Mathematics - 2011-11-18 (page 1 of 3)

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12:00 AM

user image

With that, I will leave you for the night. Ciao.

@zhenlin -Also does the AR theorem exclude the probility that a root be irrational or infinite series?

I think you need to first make sure you understand all these terms properly, before going off to think about such things.

2

I'm going to sleep now, so I'll leave you to study these words on your own.

@ZhenLin - can i ask which state you live?

@robjohn Great!!! Thanks :-).

@robjohn @jonas - are you people able to talk when robjohn is not talking?

12:07 AM

Zhen Lin lives in a state of perpetual motion. This alone proves that he is not a machine. Otherwise there would be a contradiction in the laws of thermodynamics.

@Victor I can just ping @robjohn and he will see this next time he checks this window.

@AsafKaragila What do you know about physics? You're a set theorist.

@Victor It also makes a ping on my computer.

@AsafKaragila @ZhenLin - appreciate for your teaching

@JonasTeuwen Well, I know there was this episode of the Simpsons, I think third season or so, in which Lisa built a perpetual motion machine. When Marge told that to Homer he yelled at her "Lisa! In this house we obey the laws of thermodynamics!!"

@Victor: You can set whether @Victor makes an audible ping by clicking on the speaker icon next to the "all rooms" button.

"when mentioned" is the setting I have currently.

12:13 AM

Does anybody mind if I ask a small field theory verification?

Damn, I'm exhausted :-).

Just to keep things simple: what Abel essentially said was that "there isn't a simple general formula for the roots of a quintic".

@yunone chat away :-)

@JonasTeuwen you have a right to be.

Some special quintics do have simple representations for their roots, like x^5-1...

@robjohn But I have to work in 7 hours! :-).

12:15 AM

...but most quintics, like x^5-x-1 have roots that aren't expressible as combinations of radicals.

Ok good: : ) Is the splitting field of x^8+x^6+x^4+x^2+1 just Q(\zeta_5)?

over Q that is.

@JonasTeuwen Well, you can't be just having fun after that defense, you know... :P

:P.

You know that song? "Heigh-ho, heigh-ho..."

I've had enough fun. Now I only need to work 40 hours a week if all goes well instead of much more as a student!

12:17 AM

Oh wait, nevermind, I think that's wrong.

@yunone x^10-1 = (x^2-1)(x^8+x^6+x^4+x^2+1), so does \zeta_5 handle x^10-1=0?

I know it satisfies it

but does Q[\zeta_5] have all the roots?

@robjohn So should I instead be looking at 10th roots of unity, and remove +1 and -1?

@yunone I believe so.

Ok, so the splitting field would be Q(\zeta_10) instead, with degree 8 over Q?

I'll go to bed. See you guys!

12:21 AM

Is it degree 8?

10*?

\zeta_10^5+1=0

@robjohn No it is 9 degrees centigrade.

Oh I remember, it will be \phi(10), so that means it's of degree 4?

@JonasTeuwen is it really? It is in the 50s and 60s here, but it is supposed to rain on Sunday.

@yunone That sounds better.

12:22 AM

@robjohn Celsius? I hope not? :D.

@JonasTeuwen no, Fahrenheit.

Thanks for your help, @robjohn.

@robjohn Just an hour north of you by plane it was sleeting earlier today (in fact, on me while I was running). Los Angeles and its balmy weather...

@JonasTeuwen Holy everloving... for 50 degrees Celsius, you can fry eggs on sidewalks...

@robjohn So about 15°C?

@JM Exactly.

12:24 AM

@JonasTeuwen 10-15 C, I guess

(On that note, it was 40 here around noon yesterday.)

40?!?

@MikeSpivey Perhaps we will have some weather like that later, but the ocean keeps it from getting too extreme.

@JM C or F?

@Jonas: Yep. This November is surprisingly hot...

people: is that all infinite series number are transcendental?

12:26 AM

@robjo: Celsius

Anyway! Good night!

@JM Ouch

@Victor You know the geometric series?

@JonasTeuwen Sleep well!

Thanks.

12:26 AM

@robjohn The heat ain't so bad. The humidity, on the other hand...

Good night Jonas!

@JM - i know nothing deep about the geometric series...

@JM Yes, I understand your humidity is a bit high there. When it is hot here, it is usually lower humidity.

@Victor there is not much deep to know ;-)

@Victor Nono, you were asking about whether things that can be expressed as series are transcendental. That's why I'm asking you what you know about geometric series...

