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11:00
@Soham you're around?
$0 \to A \to G \to B \to 0$ be a short exact sequence. Finding which $G$ fits into the sequence for given $A, B$ is called the "extension problem"
Now, classifying problem asks what are the groups of a given order $n$. If you can prove that for each group of order $n$, there is a known "smallish" normal subgroup $N$, you can put it in a short exact sequence like this $0 \to N \to G \to G/N \to 0$.
So that your problem is translated into the extension problem.
Finding such a smallish normal subgroup is what Sylow theory is about. Recall that Lagrange's theorem says that for any group of order $n$, a subgroup must divide $n$.
Sylow theory gives a partial converse : for any group of order $n$, if $p$ divides $n$, then there is a subgroup of order $p^k$.
sorry, I missed your message.
@BalarkaSen prime $p$, obviously? else $A_4$ is a counterexample, although I think you need Sylows to prove that.
what kind of sad person denotes $p$ as something other than a prime?
:P
you strike me as a not-always-very-nice person :P
11:06
yes, if $p$ is not a prime, then the converse is false.
that's the next chapter.
algebra is cool, man
@SohamChowdhury nah. you haven't (yet) seen me.
would I like to?
:P
@SohamChowdhury sure is. now there's something "easier" than Sylow's theorem out there.
11:08
Theorem (Cauchy) : If prime $p$ divides $|G|$, then there is an element of order $p$ in $G$.
Proof : Google McKay's proof. It's purely combinatorial.
then it generates a cyclic subgroup, na?
yes
i.e., $G$ contains an embedded copy of $\Bbb Z/p$
and Sylows upgrade this to any power of $p$?
nods back
11:10
Sylow is Cauchy with steroids.
on* steroids
ok, I should get down to studying. later.
no, with. I am referring to organic compounds, not asteroids.
:P
11:24
Hello@SohamChowdhury
@SohamChowdhury u must get ride of these reflexes
hey badass @r9m and magician @Chris'ssistheartist
@Soham what are you doing with hatcher?
So even you are badly excited about homological algebra.....
I would love to get an insight from Tobias...
@BalarkaSen hey, this is the most jora-taali-mara language ever. :P
@Rememberme having fun.
what universal do polynomial rings based on general (i.e. possibly non-commutative) rings satisfy?
11:43
@SohamChowdhury i don't even understand the question.
going to isi to meet prof this sunday. have to extract some of his geometric intuition to use it in my work.
@Agawa001 hey :-)
@BalarkaSen "work", wow. you sound grown-up.
work = fiddling
ultra-cool thing: what is $\sf{End_{Ab}}(\Bbb Z)$?
$\Bbb Z$, I think.
11:55
yes. as a ring!
right
$\mathsf{End}_{\mathsf{Ab}}(\Bbb Z^n)$ in general is $\text{M}_{n \times n}(\Bbb Z)$
this makes it seem like something's special about the ring structure on $\Bbb Z$. it's the initial object in $\sf Ring$, too.
The isomorphism you can write down by looking at where the generators are sent to.
yes.
I'll try.
what's an initial object, again?
11:57
oh, google it. I'm eating.
In category theory, an abstract branch of mathematics, an initial object of a category C is an object I in C such that for every object X in C, there exists precisely one morphism I → X. The dual notion is that of a terminal object (also called terminal element): T is terminal if for every object X in C there exists a single morphism X → T. Initial objects are also called coterminal or universal, and terminal objects are also called final. If an object is both initial and terminal, it is called a zero object or null object. A pointed category is one with a zero object. A strict initial object I...
there you are
@Balarka, is there any concept of an algebraic structure with three operations?
@SohamChowdhury dunno.
do your friends/relatives ask you if you've found $x$ yet when they see you?
