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17:00
@skullpatrol yipyipyip how are you? I'm.coooold
@Charlie Fine thanks. Where is your hot water bottle?
@skullpatrol haha not here
@Charlie :D:D
@skullpatrol :D
17:05
@skullpatrol hahaha
c: C: (:
c-: C-: (-:
@Charlie
@skullpatrol hahahahs
@Charlie Which one do like the best?
@skullpatrol first line,.second
17:13
@GustavoBandeira wtf?
@Charlie We should definitely make the Topology Suicide.
That would make more sense as "Algebraic Suicide" IMO.
@skullpatrol true
or "Topological Suicide"
@skullpatrol or.set theoretic suicide and so on
17:17
@skullpatrol Perhaps grammatic suicide, isn't it?!
Sure, why not...
...the adjective needs a suffix.
always rember for topology you need balls, but you need hairy balls for algebraic topology ;)
@skullpatrol I prefer (something) homicide but anyway
@DominicMichaelis Hairy balls with erotic oil?
17:22
Oh no
The hairy ball theorem of algebraic topology states that there is no nonvanishing continuous tangent vector field on even dimensional n-spheres. For the ordinary sphere, or 2‑sphere, if f is a continuous function that assigns a vector in R3 to every point p on a sphere such that f(p) is always tangent to the sphere at p, then there is at least one p such that f(p) = 0. In other words, whenever one attempts to comb a hairy ball flat, there will always be at least one tuft of hair at one point on the ball. The theorem was first stated by Henri Poincaré in the late 19th century. This i...
@Charlie I like it :D... Multiplication by 0 is numerical homicide. (in the reals)
@skullpatrol hehe good
@Charlie C:
@DominicMichaelis I know it.
17:24
@skullpatrol :D
@Charlie C:harlie
C:h
@skullpatrol haha cute!
@skullpatrol a hat!
@Charlie Yipyipyip
@Charlie Wassup? I was away eating ice cream.
Kulfi?
17:28
Nopes, proper butterscotch.
@JayeshBadwaik hi.
@AlexanderGruber Hey, just did a quick read of the paper
@TobiasKildetoft cool how'd it go?
@AlexanderGruber look nice. For now I skipped the proofs and mainly read the stuff that were not technical lemmas
I have two comments on some formulations that took me a few tried to read.
@TobiasKildetoft oh yeah, what are those?
17:31
on page 3, the line beginning with "we notice that the simplest..." the end of the line is a bit weird
at least it took me a while to see how the "as..." part that it ends with fitted into it
@AlexanderGruber HI GRUBBY
@TobiasKildetoft ah yes: actually that was a mistake i caught yesterday, it's a bit agrammatical. my next version will have fixed that line.
@Charlie hi cha li. feeling better?
and on page 4, the line beginning with "the preceding lemmas..." has a "for all nonedges $pq\not\in$..." and it seems unnecessary to have the word nonedges as well as the $\not\in$
i suppose that's true.
@AlexanderGruber yes, thanks for asking
17:33
ohh, it even says "nonedges in..." which is a bit strange
@Charlie good :)
I believe I obtained more votes in the ellection when I uploaded my photo with beard. Beard seems to be a legit item to obtain more votes in an ellection.
@GustavoBandeira the picture is quite seductive.
@TobiasKildetoft cool: i've reworded that now.
@AlexanderGruber great
10 persons gonna pass the preliminaries ?
17:35
@DominicMichaelis That's what I've heard.
@GustavoBandeira It gives you an aristocratic look IMO.
@AlexanderGruber I will have to see if I also have the time to read it thoroughly at some point soon
@DominicMichaelis i guess so, which seems a bit silly
maybe i shouldn't say that but i think that is kind of waste of time isn't it ?
17:36
but for now, I should probably start putting my various handwritten notes in electronic form before I go home in a few weeks
@TobiasKildetoft that's probably a good idea ;)
(it would be silly to carry like a pound or more of paper notes)
and it would be a shame to have to redo all the calculations
@skullpatrol I don't like beard
@GustavoBandeira That's too seductive. People might "forget" to vote. :P
@skullpatrol What you mean?
17:37
in Mathematics 2013 Moderator Election, May 13 at 20:39, by Lord_Farin
@MJD Such cut-off values (10 in this case) are always a little awkward when they're just surpassed. The result will not be known ahead of time because we can vote for more than three candidates, and cannot indicate a preference among our votes.
@GustavoBandeira Aristotle had a beard.
@JayeshBadwaik hahaha
it is interessting that i am in top 4 on the meta QA but much further behind in the preliminaries
@GustavoBandeira lol, how do you even find such pics?
Image search
17:39
@skullpatrol The mom of an old friend of mine had a beard. She had more beard than me.
@skullpatrol Wow, didn't know that! :-O
@DominicMichaelis Very few people read the complete questionnaire I suppose. Hence, the sampling goes all wrong.
@Parth Well, I made a guess and searched for love look
@Parth Google has it built-in.
@GustavoBandeira if you don't have time to shave your beard how do you you have time to be a mod?
17:41
@skullpatrol REALLLY???!!!!