@robjohn Yeah, taking a bath (at least) twice a day is mandatory: before going to work and after coming from work...

@JM twice a day, that sounds drying (of skin oil).

@robjohn Well, otherwise, you end up sticky...

12:32 AM

@JM I can imagine, but it still must deplete the oil from the skin.

@JM How prevalent is air conditioning?

Well, all the cool kids and establishments have it... :)

On the other hand, I can't afford a room with that, so I just had three or so fans running yesterday...

(I could've hung around at the mall until around 19:00 or so if I were really determined to have air conditioning...)

@JM: Interesting. I grew up in the US South, which is pretty hot (although nowhere near what I imagine it is in the Philippines). Virtually everything is air conditioned there - we were pretty spoiled.

@JM -Thanks, but why people doen't research about using a finite series or formula of number compose of the power of each term of polynomaial and the coeffcient of polynomial be the root that is not cannot express by radical

@Victor Again, because not all algebraic numbers can be expressed as radicals and/or combinations of them.

12:42 AM

@JM - what is a example?

The real root of x^5-x-1.

You need some unusual functions to represent the root explicitly.

@jm - is that all algebraic number could express by radical?

Again... no, not all algebraic numbers can be expressed as radicals.

I would even say there are a lot of algebraic numbers that cannot be expressed as radicals.

@MikeSpivey Even though the humidity is not that high, since the temps can be in excess of 110° F for the better part of a week at a time, we have A/C most places.

A couple of years ago, it hit 115° F in January here.

That was quite amazing.

12:57 AM

Is it ever cool in California? :)

@JM Right now the highs are around 10-15 C

It can freeze (below 0° C) over night.

@JM I would say that most cannot be expressed as radicals.

@robjohn yeah, "most" is a better way of putting it.

@robjohn Now that is cool... on the other hand, the last trip I had to California was smack dab in the middle of summer, so I guess I'm biased. :)

@JM Ah, you said California. Yes, some places in CA have a cold climate.

@JM It also depends on where in California you visited.

@robjohn San Diego and Burbank.

San Diego is moderated by the ocean and rarely goes outisde 65°-85°F

@JM Burbank can get hot.

But Burbank is right next door to Glendale, where I grew up (they share a border).

They are both in the Verdugo Valley.

1:04 AM

"Verdugo Valley" - so aptly named... :D

@JM ? Why is that?

@JM - so, why couldn't we find a method or formula for finding the roots of all higher polynomial by using the coefficient and the power of all the terms in the polynomial?Also, why couldn't the root be obtained from the infinite series that i obtain by using the powers of all the term in a polynomial and coefficent of the polynomial? does AR theorem contradict the possibility of this method

@robjohn If it gets hot enough, then the valley is your verdugo.

@Victor You're not clear to me. Of course you can express any algebraic number in terms of the coefficients of its minimal polynomial. The only thing Abel said is that for roots of quintics and higher-order polynomials, radicals are insufficient for getting explicit representations.

@JM In what language are you translating Verdugo? In Portuguese, it means "headsman", but I see in Spanish, it means "executioner". Heh.

@Victor You might want to look at this, for instance.

1:10 AM

It was named for the Verdugo family.

@robjohn Ah, sorry. :) I don't know Portuguese.

@robjohn I only realized that after looking at Wikipedia. On the other hand, when I was there, I was thinking "what a name"...

Time for the park. My dog needs to see her friends.

BBL

See you.

@JM - what does "Camille Jordan (1838-1922) shows that algebraic equations of any degree can be solved in terms of modular functions. " mean?

The modular functions are a bunch of special functions. You haven't seen them yet in high school...

Wikipedia should have something on them.

1:19 AM

@JM - Thanks, isn't that roots of any polynomial must be algebraic, not transcendental, in your website that give to me, it says, "Ferdinand von Lindemann (1852-1939) expresses the roots of an arbitrary polynomial in terms of theta functions".

@Victor Yes, theta functions are another set of special functions...

"Set of algebraic numbers expressible as radicals" is a (very small!) subset of "set of algebraic numbers", if it wasn't clear already.

@jm - isn't that e is a transdental number?

Yes, e is transcendental.

@jm - so how could a function in terms of a transcendal number be algebraic, that is , a root of a polynomial

e^{i\pi t} for any rational t is algebraic, no?

