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5:11 AM
yeah haha that's a hilarious story with sad undertones that nobody cares about bro
I too have a story alike
our primary schools, which were in the late 80's early 90's , always had special guest speakers that were accompanied by videos that emphasized how special they are as a category of guest
and one of the first was a dude in a human teddy bear costume that looks identical to the local police uniform, and the teacher said, ok kids, I want you to say a big hello to Constable Care
at question time I had my hand up really quick and when they eventually gave up and let me ask mine I asked " how do we know the constable cares about us?"
The they collaborated and responded with "because his uniform is blue and white, and so is his car, and so is his lights"
I then immediately said "well my mum always told me to look for red if I am out somewhere on the farm and im in trouble because she always wore a bright red jumper"
but yeah I don't have a photographic memory of my whole life OK if I am honest most of my early childhood education I was always sleepy or just waking up and wondering where I am
oh skull think of the children you monster
 
Little David was in his 5th grade class when the teacher asked the children what their fathers did for a living. All the typical answers came up - fireman, policeman, salesman, doctor, lawyer.
David was being uncharacteristically quiet, so the teacher asked him about his father. "My father's an exotic dancer in a gay cabaret and takes off all his clothes in front of other men and they put money in his underwear. Sometimes, if the offer's really good, he'll go home with some guy and have sex with him for money."
there^ I got the full joke :-D
it's just a joke, pal
 
Jeffory if you continue to be disruptive you will get another shot so help me god
 
The real trump is in the uk hoping to open a new casino there
 
haha um yeah I think all of this material is gold but what every play some violin woos
what ever*
 
5:27 AM
yup
 
that additional y is a career stopper
or my selection of book for story time for the kindis but hey im not paid here which is pretty much accurate for educators period
which is why I literally would not even dream of being a school teacher and cheat the welfare system repeatedly and put up with the abs min wage to avoid all forms of that example of bear trap
ok I just got an update that I am really hurting the economy we better stop
but that casino comment, you drove a samari sword through the heart of the economy and you know why you are not welcome in heaven now
that many soldiers of Christ voted our leader in skull, does statistics make you a very smart person for propagating your fake news with you and you deep state buddies?
exactly that's why redhat is there to protect freedom
and provide another desperately needed Linux open source operating system
 
 
2 hours later…
7:08 AM
Mornin'
 
Hello
 
lol
it's pronounced more like "mey", soz
 
7:25 AM
ah right I am expecting to see a removed unit in the stream but im not
 
wat
Go do some maths
4
 
We should start teaching bees the millienun prize problems
 
Biemann Hypothesis
schnarf schnarf
 
 
3 hours later…
10:13 AM
Jun 1 at 12:05, by Albas
If $\nabla$ is a connection on some manifold $M$, then we define for a vector field $Y$, $\nabla Y$ to be a $(1,1)$ tensor field such that when contracted with $X$ it gives the covariant derivative $\nabla_X Y$. I don't get the meaning of this definitio. How does acting $\nabla$ of a vector field give you a $(1,1)$ tensor field? And what does contracted with $X$ even mean?
Can someone help me with that question of mine?
 
$\nabla Y$ is an operator on the space of vector fields on $M$, i.e., $\nabla Y : \mathfrak{X}(M) \to \mathfrak{X}(M)$
$\mathfrak{X}(M)$ is the space of sections of the tangent bundle $TM$, and since $\nabla Y$ is a pointwise defined operator (i.e., $(\nabla_X Y)_p$ only depends on $X(p)$), you can redefine this R-linear operator in $\text{End}(\Gamma(TM))$ as an element of $\Gamma(TM \otimes T^*M)$
A section of $(TM)^{\otimes p} \otimes (T^*M)^{\otimes q}$ is a $(p, q)$-tensor field. In this case $p = q = 1$, which is a $(1, 1)$-tensor field.
 
10:35 AM
Ahhh jeez, why didn't I see that.
 
