Mathematics - 2018-07-12 (page 1 of 2)

ÍgjøgnumMeg

00:00

I should go to bed

I have an interview tomorrow for a job lugging corpses around a hospital

wish me luck

MatheinBoulomenos

@ÍgjøgnumMeg good luck!

I have an interview on monday for TAing intro abstract algebra

ÍgjøgnumMeg

@Mathein that sounds more fun than mine

viel glück :P

MatheinBoulomenos

it's certainly more ... lively

thanks

ÍgjøgnumMeg

h a

gotta pay off my overdraft somehow

and I don't wanna have to ask my parents for money

lol

Current student debt (according to the website): £ 42,644.31

woohoo!

MatheinBoulomenos

:O

ÍgjøgnumMeg

00:05

hahaha, well I only start paying this back when I earn a reasonable amount of money

but I also borrowed 2000 from the bank to live during my studies and I need to pay this back asap..

MatheinBoulomenos

wow

ÍgjøgnumMeg

#LoveMyCountry

MatheinBoulomenos

my parents pay for my living (and I don't feel bad about it), but also there's no tuition here

ÍgjøgnumMeg

that's cool, I live at my parents' place so I'm living from them basically

but I don't like to actually borrow from them hahaha

anywaaaay gute nacht good night god natt osv. usw. etc.

MatheinBoulomenos

@ÍgjøgnumMeg bonam noctem!

Daminark

00:13

See you @ÍgjøgnumMeg! Also good luck on student loans

MatheinBoulomenos

if I'm in the fortunate position to teach math someday, I'm going to give this as a reference, just for the reactions: amazon.com/dp/1593274130

mercio

I knew what series it was without clicking on it

Daminark

Lmaooooo

loch

wow

Balarka Sen

This is pure art

MatheinBoulomenos

00:26

@Daminark can you send me your paper when you're finished? I'd like to read it, it sounds pretty cool

Daminark

I definitely will!

Xander Henderson

@BalarkaSen That makes my skin crawl... $\ast$shudder$\ast$

anakhronizein

@MatheinBoulomenos until you yourself take a look inside and see how horrible it is.

MatheinBoulomenos

@anakhronizein it would just be for the lulz anyway

anakhronizein

As long as no one is compelled enough to buy it. :P

Daminark

01:12

Lmao, is it that bad?

I mean I probably wouldn't assign a book for a class that wasn't on libgen

(Assign meaning, if I were to have reading or problems from the book and there was no good way to do the class without having it. If it was a recommended reference but the class was self-contained maybe that's another story, though ideally still I'd have one of them be free)

MatheinBoulomenos

we don't really have classes taught from books here

everything is self-contained

in advanced lectures, sometimes results are referenced when the lecturer skips the proof

Daminark

01:55

Sorry I was out, but yeah here sometimes psets are just written as "From the book, chapter X, do problems a_1 to a_n"

Otherwise I think lectures were essentially standalone?

MatheinBoulomenos

hmm, we never have problem statements like that

loch

lol the psets for the AG class i took here was 90% "do hartshorne 2.1,2.2,2.3,2.4,3.1,3.2,... etc."

Daminark

Oh I guess also in my analysis class first quarter we had to teach ourselves LA. We had this one book which we had to read and do problems out of. Of course you could read something else but given that other books may organize very differently... You very much wanted to just read that book (also it was just a good book)

MatheinBoulomenos

the current pset in ANT is basically "prove Galois descent, all basic theorems about classical brauer groups and that it's isomorphic to some Galois cohomology" (i.e. stuff that takes up 2 chapters or more in a textbook, and the theorems of that, not the exercises)

Daminark

How long do you have to do it?

MatheinBoulomenos

02:05

2 weeks

there are some hints, so it's alright

Daminark

Ah okay that's good

I'm used to one week psets

anon

@BalarkaSen regular subgroup means acts regularly on whatever the original group was acting on

S^3 (say, left-multiplications by unit quaternions) within SO(4) acts regularly on S^3

Balarka Sen

I had to look up what a regular group action is lol i did not know that terminology. Oh but then that's just a reparse of my proof of the bundle SO(3) -> SO(4) -> S^3 degenerating

That copy of S^3 inside SO(4) acting regularly on S^3 is the section

anon

So for instance, SL(2,R) acts on upper half-plane, has subgroup H of upper triangulars which act regularly, and point-stabilizer K=SO(2), so SL(2,R) is a knit product of H and K (i.e. every elt is uniquely expressible as hk with h in H and k in K), which gives you its Iwasawa decomposition and shows SL(2,R) is R^2 x S^1

it's more or less a reparse of your statement, yes

Balarka Sen

If in general G acts on X with stabilizer K then you have a principal K-bundle K -> G -> X and if there is a regular subgroup H of G, the isomorphism X -> H by orbit-stabilizer is the section of this principal K-bundle

