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1:02 AM
what is the correct notation to insert a specific kind of an edge to a graph $g \gets (a, b, kind)$ ?
1:46 AM
So who is on the strike here?
1:56 AM
node: i don't think there is a standardized notation for it. and particularly in the world of graphs (where even the definition of 'graph' is subject to a number of common variations) i would be inclined to expressly define whatever notation you use even if it is in common use.
2:51 AM
If I have a symplectic manifold with an analytic structure (symplectic form is still smooth), and a Lagrangian inside, can I perturb the Lagrangian by a Hamiltonian isotopy to be an analytic submanifold? Can I do it in families?
"subanalytic" would be good enough for me.
If $f : L \to M$ is the Lagrangian, Weinstein's tubular neighborhood theorem gives an extension $F : T^*L \to M$. Pullback the analytic structure on $M$ by $f$, pick an analytic fiberwise ac structure on $T^*L$ to get an analytic structure on $T^*L$. $F : T^*L \to M$ does not preserve these analytic structures on the domain and codomain, so it is only a smooth embedding. Whitney approximate by an analytic map $F_1 : T^*L \to M$ and a connecting homotopy $F_t : T^*L \to M$. Then use Moser lemma?
For the families version, we can use a families Weinstein. Seems to work?
$F=F_0$ is an open symplectic embedding
The perturbed Lagrangian would be $f_1 := F_1|0_L$, $0_L \subset T^*L$ being the zero section
3 hours later…
6:14 AM
Can this sum be a prime for any $n>310$ ?
6:36 AM
Q is not locally connected.
Proof: Take any nbd. of 0 in Q, then for any nbd. of 0 in the nbd., there are infinitely many connected components (Q is totally disconnected so every singleton set in it is connected.)
Open problem: Mandelbrot set is locally connected.
Mandelbrot set is also an exponentially more complicated subset of R^2 than Q
I don't even remember the definition, honestly. Every point $c$ on it is supposed to say something about the dynamics of $f_c(z) = z^2 + c$ as a function $f_c : \Bbb C \to \Bbb C$
The set of all $c$'s such that $f_c$ has a nonempty Julia set, maybe?
No, connected Julia set. I have literally forgotten all the complex dynamics I learnt two semesters ago lol
But yeah, bottom line, who knows? The way its defined is already so bizarre.
6:58 AM
It's amazing that your institution held a complex dynamics course. We hadn't and maybe never so I need to self-study to learn complex dynamics.
It was a reading course, offered by one of the faculty here who does complex dynamics.
The subject didn't sit right with me
I'm too topology-pilled, I guess
I heard that complex dynamics is not very directly related to low dimensional topology because it's usually done between spheres. although I don't know both subjects... but anyway.
Dynamics of only maps between spheres are interesting, yes. I don't mind not every subject being low dimensional topology. It's just something about complex dynamics that was very off putting to me
I blame Milnor's notes, which is what the first half of the reading course was about.
Worst thing Milnor has ever written. It's shockingly bad.
{(x,y,z) in R^3: xyz=1} and {(x,y,z) in R^3: xyz=0} are homeom. or not?
both are connected as the former is continuous image of the connected set $R^3-$(0,0,0).
and the latter is the union of connected sets (x-axis, y-axis and z axis).
both are path connected as well.
$xyz = 1$ is not connected.
7:13 AM
Think about xy=1 and xy=0 for intuition
ohh right I erred.
Speaking of Julia sets, I think there are lots of examples of $c$'s such that the Julia set of $z^2 + c$ is a dendrite, @AlessandroCodenotti
Uh interesting
I haven't been thinking about dendrites much lately, the Menger curve is my new best friend
Menger "curve"?
7:19 AM
It is one dimensional and a continuous image of $[0,1]$ so it's a curve
But it is very much a three dimensional object in some sense, it contains no nonempty open planar subsets for example
I do agree that calling it a curve is weird, but it seems to be the standard name
7:36 AM
@AlessandroCodenotti Wild idea. I will define a tree 2-complex to be a 2 dimensional cell complex which is locally homeomorphic to one of the following: $\Bbb R^2$, $\Bbb R^2 \cup \{x = 0, z \geq 0\}$, $\{y = 0, z \leq 0\} \cup \Bbb R^2 \cup \{x = 0, z \geq 0\}$.