11:01 AM
So I guess a (k,1) tensor can be said to be a map from V×V...×V(k times) to V right ?
 
A multilinear map from the product, or a linear map from the k-fold tensor product, yes.
 
Interesting, never saw tensors like this
Then the torsion becomes a (2,1) tensor I guess.
 
That's right
That's one interesting tensor, by the way
 
Yeah, that's a dude I don't understand. Is it the same thing as the torsion defined for curves with respect to the Secret Frenet frame?
 
No, that's something intrinsic
I "understand" torsion in two ways. The first one is by a concrete example. Take the connection on $\Bbb R^3$ which satisfies $\nabla_X Y = Z = -\nabla_Y X$, $\nabla_Z X = Y = -\nabla_X Z$ and $\nabla_Y Z = X = -\nabla_Z Y$ where $X, Y, Z$ are the three coordinate vector fields.
 
11:10 AM
I see, also why "torsion" ? Anything to do with the torsion group?
 
Yea that example of a rotating American football.
 
Right, it's a Riemannian connection on $\Bbb R^3$, so geodesics are straightlines. But you'll see if you parallel transport a frame along a straightline it "twists"
Hence the name "torsion", I believe
 
If $(M,g)$ is a smooth compact Riemannian manifold of dimension $n$ what exactly does it mean to integrate against the volume form $\mathrm{dvol}_g(x)=\sqrt{\det(g)}\mathrm{d}x^1\wedge\cdots\wedge\mathrm{d}x^n$? How does it related to integrating against a measure?
 
@Alessandro A volume form gives a measure, given a ball in a chart you integrate $\sqrt{det(g)} dx^1 \cdots dx^n$ in that chart to get the Riemannian volume of that ball
This gives a Borel measure on the manifold
 
11:14 AM
@BalarkaSen Is the thing after the square root a differential form or just saying that you're integrating $n$ variables?
 
Integrating a volume form $F dx^1 \wedge \cdots \wedge dx^n$ on a domain $U$ in $\Bbb R^n$ is by definition $\int_U F dx^1 \cdots dx^n$. So they are the same thing if you pass to local coordinates
 
Oh ok, perfect, thanks
 
Artin in his book Algebra says:
An integral domain $R$ is a Euclidean Domain if there is a size function $\sigma$ on $R$ such that division with remainder is possible, in the following sense:
Let $a,b\in R$ and suppose that $a\ne0$. There are $q,r$ in $R$ such that $b=aq+r$, and either $r=0$ or else $\sigma(r)<\sigma(a)$.
 
We did a bunch of stuff today in noncommutative geometry using diffgeo and function analysis concepts I'm not familiar with...
 
(Where size function is any function from $R-\{0\}$ to $\Bbb N\cup\{0\}$.)
Then Artin says: 'The most important fact about division with remainder is that $r$ is zero iff $a$ divides $b$'.
I know that in Euclidean domain, $q$ and $r$ may not be unique. So, how to show that $r=0$ iff $a$ divides $b$? 'Only if' direction is obvious. For 'if' direction, suppose $b=aq_1$ and $b=aq_2+r$ for $r\ne0$, then since we are in integral domain, $a(q_1-q_2)=r$ implies $q_1\ne q_2$. How to proceed further?
 
11:20 AM
Remember that $\sigma(r) < \sigma(a)$
The size function is the compensation for uniqueness. It controls the remainder
@Alessandro Cool, what was the punchline
 
how many elements are there of order $p$ and $q$ in a non abelian group of order $pq$ with $p<q$?


Approach: The number of $p$-sylow subgroups cannot be 1, since the center becomes trivial. So, let the number be $s$. $s \mid q$ and $s = p k +1$ for some $k \in \mathbb{Z}$ with $s \neq 1$. Therefore, $s$ must be $q$ [$q$ is of the form $p k +1$].