That proves G is isomorphic as topological groups to K x H, and topologically that's K x X

This was your general fact

anon

02:16

mmhmm

Balarka Sen

06:06

My edited answer is getting so long that SE might crash because of it

#yolo

Fargle

06:18

@Balarka, known absolute madman

Balarka Sen

06:39

5

A: Sphere eversions

The short answer is "yes" but instead I'll explain the origin of the obstruction class in $\pi_n SO(n+1)$ that determines if a given immersion $S^n \to \Bbb R^{n+1}$ can be isotoped to the standard embedding. Suppose $M^m, N^n$ are manifolds, $M$ is closed, with $m < n$. The space of $1$-jets of...

I am become sphere eversor, destroyer of stack exchange

Hawk

06:52

is any1 here

user131753

07:06

I am currently reading category theory from the book The Joy of Cats. In exercise 3H(d) (page 45) we are required to show that an equivalence is an embedding if and only if it reflects identities. However, I have shown that a full and faithful functor is an embedding iff it reflects identities. Is it true?

user131753

@LeakyNun or @BalarkaSen any ideas?

jakes

math.stackexchange.com/questions/2848363/… Any clues?

Piyush Divyanakar

07:32

Does anyone know any continuous probability distributions that have a constant median.

and are also bounded.

mick

10:13

0

Q: $ \int_1^{\infty} \int_1^{\infty} f(x,y) \space dx \space dy $ special case?

Consider $$ A = \int_1^{\infty} \int_1^{\infty} f(x,y) \space dx \space dy $$ $$ B = \int_1^{\infty} f(x,y) \space dx $$ $$ C = \int_1^{\infty} f(x,y) \space dy $$ $$ D = \int_1^{\infty} \int_1^{\infty} f(x,y) - f(y,x) \space dx \space dy $$ Where $A,B,C$ converge to a positive real valu...

Edited

Secret

10:34

If the audience does not exist, then I don' exist

MatheinBoulomenos

12:18

@user170039 yes, that's true

user131753

@MatheinBoulomenos I see, thanks.

MatheinBoulomenos

note that every full and faithful functor is an equivalence to its essential image, so basically what you showed and the exercise statement are equivalent

user131753

@MatheinBoulomenos I have recently made a post on MSE representing my proof of it. In case you are interested, you are welcome to write an answer for it to remove it from the list of unanswered questions.

user131753

@MatheinBoulomenos What is the definition of an essential image?

MatheinBoulomenos

If $F:\mathbf{A} \to \mathbf{B}$ is a functor then the essential image is the full subcategory of $\mathbf{B}$ of all objects $B \in \mathbf{B}$ such that there exists an object $A \in \mathbf{A}$ such that $F(A) \cong B$

a functor is essentially surjective iff the essential image is $\mathbf{B}$

if you haven't covered essentially surjective functors yet, feel free to ignore my remark

user131753

12:29

@MatheinBoulomenos I haven't yet covered it because they are not introduced in The Joy of Cats at least to the point where I read.

user131753

@MatheinBoulomenos In fact I haven't met even with the notion of subcategories yet.

MatheinBoulomenos

@user170039 your proof is fine

user131753

@MatheinBoulomenos I see. Thanks for checking it.

ÍgjøgnumMeg

@Mathein hallo!

MatheinBoulomenos

@ÍgjøgnumMeg hello

ÍgjøgnumMeg

12:34

@Mathein how's it going? Did you have your interview?

MatheinBoulomenos

no, it's on monday

ÍgjøgnumMeg

oh I see lol

MatheinBoulomenos

how did your interview go?

ÍgjøgnumMeg

yeah it was alright, the job was a different one than I expected though and I'm not sure I want it

so I might just see if I can find a bar to work at

hahaha

also, on a lighter note, on the way back from the city centre I happened to notice that some of my lecturers had organised a conference on moduli spaces

so I sat in on some talks

MatheinBoulomenos

@user170039 The Joy of Cats introduces essentially surjective under a different name: they call it isomorphism-dense

user131753

12:37

@MatheinBoulomenos I see.

MatheinBoulomenos

@ÍgjøgnumMeg that sounds really cool

ÍgjøgnumMeg

@Mathein the first 4 or 5 minutes were cool since I understood what was being talked about

lol

MatheinBoulomenos

lol

I listened to colloquium talks like that

okay maybe I understood 10 minutes

ÍgjøgnumMeg

haha yeaaah

it was nice tho

there was just

a lot of weird technical restrictions placed on some integers and then a wall of homological algebra

user131753

@MatheinBoulomenos Would you mind to clarify this point?

