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Adam
Adam
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
 
skullpatrol
skullpatrol
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
 
Adam
Adam
Jeffory if you continue to be disruptive you will get another shot so help me god
 
skullpatrol
skullpatrol
The real trump is in the uk hoping to open a new casino there
 
Adam
Adam
haha um yeah I think all of this material is gold but what every play some violin woos
what ever*
 
skullpatrol
skullpatrol
5:27 AM
yup
 
Adam
Adam
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…
ÍgjøgnumMeg
ÍgjøgnumMeg
7:08 AM
Mornin'
 
eranreches
eranreches
Hello
 
ÍgjøgnumMeg
ÍgjøgnumMeg
lol
it's pronounced more like "mey", soz
 
Adam
Adam
7:25 AM
ah right I am expecting to see a removed unit in the stream but im not
 
ÍgjøgnumMeg
ÍgjøgnumMeg
wat
Go do some maths
4
 
Secret
Secret
We should start teaching bees the millienun prize problems
 
ÍgjøgnumMeg
ÍgjøgnumMeg
Biemann Hypothesis
schnarf schnarf
 
 
3 hours later…
Albas
Albas
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?
 
Balarka Sen
Balarka Sen
$\nabla Y$ 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.
 
Albas
Albas
10:35 AM
Ahhh jeez, why didn't I see that.
 
Albas
Albas
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 ?
 
Balarka Sen
Balarka Sen
A multilinear map from the product, or a linear map from the k-fold tensor product, yes.
 
Albas
Albas
Interesting, never saw tensors like this
Then the torsion becomes a (2,1) tensor I guess.
 
Balarka Sen
Balarka Sen
That's right
That's one interesting tensor, by the way
 
Albas
Albas
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?
 
Balarka Sen
Balarka Sen
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.
 
Albas
Albas
11:10 AM
I see, also why "torsion" ? Anything to do with the torsion group?
 
Subhasis Biswas
Subhasis Biswas
hello
 
Albas
Albas
Yea that example of a rotating American football.
 
Balarka Sen
Balarka Sen
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
 
Alessandro Codenotti
Alessandro Codenotti
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?
 
Balarka Sen
Balarka Sen
@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
 
Alessandro Codenotti
Alessandro Codenotti
11:14 AM
@BalarkaSen Is the thing after the square root a differential form or just saying that you're integrating $n$ variables?
 
Balarka Sen
Balarka Sen
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
 
Alessandro Codenotti
Alessandro Codenotti
Oh ok, perfect, thanks
 
Silent
Silent
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 $\sigma(r)<\sigma(a)$.
 
Alessandro Codenotti
Alessandro Codenotti
We did a bunch of stuff today in noncommutative geometry using diffgeo and function analysis concepts I'm not familiar with...
 
Silent
Silent
(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?
 
Balarka Sen
Balarka Sen
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
 
Subhasis Biswas
Subhasis Biswas
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 $n(o(q))=pq-q-1$
 
Alessandro Codenotti
Alessandro Codenotti
@BalarkaSen K-homology
 
Balarka Sen
Balarka Sen
Sounds scary
 
Alessandro Codenotti
Alessandro Codenotti
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
 
Silent
Silent
@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!
 
Alessandro Codenotti
Alessandro Codenotti
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 $M$
 
Balarka Sen
Balarka Sen
@Alessandro Thanks a lot, that'd be nice
Ah I see.
Did he tell you about the Dirac operator yet
 
Subhasis Biswas
Subhasis Biswas
@BalarkaSen , what do you think?
 
Alessandro Codenotti
Alessandro Codenotti
@BalarkaSen Nope, what is it?
 
Balarka Sen
Balarka Sen
@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.
 
Alessandro Codenotti
Alessandro Codenotti
That's way over my head
 
Balarka Sen
Balarka Sen
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
 
Subhasis Biswas
Subhasis Biswas
should I proceed with my approach?
 
Balarka Sen
Balarka Sen
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.
 
Subhasis Biswas
Subhasis Biswas
@BalarkaSen ok. Will try doing that way
 
Alessandro Codenotti
Alessandro Codenotti
Ah interesting, we started talking about spectral triples for $C^\ast$-algebras today
Which are somehow a generalization I believe
 
Balarka Sen
Balarka Sen
Oh shoot, you should teach the story to me
 
Alessandro Codenotti
Alessandro Codenotti
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)
 
Secret
Secret
This is such a weird letter
$$\Huge{𝔛}$$
 
Balarka Sen
Balarka Sen
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
 
Akiva Weinberger
Akiva Weinberger
The forward-facing diagonal stroke is very squiggly
 
Secret
Secret
Is it an "X"?
 
