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12:13 AM
@leslietownes I think we pretty much always agree
12:25 AM
There is no such thing that simultaneously generalizes homology and homotopy, for they are simply different kinds of entities. There are homotopical interpretations of homology like in terms of Eilenberg-MacLane spectra or via the Dold-Thom theorem, but that is something different.
Neither homology nor homotopy is a "random" concept. Singular homology was obtained as a generalization of simplicial homology, cause topology used to be the study of simplicial complexes/polyhedra in the early 20th century, which is very explicit geometrically. (You may also find value in https://mathoverflow.ne
@gf.c seems right
12:41 AM
@Thorgott thanks..im not knowledgeable enough to interpret these terms. i will come back to this after i learn more
thanks for the links.. lemme see
@Thorgott, but then doesn't this imply that the $\text{SE}(3)$ action on $\mathbf{R}^3 \times S^2$ is transitive (since I've realized it as a quotient)? the action being $(t, R) \cdot (x, r) = (Rx + t, Rr)$. this doesn't seem right to me
most of the technical stuff doesn't really matter, but I do recommend familiarizing yourself with the history and lineage of the definition if you feel unconvinced by them, I recommend a look into Dieudonné's A History of Algebraic and Differential Topology
@gf.c why does that not seem right?
@Thorgott sorry i shouldnt have said "random". i do find these groups very natural to consider
I see how the action is transitive but for some reason obtaining the space as a quotient seems like I'm "coupling" or "twisting" $\mathbf{R}^3$ and $S^2$ in a manner not present if I just started with the two spaces and took their product
I feel like the "twisting" coming from the semidirect product from $\text{SE}(3)$ should carry over somehow, no?
actually, what action on $\mathbb{R}^3\times S^2$ are you considering?
12:56 AM
given $(t, R) \in \text{SE}(3)$ and $(x, r) \in \mathbf{R}^3 \times S^2$, take $(t, R) \cdot (x, r) = (Rx + t, Rr)$.
oh, you had already said that, sorry for making you repeat
so suppose $(x,r),(x',r')$ are two points of $\mathbb{R}^3\times S^2$. the action of $SO(3)$ on $S^2$ is transitive, so there is an $R\in SO(3)$ s.t. $Rr=r^{\prime}$. then, set $t=x^{\prime}-Rx\in\mathbb{R}^3$ and we have $(t,R).(x,r)=(x',r')$
1:14 AM
no worries. i see now that it's transitive. so this would be an example of a klein geometry with $G = \text{SE}(3)$ and $H = \{0\} \times \text{SO}(2)$. i guess i'm confused how this differs from if i allowed a more general action by $G = \text{SE}(3) \times \text{SO}(3)$
specifically, does the space of "valid" diffeomorphisms change when i change the the group action?
I'm not sure I see how you write $\mathbb{R}^3\times S^2$ as a quotient of $SE(3)\times SO(3)$
Take $H = \text{SO}(3) \times \text{SO}(2)$ maybe?
i'm trying to get a grip on klein geometries. my main confusion is: how does the geometry of the base space change with respect to a change of the group action? or does the geometry change at all? in this example, what exactly changes when I extend from $\text{SE}(3)$ to some larger group of isometries?
2:00 AM
I'm afraid I'm not familiar with the purpose Klein geometries
2:14 AM
understood, thanks for your time!
2:35 AM
no problem
2:58 AM
@user85795 There were lots of clouds, but we did get short glimpses of the partial and total phases. I got a nice test shot of the sun the day before, but I could not get a well focused image of the total eclipse. The fuzzy image I got shows three prominences, however. here is the sun the day before the eclipse. here is the fuzzy picture of totality.
3:39 AM
what is this $\tilde{\gamma}_*$ map? it does not seem to mean what the symbols have been defined to mean
in this text $f_*$ is the pushforward of $f$
but taking this literally, the definition of $\tilde{X}$ doesn't make sense
@SoumikMukherjee yeah, but I don't have voter ID so sad. What about you?
@XanderHenderson as you wish!
4:44 AM
@AlessandroCodenotti I thought so and went to sleep, now that I see the game, the conversion was top notch from Fabi
Gukesh won, so now it's win or nothing for Fabi, Hikaru and Nepo. Though even if Fabi or Nepo wins, they still have to play tiebreaks unless Gukesh wins.