$\Bbb R^2 = \{z = 0\}$ here.
Can you define a "2-dimensional dendrite" as an appropriate inverse limit of tree 2-complexes?
Hmm for sure there's some subtlety involved, not all inverse limits of trees are dendrites, you want monotone maps
Makes sense.
But there should be some notion of a space that is locally connected and "looks like a 2-dimensional dendrite"
That would be ideal
Not quite.
Here is an example. Take $\Bbb R^2$, glue a hemisphere along the unit circle, then glue a hemisphere along a closed curve which goes a bit around the hemisphere and a bit around the plane.
So there's clearly "three layers" to this, once you fix the initial $\Bbb R^2$ to be the "root". Like depth of a rooted tree.
Another thought: Think of the second local model as "Y" x I, where "Y" is literally the letter Y. Replace Y by the a pair of real lines, glued along the open half-axis. Then it becomes a manifold, albeit non-Hausdorff. This can be done with the third model as well.
One should be able to do this consistently to all the "junctures" and make the whole thing a non-Hausdorff 2-manifold. Then, perhaps, this fellow is a (non Hausdorff) quotient of $\Sigma \times \{1, \cdots, n\}$ where $\Sigma$ is a Hausdorff surface.
Can one extract a genuine tree modelling the various layers of a tree 2-complex, out of this mechanism?
Maybe: Vertices are $\Sigma \times \{i\}$, edges are whenever they get attached along a subsurface. The equivalence relation should give the tree on the $n$ nodes.
If that works, will the "manifoldification" of this phantom inverse limit be a non-Hausdorff manifold quotient of $\Sigma \times C$, where $C$ is the Cantor set?
Also of note: every juncture has "valence 3". Only at most three sheets can come at any given point, by my definition of a tree 2-complex.
So the limit object should have the property that it has $H_1 = 0$, but removing any point gives $H_1$ rank $3$. (Appropriate notion of homology)
Someone should construct this space, @Alessandro.
8:06 AM
I can't think about this now (I'm giving a talk in 10 minutes), but we can chat later
Good luck!
Thanks! It's a seminar about continuum theory for masters students I coorganized with my advisor, I'm talking about the characterization of the Menger curve (it is the unique 1-dim Peano continuum which has no locally separating points and no nonempty open planar subsets)
Sounds fun!
The proof is some atrocious Polish topology relying heavily on papers by Bing from the 50s
Lmao of course
8:08 AM
But the result is nice
You've already sold me by mentioning the result relies on Bing's works
9:01 AM
Can we characterize all $n\in\Bbb {N}$ such that $\Bbb{Q}$ has a Galois extension of degree $n$?
should be all n, shouldn't it? can you find examples in cyclotomic extensions (or subfields thereof)?
Cyclotomic polynomial $\Phi_n$ has degree $\phi(n) $
well they give you galois extensions, right? and the galois group is cyclic a lot of the time.
the idea being something like, given n, find some (larger) m for which n divides phi(m) and for which you can say that there's a subgroup of the galois group of order n
or of order phi(m)/n
i always forget which way that duality works
For an example : For Gal(L/Q) =3 , we can take K=Q(\zeta_7) =Q[x]/<\Phi_7(x) >
Then [K:Q]=6
Gal(K/Q) has a subgroup of order $3$.
Then there is an intermediate extension of Q of degree 3.
Finte and separable
How to prove this extension is normal?
Question: Let L|K is a Galois extension and $H$ is a normal subgroup of $Gal(L/K) $.
Can we show that there exists a normal extension of K of degree |H| ?
Is there any relationship between normal subgroup of the Galois group and normal extension of the base field?
9:37 AM
What's $(2^\mathfrak{c})^{<2^\mathfrak{c}}$?