MatheinBoulomenos

12:41

would you mind if I introduce subcategories and essential images while doing so?

user131753

@MatheinBoulomenos No problem.

MatheinBoulomenos

okay, so a subcategory $\mathbf{A}$ of a category $\mathbf{B}$ is kind of what you would expect: it's a category such that $Ob(\mathbf{A}) \subset Ob(\mathbf{B})$ and for each $A,B \in Ob(\mathbf{A})$, you have $\mathrm{hom}_{\mathbf{A}}(A,B) \subset \mathrm{hom}_{\mathbf{B}}(A,B)$ and the identities in $\mathbf{A}$ are the same as the ones in $\mathbf{B}$ and composition is just restricted from the composition in $\mathbf{B}$

For that for any subclass $X \subset Ob(\mathbf{B})$ we can just define a subcategory $\mathbf{X}$ of $\mathbf{B}$ by saying that $Ob(\mathbf{X}) := X$ and $\mathrm{hom}_{\mathbf{X}}(A,B) := \mathrm{hom}_{\mathbf{B}}(A,B)$ for all $A,B \in X$

and we take the same composition as in $\mathbf{B}$ restricted to the objects in $X$

that's called the full subcategory with objects $X$

note that inclusion of objects and morphisms gives an embedding $\mathbf{A} \to \mathbf{B}$

for any subcategory $\mathbf{A}$ of $\mathbf{B}$

if that embedding is full, then $\mathbf{A}$ is called a full subcategory

that's where the name "the full subcategory with objects $X$" comes from

suppose you have a functor $F:\mathbf{A} \to \mathbf{B}$ and say that $F$ is full for simplicity then we define the essential image of $F$ to be the full subcategory with the following class of objects $X:=\{B \in Ob(\mathbf{B}) \mid \exists A \in Ob(\mathbf{A}), F(A) \cong B$

not that a full and faithful functor $F:\mathbf{A} \to \mathbf{B}$ is isomorphism-dense iff the essential image is just $\mathbf{B}$ (this is just a restatement of the definition)

are you following? @user170039

user131753

@MatheinBoulomenos Yeah. Sure.

MatheinBoulomenos

12:57

okay suppose $F$ is fully faithful and let's denote the essential image by $\mathrm{im}_{ess}(F)$, then $F$ factors over the inclusion $\mathrm{im}_{ess}(F) \hookrightarrow \mathbf{C}$

now by definition, the factorized version $F: \mathbf{A} \to \mathrm{im}_{ess}(F)$ is full and faithful and isomorphism-dense, so an equivalence

so if we suppose we know that an equivalence reflects identities iff it is an embedding, we get that $F:\mathbf{A} \to \mathrm{im}_{ess}(F)$ is an embedding iff it reflects isomorphisms

user131753

@MatheinBoulomenos Sorry to interrupt but what is $\mathbf{C}$?

MatheinBoulomenos

ah lol

that's supposed to be a $\mathbf{B}$, sorry

too late to edit

user131753

@MatheinBoulomenos I see. Then it's fine.

MatheinBoulomenos

but since the inclusion $\mathrm{im}_{ess}(F) \hookrightarrow \mathbf{B}$ is obviously an embedding and reflects identities, and the original $F$ is the composition $\mathbf{A} \to \mathrm{im}_{ess}(F) \hookrightarrow \mathbf{B}$, we can get the statement that reflecting identities is equivalent to embedding for general full and faithful functors

the other direction is of course clear since equivalences are full and faithful

that's what I meant by saying that the two versions with "equivalences" or just "full and faithful" are basically equivalent

so the idea is just to throw away all objects that prevent the full and faithful functor from being an equivalence

like you can get a surjection from every function by restricting to the image

okay? @user170039

user131753

@MatheinBoulomenos How do we get that? If we assume that an equivalence reflects identities iff it is an embedding then (since a fully faithful functor reflect isomorphisms) we can conclude that an embedding reflects isomorphisms. But how do we get the converse?

MatheinBoulomenos

13:07

ah lol

forget that I wrote isomorphisms

corrected version:
"so if we suppose we know that an equivalence reflects identities iff it is an embedding, we get that $F:\mathbf{A} \to \mathrm{im}_{ess}(F)$ is an embedding iff it reflects identities"

user131753

@MatheinBoulomenos Now it's ok.

user131753

(To me at least.)

MatheinBoulomenos

I agree that you don't use isomorphism-dense anywhere in the proof so it's bit strange that they included it in the exercise statement

user131753

@MatheinBoulomenos Yes. That's what motivated me to write the post. I thought that I must be missing something.

mick

0

Q: How to solve $ f\left(\sqrt[3]{1 - z^3}\,\right)^2 = 1 - f(z)^2 $?