12:14
do you want to be reported for unfunny trolling?
r9m
r9m
12:30
@dREaM thats an old one :P
@Agawa001 badass r9m?!! :P LOL
do algebraic topologists really use ring theory that much?
r9m
r9m
@anon rofl :P
12:49
@anon a lot of algebraic topological invariant are rings (cohomology!), so certainly some amount of ring theory is needed. I don't think they use full algebraic stuff like noetherian rings, etc, though.
didn't know cohomology was rings
cohomology can be given a ring structure. google cup product.
nah. don't want to.
ok.
so, what have you been thinking about? also, morning.
combinatorics. an interesting corollary to the resolution of the question I had is that if $X,Y$ are two $n$-sets on which $S_n$ acts, even if they're not equivariant, the sets $S_n\times X$ and $S_n\times Y$ are equivariant, even in a way that preserves the first coordinate: $(\sigma,x)\mapsto(\sigma,\sigma f\sigma^{-1}x)$ for any (barren) bijection $f:X\to Y$.
this is a generalization of the fact that if $G,H\le S_n$ have index $n$ then they're conjugate, ish
12:52
interesting.
@anon how does that follow from that statement? i think i am being silly
actually I think the implication is the other way around for $X=S_n/G$, $Y=S_n/H$.
the fact they're conjugate implies $S_n\times X\cong S_n\times Y$ are equivariant while preserving the 1st coordinate
right.
but even if $X,Y$ aren't equivariant, if they have different numbers of orbits of different orders and types etc., the products $S_n\times X$ and $S_n\times Y$ are equivariant, even whilst fixing the first coordinate
yes, true.
anyway this means all of the combinatorics I wanted to do can be translated into the language of group actions, which is very nice
13:06
Hello@BalarkaSen
Oh, I just got another awesome result ...
Mew
Mew
13:23
hi
14:03
@Semiclassical did you work on that integral of mine that you said about is scary? :-)
your series are all awfully scary in horrific way
chriss whats about your book news ?
@Agawa001 It's just an illusion, it's all about beauty, it's so much beauty in there that people are tempted to see a deformed image of the things, because of the profound beauty that mind hard accepts. I can't explain better, but I might try.
@Chris'ssistheartist beauty in the eye of beholder, i see beauty in mathematics but not in these kinds of endless series
@Agawa001 Thanks, all is fine. I have under research some very interesting integrals that I prepare for my book.
If you wanna run for 10 km, then you might fail after 1 km, less or more, without proper training, that's the same story with the mind, you need training to take a profound beauty.
good luck; i must leave now, by
@Chris'ssistheartist lol
14:19
@Agawa001 :-)
yes it happens always with me
Profound beauty is for trained minds. :-)
i fail before few steps toward the end
i hope you ll hold on
and continue til the end
14:20
Never give up. :-)
whatever costs
of course
Yes, definitely.
key of success, is keep trying whatever takes
Yes, die working, crawling, succeeding. :-)
einstein said , perseverance is priceless
14:21
The full success is the only outcome accepted. :-)
Yeap, it is.
i have some buisness to do now good luck
take care
14:34
I just realized the google calculator won't calculate 46^2
eh? it does, it just shows up below the youtube entry
same thing happens for 46+2, in fact
@Chris'ssistheartist: I just wrote this answer which draws on an answer we discussed in chat back in January. I tried to find that chat, because I knew I had computed $$\sum_{n=1}^\infty\frac{\zeta(2n)-1}{n+1}$$
@Chris'ssistheartist: I figured that it would have been you with whom I had discussed that sum.
@robjohn I don't remember exactly the series, but it might be.
@Chris'ssistheartist somewhere near this comment.
@robjohn Ah, got it. It was about that series involving a nice telescoping sum. :-)
14:59
@Chris'ssistheartist: did you see my answer to this question? I mentioned it yesterday in chat, but I was wondering if you'd seen that method for computing the square root. Jack D'Aurizio says it used to be in text books. I don't remember seeing it in text books when I was a kid.
@robjohn Yes, I saw that. :-)
@Chris'ssistheartist but had you seen that algorithm in any text books?
@robjohn No, I don't remember that.
I have a problem with PC, after a while HDD spins so fast I barely manage to do anything, and I don't know for what reason.
Hope to replace this machine soon.