Is it $2^\mathfrak{c}$?
Under GHC it is
Assuming , Goldbach's conjecture would be proven. Can we conclude that for $n>6$ , we can choose the desired primes distinct ?
10:13 AM
Fun fact: The maximum number desmos can handle for $$f\left(x\right)=x^{x^{x^{x^{x^{x^{x^{x^{x^{x^{x^{x^{x^{x^{x^{x^{x^{x^{x}}}}}}}}}}}}}}}}}}$$ is approximately $1.472286524187$.
2 hours later…
11:58 AM
Is it possible to find a real-valued continuous function $f$ on $[0,1]$ with infinitely many zeros, such that $f$ is non-constant on every sub-interval of the domain?
Yeah. You can take some kind of zig-zag passing through 1/n and 0
For the half-open interval $(0,1]$ , the function $f(x)=\sin{\frac{1}{x}}$ does the job.
Yeah. You can make it continuous on [0, 1] by adding x in front
12:02 PM
what if I add an uncountably infinite assumption?
@Jakobian Yep, that seems to do it.
Just take Cantor set instead then
@Jakobian Do you mean $f(x) = d(x,C)$?
Yeah that works
Works for any closed set as long as it doesn't contain intervals
Because on any open component (a, b) of it's complement the function will be a spike so not constant on any subinterval
This really just boils down to what closed set we choose
12:46 PM
Q: Academia.SE Moderation Strike

cag51Effective immediately, the moderators of Academia.SE (wrzlprmft, cag51, and Bryan Krause) are on strike. This is part of the network-wide action described here, and follows an Academia moderator resignation a few days ago. We will not perform any moderation functions until this situation is resol...

12:56 PM
Classic union flick.
@user858770 yeah, everyone is striking, really
mostly because stackexchange authorities are allowing AI-generated content (like ChatGPT)
there's also a post about it starred here
1:17 PM
Q: Moderation Strike: Stack Overflow, Inc. cannot consistently ignore, mistreat, and malign its volunteers

Mithical Introduction As of today, June 5th, 2023, a large number of moderators, curators, contributors, and users from around Stack Overflow and the Stack Exchange network are initiating a general moderation strike. This strike is in protest of recent and upcoming changes to policy and the platform that...

1 hour later…
2:21 PM
@robjohn hi
Actually I was passing through a video named as "Quantum internet" Can u help in getting mathematical model of it?
I mean how mathematical possible to designe a quantum computer
2 hours later…
4:26 PM
Apparently milk is better at long-term hydration than water
4:40 PM
Hey there guys! Wanna check out this proof ?
Q: If $A$ is a Dedekind cut, then we define $-A=\{r\in \Bbb Q:\exists t\notin A; t\lt -r\}.$ Let $O=\{x\in \Bbb Q:x\lt 0\}$ then show that $A+(-A)=O.$

Thomas FinleyIf $A$ is a Dedekind cut, then we define $-A=\{r\in \Bbb Q:\exists t\notin A; t\lt -r\}.$ Let $O=\{x\in \Bbb Q:x\lt 0\}$ then show that $A+(-A)=O.$ The proof, I encountered was : The general strategy is to show that $A+(-A)\subset O$ and $O\subset A+(-A).$ It is easy to show the former inclusion...

4:56 PM
@robjohn I saw some of uour answers, and it seems you have a knack for calculus. Will u mind helping me a bit, here?
5:28 PM
@Jakobian but most of us haven't evolved to drink milk
5:40 PM
I evolved. The only byproduct is the additional fins on my back
anyway, don't they produce milk for the lactose intolerant
probably should check whether long-term hydration is the same in that case
I think it was about salt and some minerals in the milk, didn't read the article in full
@Jakobian hope your hands don't evolve; it's hard to type with pectoral fins.
better eat salt and minerals instead
scoop it up
we'll leave a salt lick out for those who need it.