Akiva Weinberger
Akiva Weinberger
(Top left and bottom right)
 
Balarka Sen
Balarka Sen
11:43 AM
@Secret yeah
 
Secret
Secret
I see
 
Balarka Sen
Balarka Sen
fraktur X
@AkivaWeinberger I don't draw it with those strokes
I just do ) ( and cut it
 
Akiva Weinberger
Akiva Weinberger
-)(-
)(
Ж
Wait no
 
Balarka Sen
Balarka Sen
LOL
 
Secret
Secret
let me fix that for you:
$e^{i\frac{\pi}{2}}Ж$
 
Akiva Weinberger
Akiva Weinberger
11:45 AM
heh
 
Balarka Sen
Balarka Sen
good joke
took me a sec
 
ÍgjøgnumMeg
ÍgjøgnumMeg
$\Huge{\frak{P}}$
aw
is my favourite
 
Balarka Sen
Balarka Sen
@AlessandroCodenotti Do you listen to Sigur Ros
 
Alessandro Codenotti
Alessandro Codenotti
I haven't listen to them in a while but I do sometimes
 
Balarka Sen
Balarka Sen
Can you recommend me an album or something
I have only heard a few tracks
 
Alessandro Codenotti
Alessandro Codenotti
11:53 AM
Ágætis byrjun
 
Balarka Sen
Balarka Sen
thanks
 
ÍgjøgnumMeg
ÍgjøgnumMeg
someone's origin
I think?
Or start
 
Alessandro Codenotti
Alessandro Codenotti
@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
 
ÍgjøgnumMeg
ÍgjøgnumMeg
rofl
 
Alessandro Codenotti
Alessandro Codenotti
I now wonder, what does a $\mathfrak{B}$ look like?
Oh, that looks much more like a B I must admit
 
Balarka Sen
Balarka Sen
11:54 AM
Uncannily similar
 
ÍgjøgnumMeg
ÍgjøgnumMeg
I just write $\underline{P}$ on a board
 
Balarka Sen
Balarka Sen
I like $\wp$
All my $\mathfrak{p}$ are like that
 
ÍgjøgnumMeg
ÍgjøgnumMeg
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"
 
Balarka Sen
Balarka Sen
LOL
 
Alessandro Codenotti
Alessandro Codenotti
@ÍgjøgnumMeg That's great
I love the "if you think notation doesn't matter then let $x(f)=f^3$ and compute $\frac{\mathrm{d}}{\mathrm{d}f}x$."
 
ÍgjøgnumMeg
ÍgjøgnumMeg
12:00 PM
hahaha yeah
 
Alessandro Codenotti
Alessandro Codenotti
@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)
 
ÍgjøgnumMeg
ÍgjøgnumMeg
lolol
Lang starts going on about $\frak O$ and $\frak Q$
which are nearly indistinguishable
 
Balarka Sen
Balarka Sen
wtf whats the difference
 
ÍgjøgnumMeg
ÍgjøgnumMeg
O and Q
 
Balarka Sen
Balarka Sen
one is O one is Q
got it
sniped
 
ÍgjøgnumMeg
ÍgjøgnumMeg
12:02 PM
hahaha
 
Alessandro Codenotti
Alessandro Codenotti
@ÍgjøgnumMeg I can totally see those being used in a book with poor printing from the 70s with random ink dots here and there
 
ÍgjøgnumMeg
ÍgjøgnumMeg
y e p, he uses them in his ANT book
 
Balarka Sen
Balarka Sen
sadist man
I like Mochizuki's notations tho
 
Alessandro Codenotti
Alessandro Codenotti
That's like using $t$ and $\tau$ in the same equation while writing on a blackboard
 
ÍgjøgnumMeg
ÍgjøgnumMeg
or i and $\iota$
or $v$ and $\nu$
or $w$ and $\omega$
rofl
 
Balarka Sen
Balarka Sen
12:04 PM
$\mathscr{F}^{blah}$-prime strips, where blah is a string of symbols that you see in a 90s cassette player
 
Alessandro Codenotti
Alessandro Codenotti
$k$ and $\kappa$ are often used together in set theory
 
