have required the student to present solutions in beamer instead, I don't know.
His full name's shorter than mine. :)
@PM2Ring Yeah, hash tag me also.
But I go by "Xander".
Which is a lot shorter.
And one of the other faculty here calls me "Dr X". Which is shorter, still.
see, dr. x is the one who should have said that quote.
although i have to give tony credit, his name would certainly have been longer if he'd used his given name ("Antoine Marie Jean-Baptiste Roger") and not his title of nobility
18:54
I don't mind long name systems. I like the Hispanic system that maintains surnames from both the maternal & paternal sides. OTOH, I also think the traditional Icelandic system is cute, where a mother, father, daughter, and son all have different last names.
@leslietownes Interesting how Marie can be a boy's name in French, with no attempt to create a male gender specific form, whereas Italian has Mario. There's no equivalent in English.
I guess there's Marion.
@PM2Ring It, well, can't. Not as a first name.
The practice of having some variant of Mary as some form of middle name for boys used to be rather widespread across different languages. I don't know where it comes from, however.
@BenSteffan Ah, ok. I must admit that I've only ever seen it as a 2nd or 3rd name.
Precisely :)
19:32
test
it's the season
 
2 hours later…
21:05
@Joe this is a really good question. for reasons i cannot explain it seems clear to me that the thing in the middle of a SES is 'the star of the show' and thus worthy of first place. as to why it's M' and M'' on the other ones, maybe it's just, you've used M and now work left to right adding primes as you go
i also wasn't consciously aware of this convention although when i looked at some old notes i saw my prof using it
Of course $M$ is the star of the show: The exact sequence describes $M$ as an extension of $M''$ by $M'$! :)
It's the whole that you put together from the parts to its left and right
yeah i mean i get that, but suppose you're going to call the other ones M' and M'' and you are going to decide once and for all which is which. is the thing on the left being M' just left-to-right bias, or is there more to it
betting on left-to-right bias for that one
the case one should always have in mind is the trivial extension where $M = M' \oplus M''$, and here the roles of $M'$ and $M''$ are interchangeable
clearly, one prime looks like the start of $\hookrightarrow$ and two primes look like the tail of $\twoheadrightarrow$
2
21:32
I posted a question on the big boy site
0
Q: What is known about elementary equivalence of open set posets of topological spaces?

Akiva WeinbergerRecall from model theory that two structures are called elementarily equivalent if they satisfy the same first-order sentences. In other words, two structures $\frak A$ and $\frak B$ of the language $\cal L$ are elementarily equivalent (written $\frak A\equiv\frak B$) if $\operatorname{Th}(\frak ...

I have a question related to

https://math.stackexchange.com/questions/4520689/if-a-subset-0-1-is-strong-measure-zero-then-fa-is-strong-measure-zero-w

Suppose that $f:[0,1] \rightarrow[0,1]$ is continuous. The question above proved that $f$ maps strong measure zero sets to strong measure zero sets; that is, if $X \subseteq[0,1]$ is strong measure zero, then so is $f[X]$.

Now suppose that $f: \mathbb{R} \rightarrow \mathbb{R}$ is continuous, how do you show that the same conclusion holds, that is, if $X \subseteq \mathbb{R}$ is strong measure zero, then so is $f[X]$?
I am not sure how to handle that $f$ is not necessarily uniformly continuous
$f(X\cap[k,k+1])$ is strong measure zero by this result for all integer $k$ and the countable union of strong measure zero sets is strong measure zero
21:48
Ok, thanks.
*The countable union of strong measure zero sets is strong measure zero*
My proof: Let $(A_k)_k$ be a sequence of strong measure zero sets. Let $(\varepsilon_n)$ be a sequence of positive real numbers. Since $A_n$ is strong measure zero, there exists a sequence of intervals $(I_{n,k})_k$ such that $A_n\subseteq\bigcup_k I_{n,k}$ and $\mu(I_{n,k})\leq 2^{-k}\varepsilon_n$. Then, $\mu(\bigcup_k I_{n,k})\leq \sum_k\mu(I_{n,k})\leq \varepsilon_n$ so $\bigcup_n A_n$ is strong measure zero.
Sorry, my subscript is wrong.
[deleted]
Proof: Let $(A_k)_k$ be a sequence of strong measure zero sets. Let $(\varepsilon_n)$ be a sequence of positive real numbers. Since $A_k$ is strong measure zero, there exists a sequence of intervals $(I_{k,n})_n$ such that $A_k\subseteq\bigcup_n I_{k,n}$ and $\mu(I_{k,n})\leq \varepsilon_{f(k,n)}$ where $f(i,j)=\frac{(i+j)(i+j+1)}{2}+j$ is a bijection from N^2 to N.
Then, $\bigcup_k A_k\subseteq\bigcup_{k,n} I_{k,n}$ and $\mu(I_{k,n})\leq \varepsilon_{f(k,n)}$ so $\bigcup_k A_k$ is strong measure zero.
that does not seem to provide what the definition is asking for
22:03
Why not?
$\{I_{f^{-1}(0)},I_{f^{-1}(1)},I_{f^{-1}(2)},\ldots\}$ is a countable collection of intervals that covers $\bigcup_k A_k$ and $\mu(I_{f^{-1}(n)})\leq \varepsilon_n$ so $\bigcup_k A_k$ is strong measure zero.
yeah ok, that works
I suppose it's obvious that this should've worked anyhow
Yes, thanks.
 
1 hour later…
23:18
5
Q: No perfect non-empty set in $\mathscr{P}(\Bbb{R})$ can be strong measure zero.