5:51 PM
6:28 PM
I've just answered a 5 years old question
for no other reason than because I had another proof
6:47 PM
I've gotten my 109 accounts banned for no valid reason. Help me get unbanned. pastebin.com/raw/rzVk4rVj I've gotten my 109 accounts banned for no valid reason. Help me get unbanned. pastebin.com/raw/rzVk4rVj I've gotten my 109 accounts banned for no valid reason. Help me get unbanned. pastebin.com/raw/rzVk4rVj
is it alive, are you alive
odd things are happening
i mark this day, as the day i consider any account created after today to be suspicious
109 accounts!? XD
also old accounts with recent new spikes of activity
7:23 PM
:63742303 Hell no!
now I've gotten my 110 accounts banned for no valid reason. Help me get unbanned
I would expect even more banning for Munchkin.
as she would say: phbhtbht
7:57 PM
I was wondering, is there a fast way to see how many questions of me have answers or accepted answers ? I believe this was possible in the past... is this still so ?
Take $X=(0,1)^n.$ Fix points $p,q$ s.t. $\text{dist}_n(p,q)=\sqrt{n}.$ Construct a smooth regular foliation of $X$ with $(n-1)-$dim. leaves which are topologically $(0,\sqrt{n})\times S^{n-2} $ accumulating to $p,q.$

How would I re-write this question for a holomorphic foliation? I presume I can just trade 'smooth' for 'holomorphic' and then trade $S^{n-2}$ as a submanifold of Euclidean space, for a complex sphere as a subset of complex space. Would that be correct? Anything else I would need to revise?
Note in the smooth site, that the leaves in $X$ are all diffeomorphic to open annuli, and the open annuli are diffeomorphic to once punctured planes. This is why the problem is equivalent to asking about a foliation by once punctured planes. In the complex site I would have to think some more about whether the leaves are all holomorphic to open annuli and whether the open annuli are holomorphic to once punctured planes.
:63742303 109 seems unreal
8:27 PM
@geocalc33 This is unlikely to make any sense. Do you even know what a “complex sphere” is?
Hi @Ted
@TedShifrin it's a complex variety determined by the polynomial $x^2+y^2-1=0$
That is the $1$-dimensional case, yes. What does this look like?
Hi, a Balarka.
balarka: don't you mean e^{i S^n} ?
Something like a hyperbolic surface I think for the real surface
8:40 PM
@leslietownes good point
the real surface that corresponds to the 1 dim. complex variety
For starters, it is very non-compact.
@leslietownes Doing stationary phase or stationery phase?
already looks much different than the smooth case
Holomorphic world is very different, yes. No compact complex submanifolds of $\Bbb C^n$.
I could just ignore the beginning part and skip to "holomorphic foliation of $\Bbb C^2$ by the class $S=\Bbb C - \lbrace 0 \rbrace.$ I know this is Stein now because it's a noncompact riemann surface. But the smooth case I wrote is also dealing with noncompact surfaces
8:49 PM
Huh? That is not a Riemann surface.
Oh I thought that you had said that before in your reply - I'll have to go back and re-read it
If $(x, y) \in \Bbb R^2$, $e^{i(x, y)}$ should the notation for the point on $S^2$ of longitude $\mathrm{tan}^{-1}(y/x) \in [-\pi, \pi]$ and latitude $\sqrt{x^2+y^2} \pmod{\pi/2}$
All sorts of stuff is different. Look up Hartogs’ Theorem. You were talking about $\Bbb C-\{0\}$ last time.
Likewise in higher dimensions
Arctan has issues.
8:52 PM
Essentially trying to write the exponential map $\mathrm{exp} : T_N S^2 \to S^2$ in a more concealed manner.
$N$ = north pole
More like congealed.
It's a cool fact that Stein = domains of holomorphy
= pseudoconvex
@TedShifrin Shocking theorem: Every closed $4$-manifold is union of two Stein domains with smooth boundaries, glued along the boundary by a diffeomorphism.
I think that's probably true in all even dimensions, but I'd have to think.
even the exotic S^4s (if they exist)?
