12:05 AM
@TedShifrin you are a good mind reader.

@jacksoja that relation is far from uniquely determining r,s

12:30 AM
@anakhro LOL, gee, thanks.

2 hours later…
2:47 AM
For m even, why $$\sum_{n=1}^{m}\left|x\sin\left(\frac{2\pi n}{m}\right)+y\cos\left(\frac{2\pi n}{m}\right)\right| < 2\cot\left(\frac{\pi}{m}\right)$$ iff $(x,y)$ is inside the regular m-gon?

4 hours later…
6:37 AM
Please let me know if I am correct here: maximal ideal implies prime ideal need not hold for commutative ring without unity: Consider ring $2\Bbb Z$. Here $4\Bbb Z$ maximal ideal, but not prime ideal.

6:53 AM
@Silent That's right

7:15 AM
thanks :)

I seem to get negative entropy values for the normal distribution with sufficiently low variance. This is clearly wrong, right? But since one of the terms in the entropy calculation is is $\log{2\pi\e\sigma^2}$, it will go negative once \sigma^2 < 1/(2\pi\e) .
Please let me know what I'm missing

7:50 AM
I gather that it's just because it's differential entropy, where there isn't a bound at zero.

8:10 AM
Radoslav Rochallyi (born 1 May 1980, Bardejov , Czechoslovakia) is a Slovak philosopher, writer and poet. == Biography == Rochallyi was born in Bardejov, located in the Prešov region of what is today the Slovak Republic. The author finished his studies in Philosophy at the Faculty of Arts of the University of Prešov (1999–2005) and completed postgraduate PhD studies in pedagogy at the Faculty of Education of the University of Prešov (2005–2009). He worked as a teacher at the Pedagogical Faculty of Prešov University in Prešov. == Writer == Rochallyi is the author of eleven books. He wri...

3 hours later…
11:30 AM
@ted Ok I agree that the group action is discontinuous, but I don't understand how this doesn't disprove the statement I had about partial applications, since both $g\mapsto g\cdot f$ and $f\mapsto g\cdot f$ look very continuous to me

@Astyx what is the statement

That the action of $\Bbb R/\Bbb Z$ on $C(\Bbb R/\Bbb Z)$ by translation isn't continuous
I have a proposition that tells me that when V is a barreled space a group morphism $\pi:G\to Aut(V)$ is a continuous representation iff $g\mapsto \pi(g)(v)$ and $v\mapsto \pi(g)(v)$ are continuous
The argument is that $\sup_{\vert f\vert =1}\vert g\cdot f - f\vert >1$ whenever $g\ne id$

@Astyx in that proposition, what is the topology on $\mathrm{Aut}(V)$?
the representation $\Bbb R/\Bbb Z \to \mathrm{Aut}(C(\Bbb R/\Bbb Z))$ is discontinuous if you put the operator norm topology on the latter, but that topology makes only sense for normed spaces, so I doubt that's the topology on $\mathrm{Aut}(V)$ in the proposition

11:47 AM
$V$ is normed in the proposition, forgot to add that

@LukasHeger C(X) is a normed space for X compact, I don't get your point

So I think it is operator norm topology (it isn't specified and I don't see what else it could be)

Ah I get your point now
Anyway yes it's surely about the operator norm

I'll be right back

12:26 PM
@Lukas went in to my old Uni to say hi to my old lecturers and now I’m giving a talk about some fun basic results in the theory of modular forms for undergrads lol

@EdwardEvans nice

My lecturer said „you’re a postgrad student, you don’t get free time“

lol

So I’ll probably talk about the spaces of forms as graded algebras and their structure theorems and then talk about weird congruences in the Fourier coefficients or smth
I‘m gonna be handwaving quite intensely lol

just talk about the associated Galois representations of modular forms
and how that is a form of global Langlands for $\mathrm{GL}_2$

12:29 PM
Rofl I’m sure that’ll be fine

realistically, you can't do Hecke theory, so you can just do elementary consequences of dimension computations
i.e. comparing the Fourier coefficients in the equality $E_4^2=E_8$ or something
or you can use Theta series and talk about stuff like how you can prove the four squares theorem with modular forms
(handwaving very heavily)

Yeah that’s what I was thinking
And other things like tau(n) = sigma_11(n) mod (691?)

yeah you can do that

I don’t think any of them know what a graded algebra is though, so I guess I’ll quickly define those in the talk as well
Most of them just about know what a vector space is lol

do they know holomorphic maps, fourier series?