How to Find analytic $f(z)$ such that $$ f\left(\sqrt[3]{1 - z^3}\,\right)^2 = 1 - f(z)^2 $$ Koenigs function can not be used here So I am stuck. How does the riemann surface look like ?

dynamical-systems functional-equations riemann-surfaces complex-dynamics

user131753

13:20

@MatheinBoulomenos So if we try to flesh out the details of this comment I think what you meant is something as follows: Let $F:\mathbf{A}\to\mathbf{B}$ be a fully faithful functor then since $F$ can be written as the composition of two embeddings, i.e. $H\circ G$ where $G:\mathbf{A}\to\operatorname{im}_{ess}(F)$ and $H: \operatorname{im}_{ess}(F)\to\mathbf{B}$ and since both of them reflects identities, $F$ also reflects identities.

MatheinBoulomenos

right

user131753

@MatheinBoulomenos So the key observation is that any full and faithful functor can be written as the composition of an equivalence and an embedding.

MatheinBoulomenos

and the other direction works as well

no wait

that's not true

any full and faithful functor is the composition of an equivalence and an embedding

$\mathbf{A} \to \mathrm{im}_{ess}(F)$ need not be an embedding

user131753

@MatheinBoulomenos Yeap. Noted that.

MatheinBoulomenos

I think that's a nice statement to have in general, so it's good you pointed that out

user131753

13:25

@MatheinBoulomenos In my comment here also the edit should be done accordingly.

MatheinBoulomenos

Note that $\mathrm{im}_{ess}(F) \to \mathbf{B}$ reflects identities by construction

so we can use this decomposition to deduce the statement from the exercise for general full and faithful functors from the (a priori weaker) statement just for equivalences, that's why the formulations are "basically equivalent" in some vague sense

user131753

@MatheinBoulomenos However if $F$ is a full and faithful functor that reflects identities then can we show that $F$ is an equivalence?

user131753

Only assuming that if $F$ is an equivalence then $F$ is an embedding iff it reflects identities.

user131753

@MatheinBoulomenos I don't think they are logically equivalent.

user131753

13:45

Are you there @MatheinBoulomenos?

user193319

If $\{a_{n_k}\}$ is some subsequence of $\{a_n\} \subseteq \Bbb{R}$, is it true that $\limsup a_{n_k} = \limsup a_n$?

Alessandro Codenotti

Think about a sequence with positive limsup and a constant subsequence of zeroes

user131753

@user193319 Consider the sequence $a_n:=(-1)^n$ for all $n\in\mathbb{N}$.

user193319

Shoot...Thanks!

user131753

Anyway @MatheinBoulomenos thank you very much for your detailed explanation.

Balarka Sen

14:09

lmfao

user193319

@user170039 Is it possible to prove this if $\lim_{n \to \infty} a_n$ exists, without using the fact that this implies $\liminf a_n = \lim_{n \to \infty} = \limsup a_n$ (I am trying to prove this latter fact)?

user131753

@MatheinBoulomenos: I take back my comment. The reason is the following argument.

user131753

Let $F$ be a full and faithful functor that reflects identities. We claim that $G$ is an embedding.

user131753

Since $G$ is an equivalence, to prove this it it enough to show that $G$ reflects identities. So let $G(k)$ be an identity for some $\mathbf{A}$-morphism $k$. Since $H$ is a functor, it follows that $H(G(k))$ is an identity. Since $H\circ G=F$ it implies that $F(k)$ is an identity.

user131753

Since $F$ reflects identity, it implies that $k$ is an identity. Hence $G$ reflects identities. Consequently $G$ is an embedding. Since the composition of two embeddings is an embedding, it follows that $F$ is an embedding and we are done.

user131753

14:19

@user193319 Are you asking whether it is possible to prove that $\operatorname{limsup} a_{n_k}=\operatorname{limsup} a_{n}$ if $(a_n)$ is convergent and if we assume that $\operatorname{liminf} a_{n}=\operatorname{lim}_{n\to\infty}a_n=\operatorname{limsup} a_{n}$?

user193319

Yes to everything, except we are not assuming that $\operatorname{liminf} a_{n}=\operatorname{lim}_{n\to\infty}a_n=\operatorname{limsup} a_{n}$, because I am trying to prove this.