1:32 AM

@jm - wow , some mathmatician is trying to misleading!

No, more of you're in way over your head, if I must be blunt.

2

@jm - thanks, have a good day!

2 hours later…

3:30 AM

Ugh, help vampires...

3:42 AM

@JM:V for Vampire?

4:00 AM

I wish... :D

4:20 AM

@JM ??

@J.M. Need help with something..

Oh... you've seen the article from meta.SO I linked to, right?

Fire away.

Sunni asked a question today, remember?

Is that deleted? I am yet to afford those cool 10k+ goggles...

I can dig thru today's chat. One second.

Doesn't seem to show in the deleted questions panel. Please do dig through the transcript...

@Srivatsan Why do you want to find it?

math.stackexchange.com/questions/83095/what-is-this-maximum

4:27 AM

Yes, Sunni deleted it himself...

@MikeSpivey Wanted to take a look at the question. Have some time free...

Ok thanks, @JM.

Morning guys!

@srivatsan: Here's the LaTeX code: This is related to my previous problem: http://math.stackexchange.com/q/82442/13425.

Let $a_{j,1}\ge a_{j,2}\ge \cdots \ge a_{j,n} \gt 0$, for $j=1,2,\ldots, k$, what is

$$\max_{\sigma_j\in S_n}\;\;\prod\limits_{i=1}^n\left(\sum\limits_{j=1}^k a_{j,\sigma_j(i)}\right)?$$

$S_n$ means the set of all the permuations on $\{1,\dots,n\}$.

Sn means the set of all the permuations on {1,…,n}.

I'm still not entirely sure why self-deletions don't show up on the moderation panel...

@tb Good evening! :)

4:29 AM

Good morning, t.b.! Had coffee already?

@JM Yes, but not enough... Halfway through it, second pot is boiling :)

@MikeSpivey Oh, thanks a lot, Mike.

(I suspected that it is a direct generalization of the previous question. It is.)

Good morning, @tb and @JM.

@JM "wow , some mathmatician is trying to misleading!" - we should've seen this coming... =)

Told you... I got turned off on expanding after that.

@tb Sorry, what did you tell me? // I wasn't around all this while.

You expanded in the right direction I'd say, JM.

4:39 AM

@Sri: "Help vampire"...

Serious question: Do you folks think there are more of these "Help vampires" here than we used to have?

Definitely yes.

Yeah... it's a mix of "people who want to get it over with" and "people way over in their heads"...

Yes, just found that thread in meta.so

I don't suppose this is anything new, but the first page of the "Questions" list has only three out of fifty questions from 2K+ users. And many of those other 47 are from people with <100 rep.

4:51 AM

@MikeSpivey How did you do these two counts? Manually by looking at each question?

@Mike: When does the Fall term end there?

@Srivatsan Here. Just scroll down through the first page.

@JM Mid-December. I'm on sabbatical, though, so I can ignore these things. :)

I was asking to get an idea of when things'll slow down on math.SE ... :)

@MikeSpivey Sorry, I misread that as the main page of the site.

At least it's slower today. The last question on the front page was touched four hours ago...

4:56 AM

@JM Now I understand you. :) Different schools are on different systems (quarter system, for instance), but most of the U.S. schools will end fall term in mid-December.

@JM Perhaps that doesn't mean much. With people like me editing questions often.

Ok, I will leave you guys in peace for 9-10 hours.

Bye all.

Bye Srivatsan.

Another thing I've been wondering about lately is whether voting is down. Perhaps I've just run through a spate of low-voted answers recently, but when I look through some of the top users (e.g., Arturo, joriki, Didier) their recent answers seem to be below their averages, too.

Bye Srivatsan.

See you, @Sri.

@MikeSpivey Hmm, yes. It seems to go along with the question avalanche.

For instance, if t.b. hadn't mentioned a recent answer of his a few days ago, I wouldn't have seen and upvoted it (the question was on page two, when sorted by activity).

@MikeSpivey I had the same impression. The list this week's voters reveals that there were only 100 users voting more than 10 times this week (since Sunday morning), so the vast majority didn't even vote twice a day.

5:08 AM

I'm glad to know it wasn't just me who had this impression.

Boy, I didn't vote much this week. :-(

It troubles me. I don't think it's good for the site in the long run. Maybe it's unavoidable, though, as the rate of questions being asked increases.

I agree. But we're still better off than SO, where "Tenacious" and "Unsung Hero" badges are easier to acquire...

