Mathematics - 2018-05-15 (page 2 of 3)

Secret

10:01

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Tobias Kildetoft

@Secret Just stop that already.

Secret

blogs.ams.org/matheducation/2018/05/14/…

Mathematics is a human construct and thus it is strongly tied to culture and politics

It is interesting how everything is related to politics now

Alessandro Codenotti

10:19

Hi @Tobias

Tobias Kildetoft

@AlessandroCodenotti Hi

Secret

finally, activity, this is so boring

mercio

oO

Alessandro Codenotti

I have an algebraic number theory question you probably know the answer to

Tobias Kildetoft

You will be amazed at the many things I don't know or don't remember about algebraic number theory

Liad

10:22

Hi, i got a question:
suppose $\Bbb F \subset \Bbb K_1 , \Bbb K_2$ all fields.
denote by $\Bbb K_1\Bbb K_2$ the minimal field containing both $\Bbb K_1$ and $\Bbb K_2$ i need to prove that if $[\Bbb K_1:\Bbb F]$ and $[\Bbb K_2:\Bbb F]$ both finite then $[\Bbb K_1 \Bbb K_2 : \Bbb F]\le [\Bbb K_1:\Bbb F][\Bbb K_2:\Bbb F]$

Alessandro Codenotti

I have a number field $K$ and $h_K$ its class number. If $h_k=1$ then $\mathcal{O}_K$ is an UFD, if $h_K>1$ I can conclude that it isn't a PID. Are there examples with $h_K>1$ and thd ring of integers is an UFD nonetheless?

Liad

someone can help?

Tobias Kildetoft

@AlessandroCodenotti Hmm, I think UFD implies PID for Dedekind domains

Secret

le soneon dies not exist (Gib)

Tobias Kildetoft

right, this was mentioned in the answer to math.stackexchange.com/questions/1051366/…

@AlessandroCodenotti Anyway, feeling slightly Italian, as I have a bolognese on the stove simmering away. But also very Danish, as my lunch was potatoes on rye bread.

quallenjäger

10:30

Suppose I have a continuous path in a hausdorff space, is it possible to construct an injective version of it.

Tobias Kildetoft

@quallenjäger What would it mean to be a version of a path? You mean up to homotopy?

quallenjäger

arxiv.org/pdf/1406.7871.pdf

This Lemma 3.2, I can't understand.

I don't see what the mapping $q$ does.

Secret

(Gib and the recent instability will be explained in next downtime (except that the search is terribly not helpful as neither "question" nor "hello" capture the dozens of them)

quallenjäger

And I can't find the reference to it because the book is so old.

Alessandro Codenotti

@TobiasKildetoft ah right, that's actually not hard to prove. Pick $a\in P$ with $P$ a prime ideal, factor $a$ as $p_1\cdots p_k$, there must be a $p_i$ with $p_i\in P$, hence $(p_i)\subseteq P$ so $(p_i)=P$ because primes are maximal and $(p_i)$ is also prime, so every prime ideal is principal

@TobiasKildetoft Nice, I made it with my flatmates during the weekend too! (We call it ragù rather than bolognese in Italy usually though)

Tobias Kildetoft

10:37

@AlessandroCodenotti Interesting. We usually call it meatsauce

Also, most Danish people don't actually let it simmer at all, so it tends to be somewhat gritty in texture.

Alessandro Codenotti

The longer it simmers the better it tastes, there's no limit

Tobias Kildetoft

agreed

I was a little lazy today though, so I skipped a few ingredients. It is just onions, meat, carrots and (canned) tomatoes.

Secret

Fact #1: Liad dies not exist
Fact #2: Marystar dies not exist
Fact #3: (I forgot)

0celo7

11:24

@AlessandroCodenotti no, like you make a function that does things at the points 3.4 and 3.6

Alessandro Codenotti

I know, I was joking

Secret

12:06

in The Factory Floor, 8 mins ago, by Secret

A "world" reachable only via "zooming"

Because space is too boring and restricted, why not make things more interesting by creating a space like thing with relations instead

In the context of maths, "zooming out" from various algebraic structure will end up in category theory

Of course, "zooming" too far out and then there will be nothing in common and thus you end up in the empty set

$$\text{Zoomout}(A,B) = \min (X : A \cap B \subseteq X )$$

mercio

gives more drugs to Secret

0celo7

12:27

@AlessandroCodenotti oh woooooooooooow the guy means $\frac{d}{5-r}$ instead of $d/5-r$...how did this get published

Alessandro Codenotti

12:46

Feels like those 2-2/2 questions on facebook

Mary Star

Let $(a_n)_{n\in \mathbb{N}}$ be a sequence of numbers. I want to show the equivalence of the follwoing statements:

(1) There exists a real number $q$ ($q\neq 0$) with $\frac{a_{n+1}}{a_n}=q$ for all $n\in \mathbb{N}$.
(2) There exists a real number $q$ ($q\neq 0$) with $a_{n+1}=a_1\cdot q^n$ for all $n\in \mathbb{N}$.