Balarka Sen
Balarka Sen
page 6
 
ÍgjøgnumMeg
ÍgjøgnumMeg
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
 
Alessandro Codenotti
Alessandro Codenotti
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 $\kappa_{mn}$
 
ÍgjøgnumMeg
ÍgjøgnumMeg
which we all know is the field of rational functions in $x$ over a complete local field at the place $v$
 
Balarka Sen
Balarka Sen
12:07 PM
speaking of what's the state of art for the Scholze-Stix refutation of Mochizuki
 
ÍgjøgnumMeg
ÍgjøgnumMeg
i'unno
 
Balarka Sen
Balarka Sen
that's like the first thing Scholze did after winning the Fields lmao
 
ÍgjøgnumMeg
ÍgjøgnumMeg
rofl
 
Balarka Sen
Balarka Sen
he had it all planned
 
ÍgjøgnumMeg
ÍgjøgnumMeg
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
 
Alessandro Codenotti
Alessandro Codenotti
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"
 
Balarka Sen
Balarka Sen
ugh
 
MatheinBoulomenos
MatheinBoulomenos
@AlessandroCodenotti I have a partner for a German-Italian tandem!
 
Alessandro Codenotti
Alessandro Codenotti
Ottimo!
 
MatheinBoulomenos
MatheinBoulomenos
Lei ha detto che il mio italiano sia molto buono
studia la germanistica e la linguistica a Heidelberg, mi interessano molto
 
Alessandro Codenotti
Alessandro Codenotti
@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)
 
MatheinBoulomenos
MatheinBoulomenos
12:18 PM
@AlessandroCodenotti grazie
 
Alessandro Codenotti
Alessandro Codenotti
@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)
 
MatheinBoulomenos
MatheinBoulomenos
abbiamo scritto su Facebook, la prossima settimana mangieremo dei gelati e parleremo italiano
 
Alessandro Codenotti
Alessandro Codenotti
But apart from those small details it was very good
 
MatheinBoulomenos
MatheinBoulomenos
@AlessandroCodenotti ancora grazie
 
Alessandro Codenotti
Alessandro Codenotti
@MatheinBoulomenos This is correct, but "ci siamo scritti" (we wrote each other on fb) sounds better than "abbiamo scritto" (we wrote on fb)
 
MatheinBoulomenos
MatheinBoulomenos
12:22 PM
Sì, ho capito
 
Alessandro Codenotti
Alessandro Codenotti
"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)
 
MatheinBoulomenos
MatheinBoulomenos
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"
 
Alessandro Codenotti
Alessandro Codenotti
@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"
 
MatheinBoulomenos
MatheinBoulomenos
I'd say reflexive meaning is subjective, though
 
s.harp
s.harp
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 $g$"?
 
Balarka Sen
Balarka Sen
i use precompose and postcompose quite often
 
s.harp
s.harp
and you understand it in the same order I do?
(f postcompose g = $g \circ f$, f precompose g = $f\circ g$)
 
Balarka Sen
Balarka Sen
heh dont remember
 
s.harp
s.harp
haha
If its not clear what it means then its bad style and I will avoid it
 
MatheinBoulomenos
MatheinBoulomenos
I wouldn't said "f postcompose g", but "f postcomposed with g" is clearly $g \circ f$ to me
 
Balarka Sen
Balarka Sen
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
 
s.harp
s.harp
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 $\to_f\to_g = \to_{g\circ f}$
 
Balarka Sen
Balarka Sen
ya but that's $g \circ f$ so
 
Alessandro Codenotti
Alessandro Codenotti
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
 
Balarka Sen
Balarka Sen
how about "composing on the left" and "composing on the right"
thats clearer
 
Alessandro Codenotti
Alessandro Codenotti
(I sure hope I didn't mess it up now lol)
 
Balarka Sen
Balarka Sen
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
 
Alessandro Codenotti
Alessandro Codenotti
@BalarkaSen That works too
 
MatheinBoulomenos
MatheinBoulomenos
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
 
Alessandro Codenotti
Alessandro Codenotti
check Washington @Balarka
 
MatheinBoulomenos
MatheinBoulomenos
I pronounce "$g \circ f$" as "$f$, then $g$" which makes it easy to think in terms of temporal succession
 
s.harp
s.harp
I agree with everything you said on this @MatheinBoulomenos
 
Alessandro Codenotti
Alessandro Codenotti
1:03 PM
@MatheinBoulomenos I read it as "$g$ after $f$"
 
ÍgjøgnumMeg
ÍgjøgnumMeg
1:41 PM
g surk f
 
ÍgjøgnumMeg
ÍgjøgnumMeg
1:58 PM
A 1-simplex is just.. a line right?
 