Leaky Nun

Croatia didn't exist the last time England reached semi-finals

user131753

14:34

@user193319 The only method that I can think of it to prove the result by first proving that $\limsup a_n=\displaystyle\lim_{n\to\infty} a_n=\liminf a_n$ iff $(a_n)$ is convergent.

user193319

@user170039 That's what I am having trouble with...The best I can do is show there exists $\{a_{n_k}\}$ s.t. $a_{n_k} \le \sup_{i \ge n_k} a_i < a_{n_k} + \epsilon$, where $\epsilon > 0$ is arbitrary. Letting $k \to \infty$ and $\epsilon \to 0$ gives, if you assume $\lim_{n \to \infty} a_n$ exists, $\lim_{n \to \infty} a_n \le \limsup a_{n_k} \le \lim_{n \to \infty} a_n$. If I knew that $\limsup a_{n_k} = \limsup a_n$, then I would be done.

user131753

15:12

@user193319 Instead of trying to show that $\limsup a_{n_k}=\limsup a_n$ try to show that $\limsup a_n=\displaystyle\lim_{n\to\infty} a_n$.

user131753

Observe that $\limsup a_n\ge \displaystyle\lim_{n\to\infty} a_n$. What happens if the inequality is strict?

Abcd

15:36

$$f(x)= \int_0^1e^{|t-x|}dt$$, $0\le x \le 1$

Please explain how to find f(x).

Leaky Nun

break it into two regions

Abcd

@LeakyNun Do we have to consider x as a constant?

Leaky Nun

sure

Abcd

@LeakyNun How do I know if t is greater than x or less than it

Leaky Nun

when t is greater than x, t is greater than x

otherwise, t is less than x

$$f(x) = \int_0^1 e^{|t-x|} \mathrm dt = \int_0^x e^{|t-x|} \mathrm dt + \int_x^1 e^{|t-x|} \mathrm dt$$

that's the first step

Abcd

15:39

yes, I know that.

Leaky Nun

in the first integral $\displaystyle \int_0^x e^{|t-x|} \mathrm dt$

$t$ ranges from $0$ to $x$

so what is the sign of $t-x$?

Abcd

got it, thanks

Leaky Nun

ok

MatheinBoulomenos

15:56

@user170039 no

@user170039 yes, that's what I had in mind

Leaky Nun

@MatheinBoulomenos servus

MatheinBoulomenos

@LeakyNun moin

Perturbative

Hey everyone

MatheinBoulomenos

hey @Perturbative

Perturbative

Hey @MatheinBoulomenos :)

Okay so quick question, how can I show that the bouquet of circles, $$\bigvee_{i=1}^n \mathbb{S}_i^1$$ is Hausdorff? I know that $\bigvee_{i=1}^n \mathbb{S}_i^1$ is defined as the quotient of the disjoint union of $n$ copies of $\mathbb{S}^1$ and the disjoint union will be Haursdorff because $\mathbb{S}^1$ is, but quotienting doesn't preserve Haursdoffness so what argument could I use here?

I guess I could do it by the definition, but there must be an easier way to prove it

MatheinBoulomenos

16:11

So by induction, it's enough to treat the case $A \vee B$ for Hausdorff spaces $A$ and $B$

Leaky Nun

you don't need induction

everything is locally finite

MatheinBoulomenos

fine, treat the case $\vee_{i \in I} X_i$ for any index set $I$

Leaky Nun

it's not only locally finite, it's locally two

MatheinBoulomenos

but you either use the definition or embed into the product yeah

both is easy

Balarka Sen

for any two points away from the wedge point floop it, at the wedge point you have flower with petals, shrink shrink flip flip

ez

Leaky Nun

16:14

you only need T1 if you have two points belonging to distinct sets and are not the wedge point

Perturbative

Ahh okay, so the wedge products of Hausdorff spaces is Hausdorff, cool cool

MatheinBoulomenos

@BalarkaSen ...

Balarka Sen

@MatheinBoulomenos That's how you prove obvious facts

MatheinBoulomenos

nah

Alessandro Codenotti

@BalarkaSen lmao

Balarka Sen

16:27

Do you find the new search bar in the main site annoying?

Secret

It's manageable

rschwieb

"new" as in how new?

Balarka Sen

idk, a few days ago the search bar still vanished when i scrolled down

rschwieb

I haven't seen any differences lately..

Balarka Sen

Could be that I haven't actually been active in the past few months lol

Alessandro Codenotti

16:33

@BalarkaSen I only noticed that now that you pointed it out! But it could have been here for a while

Balarka Sen

yeah i dont remember it doing that until recently

Clarinetist

I reported a design issue here. I would probably report yours there with a screenshot if I were you

Kari

18:19

I've seen other definitions of (0,2) tensors but none involve a function multiplied by a tangent vector in the linearity bit of the definition

Is this definition even correct?

Mike Miller

Yes, that should be in the definition of an (m,n) tensor!