@Charlie People do not have a beard just because they do not have time to shave it.
@Charlie Yes. Ask that to Heineken.
For some, it actually take a lot of time and care.
Ops, it's not Heineken.
@GustavoBandeira Woot?
17:42
I don't remember his name.
He's one mse user.
YES!
Henning.
@JayeshBadwaik it's not what it seems.
@GustavoBandeira but it's blonde. Blonde beard is nice.
@Charlie only like 0,000009% of my beard is blonde.
@GustavoBandeira then shave your beard!
17:46
@Charlie My GF loves it.
@GustavoBandeira ok, good for her
@Parth Did you find it pal?
@Lord_Farin hi. :-)
And also as a quote to @GitGud "I can not be arsed with that".
@Chris'swisesister Good evening.
@Lord_Farin hello! What ya doin'? :-)
@anon I loved this question of yours.
@anon hi Ani :)
@Chris'swisesister I'm weeding a bit in the Unanswered questions department.
17:50
@Charlie In one instant you have set yourself in stone for eternity as Natalie Portman in my head.
2
@Lord_Farin have you seen the query posted on meta that gives questions with unupvoted answers?
@anon Why?
@TobiasKildetoft Yes. But I'm actually answering questions.
Thanks, though. :)
@Lord_Farin ahh, even better
@JayeshBadwaik that's what her character calls young Anakin in Star Wars
17:52
@anon Ohh, that connection. Hmm.
@anon :D
@skullpatrol No :-(
@anon I loved the question btw. I have a crush in set theory.
@MphLee hi
@Charlie , @GustavoBandeira, thanks
agree with them, very nice question, it was usefull for me too
@anon so with my question? did you give up? I wanted to give you the bounty xD
the question is nearly two years old
@MphLee Like I said yesterday, I am not familiar with anything that specifically focuses on structured collections of partial orders on a given set.
i.e. I do not have an answer to your question.
okok np ;)
@anon is it hot in there?
18:02
It is.
@anon how hot?
@Lord_Farin have you seen it? $\int_0^{\infty} \frac{x^2 \sin x}{\cos x + \cosh x} \ dx=\sum_{n=1}^{\infty} \frac{H_{2n}-H_{n}}{n^2}$
I'd say upwards of 80 in my room here.
(switching off ice packs to keep the lappy happy)
@Chris'swisesister Nope.
@anon 80 what?
18:04
degrees fahrenheit
@anon damn I hate it
@skullpatrol Do you really think I am that stupid? :-)
bye all
later pal
18:06
@MphLee bye
Yo @PeterTamaroff wazzup dawg?
@skullpatrol silly tamaroff :)
@skullpatrol Just learned the Euclid Wallis algorithm, pretty dope.
rob eh?
@anon Yeah.
18:11
take a look at the thread I linked above for some head-scratching
My professor gave it, but he didn't explain anything, so I was like "dafaq"?
He just said $$s_0=1,t_0=0,s_1=0,t_1=1$$ $$s_{i}=s_{i-2}-q_{i-1}s_{i-1}$$
$$t_{i}=t_{i-2}-q_{i-1}t_{i-1}$$
@anon Ah?
@anon $\uparrow$ that one?
yeah
@anon You mean Asaf's answer, yes?
no, the crank asaf dragged in as consequence of his most recent edit :-)
@anon "probe deep infinity" LOL!!!
@anon Why is your profile linked to phil?
18:17
people's chat profiles link all over the place for some reason
changed it back to normal
@anon Dafaq is "$\bullet$-adic"?
[maximal ideal]-adic, if you wish
given a ring R and an ideal I, you can form the inverse limit of R/I, R/I^2, R/I^3, ... and as a special case you can look at Z/pZ, Z/p^2Z, Z/p^3Z, .. and get the p-adics. You can complete F[T] for instance and get F[[T]] (whose fractions form a local function field), or complete a number field('s ring of integers) wrt a prime (ideal) and get an extension of the p-adics for some p (where the prime ideal divides p), etc.
the $\bullet$ is just a symbolic placeholder for "whatever"
@skullpatrol so funny
18:29
B-)
@skullpatrol hehe the other is funnier
@anon Way over my head =)
not as much as you may think
@anon Really?
yes
18:31
Well, I just need experience. Get used to stuff, that is all.
I hope some day we can meet and talk and stuff, really. =D
it's possible
@anon What is?
heh, you find those words so unbelievable from my mouth you have to ask for clarification
well, from my fingers
@anon say it to him: "Pedro, you so silly"
@anon I don't follow XD
18:34
@Charlie oh, I do that all the time :)
It's better to solve one question 5 ways than to solve 5 questions one way.
@anon good, thanks, it's very kind
@skullpatrol lol :-)
@Chris'swisesister :D
18:39
maybe if I can derail my prof's talk later into hopf algebra territory with sufficient momentum I can get him to shine line on some rep thry stuff I want to know about even though it's completely unrelated to his research
@anon what sort of rep theory?