Sebastián P. PincheiraHere is what to be proven: No perfect non-empty set in $\mathscr{P}(\Bbb{R})$ has strong measure zero. The textbook (Teoría de la Medida, Jaime San Martin Aristegui, section 1.6) suggests the following approach: Suppose that $f\colon [0,1]\to \Bbb{R}$ is a continuous function and $A\subset [0,...

In this answer, I don't know why "At iteration $n$, the length of each interval is $\le(2/3)^n$."
$m$ is arbitrary point in $[A_0+\frac{A_1-A_0}3,A_1-\frac{A_1-A_0}3]\setminus A$ and $(A_{\frac13}, A_{\frac23})$ is the maximal open interval containing $m$ that is disjoint from $A$, why does it follow that the length of each interval $[A_0,A_{\frac13}]$, $[A_{\frac23},A_1]$ is $\le(2/3)^1$ at iteration $n=1$?
One thing I love about stable homotopy theory is that one of its fundamental tools exists only in folklore and anecdotes.
Fantastic state of affairs.
23:38
ben is that true? what is it?
It's the Atiyah-Hirzebruch spectral sequence.
You can find definitions of the spectral sequence pretty much everywhere. Multiplicativity is generally mentioned but proofs are considerably rarer (the situation is so bad it prompted Dugger to write two papers essentially proving this from scratch).
haha i thought that was like some well established thing in the spirit of "wow looks complicated but someone surely worked out the details" and i guess we were all doing that??
haha
But there's more structure: There is a pairing between the homological and the cohomological version of the spectral sequence; I know of a single source that works this out. Most sources don't even mention this.
And then there's more. This, for instance: math.stackexchange.com/questions/5028531/cap-formula-for-ahss
oh wow, haha
good for dugger
@leslietownes I doubt he had much fun doing that
23:43
and i'm sure the professional honors just kept rolling in after he did it
it's actually a fairly well-cited paper
he did things in some generality of course
well that's something
I'm not so familiar with the generality of ring spectra, but I saw that question earlier and I was under the impression this has been worked out at least in the plain context of generalized homology theories
the paper is from 2003. the spectral sequence is from '56 :^)
well, I've personally never needed to know there is a pairing between the homology and cohomology SSes
23:46
@Thorgott it should not be too hard to work this out, but would be annoying
@Thorgott good for you. I have. :)
my advisor was always hyper cautious about "verify the basic property" stuff from some things that had happened to other people in the 1960s, and was sometimes quite annoying about it, i think because he had encountered difficulty in publishing proofs that certain things were well defined in his early career
and there were a handful of people actively marketing hocus pocus when i was a student
there's a general machinery for constructing spectral sequences and pairings between them (in particular, multiplicative structures) in terms of Cartan-Eilenberg systems
you can get the Serre-Atiyah-Hirzebruch spectral sequence like that
for me 2024 was the year of the error I think. People I know, including myself, just found so many errors in various publications
@Thorgott this sounds like you're referring to dugger's paper
I'm sure I've looked at Dugger's paper before, but I don't remember
it discusses pairings, but pairings between spectral sequences of the same variance
not quite the same thing
23:49
Cartan-Eilenberg systems go back to Cartan & Eilenberg (wow!), but they were historically somewhat overshadowed by Mackey's exact couples
but they're really the better formalism IMO
now there's a hot take
every Cartan-Eilenberg system has an underlying exact couple, which suffices to describe the spectral sequence
but you can define pairings of Cartan-Eilenberg systems that induce pairings of spectral sequences
whereas you can't define 'pairings of exact couples' in a meaningful fashion
(I think Spanier has an old paper where he tries, but it doesn't amount to much)
ah no, that was also Massey, not Spanier
nowadays we would like to derive everything from the Lurie spectral sequence anyways
perhaps these facts all exist in the highbrow literature somewhere
I don't even know what that is :)
John Rognes has a great set of notes on Spectral Sequences, where he uses Cartan-Eilenbrg systems and derives the multiplicative structure of the Serre SS
jacob lurie doesn't exist only in folklore and anecdotes, i've met him
23:54
I think the pairing between homology and cohomology should not be more complicated
the hardest part is usually identifying the E_2-page
and i've at least talked to atiyah although i did not get the opportunity to corner him about his spectral sequence
It takes a filtered object in a stable $\infty$-category and spits out a spectral sequence with values in its heart
people need to go after these folks like investigative journalists
it generalizes pretty much any spectral sequence in having to do with spectra
J.P. May once alleged that there is a homotopical method of identifying the E_2-page of the Serre SS
but I have never been able to verify this
23:56
when he's putting out his garbage someone needs to pop up out of the bushes and demand an answer from him
@leslietownes he might not have known. the spectral sequence is due to whitehead, but was first published by atiyah and hirzebruch
it's misattributed in a sense
although it's fairly obvious and really anybody could have come up with it at the time
@BenSteffan of course
@Thorgott I'm not sure I understand what this even means
I asked an MO question about it
2
Q: Identifying $d_1$ in the Atiyah-Hirzebruch-Serre spectral sequence

ThorgottIn A Primer on Spectral Sequences (also later published in More Concise Algebraic Topology), J. Peter May describes the Serre Spectral Sequence for any homology theory. To recap, suppose $p\colon E\rightarrow B$ be a fibration with fiber $F$ s.t. $\pi_1(B)$ acts trivially on $F$ (in the homotopic...

ah yes, the famous "straightforward" claims by May
23:58
I genuinely believe this one is very subtle
it's a shame cause I really like the idea of the argument
all the other arguments I've seen are rather ugly

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