02:00 - 21:0021:00 - 00:00

2:48 AM
Ok this is dumb of me but what is the name/meaning the symbol that is like "<" but with the lines concave so it ends up "pointy-er" .... e.g. in the first equation on page 6 of this PDF (princeton.edu/~aaa/Public/Teaching/ORF523/ORF523_Lec16.pdf)

first its not an equation
In mathematics, a symmetric matrix M {\displaystyle M} with real entries is positive-definite if the real number z T M z {\displaystyle z^{\operatorname {T} }Mz} is positive for every nonzero real column vector z , {\displaystyle z,} where z T {\displaystyle z^{\operatorname {T} }} is the transpose...

ty

1 hour later…
4:18 AM
What's the cure for being British?

4:29 AM
Hi @Jakobian
@Jakobian you asked me to use difference of squares but there’s no square…. And what did you mean by bring everything in?!
for context I was trying to maximise and minimise $\lfloor \sqrt{4n+1000} \rfloor - \lfloor 2\sqrt{n}\rfloor$ for $n \in \mathbb Z$ and $1\le n \le 1000$

a-b = (a^2-b^2)/(a+b)
For a = sqrt(4n+1000), b = 2sqrt(n)
Bring everything in means to write - floor(2sqrt(n)) as floor(something)

@Jakobian huh

Don't huh me
If you have a question then ask it

4:45 AM
You mean to write the difference of two floors as floor(something)?

Well that too

Alright thanks

But the point is floor(x+y) = floor(x) + floor(y)
And -floor(2sqrt(n)) = floor(-2sqr(n)) when 2sqrt(n) is an integer
If not then its floor(-2sqrt(n))+1 I think

@Jakobian I don’t think that’s generally true though. Eg floor(0.6+0.7) is not 0
But with some conditions on the fractional part sum I should be able to do it
Thanks

@Sahaj I guess
Yeah we do need those so its probably off by at most 3
But other than that its basically floor(250/(sqrt(n+250)+sqrt(n)))
So n = 1 or number near it should be the maximum
And n = 1000 the minimum

5:03 AM
yeah that’s right
thanks

@Sahaj can I ask you something

sure

What do you usually put on a sandwich

Uhh

Its a simple question. No need to be confused

5:07 AM
Mayonnaise and some vegetables
sometimes cheese

No butter?

Not usually
i don’t like butter

Margarine?

Yes

Oh okay. That's all I need to know. Now I have your full profile. I basically hacked into your brain
Yes...
You like uhhh... food!
I got it?

5:11 AM
True

Nice

By the way, much cuisine here is vegetarian. I haven’t eaten animal meat in my life. I wonder how it tastes like.

@Sahaj It can taste good if you cook it right. But if you don't cook it right? Awful to be honest with you

Makes sense. Do you eat seafood?

No
Rarely fish
Today my food was borsch
Red one
With uszka filled with cabbage and mushrooms

5:19 AM
What sorts of mushrooms do you like?

I didn't taste enough of them to really sense the differences. You can't go wrong with porcini though

Nice. Haven’t ever had them. I don’t think they are found here

6:00 AM
@BalarkaSen I just took a look at Gabai's theorem and it seems the result is not restricted to fibered face. In the case of non-fibered face, it gives depth $n$ foliation for $n\geq 1$.

2 hours later…
7:43 AM
Can someone help me here?
0

Let $s\mapsto A_s$ be an increasing right continuous function, $s\geq 0$. Then define $A_s^{-1}:=\inf\{t\geq 0: A_t>s\}$ with the convention $\inf\{\emptyset\}=\infty$. By definition the map $s\mapsto A_s^{-1}$ in increasing so one can define $A_{s-}^{-1}:=\lim_{h\rightarrow 0} A_{s-h}^{-1}$ the...

8:42 AM
5000% is 5000 divided by 100 so using a calculator we will get 50. So, if the question is what is percentage of the total miles being driven per 100 gallon, what will be your answer? Is it 50% or 5000% which is after you multiply 50 with 100?

9:26 AM
Sorry the question is what is the percentage of 5000 miles being driven per 100 gallon?