It is what lets you say that a tensor is a "smoothly varying linear operation", whose output (an m-vectoe field) at each point only depends on the input at each point (an n-vector field). For you, that says the number g(x,y) should depend only on two vectors you choose at each point, not the vector fields on a neighborhood.

Balarka Sen

I think I am finally able to draw the full evolution of the singularity locus of a sphere eversion.

Mike Miller

Contrast that with the Lie derivative operator, for which the value of [X, Y] at any point depends on a germ of Y near that point.

rschwieb

@BalarkaSen Great conversation starter! I'll have to remember it...

Balarka Sen

lmao

I was going to offer pictures for whoever interested, but maybe not... :P

Leaky Nun

18:34

hi @BalarkaSen

Balarka Sen

\o

Leaky Nun

@BalarkaSen with MSPaint or with words?

Balarka Sen

Using a pen and paper :P

Leaky Nun

that's better than words

MatheinBoulomenos

@rschwieb I did use conversation starters that are on that level on my math peers

Balarka Sen

18:37

Well, the one-line description using words seem to be "introduce a circle's worth of double points, pass that circle transversely through itself (so it makes a figure eight at some point - the figure eight point is the famous quadruple point of the sphere eversion), then kill the circle"

But that misses finer details of the eversion

rschwieb

Yeah but what do people do at bus stops when you say that

MatheinBoulomenos

I'll have to try that

Mike Miller

@BalarkaSen Here is a goal if you want to understand the inversion well. You know the standard picture of Boy's surface in R^3. If you could regularly homotope your immersion of a 2-sphere to the double cover of RP^2, then it is clearly evertible: thinking of S^2 as the unit sphere bundle of the tautological line bundle, you're just scaling by t, as t goes from 1 to -1.

Balarka Sen

I told you that story :P

MatheinBoulomenos

"Hey, do you know the proof for balancing Ext just using universal delta functors without double complexes? Oh and does this bus line go to the university square?"

Mike Miller

18:44

Oh, you did? Can you do it?

Kari

@MikeMiller I'm a bit confused as to what you mean by the input being an n-vector field and the output being an m

Balarka Sen

Ya but not with Boy's surface. There is an immersion of RP^2 in R^3 with fourfold rotational symmetry whose tautological circle bundle is Morin's surface

So that gives the standard sphere eversion

Kari

I thought the inputs to a type (m,n) tensor were both m co-vector fields and n vector fields

MatheinBoulomenos

Did you know that advanced ANT over function fields is apparently used by terrorists?

Terrorists Using Drinfeld Modules

►

Drinfeld module

In mathematics, a Drinfeld module (or elliptic module) is roughly a special kind of module over a ring of functions on a curve over a finite field, generalizing the Carlitz module. Loosely speaking, they provide a function field analogue of complex multiplication theory. A shtuka (also called F-sheaf or chtouca) is a sort of generalization of a Drinfeld module, consisting roughly of a vector bundle over a curve, together with some extra structure identifying a "Frobenius twist" of the bundle with a "modification" of it. Drinfeld modules were introduced by Drinfeld (1974), who used them to prove...

rschwieb

@MatheinBoulomenos Stay tuned for the next exciting episode of $2^3\cdot 3$.

loch

18:53

i have a possibly dumb number theory question - if I have a number fields $L/\mathbb{Q}, K/\mathbb{Q}$ where $L$ is an extension of $K$, is there anything that can be said about the class numbers of $L$ and $K$? (e.g. if the class number of $L$ should be bigger than the one of $K$)

I think the answer is no?

MatheinBoulomenos

@loch yeah the answer is no

class numbers are quite mysterious objects

Balarka Sen

@MikeMiller In any case, that eversion is theoretically nice but it's not transverse to itself at t = 0

loch

hmmmmm ok

Balarka Sen

I specifically wanted to understand the evolution of the singularity locus for a self-transverse immersion

where transversality is tricky to define when image of singular sets are itself singular, but think of it as stratified transversality

MatheinBoulomenos

@loch you can say something in very special cases via Iwasawa theory, but that's quite deep and difficult

loch

19:03

yeah i've tried to sit in a seminar on iwasawa theory once or twice (then i gave up) - but im vaguely aware of something like that in iwasawa theory :p

Mike Miller

@Kari A covector is the same as a functional on vectors: it takes a vector and spits out a number, linearly in the vector.

MatheinBoulomenos

@loch you can write down homomorphisms between class groups in both directions, induced by extension of scalars and the norm map, but these are neither injective nor surjective in general, so they're not helpful

rschwieb

@MikeMiller Ah, a crisp definition of something I see explained in four excrutiating pages of detail in a physics textbook.