1:01 PM
yeah I believe so
they have a first course in complex analysis
and they should be studying fourier series this semester

@MikeMiller Dang, nice. (Re: talk)

1:49 PM
@BalarkaSen i'm drowning in the sea of definitions
with $\overline\partial$-operators and connections and covariant derivatives

@LeakyNun Then try to organize those definitions and then push forward to see what they're good for
@BalarkaSen vOv

push forward?

like a sheaf

2:20 PM
If $0\to A\to B\to C\to 0$ is exact, what are conditions to guarantee that $0\to A/D\to B/f(D)\to C\to 0$ is exact, where $D\subseteq A$ is a submodule and $f:A\to B$?

if I know $log_2(a)$ and $log_2(b)$ can I calculate, or put bound on $log_2(a+b)$ ?

I was thinking about proving the exactness of the sequence in reduced homology, but the general case seems interesting

2:43 PM
if you know $\log_2(a)$ and $\log_2(b)$, you know $a$ and $b$, so you also know $\log_2(a+b)$

1 hour later…
3:57 PM
@Thorgott consider $a$ and $b$ as arbitrary size binary numbers. I know number of binary digits for both numbers (hence $log_2(a)$ and $log_2(b)$ is known). I am trying to put a maximum size on $log_2(a+b)$ so that I can know how large the resulting number can be. I plan to use this result to allocate space in computer program for the result of summation.
my guess is that $log_2(a + b) <= log_2(a) + log_2(b)$

consider $a=b=1$

if both of a, b are greater than 2 then $a+b≤ a\cdot b$ and so $\log(a+b) ≤ \log(a)+\log(b)$ by monotonicity

1 hour later…
5:06 PM
If $a,b$ are commuting elements, and neither $a \in \langle b \rangle$ nor $b \in \langle a \rangle$, then may I conclude that $\langle a,b \rangle \cong \langle a \rangle \times \langle b \rangle$.

How does $ln(2x^5)$ turn into $5ln(2^{1/5}x)$

Hello please does $(A_1\cup A_2) \times (B_1\cup B_2) = (A_1\times B_1 ) \cup( A_1\times B_2)\cup (A_2\times B_1 ) \cup (A_2\times B_2 )$
someone here ?

Oh i figured it out, nvm

5:21 PM
????

ok
@Astyx hello

hi

I hope that you are fine
please do you remember this result of product of sets

@user193319 Fundamental Theorem of finitely generated abelian groups might be your friend, there

5:25 PM
This would follow from $(A_1\cup A_2)\times B = A_1\times B\cup A_2\times B$
Do you think that's true ?

yes I think

@user193319 take $2,3\in\mathbb{Z}$, these commute and $2\not\in3\mathbb{Z}$,$3\not\in2\mathbb{Z}$, but $\langle2,3\rangle\cong\mathbb{Z}$ by Bézout, yet $\langle2\rangle\times\langle3\rangle\cong\mathbb{Z}^2$

@Thorgott Oh, very nice!

your conclusion follows under the stronger hypothesis that $\langle a\rangle\cap\langle b\rangle=\{e\}$

Ah! I think that might be the case for this proof I'm working through.

5:32 PM
@Astyx ce n'EST pas in produit cartesian

Comment ça ?

$x\in AB$ veut dire que x=ab

???? On parle des ensembles ou plutôt des groupes?

des ensemble de R

mais $A\times B\subset\Bbb R\times\Bbb R$?

5:37 PM
$AB=\{ab, a\in A, b\in B \}$

sans $\times$

you have $\langle a\rangle\langle b\rangle=\langle a,b\rangle$ and $\langle a\rangle,\langle b\rangle\triangleleft\langle a,b\rangle$ since the elements commute, so if $\langle a\rangle\cap\langle b\rangle=\{e\}$, $\langle a,b\rangle$ is an internal direct product of $\langle a\rangle$ and $\langle b\rangle$

non c'est x du produit cartésien on reste dans R

alors, ce n'est pas un produit cartésien.
Ce sont deux choses complètement différentes.
@LeakyNun What course are you taking?

est ce que je peux dire que
$(A_1\cup A_2) . (B_1\cup B_2) = (A_1. B_1 ) \cup( A_1. B_2)\cup (A_2 . B_1 ) \cup (A_2. B_2 )$
@TedShifrin

5:45 PM
Hi @Ted

Il ne faut que prendre la définition de $\cup$
hi, demonic @Alessandro

How did the discussion about the action of R/Z on C(S^1) not being continuous end? I'm dealing with a very similar problem today...