9:05 PM
That is easy I think
Every exotic S^4 is union of two standard D^4s glued along the boundary by a diffeo
Give D^4 the structure of a polydisk D^2 x D^2 in C^2
That's Stein
@BalarkaSen oh right
Polydisk is too hard, of course, you can just take an open ball in C^2.
That's Stein because its a sub-level set of the plurisubharmonic function $f(z, w) = |z|^2 + |w|^2$
I don't know how to directly show an open ball in C^n is a domain of holomorphy though
Some kind of lacunary nonsense has to be cooked up
9:22 PM
I knew this 50 years ago.
i signed the petition but do not see my name
have you tried expanding the list and ctrl+f your name
@BalarkaSen Nothing personal about you, but since last year I've heard everybody in my institution use the phrase "cook up" to refer to math basically nonstop. I hated this metaphor already the first time I heard it.
why do you hate it
i've been using it since eternity
i have no idea how often i would have to hear it for it to bother me, although "nonstop" would definitely be over my line. but it's certainly not that uncommon, or so common that i notice it.
9:29 PM
I don't know, it doesn't fit at all. Maybe its because I love cooking and I love math, but the way I think and work with both is so different.
math has a shortage of words for things that mean "make up, but with more intentionality and purpose than the way you might 'make up' a children's story"
Its like saying you are going to perform a function-solo as a musical metaphor when a function plays an important role in your proof, it just feels bad.
when you "cook up" an apparatus in math, it's less like cooking food and more like cooking meth
mm. it would definitely bother me in a textbook or article, for that reason.
i would never use it in formal writing
9:32 PM
Yes, I only hear it verbally
people should lean more heavily into terms that literally mean 'create' but also often have the connotation of falsity. "cook up" sort of has this, but "fabricate" would be even better.
fabricate sounds like youre about to give a wrong proof lol
if you don't buy that argument, i have others. just tell me what works for you.
we can whip up a little something special, off-menu, just for you
i dont see a problem with "cook up". i agree its meant to evoke "create" in a situation where the created object is not that important, only created to close the proof
eg its not important to create a holomorphic function with domain of holomorphy D^n to demonstrate why D^n is stein
but you might have to, if youre going to explain someone from first principles
We will now perform a deception by assuming neg(A). Stitching this assumption into the fabric of the X-theory then leads to a shirt without an opening - an absurdity.
9:36 PM
it's like "its unfortunate we have to go this route, but lets proceed..."
"we're about to make a bunch of choices that maybe don't matter too much, and aren't logically necessary, only because you're being so irritating about it"
Somehow when I see "cook up" it seems like a more polite way of saying "I'm going to do a big mathematical poo, then smear the result on the blackboard and hope it solves the problem"
i would never use it for constructions i like and think are insightful
I do not think of it as having negative/pejorative connotations at all.
not all of us are award-winning chefs, ted.
9:46 PM
Ah, my star chefdom must explain it, then.
You might use "unearth" instead of "cook up," but then some grave-robbers would get ideas.
just disinter the following auxiliary function
Maybe it's cloaked in black.
@Shaun I just saw this in the starred messages. I hope you're feeling better and having a quick version of it. Many of us who stayed healthy for 3 years have finally succumbed during the last month or two (me included).
Conjure is a good one, but that one has a positive connotation to me.
A particularly slick trick
That suggests sorcery, so I'm not sure about the positive connotation.
Sorcery is positive for me.
10:01 PM
That figures.
Of course, I did always love Elizabeth Montgomery.
How about the pretentious, "at this point in the proof, we demonstrate a particularly extraordinary piece of legerdemain with the following chain of inequalities"
de main versus demain?
Of hand (not offhand) versus tomorrow.
Oh, interesting. The word is "legerdemain", and means "sleigh of hand", so "de main" should be the correct thing.
Always miss the t in sleight.
As long as I don't slight the hand, I should be fine.
10:05 PM
I'm just here to be the pompous/pretentious linguist.
Hand-waving is used in physics.