That's why you should use solid state
definitely. that's part of why i like this new laptop.
no hard drive spinning, and it boots soooo much faster
15:17
Yeah, SSD is the future, bye HDD.
15:43
@Chris'ssistheartist what do you mean it spins so fast? It used to be that some drives spun at different rates while accessing inner and outer tracks, but I think that nowadays, most drives spin at a constant rate. In any case, the rate is the same when reading the same tracks.
@robjohn no one likes a party pooper
The party's over pal.
Ain't no party like a Vegas Pary coz Vegas parties don't stop
oh apparently the original version is with west coast pary
How would I go about this problem?
1
Q: Help to Prove Convex Quadrilateral Problem

FlyAwayBirdieI'd appreciate help with solving this convex quadrilateral riddle. It appears to hold true in all of my test cases, but I'm not sure how I would go about writing a proof for it. Let $UVWX$ be a convex quadrilateral. Let $Y$ be the midpoint of $UX$, and let $Z$ be the midpoint of $VY$. ...

Oops I broke chat rules #1 and #3, sorry.
:(
16:03
@robjohn I mean that after a while I see the HDD red light accompanied by noise, and the whole computer works very slowly.
We have chat rules?
Are there theorems particularly useful for trigonometric diophantine equations?
Of course most of these are not solvable, but are there specific ones that we can deal with? I attacked $$\cos \frac{a}{b}\pi = \phi \cos \frac{c}{d} \pi $$ using infinite product representations, but that did not help.
67
Q: Main Chatroom Etiquette Rules

robjohnThe previous owner of the Mathematics chat Asaf Karagila drafted some etiquette rules for the chat. They were deleted, so I am reposting them so that they might be observed. It's nice of you to drop by, after saying hello please spend a minute reading the transcript to see if there is an active...

Oh, I think I have broken all of them
except 6
sounds good
Wow, you left my comment hanging there dude
17:03
hi @Ted
Hello Professor @TedShifrin
17:20
@DanielFischer interesting to note that this is false when considered $\Bbb R^3$.
$K$ be the whitehead solenoid. solenoids are compact. but complement is the whitehead manifold, which is simply connected
@BalarkaSen $K = \{0\}$ would already do it in $\mathbb{R}^3$.
oh, fair enough. meh.
But if we take an Alexander horned ball ...
if you want a non-simply connected example, just take a knot.
the whitehead solenoid example above is interesting because it's a 1-dimensional (in the sense of covering dimension) subset of R^3. is there a simpler 1-dimensional cpt subset of R^3 s.t. complement is simply connected?
hmm
ok, i don't think covering dimension is the right thing. i think we need inductive dimension.
17:38
lollll, I have so much fun here.
@robjohn no one calculated that integral I showed you some days ago referring to some I gave it to.
@Chris'ssistheartist I should look back at it...
@robjohn OK, take it when you have time. :-)
@robjohn I should propose it to some magazine. Actually I'm going to do that.
17:53
@Chris'ssistheartist was it this one, or something earlier?
@robjohn that one :-)
18:11
Hi @robjohn
Can you give me a reference for "Hilbert space"
I was reading a quantum mechanics text and it came up
$$\frac{\Bbb R[x]}{(x^2+1)}\cong\Bbb C$$
wow.
@SohamChowdhury You can similarly construct the dual numbers and the split-complex numbers, respectively with $(x^2)$ and $(x^2 - 1)$.
yes.
the idea is very well explained here:
this helped me.
@SohamChowdhury this is covered in the second semester of most algebra courses
is that relevant?
18:26
what do you mean?
Yeah. Basically, making the polynomial ring adds an indeterminate $x$, and taking its quotient by the ideal generated by $x^2 + 1$ essentially means you're setting all instances of $x^2 + 1$ equal to zero--so it's literally the same as adjoining to the reals a solution $x = i$ of $x^2 + 1 = 0$.
yeah, that's pretty cool.
@Fargle exactly, that captures the intuition I have.
Is that Aluffi?