3 hours later…
12:13 PM
@onepotatotwopotato Ah, yeah, sure, because the norm minimizers will appear as leaves of a taut foliation which is not fibered

12:54 PM
@onepotatotwopotato Don't read the original paper, just learn from Calegari's book on foliations (cambridge uni press)
I was instead taking a look at Gabai-Oertel, which is about essential laminations

@BalarkaSen why? is it hard to understand?
I don't actually know the definition of essential lamination and why people care. It seems quite important though.

@onepotatotwopotato I think the book will do a better job
Better job as in easier to read from a book.
The proof in Calegari seems pretty short too
@onepotatotwopotato It's like a taut foliation of a 3-manifold but you only have a closed union of a bunch of leaves, instead of the full taut foliation
Indeed, I think "sublaminations of taut foliations" are examples.
It seems useful to me because not all (say, hyperbolic) 3-manifolds admit taut foliations. Admitting an essential lamination is a weaker notion, and one can (I think) write down precise conditions for when an essential lamination extends to a taut foliation
Think of it as a middle-ground between an incompressible surface and a taut foliation. These are both two extremes of an essential lamination.
Since there are many non-Haken hyperbolic manifolds, you don't always get embedded incompressible surfaces. Moreover, many hyperbolic manifolds which are so-called "L-spaces" do not admit taut foliations (this is the case if the hyperbolic manifold is surgery on a hyperbolic knot which admits an exceptional Dehn surgery that is an elliptic manifold).

1 hour later…
2:28 PM
@BalarkaSen there's L-spaces in topology

2:38 PM
@Jakobian Awful

i prefer L-groups

3:17 PM
@BalarkaSen are you saying that because you googled?

i didnt, i guessed it must be some point set topology thing
dont want to know what ti is

it is
its about countability properties (Lindelof, separable, and so on)

hi jakobian
Did you ever participate in math contests in school or college

I have browsed through the lecture notes, but nowhere can I find anywhere why the author insists on writing "$f=g$ pointwise $\mu$-a.e."? Why not just write "$f=g$ $\mu$-a.e.", why the word pointwise? What difference does it make?

@Thorgott do you know if the Berkovich space has an interpretation using prime ideals?
I mean, I saw pictures of it and I was wondering if they correspond to prime ideals
@Sahaj please stop asking me those kinds of questions

3:27 PM
the answer ought to be yes in some sense, but this is a question for Lukas rather than me

alright
@LukasHeger Do you know if the pictures of Berkovich space correspond to partial orders of prime ideals in some sense?
@psie it makes no difference

that's good to hear :)

@Jakobian why?
What do you mean by those kinds :c
I only asked you once..

I was thinking to myself, is there some other way in which two functions could equal each other if not pointwise? Just out of curiosity. Maybe the author wants to distinguish it from some other form of equality between functions. I'm just speculating...

@psie not that I know

3:39 PM
jakobian why do you ignore my question!!!
2
I guess I should quit asking it

ok, I will just accept it

in probability you have convergence in measure which means $\mu(\{|f-f_n|\geq\varepsilon\})\to 0$ for all $\varepsilon > 0$
so "equality in measure" would mean $\mu(\{|f-g|\geq \varepsilon\}) = 0$ for all $\varepsilon > 0$, I guess
but then we just get back equality a.e.

@Thorgott i sent you a message elsewhere

interesting

For $L^p$ you have $\int |f-g|^p = 0$
but this also, again, equality in $L^p$ just means equality a.e. back
what the author could have in mind when writing that? I have no idea

3:42 PM
functions in $L^p$ are already equivalence classes of functions upto a.e.

$\mathcal{L}^p$ then
pedant

wth is $\mathcal{L}^p$ lol

set of all measurable functions with finite $p$-th integral
$p$-th power. Like $\int |f|^p$

gotcha

So $L^p = \mathcal{L}^p/\sim$ where $f\sim g$ iff $\int |f-g|^p = 0$ iff $f = g$ a.e.
the point I am making here is the last iff

3:47 PM
yeah i like this notation

1 hour later…
5:07 PM
@Sahaj why not

Oh how very rude of you!
Actually, I’d take that off because the word “n**b” might be rude.