Competence absolves pretention.
remembering the faculty member who would constantly use both "there's no trick to this" [in high level explanations] and "it's just this one little trick" [in the middle of more technical arguments]
Uh, I mean the guilt of pretention.
Or blame thereof
Can someone explain this example (https://en.wikipedia.org/wiki/Bayes_estimator#Practical_example_of_Bayes_estimators)? I get the idea that $W$ is a weighted average of the movie's average rating and the global average rating, but not how the formula relates to the broader article.
Which piece in the example is the loss function and which piece is the Bayes Estimator?
10:08 PM
@leslietownes Probably not a unique person, but who?
is using the word "trick" better than "cook."
Apparently, for any Tychonoff space $X$, there is a pseudocompact Tychonoff space $Y$ such that $\beta Y\setminus Y\cong X$. Isn't that pretty cool?
Meh, @Jakobian
ted: way after your time. mostly a topologist.
10:11 PM
Point set drivel.
\beta Y is the biggest trick, illusion, fraud, perpetrated on the mathematical public since negative numbers were invented.
Still curious, @leslie.
Maybe this will be more interesting for you. If $X, Y$ are metric spaces, and the rings $C^*(X)$ and $C^*(Y)$ of bounded continuous functions into $\mathbb{R}$ are isomorphic, then $X$ and $Y$ are homeomorphic
This one is a classic.
@leslietownes they weren't invented, they were created
10:19 PM
confected. concocted.
Oh right, "concoct" is a good word
@Jakobian this means $\beta X$, $\beta Y$ being homeomorphic $\implies$ X and Y homeomorphic?
in case of metric spaces, yes
That which I cannot create, I donot understand.
Is the point that $\beta X$ is the bordification of $X$ by embedding it in $\prod_{I \, bdd} I$ using a countable many bounded continuous functions on $X$?
10:22 PM
what's a bordification?
wasa bordification
Take closure
Compactification for the atheists
the construction you use doesn't really matter that much
Well, neither does $\beta X$, so...
why so rude
10:23 PM
truth hurts
that's his style
not really, it's just unnecessary
dont the stone type spaces appear as those dust things in condensed math
we used to call it "terse"
so they literally solved langlands project
10:24 PM
lol sure
@s.harp well, someone mentioned $\beta X$ for discrete $X$ does appear there
its ok i like useless things
profinte sets, thats what they were called
For me $\beta X$ isn't an useless thing. I can apply it in topology
i think we mean vastly different things by "topology"
but its alright, i was just joking
10:27 PM
if its not related to breakfast is it really topology?
no need to take slight against the stone-cech compactification so personally!
I mostly study point-set topology
I even know what book I'll read next after this one
A good classic foundation.
I have free time so I read what I like
For me $\beta X$ is just a special compact space that tells us things about $X$, and which we can use to say things about $X$.
i only like point set topology if they can help me understand nicer spaces than the point set topologists care about
10:35 PM
so you care about Stein spaces but not about Stonean spaces? what an alemanophile...
10:58 PM
@TedShifrin Thank you. It's rough going. I haven't slept since Saturday. I'm sorry you got it recently.
Hmm ... Have you talked with your physician about getting Paxlovid? @Shaun
I had headache, sneezing, some sinus issues, but no fever. I was quite tired for a few days, though, and I am not sure I've completely lost that symptom ... but I'm antique.
yikes. it messed with my sleep, too, although maybe not quite that badly. for about three days, i remember trying to sleep extra because i didn't enjoy being awake and couldn't do very much while awake, but only partially succeeding with getting rest. i also had no appetite.
get well soon
sleeping was the only thing that helped when i got it.
11:23 PM
@TedShifrin Not yet. I told the NHS on Sunday my test result. That's about it.
I have a headache, sinus issues, fatigue, a change in my sense of taste, occasional sneezing, and shortness of breath.
I try to sleep. Often, I just lay down with my eyes closed, almost pretending to sleep.
@BalarkaSen Thank you.
Paxlovid has to be taken by the sixth day, I think, and it starts counting before you tested.

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