18:31
I tried reading it once while watching the whole one piece series at the same time
It didn't work out
Hmm. It seems like if you let consider $\Bbb R[x,y,z]$, letting $I$ be the ideal generated by $x^2 + 1$, $y^2 + 1$, $z^2 + 1$, and $xyz + 1$, it should stand to reason that $$\Bbb R[x,y,z]/I \cong \Bbb H.$$
(Letting $\pm 1 \mapsto \pm 1$, $\pm x \mapsto \pm i$, $\pm y \mapsto \pm j$, $\pm z \mapsto \pm k$.)
hmm, yes.
this stuff is very interesting indeed.
Actually, I just looked this up. Consider $\Bbb R\langle x,y \rangle$, the free algebra on $\Bbb R$ with two indeterminates (that is, like a polynomial ring, but x and y may not commute.)
Then consider $I$, the ideal generated by $x^2 + 1$, $y^2 + 1$, and $xy + yx$. Then $$\Bbb R \langle x,y \rangle / I \cong \Bbb H.$$
$x$ is $i$, $y$ is $j$, and $xy$ is $k$.
Anyway, I'm hungry.
Hello!!! :)
hi, @evinda, wie geht's?
18:45
@SohamChowdhury Gut.. Wie geht es dir?
Sehr gut. Ich studiere Algebra im Moment. :)
@SohamChowdhury Gut!!! Macht dir Algebra Spaß?
Ja, sehr.
@SohamChowdhury Schön!!! Was gibt es sonst so neues bei dir?
Zahlringen sind in Zahlentheorie sehr wichtig, und ich mag Zahlentheorie, so . . .
18:47
@SohamChowdhury Ich mag auch Zahlentheorie!!!
can some one help me through what it means to "take logs" to solve t for something like this 450×1.05 t  =  1325

I've no idea what it means to "take logs" but am trying to teach myself it
Wusstest du das "Weihnachten-Theorem"?
@Dave do you know what a logarithm is?
vaguely - but the logic of them was not totally easy to understand
and I guess that was $450\times 1.05^t=1325$?
should've been a power.
yes sorry ^t
my book on it is 7 pages rammed full of blocks of text, rather uninspiring way to understanding it, doesn't seem to simplify how to understand it very well for me
18:50
@Dave go learn what logs are, properly. link 1, and link 2
this website gives good intuition.
ok thanks ill try it out !
@SohamChowdhury Nein. Worum geht es?
Are logarithms used a lot in science by the way?
i see they use powers a lot for shortening large numbers, but not often seen logarithms
@Dave Use the fact that $\log_a a=1$
@Dave I think that they ae used a lot
@Dave What do you study?
i plan to do physics at university but want to brush up on maths this year first
got bored of unemployment :P
18:56
@Dave Have you studied before something else?
Not at university no. I didn't do well in my youth. Also had no care for it - but now i love learning
es ist auf Deutsch schwer. :P
wenn $p$ eine ungerade Primzahl ist, sie ist Sum (?) von zwei Quadratzahlen nur wenn $p\equiv 1 (\rm{mod}\; 4)$.
@Dave How old are you? If I am allowed to ask...
27
@Dave A ok.. And is the undergraduare program 4 years?
18:58
yup
It's very brave of you to actually want to study physics now. Good luck!
@SohamChowdhury Interessant... I have seen the theorem at the university...
Yeah i know ! Bit scary with the maths
And logs are everywhere in physics!
@Dave Yes, I think so too...
18:59
@Dave ah, sure you'll do fine, considering you seem to be someone who can work very hard!
also, do give the BetterExplained articles a read.
they're fantastic.
Will do ! Thanks :)
Where are you from? @Dave
UK
Try the Khan Academy on YouTube for math
@evinda Ah. Korrigier (?) mein Satz, bitte. :)
@skillpatrol yeah, seconded. the calculus videos, in particular, are nice. I learned from there.
19:04
Wenn $p$ eine ungerade Primzahl ist, ist sie die Summe von zwei Quadratzahlen nur wenn $p\equiv 1 (\rm{mod}\; 4)$. @SohamChowdhury
Hi @zed111
@evinda ah, fast richtig!