Alessandro Codenotti
Alessandro Codenotti
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
 
ÍgjøgnumMeg
ÍgjøgnumMeg
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
 
Balarka Sen
Balarka Sen
@RyanUnger He didn't respond to my mail. Tell that fucker to get back I have something important to tell him!
 
Ryan Unger
Ryan Unger
2:17 PM
@BalarkaSen look he told me he's waffling
I know what the issue is but I can't tell you
 
Balarka Sen
Balarka Sen
OK I will send another email telling my important thing
 
Ryan Unger
Ryan Unger
2:30 PM
He says he’ll email back soon he says
 
Subhasis Biswas
Subhasis Biswas
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
 
Ryan Unger
Ryan Unger
seems legit
 
Subhasis Biswas
Subhasis Biswas
which one?
 
Balarka Sen
Balarka Sen
@SubhasisBiswas You need to invoke the spectral theorem here; all the eigenvalues are 0 and it's diagonalizable.
But yes, that's correct
 
Subhasis Biswas
Subhasis Biswas
so exactly where the spectral theorem comes in?
 
Balarka Sen
Balarka Sen
You said "$t = 0$ is an eigenvalues of $M$". Why does that mean $M = 0$?
 
Subhasis Biswas
Subhasis Biswas
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
 
Balarka Sen
Balarka Sen
So what?
The constant term in the characteristic polynomial is determinant
 
Subhasis Biswas
Subhasis Biswas
@SubhasisBiswas i was talking about the spectral theorem. Never heard of it.
 
Balarka Sen
Balarka Sen
Lots of symmetric matrices with zero determinant
 
Ryan Unger
Ryan Unger
oof then pretty hard to do this exercise
 
Subhasis Biswas
Subhasis Biswas
is it because every eigenvalue is zero?
 
Balarka Sen
Balarka Sen
3:36 PM
Yes.
 
Ryan Unger
Ryan Unger
@SubhasisBiswas you need to know the spectral theorem
 
Balarka Sen
Balarka Sen
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 $0$!
 
Subhasis Biswas
Subhasis Biswas
@RyanUnger I will learn about it.
@BalarkaSen wow.
 
Ryan Unger
Ryan Unger
@BalarkaSen do you know why Mike is writing a book for his first paper o.O
 
Balarka Sen
Balarka Sen
Why are you writing a book for your first paper? :P
 
Ryan Unger
Ryan Unger
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
 
Balarka Sen
Balarka Sen
like about 200 pages :P
 
Ryan Unger
Ryan Unger
my thesis isn't a paper
 
Balarka Sen
Balarka Sen
this is his thesis as well
 
Ryan Unger
Ryan Unger
yeah but I don't think he has preprints
most people break the thesis up
 
Balarka Sen
Balarka Sen
idk the formalities
its good tho
i have to read bits of it
 
Miksu
Miksu
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!
 
Subhasis Biswas
Subhasis Biswas
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, $P^tDP=M=0$
 
Balarka Sen
Balarka Sen
Right.
 
Ryan Unger
Ryan Unger
@BalarkaSen You'll like the current one. Main theorem 1 has a constant depending on 7 parameters
the theorem itself involves 10 parameters
 
Balarka Sen
Balarka Sen
jesus
 
Miksu
Miksu
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..
 
Ryan Unger
Ryan Unger
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"
 
Mats Granvik
Mats Granvik
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.
 
ÍgjøgnumMeg
ÍgjøgnumMeg
@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
 
Ajay Mishra
Ajay Mishra
5:07 PM
What does it mean to integrate $f(x,y,z) = xy$ over a unit sphere centered at origin?
Anyone? @BalarkaSen ?
 