@LuckyChouhan Oh! I can and probably will vote. Anyway I was going to say that don't vote for bjp.
1 hour later…
6:14 AM
They don't make music like that anymore.
2 hours later…
7:58 AM
We'll have fireworks today @SoumikMukherjee
@SoumikMukherjee Please avoid political discussions. It will result in endless inconclusive discussion.
For every f in Schwartz class function on R^n, we define its Fourier transformation. Now suppose that g is in L^1.
We know that for every f in the Schwartz class, |f hat|_oo <= |f|_1. (A)
So we can densely define g hat as follows:
Let f_n be a sequence of Schwartz class functions converging in L^1 to g.
By (A), |f_n hat - f_m hat|_oo <= |f_m- f_n|_1 so the sequence f_n hat is a Cauchy sequence. L^oo is complete so f_n hat must converge to some G, which we define as g hat.
Je veux savoir why this g hat is well defined.
@XanderHenderson are political discussions like 'don't vote for political party ___' allowed here?
8:46 AM
@Koro I don't think so
8:59 AM
@SoumikMukherjee 'I don't think so' is irrelevant here as you are advising don't vote for...
"user profiles and chat rooms allow for more subtle and nuanced conversations where pointed or harsh criticism of political figures, their policies, and governments (including satirical statements) can be allowed as long as they do not otherwise violate the Code of Conduct and do not contain insulting language directed at individuals."
@robjohn thanks for sharing.
don't vote for ... is not subtle
besides, you are advising 'don't vote for...'
@Koro I don't think so as a reply to your 2nd sentence, not 1st
relax guys :-)
9:03 AM
@Koro Yes, is there anything wrong with that?
Didn't we learn anything from the pandemic politics?
@SoumikMukherjee I'm just replying in good faith. I consider at least this place to be for non-political discussion. There is already too much of politics outside here. And your bringing politics here is annoying.
'don't vote this, vote that'. Aren't media houses already enough for this?
You can very easily make another chatroom.
@Koro You can always ignore a user if you don't like their comments
9:09 AM
done already.
@CowperKettle and some of us are very happy that they don't :-)
9:27 AM
There's a storm of such songs, I think it started literally weeks ago with the advent of Suno and Udio.
I'm afraid it will get hard to distinguish a real song from an AI product
and thus taking the meaning of "techno" to a whole new level
Enshittification of musical media in the digital sphere: a review of the issue
10:10 AM
Let X be a random variable with continuous density. I want to solve this equation for c.

E(X-c | X > c) = E(c-X | X < c)

How should I go about doing this?
For every f in Schwartz class function on R^n, we define its Fourier transformation. Now suppose that g is in L^1.
We know that for every f in the Schwartz class, |f hat|_oo <= |f|_1. (A)
So we can densely define g hat as follows:
Let f_n be a sequence of Schwartz class functions converging in L^1 to g.
By (A), |f_n hat - f_m hat|_oo <= |f_m- f_n|_1 so the sequence f_n hat is a Cauchy sequence. L^oo is complete so f_n hat must converge to some G, which we define as g hat. why is this g hat well defined.
10:26 AM
A: extending a bounded linear operator

Giuseppe NegroHINT: Imagine for a moment that an extension $T$ of $T_0$ exists and take $x\in E\setminus E_0$. Since $E_0$ is dense, you can approximate $x$ with a sequence $x_n\in E_0$. Since our "imaginary" operator $T$ is continuous, it must hold that $$\tag{1}Tx=\lim_{n\to \infty} T_0 x_n.$$ Now go back to...

1 hour later…
11:28 AM
@Koro curious that you've never taken such offence the many times in the past that Ted, Xander or robjohn criticized or explicitly talked about voting against Trump
@Koro There are no broad topics which are explicitly disallowed, but if it is disruptive, it is not going to be tolerated.
2 hours later…
1:33 PM
@Thorgott I think we all should be taking offense to everything. For example, the letter L greatly offends me. Unacceptable
and if you're not with me then you're against me
Subscribe to my email for more bad advice
@Jakobian |_()|_
1 hour later…
2:41 PM
@psie @Obliv apparently you use a hot egg to massage the stye
Because it keeps heat
3:10 PM
@SoumikMukherjee why? People are left with one option which is BJP because Congress and AAP suck.