??? je ne comprends pas \cup c'est l'union

oui, évidemment. Est quelle est la définition de $C\cup D$?

@Alessandro I asked my prof and it turns out the topology for being a continuous representation isn't the same as the one for being a continuous action, which is what was bugging me

5:48 PM
Namely that a C0-semigroup is not necessarily continuous in norm

@Astyx: Yeah, it was that operator norm thing .... So Demonark's and my solution were right, right?

@Astyx that's exactly what I said
that you have different topologies on $\mathrm{Aut}(V)$

Yup, Lukas gets credit, too. :D

@TedShifrin c'est la réunion des éléments des deux ensembles

19 hours ago, by Lukas Heger
I don't think putting the operator norm on $\mathrm{Hom}(V,V)$ gives you the right notion for a continuous action on a normed space $V$. In general a continuous action corresponds to a continuous map $G \times V \to V$ which is equivalent to a continuous map $G \to C(V,V)$ where we put the compact-open topology on the latter. Now I don't think the compact-open topology restricts to the operator norm

5:51 PM
Donc? $(A_1\cup A_2).B =$?

A_1.B \cup A_2 . B

Alors, je ne comprends plus votre question.

je voulais juste mon assurer

Yes you were right. I failed to see that being continuous $G\to Aut(V)$ and being continuous $G\times V\to V$ is not the same thing

@TedShifrin Riemann surfaces

5:56 PM
@Leaky: Pretty fancy with covariant derivatives. Really? What context?

@TedShifrin ^
I'm a bit confused
they just called it a connection
now they call it a covariant derivative

Yeah, he's talking about connections on holomorphic line bundles.
Same thing.
A connection is given by a (specific) covariant derivative, and vice versa.
The point is that there's no canonical way to differentiate sections of a line bundle unless you have a global trivialization thereof.
Is this a second course in Riemann surfaces? Things have gone awfully fast for just a few weeks.

this is a first course I think... yeah it's a bit fast
It's taught by Tomasz Mrowka

Yeah, it's super fast. Ah, he was an MIT undergrad when I was a postdoc there. Say hi :P

6:03 PM
Is he following a text?

I don't know; there are not a lot of information about this course that I can find
every link on the stellar page redirects me to the l-mod page
and there are only three tabs on the l-mod page
calender, assignment, material
and there are only lecture notes in the material tab

Hmm, there's no syllabus or anything. Weird. Times have changed :P

and his personal page doesn't even mention this course so...

Anyway, it's reasonable to get to such matters later in the course. You certainly need to know differential forms and basic differentiable manifolds.
I usually taught Riemann surfaces as the end of my complex analysis course, so I didn't presume that background.

Are you still at Imperial College @Leaky ?

6:06 PM
I'm in MIT now
@TedShifrin I'll send you an email

Ok right
Makes sense

Anyhow, @Leaky, if i can I'm happy to answer questions for you.
@Astyx: If you're on vacation, how come you're bugging your professor ? :D

I'm mean >:)

Well, we knew that!

sent

6:08 PM
Have you heard anything from the rest of the Paris gang we got together with a few years ago?

Nope, not at all

Hmm, I guess I shouldn't visit again. :(

Well to be fair I didn't know about them before either

@TedShifrin should an undegrad thesis be self-contained?
or can I just cite a paper

Of course you can cite papers, books, etc. But you should write more expository stuff than would be appropriate for a published research paper, of course.

6:12 PM
hell, one exercise wrote "the space of $m \times n$ matrices, $\operatorname{Hom}(\Bbb R^m, \Bbb R^n)$"
@TedShifrin have you received my email?

So the letters are reversed.
The point is that the bundle is smoothly but not holomorphically trivial.
What does "difference $(0,1)$ form" mean?

exactly
two $\overline \partial$ operators differ by a (0,1) form, I've been told

(What's a $\overline\partial$ operator ?)