5:25 PM
@Sahaj why is that rude. I can choose to not respond to anything I like. You are rude for expecting me to respond

I was joking
its obviously not rude

@Jakobian point-set topology is for losers

I see

^ this is how you should actually be rude to Jakobian

well, yes. If you're going to be this direct then I won't take you seriously

5:28 PM
oh, so you're saying i should state the same but more indirectly

well. If you want me to be hurt

i'll scribble that on my excel sheet on how to hurt people

D:

@BalarkaSen What do you think the L in L-spaces stands for to go back to the previous topic

lool

5:30 PM
jakobian what does one study in topology in simple terms?

brutal

@Sahaj continuity

Of?

maps

Does it have any simple real life applications?

5:31 PM
and structures for which you can talk about continuity of maps
this is all that topology is

"structures" of course meaning "(infty,1)-categories"

@Sahaj yes. You can pursue other branches in mathematics

No that’s not what I meant

topology is not a field of applied mathematics

oh

5:33 PM
there are real life applications, supposedly
but I don't care
you do mathematics for fun, not for applications

True mathematician

@AlessandroCodenotti it stands for Lindelof

I know

I got the joke but not at first

#IStandWithLindelöf

5:36 PM
Do mathematicians even understand jokes like normal humans?

I just thought you're trying to engage in a discussion

@Sahaj suppose you were running at the velocity of 5 km/h and 10 minutes later you sped up and started running at the velocity of 8 km/h. Then there must have been a point in these 10 minutes where you achieved the velocity of 6 km/h
this is a simple application of topology

@BalarkaSen oh. But isn’t that obvious from basic calculus too since my velocity is continuous and intermediate value theorem

@Sahaj Just like there is good and bad people in academics, mathematicians also come in various flavours

@Sahaj that is indeed in fact a topological theorem
its easy enough that it is done in a calculus course typically

5:38 PM
Huh!?
Wow.
I’m amazed. Thanks.

But I wouldn't call it an application of topology
its more that in topology you look at intermediate value theorem and say "aha! This is just a topological fact"

@Jakobian in some sense yeah
not exactly "correspond" because they are a lot more points in the Berkovich spectrum than there are prime ideals

I was asking, because as far as I understand, the prime ideals of $C(X)$ look like forests
so the picture should be similar

Spec(A) for A an Banach ring embeds into the Berkovich spectrum of A and there's a continuous retraction from the Berkovich spectrum to this subspace homeomorphic to Spec(A)
that's the relation between prime ideals and points in the Berkovich spectrum

Oh, at least when $X$ is compact, then $C(X)$ should be a Banach ring
so this is even related, more closely than I expected

5:43 PM
@Jakobian I wonder if there’s any generalisation to the theorem in topology

@LukasHeger is this embedding order-preserving?

@Jakobian what is the order in the Berkovich spectrum?

@Sahaj here's another application. suppose you have a road and there's three homes on the left side of the road and three factories to the right side of the road which distribute three different utilities (lets say, water, food, air). each home has a demand for each of the utilities, and the products will be delivered by pipelines. the companies are having a dispute regarding the pipeline configurations; they'll sue each other if the pipelines cross.
you can prove using topology that this dispute cannot be settled without government intervention forcing one of the companies to eat the short end of the stick

@LukasHeger right. I've assumed there would be one. So there's no canonical order?

@BalarkaSen That’s very interesting. Is there any elementary way to work on this problem?

5:46 PM
Oh sorry now I remember an important fact
the "embedding" is not even continuous
it's just an injection
but the retraction is continuous
the point is that the Berkovich spectrum is always Hausdorff
and Spec(A) often isn't
@Jakobian if you have a complex Banach algebra, then the Berkovich spectrum is just isomorphic to the maximal spectrum
I was thinking of other examples where this isn't true

so $A$ here is complex?

yeah if it's a complex Banach algebra that is true

by $C(X)$ I meant functions from $X$ to $\mathbb{R}$

yeah, but the complex case tells us something about that as well, using Galois descent

well, if the Berkovich spectrum and partial order of prime ideals of $C(X)$ are related, then this would explain why the picture of them should be pretty similar
is Berkovich spectrum enormous in some sense?