Hello @XanderHenderson how is life?
3:24 PM
Okay, it has crossed into disruptive.
Knock it off.
What happened? which parts of my texts were disruptive?
Did I say "You, @SoumikMukherjee, are being disruptive"?
You didn't say those words verbatim
The tone of the room had become heated. It needed to stop. You can either keep arguing with me, or move on.
@XanderHenderson No, but then why my messages got deleted?
3:30 PM
38 secs ago, by Xander Henderson
The tone of the room had become heated. It needed to stop. You can either keep arguing with me, or move on.
3:40 PM
@XanderHenderson do you read kafka's books?
Is the word fascist banned here?
@LuckyChouhan I've read a very little of Kafka.
@XanderHenderson Metamorphosis?
@SoumikMukherjee I don't know how to be more clear about this: the TONE of the room had become heated. This is not about content. It is about the hostility and acrimony which was developing.
@LuckyChouhan Yeah, that one, and a one about a guy who performed by not eating.
@XanderHenderson great, do you Sherlock Holmes's stories? A Scandal in Bohemia, Red-Headed League and A Case of Identity are good read :')
Anyone Red-Headed here?
3:45 PM
@LuckyChouhan In the third or fourth grade, I went through a Conan-Doyle phase, and read a bunch of Holmes. But I don't remember much of it. My strongest memory is a locked room mystery where the culprit is a chimpanzee.
@SoumikMukherjee The conversation is over. Enough.
xander: ... which is poe, isn't it?
@leslietownes Is it?
@XanderHenderson I see, I haven't read that Chimpanzee story.
"The Murders in the Rue Morgue" is a short story by Edgar Allan Poe published in Graham's Magazine in 1841. It has been described as the first modern detective story; Poe referred to it as one of his "tales of ratiocination". C. Auguste Dupin is a man in Paris who solves the mystery of the brutal murder of two women. Numerous witnesses heard a suspect, though no one agrees on what language was spoken. At the murder scene, Dupin finds a hair that does not appear to be human. As the first fictional detective, Poe's Dupin displays many traits which became literary conventions in subsequent fictional...
spoiler alert, i guess
Well, there you go. I am clearly misremembering the Holmes that I have read.
3:48 PM
@leslietownes are you also a avid reader of english literature?
that's actually a pretty good diss. "the best conan doyle i ever read was a poe story"
lucky: not really, no, although like xander i had a phase
hard to not read English literature if you're English
@XanderHenderson "I confess that I have been blind as a mole, but it is better to learn wisdom late than never to learn it at all." ~ Sherlock Holmes
@leslietownes Are you really a lawyer?
@XanderHenderson you know moles are blind then I'm not getting how they live their life? How they eat and do other stuff?
Moles have excellent smell. You're an ape so you developed visual senses more, moles developed smell
lucky: is anyone really a lawyer?
3:58 PM
Evolution causes animals to fit in life into distinct ways, its not susprising that distinct animals evolved to fit different purposes
@leslietownes oh sorry, May I ask "what is your profession?"?
Just like human is no greater than a gorilla, there is no one path to evolution
@Jakobian Now I see, they are attracted to each other by their smell. Thanks Jako
@leslietownes My brother is. He works for the state.
He passed the bar and everything!
Apparently, in the bible nowhere does it disallow polygamous relationships
4:09 PM
@Jakobian No. Indeed, polygamy was likely something of the ideal across many societies throughout history.
I.e. you knew you were successful if you could afford multiple wives.
I'm watching videos by a former pastor, its interesting to hear what made him atheist and his analysis of stuff in the bible. He said that the only things that the bible prohibits is bestiality and homosexuality
I think he was talking specifically about sexual relationships
xander's brother prosecutes them and i spring 'em out of jail, or vice versa
well, I'm of pretty strong opinion that bible makes no sense
@leslietownes Vice versa. He's a public defender.
In Arizona. Where everyone is guilty!
4:29 PM
I learned a while ago that I have a distant relative who is only 26 and already a criminal judge in court
@leslietownes Do you think USA is a country of freedom? Or is it the opposite, a country of lack of freedom?