I guess he wants you to see how the $\bar\partial$ operator varies with $\sigma$, since you get different holomorphic structures on the line bundle.

it's an operator from the space of (0,0)-sections to the space of (0,1)-sections satisfying the Leibniz rule

6:17 PM
If you have $\Bbb C$ you can split $d=\partial + \bar\partial$, where $$df = \frac{\partial f}{\partial z}dz + \frac{\partial f}{\partial\bar z}d\bar z.$$

Ah, ok

and I don't know which two $\overline\partial$ operators I'm supposed to consider

On a complex manifold the complex structure gives you a similar splitting.
The point is that on a holomorphic line bundle $\mathscr L$, $\bar\partial$ is uniquely defined, no choices, because the transition functions of the bundle are holomorphic. So when you do a change of trivialization, you have $$s_\alpha = \phi_{\alpha\beta}s_\beta.$$ Apply $\bar\partial$ and you get $$\bar\partial s_\alpha = \phi_{\alpha\beta}\bar\partial s_\beta,$$
so you have a well-defined $(0,1)$-form with values in $\mathscr L$.

@Ted is Dolbeault cohomology sheaf cohomology of the sheaf of holomorphic $p$-forms? I think this should be obvious from the $\overline{\partial}$ Poincaré lemma and the fact that sheaf of smooth $p,q$-forms is fine

@TedShifrin right
they call it $\Lambda^{1,0} \otimes \mathcal L$

6:21 PM
so basically the same proof that De Rham cohomology is sheaf cohomology of the constant sheaf

yeah, @Lukas, that's very useful, but still important to understand Dolbeault cohomology as it is :P
@Leaky, no $\Lambda^{0,1}\otimes\mathscr L$.

right
now what is the second $\overline\partial$ operator?

Oh, I see what he's saying. It's poorly written.

ugh

He wants the $\bar\partial$ operator for the bundle versus the $\bar\partial$ operator for the (smoothly) trivial bundle.

6:23 PM
I don't know what the latter is (read: I don't think he defined it)

Well, how did you show the bundle was smoothly trivial?

I constructed a nonvanishing smooth section

OK, so the trivial connection comes from just declaring that $\bar\partial$ of that section is $0$ (and then use the Leibniz rule in general).

ok...

So what stops your smooth section from being holomorphic?

6:26 PM
the fact that it is not, lol
or rather

How do I know? :D

the fact that its $\overline\partial$ is not zero?

@TedShifrin does the Frölicher spectral sequence work for local systems?

the happy spectral sequence?

Agh, @Lukas, I don't berember.
So @Leaky did you use bump functions or something?

6:28 PM
no
fortunately

I guess I'm not guessing how you did the exercise. What is the section?

ok I'll send you

LOL

sent
ich bin ein froehlicher Spectralsequenz

6:47 PM
@TedShifrin did you receive my email?

Yeah, it seems right. Of course, you could write the section in terms of $z$ and $\bar z$ to continue ...

1 hour later…
8:01 PM
tom is not going to write a detailed summary of his course lol

8:52 PM
@TedShifrin sanity check: the quotient of a (compact) Riemann surface by a finite group of automorphisms is again a Riemann surface, right?

If it acts freely and properly discontinuously.
Otherwise, you may have singularities in the quotient.

hmm

Think about acting on the disk by rotation by $\pi$.
You have so-called orbifold singularities.

@TedShifrin Yeah but you still get a perfectly good complex structure don't you

Not if it's not a manifold?

8:56 PM
C mod the action of z mapsto z^k is still homeomorphic to C

I don't know what that means.

Which is what the group action looks like at a fixed point
It's the same logic as the symmetric products of a surface being complex manifolds

Curve :)

fair nuff
I could be being stupid right now but I think Mathein is right

Yeah, I guess you're probably right.
<--- hides in corner
And I wasn't asking for a detailed summary of Tom's course. Just a course outline or vague list of topics/text/etc. I guess I at least put such things on the board when I taught advanced graduate courses.

8:59 PM
at the level of algebra, this is saying that C[z] is isomorphic as algebras to C[z^k], I think, if I remember my GIT
@TedShifrin Oh I'm not saying that's unreasonable

Yeah, you're right that things are special in dimension 1.

Just that Tom won't do it

LOL, yeah.
I actually have no idea what kind of lecturer/teacher he turned out to be.