@MikeSpivey It may have to do with the fact that the vast majority of the recent questions are just plain routine (or silliness if you forgive that blunt assessment). This doesn't exactly inspire well-crafted informative answers, so the regulars aren't impressed or excited.

@JM True; I just wonder if we're heading in that direction.

5:16 AM

The day one of us gets an "Unsung Hero" would be likely be the point where we have too many questions...

@tb Do you think it's a trend or just one of those random events that pops up? (And perhaps you're right that that's a partial for recent lack of voting.)

tb's now-starred comment is cut off at a suboptimal location on the sidebar... :)

:D

I think we can start expecting this every September...

What JM said.

@JM Maybe so. And fortunately we appear to be a long way from anyone getting Unsung Hero - only one person with Tenacious, and even he wouldn't qualify for it anymore.

5:24 AM

@MikeSpivey My impression is that the quality of the questions dropped dramatically sometime between end of August-mid September, so exactly when the new academic year started. When I joined the site sometime end of January, half the academic year was over, people had the time to review what they just learned during the Christmas break. The level of the exercises given to the students in the second half of the year is usually more than just plain definition-checking, as it is at the beginning.

5

So this could account for the better questions during spring/summer

@tb Makes sense.

t.b. is spot on, I would say.

Well, it's just another variation on the eternal september theme that JM already alluded to above.

September 1993... wow, I was barely using the internet back then.

Maybe I should dig through some of my old stuff to find some questions to ask on the site - help improve the quality.

On that note, maybe we shouldn't have run off the question-asking competition; if the more regular users asked more questions then the quality would go up.

It kind of feel forced if there's a competition looming, I'd say. That was what Henning and Asaf were driving at in meta.

(On that note, the questions I haven't yet posted seem to be both MO- and m.SE-appropriate, so there's that...)

5:35 AM

@JM Overall, I was not in favor of the competition. But it probably still would have improved the question quality.

I think I am going to see if I can find some old questions to ask.

6:09 AM

Good night, folks.

Good night!

Alexei Averchenko

6:30 AM

@tb heh, I guess SE is truly the new Usenet then :)

@Alexei: you're aware that that was Jeff and Joel's precise intention for SE? :)

Alexei Averchenko

yes :)

I know very well how he feels...

Alexei Averchenko

6:49 AM

Hello @Rob, are you interested in an intelligent discussion today?

@AlexeiAverchenko always Sir

Do we have any algebraic geometers in the house?

7:09 AM

I just watched a video from the University of Sidney about the influence of Godel on maths foundations

He also worked with Einstein and wrote a paper on General relativity

Alexei Averchenko

@Daniil, hi, are you from Omsk? :)

In is paper he proved the possibility of time travel according to Einstien's Field equations

The lecturer said "proving it physically possible is much stronger that proving it logically"

Alexei Averchenko

@Rob what do you mean by that? the possibility that space-time is not simply connected?

@AlexeiAverchenko The lecturer said that Godel proved that if the Universe was disk shaped that a path in space-time could loop back on itself

thus arriving before it left

Alexei Averchenko

@Rob doesn't sound credible. did he prove it?

7:18 AM

I have not seen the paper of Godel

Alexei Averchenko

or maybe by 'disk-shaped' you mean something other than a en.wikipedia.org/wiki/Disk_%28mathematics%29 ?

because disk is contractible

i wouldn't be surprised to see a wormhole-type thingy :)

"proving it physically possible is much stronger that proving it logically possible" seemed like an odd statement ... what do you think???

Alexei Averchenko

nothing :)

oh

Alexei Averchenko

i'm boring that way :)

7:22 AM

He probably means the Gödel spacetime, which has some rather unusual properties (including, yes, closed timelike curves).

hello

Alexei Averchenko

@ZhenLin is it simply connected?

It's even diffeomorphic to R^4.

It's funny how we classify internet personalities as trolls, vampires, .. geeky D&D influence

@QED Hello QED shouldn't logically possible be stronger than physically possible?

7:25 AM

physically possible is stronger

Alexei Averchenko

@ZhenLin now THAT's interesting 8)

physics is just one logically possible world

@QED The world that we live in

is that what makes it "stonger"

oops "stronger"

if something is physically possible it is logically possible

if something is logically possible, it may or may not be physically possible

I say that if A implies B but B does not necessarily imply B then A is stronger than B

@ZhenLin I find it very interesting that Godel did work outside of pure math

7:32 AM

General relativity is just geometry really. :p

that should be

@QED " B does not necessarily imply B"???