I heard USA has the record for people serving jail time for example (not sure the exact statement)
jakobian: we do incarcerate considerably more than most civilized countries, but, that level of state involvement in the lives of citizens is very far from uniformly distributed
in conclusion, america is a land of contrasts
Suppose that there is a cat, which gives birth to three kittens A,B et C. C is lost (perhaps someone took it away) and returns after 2 months say.
C is aggressive to A and B and to its mother as well.
Looking at the fur patterns of C, it seems that it is indeed the C which was lost.
But the growth in returned C (let's call it C tilde) has been more than the growth in A and B.
How to ensure that it is indeed the cat?
is C tilde same as C?
In a real situation?
4:43 PM
I'm not sure, maybe there is a method, but I feel like you can't be 100% sure unless you do some kind of genetics test
the one on the extreme right is C.
This is C tilde.
This is how the middle one looks like now (still looks like a baby).
How in the world C grew so big!!
are you sure $C$ tilde has the same coloration as $C$?
yes. even the fur pattern is same.
on the head, from the pictures you posted, the coloration looks a bit different to me. But perhaps it really is the same
C tilde was found near a pond today. So I suppose that it actually had to learn hunting. It probably ate only fish. While the other two never had any food problem. I doubt these two can hunt.
4:50 PM
yeah it's basically impossible to tell
as in the colored areas look to be different shapes
with older cats in this country, this would sometimes be solved by microchipping the cats near enough to birth that if you lost them later you could ID them again, but that requires you to have microchipped them before one ran away
which is odd to me because B did get the same coloration patterns
i'm a cat person, if it doesn't matter to anyone else whether it's the same cat, i would just enjoy the potentially new cat :)
4:52 PM
@Jakobian From a different angle.
@leslietownes that's encouraging :)
two different cats can have identical fur that's why I am not 100% sure if C tilde and C are the same.
Diet – The kitten phase is the time cats develop the fastest, which is why it is so essential to make sure they’re getting the nutrition they need. Kittens that aren’t getting the essential minerals and nutrients they need to grow will have stunted growth and reach their full size prematurely.

Abandonment – kittens or young cats that have been abandoned are usually much smaller than they should be. This has a lot to do with their diet and a lack of nutrition. To help feral or stray kittens survive on minimum food and water, their bodies will shut down, and their growth will be stunted.
those are ^ things that affect growth of cats
So I know questions along the lines of "what did the author mean here" are potentially fraught so forgive me...
EE18: how dare you! :)
This is from Hoffman and Kunze...what do they mean with the last aside in parentheses? Why would $V$ being finite-dimensional matter at all?
@leslietownes I've learned my lesson!
This sounds a bit weird to me because it seems like an abandoned cat would be smaller than it should be
4:59 PM
(That is, it feels easy to see irrespective of the dimension of $V$: $f \in S^0 \implies \forall v \in V, f(v) = 0 \implies f = 0$ by definition of the 0 map in $V^*$)
EE18: yes, you're right. the author probably didn't think it through
the existence of nonzero linear functionals on an arbitrary vector space is a subtler question and maybe what they were nervous about
Possible to say more there? Is that because we confirm the existence in finite spaces by using the theorem that says that "the action of a linear map on a basis of a finite dim space fully defines it"?
the author's thinking "maybe this is something where the easiest way to see it is to choose a basis," where finite dimensional vector spaces are indeed nicer than arbitrary ones
but it isn't, as you say, if something is 0 on all of V, it's the 0 functional. i don't know there is more to say about that specific remark
Never thought asking someone if they are an f supporter would win me getting banned speedrun.(f for football, not fascist)
@Thorgott Rookie numbers, my relative is 16 and already a criminal
5:08 PM
@leslietownes I am not sure what argument would be made here by picking a basis? But if I'm understanding you you're saying your first instinct would be to argue like i did?
that's my point. you're right and the author is confused
they're thinking about this other world, not the world of that remark
Hey @AlessandroCodenotti any final prediction?