Each such subgroup is the cyclic subgroup generated by the elements. So, $n(o(p))=q$ and $n(o(q))=pq-q-1$
 
@BalarkaSen K-homology
 
Sounds scary
 
It is, I can't say I understand it very well
I'm TeXing today's notes, I'll send you a link later if you want! I'll probably have more questions too
 
@BalarkaSen I think I should apply $\sigma(ab)=\sigma(a)\sigma(b)$ and $\sigma(r) < \sigma(a)$, but $\sigma$ can take zero value as well, I think!
 
11:25 AM
The integration thing I asked you about came up in defining $L^2(E)$ where $E\to M$ is an Hermitian bundle over $M$
 
@Alessandro Thanks a lot, that'd be nice
Ah I see.
Did he tell you about the Dirac operator yet
 
@BalarkaSen , what do you think?
 
@BalarkaSen Nope, what is it?
 
@SubhasisBiswas It looks off. The Sylow $q$-subgroup is normal, so there are $q - 1$ elements of order $q$.
@Alessandro I don't understand it very well. It's the "square root" of the Laplacian operator on a Riemannian manifold, but to make sense of that you need to pass to an appropriate Clifford algebra
I think you require a spin structure for that.
 
That's way over my head
 
11:32 AM
The point, there is some such operator $D$, and associated to each Riemannian manifold $M$ there is a "spectral triple" $(M, L^2(M), D)$ where $L^2(M)$ is the space of $L^2$ functions on the manifold, and $D$ is some operator on a Clifford algebra? of $L^2(M)$.
This is enough to recover the isometry type of $M$. Two spectral triples are isomorphic if and only if the corresponding Riemannian manifolds are isomorphic
 
should I proceed with my approach?
 
I think this is (1) related to "can you hear the shape of a drum?", which says the isometry type of the Riemannian manifold can be recovered from the spectrum of the Laplacian (2) a Gelfand-Naimark theorem in the Riemannian setup, recovering the manifold from the function space on the manifold
Connes has defined noncommutative Riemannian manifolds using this, as a spectral triple
@Subhasis I'm not thinking too hard about it right now. I'd do it by understanding it as a semidirect product first.
 
@BalarkaSen ok. Will try doing that way
 
Ah interesting, we started talking about spectral triples for $C^\ast$-algebras today
Which are somehow a generalization I believe
 
Oh shoot, you should teach the story to me
 
11:39 AM
I'm writing it down
We only defined a spectral triple for a $C^\ast$-algebra today, we will talk about those in two weeks (there are no classes next week)
 
This is such a weird letter
$$\Huge{𝔛}$$
 
I like it a lot
Drew a couple of those when teaching a differential geometry class, it feels very nice to write it on the board lol
 
The forward-facing diagonal stroke is very squiggly
 
Is it an "X"?
 
(Top left and bottom right)
 
11:43 AM
@Secret yeah
 
I see
 
fraktur X
@AkivaWeinberger I don't draw it with those strokes
I just do ) ( and cut it
 
-)(-
)(
Ж
Wait no
 
LOL
 
let me fix that for you:
$e^{i\frac{\pi}{2}}Ж$
 
11:45 AM
heh
 
good joke
took me a sec
 
$\Huge{\frak{P}}$
aw
is my favourite
 
@AlessandroCodenotti Do you listen to Sigur Ros
 
I haven't listen to them in a while but I do sometimes
 
Can you recommend me an album or something
I have only heard a few tracks
 
11:53 AM
Ágætis byrjun
 
thanks
 
someone's origin
I think?
Or start
 
@ÍgjøgnumMeg When I read chapter 1 of Neukirch I believed $\mathfrak{P}$ to be a weird B, so I did a presentation in my ANT class which was full of Bs while the book used Ps everywhere
 
I now wonder, what does a $\mathfrak{B}$ look like?
Oh, that looks much more like a B I must admit
 
11:54 AM
Uncannily similar
 
I just write $\underline{P}$ on a board
 
I like $\wp$
All my $\mathfrak{p}$ are like that
 
lol
Weierstrass elliptic function
$\frak p$ is a prime in an extension of $\Bbb Q$
and $\frak P$ is a prime lying above such a prime
as we ALL know
I saw a video on facebook like "every math/physics professor ever"
and one was "We'll be studying the formula D = AB, usually you'll find this written in textbooks as something like F = ma, but we'll be using this notation"
 
LOL
 
@ÍgjøgnumMeg That's great
I love the "if you think notation doesn't matter then let $x(f)=f^3$ and compute $\frac{\mathrm{d}}{\mathrm{d}f}x$."
 