Ryan Unger
Ryan Unger
@BalarkaSen Waka Flocka released an album
it has a big room track ahha
 
Miksu
Miksu
@ÍgjøgnumMeg Okay, thanks. So maybe I should use my money on some other book then :D
 
s.harp
s.harp
5:26 PM
If I may make you aware of the following image:
aka "Hamster geometry"
 
ÍgjøgnumMeg
ÍgjøgnumMeg
@Miksu I swear by Dummit and Foote (there are others in this room who strongly disagree but they are nerds so ignore them)
 
Ryan Unger
Ryan Unger
why would someone disagree
 
ÍgjøgnumMeg
ÍgjøgnumMeg
@Ryan @Daminark certainly dislikes D&F
lol
 
Subhasis Biswas
Subhasis Biswas
@RyanUnger because they are nerds
 
Ryan Unger
Ryan Unger
why does he dislike it
 
ÍgjøgnumMeg
ÍgjøgnumMeg
5:37 PM
His opinion is that the exposition is boring, if I remember rightly
@Daminark speak of the devil
 
Daminark
Daminark
It's so boring
 
Ryan Unger
Ryan Unger
but that's algebra
is there a legitimate reason
 
Daminark
Daminark
I went to sleep reading it
 
ÍgjøgnumMeg
ÍgjøgnumMeg
blocked
 
Subhasis Biswas
Subhasis Biswas
¯_(ツ)_/¯
 
Daminark
Daminark
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
 
ÍgjøgnumMeg
ÍgjøgnumMeg
yeah it's pretty comprehensive and has some nice problems
but the group theory section is pretty dry
 
Ryan Unger
Ryan Unger
ok nerds am I stupid
is there any simple inequality between $(x-y)^2$ and $x^2-y^2$
$x>y>0$
 
Subhasis Biswas
Subhasis Biswas
hmmm
 
Daminark
Daminark
Hmm, $x-y < x+y$ so $(x-y)^2 < x^2 - y^2$ I think
 
ÍgjøgnumMeg
ÍgjøgnumMeg
I guess $x + y > x - y$ so $(x - y)^2 < x^2 - y^2$
lol sniped
 
Daminark
Daminark
5:43 PM
Lmaoooooo
 
Subhasis Biswas
Subhasis Biswas
$2y^2-2xy< 0$
 
ÍgjøgnumMeg
ÍgjøgnumMeg
shouldn't have typed "I guess"
 
Ryan Unger
Ryan Unger
ah yes thanks
 
Miksu
Miksu
@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..
 
Daminark
Daminark
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
 
Miksu
Miksu
5:49 PM
@Daminark Oh, okay. Thanks. So maybe I should consider buying Lang?
xD
 
ÍgjøgnumMeg
ÍgjøgnumMeg
test $H^{\bullet}(G, M)$
is \bullet the usual TeX for this?
lol
 
Ryan Unger
Ryan Unger
bullet is for try hards
 
ÍgjøgnumMeg
ÍgjøgnumMeg
rofl
what do you usually use?
$H^\cdot$
ew
 
Ryan Unger
Ryan Unger
$H^*$
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
 
ÍgjøgnumMeg
ÍgjøgnumMeg
yeah Weibel is using $\cdot$
also alright hahaha
 
Ryan Unger
Ryan Unger
5:52 PM
Weibel is literally a try hard book
 
ÍgjøgnumMeg
ÍgjøgnumMeg
lmfao what does that even mean
so homology is $\ker d_n / \operatorname{im} d_{n+1}$ ye
 
Ryan Unger
Ryan Unger
are abelian categories even real
 
ÍgjøgnumMeg
ÍgjøgnumMeg
lol
 
Ryan Unger
Ryan Unger
I can take a homological algebra course in the fall from Weibel
from the book
not the person
haha
 
ÍgjøgnumMeg
ÍgjøgnumMeg
Lol fair, I actually don't know what courses I can take yet
 
Ryan Unger
Ryan Unger
5:55 PM
oh he's at rutgers
ok $\mathcal X$ or $\mathfrak X$
 
ÍgjøgnumMeg
ÍgjøgnumMeg
$\frak X$
we had this discussion earlier actually lol
 
Ryan Unger
Ryan Unger
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
 
ÍgjøgnumMeg
ÍgjøgnumMeg
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?)
 
ÍgjøgnumMeg
ÍgjøgnumMeg
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)
 
Tobias Kildetoft
Tobias Kildetoft
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.
 
mlane
mlane
6:44 PM
People talking in LateX. What is this wonderful place?
 
Daminark
Daminark
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)
 
Astyx
Astyx
Hullo chat
 
Ryan Unger
Ryan Unger
7:46 PM
@BalarkaSen @ÉricoMeloSilva Sinai is teaching ergodic theory in the fall
 
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