5:52 PM
depends, in some sense yeah

because the poset of prime ideals of $C(X)$, is set-theoretically very large

Let $G=\mathrm{Gal}(\Bbb C/\Bbb R)$, then $C(X)=(\Bbb C \otimes_{\Bbb R} C(X))^G$. This implies $\mathcal{M}(C(X))=\mathcal{M}(\Bbb C \otimes_{\Bbb R} C(X)))^G=\mathbf{Max}(C(X))^G$

Here $\mathcal{M}$ is the Berkovich spectrum?

yeah
So in some sense we only have the maximal ideals here
which is weird given that the prime spectrum embeds...

well, if you only have maximal ideals then you only have $\beta X$
which is still a lot but it misses all the poset structure
the elements of $\beta X$ are only the roots for the trees I'm talking about

5:57 PM
hmm

what exactly do you mean by $B^G$ here?

oh no wait
the last term should be $\mathbf{Max}(\Bbb C \otimes_\Bbb R C(X))^G$
I mean invariants under the action of $G$
i.e. fixed points

I'm not sure how a maximal ideal of a tensor product like this looks like

@Sahaj It's about planarity of a graph

By Gelfand-Mazur and the fact that every maximal ideal is closed, we get that $\mathbf{Max}(\Bbb C \otimes_\Bbb R C(X))^G=\mathbf{Spec}(\Bbb C \otimes_{\Bbb R} C(X))^G(\Bbb C)=\mathbf{Spec}(\Bbb C \otimes_{\Bbb R} C(X))(\Bbb C)^G=(\mathbf{Spec}(\Bbb C)(\Bbb C) \times_{\mathbf{Spec}(\Bbb R)(\Bbb C)}(\mathbf{Spec}(C(X))(\Bbb C))^G$
hmm
$\mathbf{Spec}(\Bbb C)(\Bbb C)$ is just a point. $\mathrm{Spec}(\Bbb R)(\Bbb C)$ is also just a point
so this is just $(\mathbf{Spec}(C(X))(\Bbb C))^G$
again using Gelfand-Mazur and the fact that maximal ideals are closed this should just be $\mathbf{Max}(C(X))$!
so the result also holds for real Banach algebras, if I didn't mess up
but this is not what you wanted, because it doesn't include information on the prime ideals that are not maximal...

6:16 PM
@BalarkaSen right. Graph theory is on my bucket list. I’m going to study it after my exams are over.

@LukasHeger yeah.

@Jakobian wait I have another idea! What if we don't endow $C(X)$ with the sup norm, but if we take the trivial norm. The trivial norm makes every ring into a Banach ring
I made an error earlier. You only have a continuous surjection from the Berkovich spectrum to Spec(A) if you endow A with the trivial norm
In this case at least the Berkovich spectrum "sees" the prime ideals

oh okay

If $U$ is open in metric space must the closure of the interior of its closure be equal to itself?

@monoidaltransform no
wait closure of interior of closure

6:25 PM
yes
closure of interior of closure

thats a closed set

Sorry, I meant to add U is connected

$cl(int(cl(U)))$ is closed, $U$ is open
thats not true even for $U = (-1, 1)\subseteq\mathbb{R}$

Better yet, think about a punctured disk. The interior of the closure strictly contains the original open set. What you’re writing makes no sense.

I thinik monoidal's question will converge to regular open sets along the way

6:30 PM
What are those?

open sets $U$ such that $\overline{U}^\circ = U$

I've given up on Halmos for right now, such a darn grind. Maybe it'll go smoother down the line...Back to strictly analysis for now
2

Oh, precisely what I addressed. Yet another point set term unknown to me.

When does interior of closure = the open set?

LOL

6:32 PM
when... what do you mean when
sometimes it does
for example for convex sets

Or homeo to convex …

@TedShifrin that's not true I believe

@Jakobian I can't say much about the trivially-valued case unfortunately. But I'm also not an expert on Berkovich geometry

Hmm, I might need ambiently homeo, yes

unless you mean a global homeomorphism
yeah, you call it ambient

6:34 PM
I mean I can't say much other than the Berkovich spectrum surjects onto the prime spectrum in that case

hmmm interesting. Can this be generalized to whatever convex sets mean in riemannian manifolds?

The homeo needn’t be global, but it needs to be on a nbhd.
Geodesically convex is a thing.

@LukasHeger I see. That's pretty disappointing from the previous results you were telling me about, but its alright. Thanks for the discussion

In case we have geodesically convex, then what Jakobian said is true?

@monoidaltransform I have no idea
I only know that this holds for topological vector spaces

6:37 PM
Prove it.

what's the question, if geodesically convex opens/closeds are regular?