12:00 PM
hahaha yeah
 
@ÍgjøgnumMeg That's exactly how Neukirch uses them
(Yes I should have guessed that above a p there's a big p rather than a big b)
 
lolol
Lang starts going on about $\frak O$ and $\frak Q$
which are nearly indistinguishable
 
wtf whats the difference
 
O and Q
 
one is O one is Q
got it
sniped
 
12:02 PM
hahaha
 
@ÍgjøgnumMeg I can totally see those being used in a book with poor printing from the 70s with random ink dots here and there
 
y e p, he uses them in his ANT book
 
sadist man
I like Mochizuki's notations tho
 
That's like using $t$ and $\tau$ in the same equation while writing on a blackboard
 
or i and $\iota$
or $v$ and $\nu$
or $w$ and $\omega$
rofl
 
12:04 PM
$\mathscr{F}^{blah}$-prime strips, where blah is a string of symbols that you see in a 90s cassette player
 
$k$ and $\kappa$ are often used together in set theory
 
unfortunately I was taking a class on PDEs during my undergrad while I was also reading about local fields and $K_v(x)$ is the notation for a Bessel function of some kind
 
There's an awful proof in Kanamori that I had to go through for my seminar talk where he has a sequence of elementary embedding $k_n$ with a sequence of critical points $\kappa_n$. Later in the proof he also introduces $k_{mn}$ and $\kappa_{mn}$
 
which we all know is the field of rational functions in $x$ over a complete local field at the place $v$
 
12:07 PM
speaking of what's the state of art for the Scholze-Stix refutation of Mochizuki
 
i'unno
 
that's like the first thing Scholze did after winning the Fields lmao
 
he had it all planned
 
win the fields medal just to sh*t on a japanese dude
Right I should probably do my job, or some loser crap like that
 
12:11 PM
@BalarkaSen I think they're still at "Scholze-Stix: this doesn't work" "Mochizuki: lmao even an undergrad would not make such mistakes like you two"
 
ugh
 
@AlessandroCodenotti I have a partner for a German-Italian tandem!
 
Lei ha detto che il mio italiano sia molto buono
studia la germanistica e la linguistica a Heidelberg, mi interessano molto
 
@MatheinBoulomenos è* (dire and affermare are more of a certainty. However "lei crede che il mio Italiano sia molto buono" would definitely be correct, while the same sentence with è would be wrong)
 
12:18 PM
@AlessandroCodenotti grazie
 
@MatheinBoulomenos Without both "la" (I can't really explain why, we usually use articles before nouns even when English doesn't, but they sound off in this sentence)
 
abbiamo scritto su Facebook, la prossima settimana mangieremo dei gelati e parleremo italiano
 
But apart from those small details it was very good
 
@AlessandroCodenotti ancora grazie
 
@MatheinBoulomenos This is correct, but "ci siamo scritti" (we wrote each other on fb) sounds better than "abbiamo scritto" (we wrote on fb)
 
12:22 PM
Sì, ho capito
 
"ci siamo scritti" is a funny construction, it looks like a reflexive verb but it doesn't mean "we wrote ourselves" (well it could also mean that, but that's unusual)
 
it's like in German where you can use reflexive pronouns to mean "each other" (it would be better to use a reprocal pronoun like "einander" though)
English is more consistent in the distinction between "each other" and "themselves"
 