Mike Miller

A covector field is the same as a 1-form. You plug in a vector field X and it spits out a function. And because that operation is computed pointwise, and pointwise you know that $f(\lambda v) = \lambda f(v)$, then you agree that the 1-form has $\omega(fX) = f \omega(X)$.

loch

hmm i see. for a moment i thought maybe there's something to be said (perhaps in nice cases) because i think we can identify the class group with the galois group of the hilbert class field, but i think in general the hilbert class fields of L and K are not going to be related (?)

Mike Miller

19:07

(Just applying the same result pointwise.)

MatheinBoulomenos

@loch yeah, there's no relation really I think. You can't go from an unramified abelian extension of $L$ to one of $K$ or the other way around. Unramified behaves nicely in towers, so if $L/K$ is unramified, there might be chance, but towers of abelian extensions need not be abelian (even Galois)

Well, wait if $L/K$ is unramified abelian, then there's a relation

Mike Miller

@BalarkaSen Very cool. Emmy asked me the same and I forgot you'd once told me the answer

MatheinBoulomenos

at least then the Hilbert class field of $K$ is an unramified abelian extension of $L$, so contained in the Hilbert class field of $L$

Balarka Sen

@MikeMiller i read that Tony Phillips knew this story before Smale wrote his paper

Mike Miller

Hehe

MatheinBoulomenos

19:16

Let $H_K$ be the Hilbert class field for $K$, then $c_K=[H_K:K]$ and we get that $[H_L:L]=[H_L:H_K][H_K:L]$, so $c_L[L:K]=[H_L:L][L:K]=[H_L:H_K][H_K:K]=[H_K:H_L]c_K$

that's some relation I guess

loch

ah yes

Mike Miller

@rschwieb well, I have skipped the laborious definition of vectors..

loch

interesting nonetheless :p

MatheinBoulomenos

$c_K$ divides $c_L [L:K]$

Kari

@MikeMiller I thought that a 1 form took a vector field in and spat out a scalar $$ T^{\star}\mathcal{M} \ni \omega : T \mathcal{M} \to \mathbb{K} $$

MatheinBoulomenos

19:17

but assuming that $L/K$ is abelian unramified is quite specialized, of course

Balarka Sen

I appended my answer with more stuff. Apparently the regular homotopy classes of immersions of $S^3$ in $\Bbb R^4$ is generated by the movie of the 2-sphere eversion in R^3, and twice that movie.

I can prove the first dude is a nontrivial class but I don't know a good way to see the latter is what it is.

(This is partly why I was asking about $SO(4) \simeq SO(3) \times S^3$)

Mike Miller

@Kari a 1-form defines a map of vector bundles $TM \to \Bbb R$, which means at every point you have a map $TM_x \to \Bbb R$. A section of the first thing is a vector field, a section of the second thing (the trivial bundle) is a function. Applying a 1-form to a vector field gives you a function.

Put differently: if $dx$ is the usual 1-form on $\Bbb R$ that sends $\partial/\partial x$ to $1$ everywhere, what does $dx$ send $x \partial/\partial x$?

Abcd

How do I compute $$\lim_{n \to \infty}\displaystyle \sum_{r=0}^n\left(\dfrac{1}{(r+1)(1+2r)}\right)$$

MatheinBoulomenos

@loch if $L/K$ is Galois, then it's a theorem that the quotient of Dedekind zeta functions $\zeta_L/\zeta_K$ is entire (this actually follows from some not that difficult representation theory of finite groups, I think the rep theory part is an exercise in Serre, the case that $L/K$ is not Galois is an open conjecture), so maybe one can do something with the class number formula in that case

but there are so many parameters in the class number formula that a simple relation seems hopeless

iirc there's an expression for $\zeta_L/\zeta_K$ as a product of some simpler stuff, but I don't know the formula right now and maybe it's hard to evaluate at $1$ (special values of L-functions are not exactly easy in general)

That seems interesting actually, maybe I should write out the details for that

loch

hmm that's interesting i didn't know about the quotient of zeta functions result.

Abcd

19:34

Never mind, asking on main.

MatheinBoulomenos

@loch if the values of the L-functions whose product gives $\zeta_L/\zeta_K$ at $1$ are known, then this implies some formula that relates $c_L$ and $c_K$, but it's probably going to be at least as complicated as the class number formula itself (if not more complicated), unless there are some magical cancellations

Kari

19:51

@MikeMiller Ah, according to the $\omega(fX) = f\omega(X)$, we'd have that $\text{d}x (x \partial_{x}) = x$

Mike Miller

@Kari Yeah, but even better, we only need to apply that pointwise! At $x_0 = \pi$, say, the vector field is $\pi \partial/\partial x$, and because $(dx)_{\pi}$ is linear and eats $\partial/\partial x$ to spit out 1, this should spit out $\pi$.