@AlessandroCodenotti ha
something related to this idea (but not closely related to this remark) is: given nonzero v in V, is there f in V* with f(v) nonzero, i.e., is the annihilator of {v} not V* when v is not zero
I feel like Hikaru will win and Fabi Nepo draw
5:09 PM
that's where choosing a basis helps and maybe what they're thinking about
yes, leslie, I also think they may have been thinking of that
but you don't need subtle thought to figure out what the annihilator of {0} or V is
related: the entire space is the only subspace with trivial annihilator
yeah and by this point they perhaps have not established that any vector space has a basis.
god, does it really all come back to set theory whether we like it or not
5:12 PM
ahem ahem
@leslietownes Wait, should I be keeping along with my set theory studies then :p
clearly :)
I am just seeing that it's open whether existence of nonzero linear functionals implies AC?
Q: Existence of non-trivial linear functional on any vector space

Souvik DeyFor every vector space $V$ does there exist a linear functional $f$ ( a linear map from $V$ to $F$ the underlying field ) such that for some $ \vec v \in V$ , $f(\vec v) \ne 0$ ? If it does exist , can we prove the existence without the "axiom of choice " ? Is the existence equivalent to axiom of...

Is that true?
if asaf says it, it's true
well, if he says it about set theory. maybe we do not see eye to eye on other things, but if he says something about set theory, it's true
5:28 PM
@EE18 it was in 2014 so it might have changed
@leslietownes In finite-dimensions the answer is positive because I can choose to map f(v) to $a \neq 0 \in F$. Is the answer difficult in infinite dim or should I be able to see that?
EE18: well, your intuition around "i can choose to map [v] to a in F" probably relies in part on the fact that you can define a functional arbitrarily on a basis, and that any nonzero vector can be extended to a basis
i mean, dig into how you define f for a minute. you want f(v) to be a, OK, but what is f(x) for x that isn't v? how do you answer that question
bear in mind that "you can define a functional arbitrarily on a basis and any nonzero vector can be extended to a basis" is still true [assuming AC] in infinite dimensions
it's just harder to "see" in concrete terms because you do need something like AC to get it
Ah OK, so ya I wasa using that/those theorems exactly in the finite case
Didn't realize the define a function arbitrarily on a basis was true in infinite dim too
I was familiar with extend to a basis, but not that
hi guys
alongside all of this, defining things in terms of linear algebraic bases is not really 'the done thing' in infinite dimensions, but as an abstract thing that one can do, it certainly works just as well there
5:43 PM
the fact that a linear function can be defined arbitrarily on a basis is always true and easy
the AC comes in when we claim that every vector space has a basis
I never turn on the AC
how eco friendly of you
mathematically, I never turn it off
That is my personal choice
@Thorgott Probably a better way to argue but does one say some thing like if $V \ S \neq \emptyset$ then taking $v_1 \notin S$, extending to a basis with $v_2,...$, and defining $f(v_1) = a \neq 0$ and $f(v_j) = 0$for all other $v_j$ gives us $f \neq 0 \in S^0$?
5:44 PM
Ok, I lied. I have created two oranges out of one using non measurability.
@leslietownes Hamel and Schauder bases being alluded to here? Vaguely recall from some QM book I read at some point, can't recall which
But hopefully I'll get there rigorously in my analysis book(s) at some point
Thorgott, I think my argument above relies on the sorts of AC consequences Leslie was mentioning but hopefully it's what you had in mind
Asaf is sooo depressingly smart.
There's something philosophical about ZF being neither pro nor against AC
you're missing a symbol there
Ah you're right, tried to write set difference but forgot \ is an escape character
5:49 PM
If you're using ZF would you say you turned AC off? Or is it in some kind of middle state?
the extruded middle
the dearth of convex problems suggests that ai is dead
@copper.hat Asaf is good at what he specializes in. I think. I assume. I can't check that because my expertise in set theory is only so limited
@EE18 sort of, but also a broader point, the idea of "do X in terms of a basis" is kind of a finite dimensional one. like, it's a default option that always exists in the finite dimensional setting, but it's often not how you want to approach an infinite dimensional problem even if you have the option via some subtler notion of "basis"
unless you're a physicist in which case yes, spew formulas until you run out of ink
@Jakobian I am sufficiently intimidated by his writings. Plus we have had some brief exchanges during which my lack of knowledge became very apparent.
I would love to be able to think abstractly like he seems to do.