@MatheinBoulomenos true
We also have some verbs that are usually used reflexively with no reflexive meaning
You have those in German too though "Du brauchst dich nicht schämen" does not mean "you don't need to embarrass yourself" rather "you don't need to be embarrassed"
 
I'd say reflexive meaning is subjective, though
 
12:43 PM
When I say "composing function $f$ with function $g$" do I mean $f\circ g$ or $g\circ f$ or is the order not really specified? If I mean $g\circ f$ does it make sense (or is it shit style) to say "post-composing $f$ with $g$"?
 
i use precompose and postcompose quite often
 
and you understand it in the same order I do?
(f postcompose g = $g \circ f$, f precompose g = $f\circ g$)
 
heh dont remember
 
haha
If its not clear what it means then its bad style and I will avoid it
 
I wouldn't said "f postcompose g", but "f postcomposed with g" is clearly $g \circ f$ to me
 
12:47 PM
just where the arrow is put, right?
before or after
no thats reverse
i usually think in terms of arrows so my conventions are reversed i guess
 
If you have $\to_f$ and want to "post-compose" with $\to_g$ then I guess $\to_f \to_g$ makes the most sense, because you first move along the arrow $f$ then the arrow $g$
but $\to_f\to_g = \to_{g\circ f}$
 
ya but that's $g \circ f$ so
 
Pre-composition with $f$ means $-\circ f$ to me
And I remember that because my AG professor always said that the induced map on coordinates ring is given by precomposing
 
how about "composing on the left" and "composing on the right"
thats clearer
 
(I sure hope I didn't mess it up now lol)
 
12:52 PM
clearly you mean the algebraic composition when you're saying something like that instead of arrows
@Alessandro $f \circ g$ is pullback of $f$ along $g$ :3
 
@BalarkaSen That works too
 
I think the temporal terminology "pre-" or "post-" is clearer than the spatial terminology "on the left/right". If you precompose $g$ with $f$, then you have to do $f$ first, so it's $g \circ f$, I don't see any ambiguity here
if you say on the left and then think of arrows, it goes wrong
 
check Washington @Balarka
 
I pronounce "$g \circ f$" as "$f$, then $g$" which makes it easy to think in terms of temporal succession
 
I agree with everything you said on this @MatheinBoulomenos
 
1:03 PM
@MatheinBoulomenos I read it as "$g$ after $f$"
 
1:41 PM
g surk f
 
1:58 PM
A 1-simplex is just.. a line right?
 
The standard 1-symplex is the segment between $(0,1)$ and $(1,0)$ in $\Bbb R^2$ (because of reasons the standard $n$-simplex $\Delta^n$ lives in $\Bbb R^{n+1}$)
A singular $n$-simplex in a topological space $X$ is a continuous function $f\colon \Delta^n\to X$, which can be quite messed up even for $n=1$ I'd say
 
bleh
This text doesn't really do a good job of explaining that, it just gives some hand wavy explanations and the defines simplicial complexes
 
@RyanUnger He didn't respond to my mail. Tell that fucker to get back I have something important to tell him!
 
2:17 PM
@BalarkaSen look he told me he's waffling
I know what the issue is but I can't tell you
 
OK I will send another email telling my important thing
 
2:30 PM
He says he’ll email back soon he says
 
3:21 PM
let $M$ be a real symmetric matrix such that for some $k \in \mathbb{N}$, $M^k=0$. Show that $M=0$
We know that if $t$ is an eigen value of $M$, then $t^p$ is an eigen value of $M^p$. Now, we know that a real symmetric matrix has only real eigen values. So, we must have $t^k=0$ as an eigen value. Which implies that $t=0$ is an eigen value of $M$.
@BalarkaSen
 
seems legit
 
which one?
 
@SubhasisBiswas You need to invoke the spectral theorem here; all the eigenvalues are 0 and it's diagonalizable.
But yes, that's correct
 
so exactly where the spectral theorem comes in?
 
You said "$t = 0$ is an eigenvalues of $M$". Why does that mean $M = 0$?
 