Kari

Yep, I noticed that!

What did you mean by the Lie derivative comment, @MikeMiller?

1 hour ago, by Mike Miller

It is what lets you say that a tensor is a "smoothly varying linear operation", whose output (an m-vectoe field) at each point only depends on the input at each point (an n-vector field). For you, that says the number g(x,y) should depend only on two vectors you choose at each point, not the vector fields on a neighborhood.

1 hour ago, by Mike Miller

Contrast that with the Lie derivative operator, for which the value of [X, Y] at any point depends on a germ of Y near that point.

Mike Miller

That's what I wanted to say about the tensors earlier - pointwise linearity implies that we can pull functions out as scalars, like in your definition of a metric

Well the point for covector fields (or tensors more generally) was that we eat the vector fields pointwise, right? $\omega(V) = \omega_x(V_x)$.

But there are other kinds of operators that depend on more data than that, for instance taking the derivative of a function in the direction of a vector field

(Vf)(x) depends only on $V_x$ and the value of $f$ in a neighborhood of $x$, but just knowing $f(x)$ is not enough

Kari

What does $\omega(V) = \omega_{x}(V_x)$ mean?

Mike Miller

But the formula for $V(fg)$ is correspondingly more complicated

I have to go and don't have time to continue this right now, sorry.

Kari

20:04

Ah right, I'll try do some more reading to figure it out!

loch

@MatheinBoulomenos yeah i imagine it'd be complicated

number theory is hard

Kari

Thanks for all the help, @MikeMiller!

Really appreciate it :-)

Leaky Nun

20:20

@loch why is (x,y) a subvariety of (x^2,xy) but (x,y-1) is not a subvariety?

loch

1) it's probably a better idea to say something like $V(x,y)$ instead of just writing the ideal
2) it is!

Leaky Nun

I see

loch

something that's not going to be a subvariety is the closed subscheme defined by the ideal $(y)$

Leaky Nun

but that's no subscheme

loch

?

Leaky Nun

20:25

why is that a subscheme?

loch

oh, I meant the ideal $(y) k[x,y]/(x^2,xy)$ in $k[x,y]/(x^2,xy)$

MatheinBoulomenos

closed subschemes of an affine scheme correspond to ideals

Leaky Nun

oh ok

rschwieb

20:44

@LeakyNun Afternoon... I felt like I didn't see you all day!

Sam

Evening all

Or morning... or afternoon...

Can someone please help me understand the following derivative formula screenshots.firefox.com/vTX6vloz09twVMgT/cs231n.github.io.

ÍgjøgnumMeg

Officially have a first class honours degree in maths!

11

dabbing

Daminark

Hey everyone!

Congrats @ÍgjøgnumMeg!

ÍgjøgnumMeg

@Daminark hey man

Sam

I'm not sure how the >= is used or what is returned

@ÍgjøgnumMeg Congrats dude

ÍgjøgnumMeg

20:47

@Sam @Daminark thanks :)

Sam

Where did you study?

ÍgjøgnumMeg

@Sam Plymouth! Are you at uni in the UK?

Sam

I graduated last year :D

ÍgjøgnumMeg

oooh nice, where from?

Sam

Swansea

ÍgjøgnumMeg

20:48

Nise, I've not been

Sam

In computer science not Maths. Otherwise I'd know that derivative formula ;)

ÍgjøgnumMeg

Ha fair, I'm unfamiliar with the notation

it's probably

Sam

Yeah I'm a bit confused with the logical notation

ÍgjøgnumMeg

$\frac{\partial f}{\partial x} = 1$ if $x \geq y$

Sam

else 0?

In that case, I'd assume the logical returns a 1 if true and 0 if not true, then the 1 is multiplied by its result?

MatheinBoulomenos

20:52

@ÍgjøgnumMeg congrats!

Hi @Daminark

ÍgjøgnumMeg

@Mathein thanks man!

@Sam I think the (x >= y) is a condition

@Sam as in, if $x \geq y$ then $\partial f/\partial x = 1$

Sam

what if y >= x. What would the deriviative of the function w.r.t. x be then?

ÍgjøgnumMeg

@Sam because if $x > y$ then $f(x , y) = x$

Sam

ahhh

ÍgjøgnumMeg

@Sam if $y > x$ then $f(x, y) = y$ so $\partial f/\partial y = 1$

and in the cases where $x = y$ you just have $f(x,y) = x = y$ so both derivatives are $1$

Sam

20:57

So if (y>x) is df/dx = 0 ?

ÍgjøgnumMeg

@Sam well if $y > x$ then $f(x,y) = y$ so yeah

because there is no $x$ dependence

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