@SoumikMukherjee Hmm it's very hard to predict. I think both games will have decisive results, but which way I have no idea. If Naka's prep works he might win, but if he has to push for a win in a balanced position he might get punished
5:54 PM
@leslietownes invent ink which never ends and become an infinity physicist
Similarly in Fabi-Nepo I think we will have to see whether Caruana has something spicy prepared
@copper.hat I can't relate to that experience as it never happened to me
@leslietownes LOL :) many bad habits were formed...anyway thank you as always for your help here!
@AlessandroCodenotti Yeah, though Gukesh has been really solid throughout the tournament and the only time he lost, he was in time scramble. So maybe Naka will unbalance the game enough that it reaches a crazy position in time scramble.
@AlessandroCodenotti I hope so, Nepo may go for a french.
@copper.hat Here's an experience that maybe has happened to you before though: I ran out of milk for my coffee :(
6:01 PM
you want $S$ to span a proper subspace of $V$ and then you need to take a basis of this subspace and extend it to a basis of $V$ for everything to work out
@Thorgott Ah yes, I see now that some of what I said wasn't exactly right, but this was vaguely what I had in mind. appreciate you clarifying :)
@EE18 Sounds a bit surprising to me but could very well be. It is known that the existence of a nontrivial continuous linear functional on every Banach space is equivalent to Hahn-Banach over ZF (which is strictly weaker than AC)
@AlessandroCodenotti Then existince of non-zero functional on every vector space should be implied by this weaker version of AC, no
did hahn ever do anything that was just him
there's hahn-banach this and hahn-hellinger that and jordan-hahn the other
Ah yes of course, I wasn't thinking
6:04 PM
No no
@leslietownes Hahn-Mazurkiewicz is another famous one
Banach spaces are over real or complex fields
Here we are probably thinking about arbitrary fields
> Within the Vienna Circle, Hahn was also known (and controversial) for using his mathematical and philosophical work to study psychic phenomena; according to Karl Menger he sometimes openly advocated further research into extrasensory perception while lecturing.
when you write each vector space has a basis is equivalent to AC you also need to take arbitrary fields right
hahn solo stuff must be wild
6:05 PM
@SoumikMukherjee Nepo might go for a super solid Petrov and force Fabi to take risks. But this is also risky because Fabi's team will surely have something ready for the Petrov. So many mindgames
@leslietownes Hahn theorem from measure theory?
In mathematics, the Hahn decomposition theorem, named after the Austrian mathematician Hans Hahn, states that for any measurable space ( X , Σ ) {\displaystyle (X,\Sigma )} and any signed measure μ {\displaystyle \mu } defined on the σ {\displaystyle \sigma } -algebra Σ {\displaystyle \Sigma } , there exist two Σ {\displaystyle \Sigma } -measurable sets...
oh that's what i think of as hahn-jordan
according to wiki those refer to slightly different versions of whats basically the same thing
but yeah it seems unfair to credit jordan with anything to do with abstract measure theory
@AlessandroCodenotti Yeah many mind games as both need to win, but Fabi is a Petrov expert as well so not sure if this will be a valid strategy for nepo
6:08 PM
let's give this one to hahn
@Jakobian I'm not sure, I don't remember if I've ever seen a proof of (the hard direction of) the equivalence
Hahn needs a win.
Personally I wish they go for a Benoni, anything but a draw
@AlessandroCodenotti I was pretty sure this fails if you fix your field
i wonder if anyone has published hahn's lectures or table talk on psychic phenomena
6:10 PM
@SoumikMukherjee it feels like they played so few d4 games this candidates
Yes, also most of the e4 games were Italian
Very few Ruy Lopez (only 2 I think)
Were there any c4 games?
Maybe they'll go crazy and play a KID or something like that
@SoumikMukherjee hmm not sure. Also not sure about Nf3
@AlessandroCodenotti That'll be amazing
@AlessandroCodenotti Nf3 happened in round 11 between Firo and Abasov
@Jakobian I just ran to the local shop to pick up some milk for my tea.
Why anyone would drink coffee is beyond me.
(I'm kidding.)
6:28 PM
@SoumikMukherjee Ah nice, did it transpose into an english or some d4 line?
What is a c4 game? When someone explodes?