I say that if A implies B but B does not necessarily imply A then A is stronger than B

@QED thank you

@ZhenLin geometry without the "geo"

Alexei Averchenko

@ZhenLin, yes, this is an extremely bizarre solution indeed 8)

7:38 AM

@QED so physics is stronger than logic?

Alexei Averchenko

btw, yesterday i helped a guy on one forum to better understand affine schemes

i'm proud of myself, especially since i haven't studied them yet

reading MO pays off 8)

@Rob, the set of logical truths contains the set of physical truths but not vice versa

Alexei Averchenko

@QED are you sure it's not a proper class? :P

@QED and logic is stronger than math

well this is just philosophy

I don't actually mean to make formal statements

Alexei Averchenko

7:40 AM

i'm just kidding :)

but it's a nice way to classify your views and thoughts

I'm lost

I think logic and math are (different parts of) the same thing

@Rob, well why do you ask? I wonder what context this came up in

@QED what is the same thing that both math and logic are apart of?

just mathematics

7:46 AM

@JM: I've now commented on KCd's thread. I don't think I can do any better than that.

@Alexei: What's there to understand? An affine scheme is simply an object of CRing^op. :p (At least, the way my commutative algebra lecturer teaches it...)

The context was: I just watched a video from the University of Sidney about the influence of Godel on maths foundations
He also worked with Einstein and wrote a paper on General relativity n is paper he proved the possibility of time travel according to Einstien's Field equations
The lecturer said "proving it physically possible is much stronger that proving it logically" The lecturer said that Godel proved that if the Universe was disk shaped that a path in space-time could loop back on itself

ah that's just silly

@Rob: That's incorrect. What Gödel proved was that there is a contractible spacetime in which there are closed timelike curves.

sorry I misunderstood

Godel proved that the abstract theory of GR admits a certain very odd universe

I guess the professor is saying.. that's fine in theory but in reality there's no such thing

7:51 AM

Now, I wish I knew enough generalised geometry to make sense of this page...

@tb: I got Kashiwara and Schapira out of the library yesterday. It's a scary book. I might have to return it for something easier...

Well thank you all for the very interesting and intelligent discussion today?

hehe

;-)

@ZhenLin Which one? Well, doesn't really matter as both are of the same degree of readability. Did I seriously recommend it? I'd imagine I'd have passed a word of warning along...

@tb: Categories and sheaves (2006). Yes, I suppose you did, but Iversen was out on loan...

7:59 AM

So, what are you trying to learn/look up?

Sheaf cohomology (in Grothendieck toposes).

I need to (try to) learn a lot of material very quickly so that I can decide whether or not I want to write an essay on "Étale cohomology and Galois representations".

Alexei Averchenko

can i ask a silly question?

You just did...

Alexei Averchenko

can i ask another one? :)

that's another one :p

8:03 AM

@Zhen: There's Artin's notes now available electronically (didn't know that!). Of course also Milne's notes would be a place to look.

2

Alexei Averchenko

how do i show that a set having a countable subset of [0, 1] has an accumulation point?

Uhu, Artin. I'll have a look.

@Alexei: [0, 1] is compact.

Alexei Averchenko

or (what i really want to prove) that it's not closed

@ZhenLin yes, i remember that any infinite subset of a compact space has an accumulation point, but i don't remember why

@tb: The trouble is, I haven't actually learned sheaf cohomology properly. I don't know how to compute it at all. So I was hoping for a nice introductory text on the subject...

@Zhen: since you like differential geometry: did you have a look at Bott-Tu? That should provide some basic intuition.

8:06 AM

@Alexei: Cover your infinite set with open sets. By compactness, you only need finitely many. So one of the open sets contains infinitely many points.

Alexei Averchenko

oh

so easy :)

@tb: Someone's taken that out of the library too. It seems the good books are gone, hah.

Alexei Averchenko

now that's unfair!

@ZhenLin You can't reserve books there?

I can't really read at a computer...

@JM: I have to fight with about 100 faculty, 200 or so Ph.D. students, and 300 or so graduate students...

8:10 AM

Bribe the librarian... >:)

Actually, 200 Ph.D. students doesn't sound right. But you get the idea.