3:29 PM
i am reading about it right now. But.. still it's a bit fuzzy
because the constant term must be zero in the characteristic polynomial
 
So what?
The constant term in the characteristic polynomial is determinant
 
@SubhasisBiswas i was talking about the spectral theorem. Never heard of it.
 
Lots of symmetric matrices with zero determinant
 
oof then pretty hard to do this exercise
 
is it because every eigenvalue is zero?
 
3:36 PM
Yes.
 
@SubhasisBiswas you need to know the spectral theorem
 
Every real symmetric matrix $M$ is conjugate to a real diagonal matrix, and the entries of the diagonal are the eigenvalues.
If all the eigenvalues are zero, that guy is the zero matrix
Conjugate to $0$ implies $0$!
 
@RyanUnger I will learn about it.
@BalarkaSen wow.
 
@BalarkaSen do you know why Mike is writing a book for his first paper o.O
 
Why are you writing a book for your first paper? :P
 
3:48 PM
my first paper is like 6 pages...too short to actually do anything with at the momenyt
paper 2 is going to be longer
 
like about 200 pages :P
 
my thesis isn't a paper
 
this is his thesis as well
 
yeah but I don't think he has preprints
most people break the thesis up
 
idk the formalities
its good tho
i have to read bits of it
 
3:52 PM
Hi all! Just a quick question: I have Michael Artin's book Algebra and am wondering whether I should by Lang's algebra to go deeper in the subject. Or does Artin cover much of the same things as Lang does? What are your thoughts? Thanks!
 
So, i ended up with this: Since $M$ is a symmetric matrix, there is an orthogonal matrix $P$ such that $PMP^{-1}=D$, $D$ being a diagonal matrix. Now, both $D$ and $M$ has the same eigen values. All eigenvalues of $M$ must be zero, hence, the diagonal (and the entire matrix) of $D$ consists only of zero. So, $P^tDP=M=0$
 
Right.
 
@BalarkaSen You'll like the current one. Main theorem 1 has a constant depending on 7 parameters
the theorem itself involves 10 parameters
 
jesus
 
And while I'm here, I might as well ask another question that has been in my mind. I graduated from upper secondary school this spring and I wrote a so called mathematics diploma work. It's not very special, I just solved a problem I came up with myself. The paper is around 20 pages long and I'd be eager to publish it somewhere so that my work wouldn't be wasted. Do you have any suggestions where I could publish it? I wonder if I could send it to arXiv..
 
4:01 PM
@BalarkaSen these are the kinds of stupid things you have to prove to do surgery i.gyazo.com/0e754410958955ace98a64469f05cb65.png
proof "immediate"
 
4:15 PM
What is the point of duality in linear programming? What is it good for?
"The importance of duality is twofold. First, fully understanding the shadow-price interpretation
of the optimal simplex multipliers can prove very useful in understanding the implications of a particular
linear-programming model. Second, it is often possible to solve the related linear program with the shadow
prices as the variables in place of, or in conjunction with, the original linear program, thereby taking advantage
of some computational efficiencies. The importance of duality for computational procedures will become
I found this now.
 
@Miksu Lang's algebra is more of a reference text I would say
if you're learning algebra for the first time I'd steer well clear of Lang
lol
 
5:07 PM
What does it mean to integrate $f(x,y,z) = xy$ over a unit sphere centered at origin?
Anyone? @BalarkaSen ?
 