Q: Axiom of choice and vector spaces over a given field

JohnIt is my impression that the following question is open: Does the existence of a basis for every vector space over the field K = the reals having a basis imply the axiom of choice? I saw an answer from several years ago that indicated it was open. There was also a somewhat vague comment about t...

Ah that's very interesting
@SoumikMukherjee I'm not sure why the KID is so rare nowadays, it used to be played a lot in the Karpov/Kasparov days
I see lots of discussion about chess...do any of you play Go too?
i thought go to was considered harmful?
6:34 PM
@AlessandroCodenotti It went b3 Bb2 and eventually d4 c4
uh what a weird line
@copper.hat A chess game where white plays 1. c4( white pushes their c pawn 2 squares)
@AlessandroCodenotti Maybe due to engines?
@EE18 I tried once. I played 3 games. Lost my first game in 1-20ish margin.
That was my best performance, i lost the other 2 without scoring a point:"
But surely that's how all games go in any game for a beginner, no? :)
6:38 PM
Or maybe there are not many top level Ideas in KID to play for
I was also surprised to see so few catalan setups from the players with white
Grünfeld is the final boss of top level d4 openings
Yeah I have to study it a bit at some point
Yeah, very few Catalan setups
Oh man :O Thursday is the final :O
6:43 PM
There was a beautiful game in the closed catalan in the women section
Honestly I never understood Grünfeld, seems so easy yet so hard
@SoumikMukherjee Would love a chance to play some chess
@AlessandroCodenotti The women section had more fireworks overall, they were going for crazy setups
@YourLordJoyBoy wanna play now?
Can't, gotta study, its why I
m on now
Cool, all the best for your study session
6:47 PM
@SoumikMukherjee yeah, I also feel like their games are much better for us to learn from instead of the absurd prep straight from 3600 elo engines that they're playing in the open section
Yeah right
QGA in Naka-Gukesh is also surprising
5... Be7 is interesting, iirc 5... c5 is virtually the only move played in the QGA with e3
Vidit and Alireza already drew 😂
Ah no there's also the 5... a6 modern stuff
@SoumikMukherjee Well they have nothing to play for
I would also take 3500$ to show up five minutes and play the Berlin draw
@AlessandroCodenotti Slight change in move order, also Qga is interesting
6:51 PM
exchange QGD in Fabi-Nepo
Was not expecting Qga
@SoumikMukherjee Thanks!
in Set theory, 12 mins ago, by Jakobian
@MartinSleziak Do we know if there can be a model of ZF with $AC$ false but $B_K$ true for some field $K$?
@SoumikMukherjee Appreciated. And here's to getting through that final. Omgosh I have got to pass. I just need a 70, though I'd take an 80.
@SoumikMukherjee Indeed
QGA is also not played so often but I think that objectively there is nothing wrong with it
6:55 PM
Qga can be very risky sometimes, one wrong move and game over
Maybe Gukesh is betting on Naka not expecting the QGA and so not being as well prepared with weird novelties as he might be in other openings
Probably, who would expect QGA in such a game. Also Gukesh can still have chances if he gets a draw, so not necessary for him to go all out. Surprising that he played QGA
Naka probably thinking whether to go for IQP or not
Probably he will go for IQP, as he often says in his videos that this unbalances the position
2 hours later…
8:34 PM
There are two definitions of monotone class; the above one and another, probably more common one, which says that a monotone class is a collection of sets such that it is closed under monotone countable unions and monotone countable intersections. Are they equivalent? I struggle with verifying that they are. The above one seems to say that a monotone class is closed under complements, which this answer seems to contradict.
Maybe I should make a post about it on the main site.
This is from Le Gall's Measure theory, Probability and Stochastic Processes.
ok, I think I've made some progress in this jungle of terminology; apparently the above is a so-called $\lambda$-system, but a $\lambda$-system is a monotone class
9:42 PM
I Love Maths so damn much, it hurts. That is all.
10:26 PM
Is it correct to say that a linear algebra over a field can be thought of as a field and vice versa?
@EE18 What is your definition of "a linear algebra"?
My instinct is to say "No, the words that you are saying don't make sense", but maybe you mean something other than what I infer you mean?
This is what I had in mind Xander: en.wikipedia.org/wiki/….