I find him enlightening but I doubt he's ever had notes prepared more than a day in advance

Forgive me my algebraic nonsense argument, but my motivation behind asking this was this idea: the category of compact Riemann surfaces is equivalent to the opposite category f.g. field extensions of $\Bbb C$. Now by Artin's lemma from intro Galois theory, if $L$ is a field and $G$ is a finite group of automorphisms, then $L/L^G$ is a finite Galois extension, so the category admits categorical quotients by actions of finite groups.
Now I find it unlikely that the Riemann surface associated to $L^G$ is anything other than the quotient of the Riemann surface associated to $L$ by the correspo

9:01 PM
jesus

I knew you'd like that :P

On that note, bye.

as far as i can tell you are just rediscovering GIT but for the easy case of curves

I mean $\mathcal{M}(X)^G=\mathcal{M}(X/G)$ seems like it should be true
what is GIT?

for an affine scheme X with ring of function O(X), the moral of GIT is that "O(X/G)" should be precisely O(X)^G
geometric invariant theory

9:03 PM
ah I see

but what X/G is for non-affine guys takes some work

for projective curves over any field you can just argue with their function field
so yeah this is just a really easy case
bye @Ted
@MikeMiller I'm really trying to find a conceptual proof that modular curves are Riemann surfaces without explicitly looking at charts around cusps

9:37 PM
0

After reading about Ackermann functions , tetration and similar, I considered the commutative following hyperoperator ? $$F(0,a,b) = a + b$$ $$F(n,c,0) = F(n,0,c) = c$$ $$F(n,a,b) = F(n-1,F(n,a-1,b),F(n,a,b-1))$$ I have not seen this one before in any official papers. Why is this not con...

Any ideas ?

what is an example where one can understand the advantage of the median over the mean(average). I read the wikipedia article and it says, that the average is skewed by extremely large(or small) values, but I don't understand why I should prefer the median over the mean.

Say you're in a country of 1000 citizens where everyone has 1 dollar except one person that has a billion dollar. The average wealth is approximately 1 million, but that's way higher than the median which is 1 dollar.

@Astyx merci beaucoup :)

De rien
La mediane et la moyenne apportent toutes les deux une information sur les données, mais elles suffisent rarement à en avoir un bon aperçu
Typiquement pour présenter les données on va utiliser des diagrammes en boite à moustache où on représente les quartiles, et la médiane ce qui donne déjà un bon aperçu de la répartition des données

9:54 PM
@Astyx Les deux sont donc nécessaires pour obtenir un aperçu?

Ce qu'il y a de dangereux c'est qu'on peut présenter la moyenne pour dire "ce pays est vraiment riche" alors qu'en réalité il y a une énorme inégalité sociale et la majeure partie de la population est très pauvre
@SAJW je suppose que mon message plus haut répond (du moins partiellement) à ta question ?

@Astyx communism intensifies

@Astyx yes it answers my question! If 1000 citizens have 1 dollar except one who has 1 million than that would be for "average" a rich population, where in reality only one guy is rich and 999 are poor.

@LeakyNun I believe in socialism

@Astyx I don't speak french fluently but I remember you being french :D

10:04 PM
Oh sorry :p did you understand what I said or would you like me to translate it ?

@Astyx it was easy enough for gtranslate ;) no worries, i think I got it

Cool !

That leaves the question, when to use the mean then?
it would be of no advantage to say the average person has 1.2 babies...

Well it kind of would. If the average number of children a woman has in a country is below 2, you know that this country's population is decreasing

ah!

10:10 PM
For instance in Africa I think the average woman has 4.9 children, while it is at 1.6 in Europe

So one has to observe both, the mean and the median to come to a good conclusion?

It depends how relevant both are. For birthrate, it's ok to take the mean because a woman can only ever have so many children. Other data sets are more complicated. It's not an easy task to analyze data, which is why the job "data analyst" exists

So: the statement "the average(read mean) winning hand of poker is two pair" is to be taken with a grain of salt?

10:26 PM
I'm not sure what the context is, but I guess this means that if you have two pairs you can consider your hand to have a chance to win

1 hour later…
11:30 PM
@Astyx bonne nuit

À toi aussi

11:44 PM
I'm reading something and they talk about $Rt^n$ being the standard grading of $R[t]$, what does that mean ?

The ring is graded by the degree of the polynomial.

Why is it called standard ?

A ring $R=\oplus R_m$ is graded when $R_m\cdot R_n\subset R_{m+n}$.

Right

Because someone might decide to do it differently.

11:46 PM
Ok, it's that simple
Thank you

LOL, sure.
You could lump all even stuff together and all odd stuff together, for example.

Ah ok, like a superalgebra

Yeah.