Alexei Averchenko

can you check my reasoning please?

proposition: let X be a graph with no loops. then every closed path f: I \to X satisfying f(0) = f(1) = x_0 (where x_0 is a vertex) visits finite number of vertices finite number of times

Why cant there be system which shoots an e-mail to the user upon any new notifications (posts/comments/edits) ?

8:25 AM

Huh?

@AlexeiAverchenko Can't you oscillate around a vertex using a variation on the x\sin{1/x}-theme?

Alexei Averchenko

@tb i miss-staed the proposition, sorry :)

as you mispeled the apology? :)

So, what do you really want to prove?

Alexei Averchenko

19-th exercise in Hatcher, about loops on connected 1-dimensional CWs

i dread his proof that \pi_1(S^1) = Z, and he promised this will make it easier :D

so i have to do it!

you dread it? :)

It's pretty horrible

you can find better proofs

Alexei Averchenko

8:34 AM

yeah, he basically proved the entire homotopy extension property there :D

Alexei Averchenko

9:06 AM

homotopy lifting

As May put it: HELP

Alexei Averchenko

@QED i actually traced it with pen and paper, working everything out to the smallest detail. and yet i can't remember how the last part of it goes

yeah I am not a fan of this proof

I don't know the more conceptual category theory way to do it though :/

Alexei Averchenko

van Kampen theorem also has a difficult part

although it's easier conceptually, it's still a technical nightmare, at least for me

But what's wrong with the proof?

Alexei Averchenko

9:10 AM

what proof?

Hatcher's proof looks like a perfectly natural way of doing it to me and it involves important ideas to be used later on.

I mean, you can also cheat your way around the issues and define the problem away: t -> exp(2 pi i t) is a universal covering map R -> S^1, obviously...

Top of the morrow.

good morning Asaf! How are the inaccessibles doing? Still non-existent?

Alexei Averchenko

the only part of it that's really hard for me is proving that any F: Y \times I \to S^1 lifts to \tilde F: Y \times I \to R for any given \tilde F: Y \times {0} \to R.

@tb If I tell you, I just might have to kill you.

Alexei Averchenko

9:15 AM

mornin, Asaf

I like that new XKCD strip. The tooltip is great.

@Alexei: look at it this way: as soon as you have the fundamental group of S^1 you have the fundamental theorem of algebra for free, so something should be non-trivial...

Alexei Averchenko

i'm perfectly fine with it

it's just that it's all cramed in two pages that i don't like

i think i know how to prove 19th

@Alexei: Even May spends an entire page...

(page 9)

Alexei Averchenko

maybe there's a way to grothendieck this proof, i hope :)

ok, 19th:

first, wlg we can assume that every edge is glued to two different vertices

9:26 AM

let a group G act freely and properly discontinuously on a simply connected space. Then the quotient map X -> X/G is a universal covering map.

Alexei Averchenko

construct a cover of f(I) that has no non-trivial subcover. it can be done, just take each edge's small neighborhood. thus there's only finite number of edges

so if we take preimages of all vertices we get a partiion 0 = t_1 < ... < t_m = 1

suffice to show that we can homotope f so that it's finite

indeed, if {t_i} is infinite, then it has a cluster point, and for {t_i} to be closed this point has to be in {t_i}. But then by continuity all points of the subsequence converging to this point (except for a finite number of them) must correspond to the same vertex

The sky looks somber and grim today. I hope it will rain.

Alexei Averchenko

since edges are glued to different vertices consider the vertex that is the image of the cluster point together with all edges attached to it

it's compact so there are only finite number of times that f passes through it

and it's contractible so we can homotope f such that every time it passes through this 'star' it only passes through its center once

thus wlg we can assume that 0 = t_1 < ... < t_m = 1 is finite

but then we can simply reparametrize each f|_{[t_i, t_{i+1}]} and obtain the desired homotopy

is this correct?

Good morning @Alex, @Asaf, @tb, @QED

hello

Alexei Averchenko

9:35 AM

morning

Hi Gorta.

Gorta reminds me names of creatures from Godzilla )

That'd be Mothra...

Or Ghidorah. Or whatever. :P

@JM Good afternoon)

Gonorrhea?

9:37 AM

That monster's sticky...

@AsafKaragila should screw a bit your association centre, shouldn't you? )

I thought Gonorrhea messes with the heart, syphilis is what eats the brain.

More or less right.

But both will certainly sear you...

Obviously.

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