Yes, now that Jakobian taught me yet another meaning of regular.
It’s almost on a par with normal.

I also learned the term regular closed from him few days ago.

hmm, not sure if this question is subtle or not

6:52 PM
I thought they are only important in set-theoretic topology when deducing some estimates. Apparently they must be important elsewhere too, since many people ask about regular open and regular closed sets

ill be honest, ive never needed the notion

Not enough people that I’ve ever heard it.
Maybe I spent my career in a cave and read only the wrong articles and books.

"many" here means, just enough for me to notice

I know it — with no name — only as a question in Munkres and hence in my own book.

Oh - here's a fun thing one might want to toy with
The family of regular open sets forms a complete Boolean algebra. Thus to any topological space there corresponds a certain Stone space
oh those correspond to so called "Stonean spaces"
compact, Hausdorff, extremally disconnected space
@AlessandroCodenotti what's a characterization of Stonean spaces in terms of subspaces of the Cantor's cube?
do you know any?
@Thorgott is there a way, in category theory, to obtain a projective object from an object?
or some way in which an object corresponds to a projective object
maybe I'm just describing a functor
well anyway, I am thinking of mapping a topological space $X$ into its complete Boolean algebra $RO(X)$ of regular open sets and then this into its corresponding Stonean space, say $S(RO(X))$
I'm wondering how interesting this is - I read that Stonean spaces are projective objects in category of compact Hausdorff spaces

7:15 PM
in this generality, no. usually one works in a model-theoretic context (like chain complexes on a nice abelian category) where the projective objects are the cofibrant ones and then we're talking about cofibrant replacement, but that's a very different thing.
my best guess regarding your construction is that it could be an adjoint, potentially exhibiting a reflective subcategory

If $X$ is a Stonean space, then $RO(X)$ coincides with the Boolean algebra of clopen sets
so $S(RO(X))$ gives you $X$ back
@Thorgott This should indicate its reflective, right?
its not a proof but it suggests that

it's consistent therewith, yeah
a left-adjoint to a fully faithful functor is automatically left-inverse to it by abstract nonsense

7:40 PM
oh so we need a morphism $X\to S(RO(X))$

in the other direction

but wikipedia says I need $X\to S(RO(X))$
In mathematics, a full subcategory A of a category B is said to be reflective in B when the inclusion functor from A to B has a left adjoint.: 91  This adjoint is sometimes called a reflector, or localization. Dually, A is said to be coreflective in B when the inclusion functor has a right adjoint. Informally, a reflector acts as a kind of completion operation. It adds in any "missing" pieces of the structure in such a way that reflecting it again has no further effect. == Definition == A full subcategory A of a category B is said to be reflective in B if for each B-object B there exists ...
$r_B$

Would it be possible to enlist your help with understanding how (2.6) obtains? I follow up to there but I can't see how he uses (2.5). I thought to define $y := x - \alpha_pq^p$ but that is only nonnegative and not positive.

oops, i did the categorical equivalent of a sign error, sorry
@EE18 this is correct, i dont see why youd have to care about positivity
if $y=0$, you simply get $\alpha_{p-1}=0$

going to go back over it right now, i think i got spooked by the cases since he didn't allude to it and just said to use (2.5) (which required positivity)
gonna treat the cases right now to confirm but i think i see it

7:56 PM
@Jakobian no clue. Why do you expect such a characterization?

@AlessandroCodenotti Because I know there is one for zero-dimensional spaces

8:10 PM
I've scored 10 on an AQ-10 test. 0_0

@Thorgott I'm struggling to stitch up the argument. Would it be possible to ask for help?
it's kinda nitty gritty ugly so no worries if not

textbooks should be forbidden from referring to anything as "the principle of Archimedes"

lol
he gives his statement of it
and various others

on the grounds of how silly that would sound to anyone who overheard it. it sounds like a TV writer who doesn't know math writing "mathy" dialogue

feel free to ask
@leslietownes what about the banach contraction principle

8:22 PM
that one is OK
sounds a bit like a planet killing weapon from the Dune universe

It's very silly but I'll write what I have so far. I've proved the $y = 0$ case meets the criteria (of a unique $\alpha_{p-1}$ (obviously equal to 0) such that (2.6) holds). Thus I'm now examining the $y > 0$ case and trying to show that $q^{p-1} \leq y < q^{(p-1)+1} = q^p$ so that I can apply what's already been developed in (2.5) to $y$ (and then I'll be done).
Embarrasingly, I can't seem to show that $q^{p-1} \leq y := x - \alpha_pq^p$
Once I have that I can push the proof through, but I'm struggling to derive a contradiction from $q^{p-1} \geq y$

I have a basic question. Why does it follow from density that $L^p$ space is the completion of the space of continuous, compactly supported functions? Just because a set has a dense subset doesn't mean it is the completion of that subset, right? It is stated earlier in the text that $L^p$ is complete, but I still do not understand.