Ja :) @SohamChowdhury
Ich muss ins Bett gehen.
Good night, chat.
19:06
@SohamChowdhury Gute Nacht!!!
Later pal
19:27
Hello @evinda
I love when edits I make to an answer bring 3 upvotes to another answer and nothing for mine >8(
Are you a student? @zed111
Yeah, in engineering. you?
@robjohn I am sure you can fetch many upvotes on my easy question
0
Q: Integrating $\prod_{i=k+1}^{kN} \left(\int_{-\infty}^{\infty}dp_i\right)\times$ with conditions on $p_i$

zed111I am trying to integrate this expression which came up in a derivation of the momentum distribution function for an ideal gas. $\Theta(x)$ is the Heaviside step function which is $1$ when $x$ is positive and $0$ if negative. $$\prod_{i=k+1}^{kN} \left(\int_{-\infty}^{\infty}dp_i\right)\times\...

19:36
@zed111 In applied mathematics
@zed111 In which semester are you?
@evinda Just finished my BS, starting my PhD in Aug
which country @zed111 ?
@zed111 Congratulations!
@KarimMansour india
congratulations @zed111
19:39
@evinda I love pure math btw. Just keep reading it all the time even though it has nothing to do with my PhD/research
@zed111 In which field will your phd be?
Mechanical Engineering
On energy systems
@evinda @KarimMansour which sem are you in
@zed111 I have finished the 6th semester
summer semester I have 3 semester left then I should graduate
and hopefully planning to go to waterloo
@zed111 ahh, the joys of stat mech
19:42
cool
i may take a crack at that myself, depending how productive i'm being otherwise
@Semiclassical yay. It feels awesome to derive everything from stat mech
aye. though, are you doing classical stat mech or quantum?
I am currently on classical part
@KarimMansour which country are you from?
19:44
canada @zed111
@evinda ever read quantum?
sorry this was for @Semiclassical
i know about quantum stat mech, yeah.
at the intro level that's just a matter of including fermi-dirac and bose-einstein distributions functions
@zed111 No, I haven't
the higher level stuff is formulated in terms of statistical field theory, since that allows you to use diagrammatic techniques
k
@evinda I am planning to read about "Hilbert space". It came up in a lecture on quantum.
Which math area does it belong to?
19:50
quantum mechanics you need to know functional analysis
and C* algebra
for the mathematical side of QM
depends on if you're learning it as pure math or as physics
@zed111 I don't really know
@zed111 I haven't seen it
on the physics side, you really don't see it talked about in terms of C* algebras
and we're usually not terribly rigorous about the functional analysis stuff either.
Whats the C* algebras?
thanks @KarimMansour
its an algebra with some properties
19:53
where do people read it from @KarimMansour
functional analysis and C* algebras
as for your question, at least at the level of intuition: when you integrate over all $kN$ coordinates, you're getting the total volume of the allowed phase space
right @Semiclassical
if you drop the first $k$ integrations, that amounts to leaving one of the particles in the gas out
heeeeellllllllllllllllllllooo
19:54
I heard this book is quite good too from my prof
yes, thats because we already fix $p_i$ of one particle @Semiclassical
did you ever see it @Semiclassical ?
never read von neumann, no
@Karim: It is by John von Neumann. ;-D. So it has to be good
right. you're treating that momentum, for the moment, as a controlled quantity (i.e. not integrated over)
19:56
yeah
if your original integral was the volume in the total phase space, the new one is the volume in the 1-particle-reduced phase space
hello. what is the job of you guys?
another way to make the same point is that you can get your second integral from the original one if you include $k$-dimensional Dirac delta function $\delta^{k}(\mathbf{p}-\mathbf{p}')$ in the integrand
I have this book @KarimMansour. Chapter 2 is exactly on Hilbert space
Will read it soon
hilbert space normally appear in functional analysis
a hilbert space is essentially a complex vector space that is inner product space
19:59
that's where it's rigorously presented, yeah
and it is also complete
in a sense that every cauchy sequence is convergent

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