@BalarkaSen Waka Flocka released an album
it has a big room track ahha
 
@ÍgjøgnumMeg Okay, thanks. So maybe I should use my money on some other book then :D
 
5:26 PM
If I may make you aware of the following image:
aka "Hamster geometry"
 
@Miksu I swear by Dummit and Foote (there are others in this room who strongly disagree but they are nerds so ignore them)
 
why would someone disagree
 
@Ryan @Daminark certainly dislikes D&F
lol
 
@RyanUnger because they are nerds
 
why does he dislike it
 
5:37 PM
His opinion is that the exposition is boring, if I remember rightly
@Daminark speak of the devil
 
It's so boring
 
but that's algebra
is there a legitimate reason
 
I went to sleep reading it
 
blocked
 
¯_(ツ)_/¯
 
5:38 PM
But to be fair I've heard it has good problems (my professor first two quarters mostly wrote his own) and it has the content
Just wish it could take less than 6 years to get to it
 
yeah it's pretty comprehensive and has some nice problems
but the group theory section is pretty dry
 
ok nerds am I stupid
is there any simple inequality between $(x-y)^2$ and $x^2-y^2$
$x>y>0$
 
Hmm, $x-y < x+y$ so $(x-y)^2 < x^2 - y^2$ I think
 
I guess $x + y > x - y$ so $(x - y)^2 < x^2 - y^2$
lol sniped
 
5:43 PM
Lmaoooooo
 
$2y^2-2xy< 0$
 
shouldn't have typed "I guess"
 
ah yes thanks
 
@ÍgjøgnumMeg Well I already have that Algebra by Artin and I think I'm just gonna stick with it. As I've understood, D&F teaches pretty much the same stuff. Though Artin has a lot of linear algebra woven into it..
 
D&F has a bunch more stuff but yeah the LA isn't bad. Some folk seem to say that the correct thing to do is Artin -> Lang
 
5:49 PM
@Daminark Oh, okay. Thanks. So maybe I should consider buying Lang?
xD
 
test $H^{\bullet}(G, M)$
is \bullet the usual TeX for this?
lol
 
bullet is for try hards
 
rofl
what do you usually use?
$H^\cdot$
ew
 
$H^*$
no, bullet is the right thing if you want to do a dot
I'm saying doing a dot and not a star is try-hard
 
yeah Weibel is using $\cdot$
also alright hahaha
 
5:52 PM
Weibel is literally a try hard book
 
lmfao what does that even mean
so homology is $\ker d_n / \operatorname{im} d_{n+1}$ ye
 
are abelian categories even real
 
I can take a homological algebra course in the fall from Weibel
from the book
not the person
haha
 
Lol fair, I actually don't know what courses I can take yet
 
5:55 PM
oh he's at rutgers
ok $\mathcal X$ or $\mathfrak X$
 
$\frak X$
we had this discussion earlier actually lol
 
I can see on the star board
$\mathcal X$ looks too much like a swastika
I think do Carmo (?) has a $\mathcal X$ with a bar through it
or at least someone does
not mathfrak
the mathcal
 
so let $C_n = \Bbb Z/(8)$ for $n \geq 0$ and $0$ otherwise and define $d_n : C_n \to C_{n-1}$ by $x \bmod 8 \mapsto 4x \bmod 8$. Then you have a chain complex because the $d_n$ always compose to $0 \bmod 8$ and the homology at any $n$ is $\Bbb Z/(2)$ because $\ker d_n = \Bbb Z/(4)$ and $\operatorname{im} d_{n + 1} = \Bbb Z/(2)$ at any $n$... right?
(tell me if the language is wrong; does one talk about the homology at n?)
 
6:18 PM
(I suppose one should also show that the $d_n$ are $\Bbb Z/(8)$-module homomorphisms but it's obvious in this case)
 
I learned to day that the English names for the possible security clearances in Denmark are "restricted", "classified", "NATO secret" and "Cosmic top secret" (no joke, the last one is the actual official name for the top level of security clearance).
A shame I only need the second highest. Would have been cool be be able to call myself cosmic.
 
6:44 PM
People talking in LateX. What is this wonderful place?
 
7:20 PM
@Miksu sorry I was out. I wouldn't necessarily suggest buying Lang, this is hearsay and also the person probably had the idea in mind of using free pdfs (either through Springer which has many free ones up or possibly using other methods wink wink)
 
Hullo chat
 
7:46 PM
@BalarkaSen @ÉricoMeloSilva Sinai is teaching ergodic theory in the fall
 

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