10:48 PM
there is not really such thing as "a linear algebra"
the adjective "linear" does not really make sense there
and not every algebra over a field is a field
Q: Local embedding and disk in domain perturbation

monoidaltransformConsider say $M=(\mathbb{S}^1\times\dotsb\times \mathbb{S}^1)-q$ ($n$-times). Assume that $B$ is an $n$ disk in $M$ (for instance, thinking of $\mathbb{S}^1$ as gluing $-1$ and $1$, the cube $B=[-\frac{1}{2},\frac{1}{2}]\times \dotsb\times [-\frac{1}{2},\frac{1}{2}]$ is in $M$ provided $q$ is not...

How does one see that two orientation preserving embeddings of a disk $D^n$ into $\mathbb{R}^n$ are equivalent?
11:08 PM
you have linked a question that explains exactly that, have you not
No @Thorgott I don't see how to use the linked question to my question
not clear to me
it is literally the same statement
But in linked question he sais 'not ambient isotopy' @Thorgott
"Since the boundary of a collared ball is always a bicollared sphere, it follows that every collared $n$-ball can be mapped to $B^n$ by a homeomorphism of $\mathbb{R}^n$."
Here he means the homeomorphism takes the image of the 'collaring' map?
+ How does that imply the existence of local embedding in my question? @Thorgott
11:22 PM
@Thorgott Ah OK, I will hold off then on asking too much more until I've studied these things formally. It was just something I was vaguely familiar with from before. But I guess there exist conditions under which an algebra over a field is a field?
@monoidaltransform I'm not sure what you mean by this. The homeomorphism transforms the given embedding of $B^n$ into the standard embedding.
@monoidaltransform I have not thought about that. I'm merely answering the very question that you posed in this chat.
@EE18 you already know the relevant example: the polynomial ring over a field is an algebra over that field and not itself a field
there are useful general theorems: any finite-dimensional algebra over a field is Artinian, hence decomposes as a product of the localizations at their finitely many prime ideals and each such factor is $0$-dimensional and local, hence a field if and only if it is reduced
(I don't expect you to know what those words mean, so just trust me it's meaningful)
@Thorgott Touche! To be left for an algebra book I guess...thank you as always Thorgott!
@Thorgott He's saying if $\Sigma$ is a collared $n$-ball, then there exists a homeomorphism $h:\mathbb{R}^n\rightarrow \mathbb{R}^n$ such that $h(\Sigma)=B^n$. I don't see why that mean if $f:\Sigma\rightarrow \mathbb{R}^n$ is an embedding then $h$ takes $f$ to the standard embedding
11:49 PM
ah, sorry
I missed that Jim Belk's answer does not prove the stronger claim
you should read Lee Mosher's MO answer linked there instead, that has the right claim
@EE18 here is one for your self study
In operator theory, the Gelfand–Mazur theorem is a theorem named after Israel Gelfand and Stanisław Mazur which states that a Banach algebra with unit over the complex numbers in which every nonzero element is invertible is isometrically isomorphic to the complex numbers, i. e., the only complex Banach algebra that is a division algebra is the complex numbers C. The theorem follows from the fact that the spectrum of any element of a complex Banach algebra is nonempty: for every element a of a complex Banach algebra A there is some complex number λ such that λ1 − a is not invertible. This is...
this is a lot easier if you can afford to work in the smooth category, for the record
@Thorgott I did. Mosher's answer lacks details and he talks about the fact that nicely embedding balls are equivalent. This, by Belk's answer, easy to see.
@Thorgott how?
@leslietownes First of all lol at the nugget at the end of this: "This result was proved first by Stanisław Mazur alone, but it was published in France without a proof, when the author refused the editor's request to shorten his proof. Gelfand (independently) published a proof of the complex case a few years later."
I am vaguely familiar with this theorem from a math for physicists book but i don't think i ever saw the proof. very cool
i think it was introduced in order to make precise the vague notion that one here that after the reals, complex numbers, quaternions, and octonions, there is "nowhere else to go"
EE18 for the most general version of that, i think you really need algebraic topology, but the gelfand mazur theorem is i guess a version of that
that is very modern math
oddly enough, i had a legal job interview once where someone asked about dimensions of real division algebras

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