(I assume rep here is representation, not reputation. Since clearly reputation theory is way beyond the current state of math)
@TobiasKildetoft Mariano responded to this question of mine yesterday. (My instructor's area is hopf-galois module theory.) So here I'd like to know more about how group algebras are given co-operations
@anon in what part of that question?
second paragraph of Q's answer
(I am only aware of one way one usually puts a coproduct on a group algebra)
18:45
oh?
and it is the "stupid" one where one simply doubles each group element
oh, it's right there on wikipedia
and in some sense, this is the "correct" one, since we would of course like the group elements to be precisely the group-like elements
Hi!
planetmath says g is grouplike if $\Delta(g)=g\otimes g$, what's the motivation for that definition?
18:47
Hi!
@skullpatrol good day!
@anon if a hopf-algebra is the group algebra of some group, then the group can be recovered as the set of group-like elements
@Nimza good day
Help me please, I'm confused looking on ODE $y' = \frac{y-x}{y+x}$. If I rewrite it in form $Pdx + Qdy = 0$ then $P_y \neq Q_x$, what does it mean for the initial ODE?
@anon I am not sure if there is some other good reason for the name
@anon I usually don't work much with arbitrary finite groups (but rather algebraic groups, where the picture is in some sense dual)
18:51
so $\Delta(g)=g\otimes g$ is the 'correct' coproduct on a group algebra since then the group elements are precisely the grouplike elements, a preexisting notion, and at the same time "grouplike elements" is defined as $g$ s.t. $\Delta(g)=g\otimes g$ so that group elements = grouplike elements under the coproduct on a hopf algebra (the coproduct a preexisting notion). this seems circular: which comes first?
@anon I am not actually sure
I guess one might be able to recover the coproduct on the group algebra from the product on its algebra of global sections when seen as an algebraic group
(and taking the dual)
but again, the way one constructs an algebraic group corresponding to a finite group is as far as I know essentially motivated to make the group algebra be the dual of the algebra of regular functions
one just ends up with a more general notion, namely that of a finite algebraic group
@anon Do you compete in Putnam? Are the (past) papers for the same available otherwise?
I competed once and did poorly. What papers are you referring to?
@anon I wanted to ask for the questions. I wanted to see what kind of questions come on it.
18:56
Ahh cool, thanks.
@TobiasKildetoft I think Q's answer here is pretty satisfactory
@anon yeah, that seems to sum up things nicely
and the one by zyx is also good to keep in mind, and when it comes to quantum groups, the way to define the hopf-algebra structure is a lot less obvious
19:36
Hi @JohnWordsworth how are you?
@skullpatrol Hi - fine, thanks - and you?
@JohnWordsworth hi, J, how was your meal?
@JohnWordsworth Fine thank you.
@Charlie It was good, thanks - just been for a 4 mile walk to burn off some calories
@JohnWordsworth heheh good
19:39
For those noticing the disappearance of the links behind one's username on main, see this MSO thread.
@Lord_Farin It never happened to me
@JohnWordsworth, just had a meal by the pond, on my folding chair. Unfolded
@Lord_Farin Thanks, just came here to ask about that.
i didn't found the chat anymore
Both me and the chair, unfolded
19:47
@WillJagy You must have better weather than we have!
@WillJagy good!
@JohnWordsworth, about the mildest weather anywhere. And not as many mosquitoes as where I grew up
@skullpatrol why hhehehe?
Primary proceeds to election in < 10 minutes! \O/
19:49
oh my Gd it will change my life
2
@WillJagy about 11C here - needed a fleece for my walk :(
It is supposed to be early summer here ...
@Charlie, went and met a sort of grey mix, who was very interested in the smell of Bernese Mountain dog on me from earlier scratching.
@WillJagy Berenese mountain are so cuuuuuuuuuuuute
1.8 * 11 + 32 = 51.8, a bit cold I would say.
hahahahhaha
19:50
let's see if there are dramatic turns in the last minutes of the primary...
@John, you could move to southeast Brazil
@WillJagy Hmm - and learn Portugese as welll as Persian?
@WillJagy :D hehehe it's the same thing here, as of now
@JohnWordsworth YES!
Oh - and a bit of Polish - I only like languages that start with a "P"
great!
19:52
@John Charlie could give you individual lessons, you could assist with mathematics
@WillJagy Of course!
The "meta link back please" thread has the quickest 50 upvotes I've ever seen (17 minutes).
Excellent @will
Apparently it is Portuguese. Noone knows why they put in the extra letter u
@WillJagy in portuGUEse?
because in portuguese g has the sound of j without u
19:55
@JohnWordsworth Would that make "Physics" a language you like?
@Charlie so it would be Gohn Wordsworth
@WillJagy no
}:)
@skullpatrol I used to enjoy physics - but that would have been 40 years ago ...
@WillJagy to transliterate John it would be like gion uôrdsuort
@Charlie, you can't keep changing the rules just because you don't like the results
19:57
@WillJagy look: ga gue gui go gu
Alright, gion uordsuort it is, with a little thing on one o
and then Ja je ji jo ju, but it's not J as in english
it sounds different
gue is like in the word Gay
go like in gossip
T-minus <1min
gui like gee
Alright, what is the word for glue?
19:59
@WillJagy glue, the translation of glue?
cola

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