8:39 PM
Sorry, I should have said $q^{p-1}>y$ in my last sentence above @Thorgott

that does not need to hold, nor is it necessary
this would force $\alpha_{p-1}=0$

I don't follow. You're saying $q^{p-1} \leq y < q^{(p-1)+1} = q^p$ does not hold in general?
Ah I think I see
OK, will need to rethink my approach then

psie: i really want to add "in the L^p norm" to your phrase "the completion of the space of continuous, compactly supported functions" because the norm is needed for the statement to have meaning. to really get into it, it might help to review any general construction for the completion of a normed vector space, or any abstract discussion of what a 'completion' of a normed vector space is and why it is essentially unique

@psie if $X$ is a complete metric space and $A\subseteq X$ is dense, then $X$ is the completion of $A$, yes
@EE18 now your inequalities are the other way round again, but yes, this need not hold
you don't have to "recreate" (2.4) at every step
you just wanna apply Archimedes repeatedly

@Thorgott correct me if I'm wrong, but didn't FShrike completely misunderstood my question?
0

$\newcommand{\top}{\mathsf{Top}}\newcommand{\stone}{\mathsf{Stone}}\newcommand{\bool}{\mathsf{BoolAlg}}\newcommand{\op}{{^\mathsf{op}}}\newcommand{\cl}{\operatorname{Cl}}\newcommand{\S}{\mathfrak{S}}$I'm not sure why we're focusing on regular open sets here. Stone duality as I know it works throu...

8:45 PM
psie: but basically thorgott's comment. any complete space that contains an isometric copy of A as a dense subset can be thought of as "the completion" of A

@Jakobian i think so, yes

ok, I didn't know this. I will have to look this up, but it's pretty hard to find via google so far.

this is something like the fourier transform where patching together differently oriented web references will not be as helpful as consulting any one book treatment of the subject
not because books are better than wikipedia but because the details (if there are any) depend on having a consistent set of definitions, which disparate web resources cannot have

the reasoning depends on how you define "completion"

i'm using scare quotes around "the completion" above because there may be different ways of realizing a completion literally as a set of things
this might not be the right time to say it, but depending on your source, you may already have this indeterminacy, or be encountering choices about it, in your definition of what "L^p(Omega)" is

8:49 PM
the spaghetti I just ate was good

@Thorgott OK, reorienting to that. I have $y:= x - \alpha_pq^p$ and I fix $q^{p-1}$. Archimedes principle guarantees me a unique integer $\alpha_{p-1}$ such that $\alpha_{p-1}q^{p-1} \leq y <\alpha_{p-1}q^{p-1} + q^{p-1}$. I just now need to confirm that that integer must be in the set $\{0,1,...,q-1\}$

almost, but not quite
$\alpha_{p-1}=0$ is allowed, even if $y>0$
in my opinion, there really is no point in separating the cases $y=0$ and $y>0$

ah of course, because (intuitively) subsequent digits might be nonzero
you are right
OK, I'll edit the above briefly

@EE18 yup

Thank you! Just trying to prove it has to be in that set of digits now
will report back

8:55 PM
yeah, the author doesn't define completion anywhere (these are just brief, introductory notes on measure theory). Seems to be a pretty intricate matter

psie: FWIW you could take that offhand comment (with a small amount of surrounding context) to be a definition of 'completion.' but if it sounds like the book will not be using this remark anywhere, maybe you are free to ignore it
we can all ruminate privately on whether it's a good idea or a bad idea for a textbook to include remarks relating to concepts that aren't elsewhere in the book, and would maybe only be familiar to someone who already knew the subject matter of the book, and thus wasn't in the likely audience of the book

02:00 - 21:0021:00 - 00:00