To show the equivalence we have to show that $(1)\rightarrow (2)$ and $(2)\rightarrow (1)$, or not?
But how? Using induction?

0celo7

@AlessandroCodenotti this is such a strange paper

there's a random 3 now that I don't know what's with it

that's not english but you get the point

Alessandro Codenotti

Has the result been published somewhere else or are you stuck with this paper?

0celo7

In some notes but they're worse

completely different proof too

this lemma is completely ridiculous too

Alessandro Codenotti

I can assure you that kind of stuff doesn't happen outside analysis

(Blatant lies)

0celo7

13:03

the proof has like 4 typosd

Secret

$lies^{lies^{lies}}$

(that makes a good echo)

Mancala

What is the boundary asymptotic of a surface?

GFauxPas

13:30

Yay I got a B in the class I was worried about, complex ii :D

skull

congrats

GFauxPas

It was funny, when my lin alg ii prof walked into the room holding the exams he was visibly embarrassed, he explained that

He had so much fun designing problems that when he finished making the exam he realised it was too long

So he said just work on it for the entire alloted time and he'll grade what we do, we don't need to try to do all of them lol

skull

-_-

GFauxPas

I was so happy one of the problems was to prove the Riesz r

Representation thm because i love that thm and its proof

Jake Rose

14:47

Can somebody explain what Im doing wrong here?

Trying to prove that if A and B are orthogonal then AB is also orthogonal

I know the conventional proof

But here is my attempt at one

Nevermind I figured it out

Swapnil Das

15:14

Any hints for this? I'm really stuck.

ffahim

15:27

In how many possible ways we can Rearrange the word "" MATHEMATICS """" such that all the vowels stay separately?

GFauxPas

15:43

@SwapnilDas I didn't work out the whole thing but it may help to note that (1+x)^1/x converges to e as x goes to zero from the right

Because (1+1/n)^n converges to e as n goes to infinity

Anshuman Dwivedi

16:12

Taking help from math.stackexchange.com/questions/1974159/… and applying L' Hospital rule, the answer comes = e/2

GFauxPas

16:38

I'd say that (1+1/n)^n goes to exp(1) is common knowledge and you'd just have to justify changing the limit from 1/x to 0 to x to +infinity

philmcole

17:04

Hi, if I look at the solution set $M$ of the set of non-linear equations

$$\begin{cases} y_1&=x^2 \\ y_2&=x^3\end{cases}$$

this is a manifold. I can check this using the implicit function theorem by defining $f: \Bbb R^3 \to \Bbb R^2$ and

$$f(x,y_1,y_2) = \begin{bmatrix} y_1-x^2 \\ y_2-x^3\end{bmatrix}$$

and checking that the $2 \times 2$ matrix

$$\partial_{y_1,y_2}f = \begin{bmatrix} 1&0 \\ 0&1 \end{bmatrix}$$

is indeed non-singular. My question is if I don't use the suggestive notation $y_1,y_2$, but for example $y$ and $z$, how do I know which part of the total derivative of $f$ I …

Anshuman Dwivedi

Hello Phimcole, if you are interested in this question ->

https://math.stackexchange.com/questions/135766/average-distance-between-two-points-in-a-circular-disk?utm_medium=organic&utm_source=google_rich_qa&utm_campaign=google_rich_qa,

then please help in verifying whether on applying the result of the thread
https://math.stackexchange.com/questions/2386926/average-distance-to-a-non-central-point-in-a-circle

to the former problem, the answer comes 0.82*r as against 145*r/28*pi

Ted Shifrin

17:19

@philmcole: You write down the entire derivative matrix and see which minor (if any) has nonzero determinant. ... But, again, this is back to a graph $y=\phi(x)$, and graphs of smooth functions are always manifolds ($x\in\Bbb R^k$, $y\in\Bbb R^\ell$).

philmcole

@Ted This would be a lot of minors to check in general. Is there no way to see which variables to name $y_1,y_2,y_3$ and so on from a plot?

This was in fact the first of the examples you did in the lecture on YT :) You said something like "I'm gonna name the variables according to the theorem to help you apply it" and my question is how would we have figured out which variables are which if you hadn't named them like that?

Ted Shifrin

If it's really a graph, you should see an identity matrix in the derivative matrix. But if it's not, you actually do have to hunt to find a nonsingular submatrix (locally).

GFauxPas

I'm examining a sequence of maps on an infinite-dimensional Hilbert spaces that converges pointwise but not uniformly, I don't know how to prove this step and would like a hint

philmcole

17:35

Okay thanks!

Ted Shifrin

It all depends on the maps, @GFauxPas. I don't think I've ever thought about this in an infinite-dimensional setting.

GFauxPas

right well it gives an example but the exercise is to prove it converges pointwise and jnot uniformly

i'm stuck on this part

Ted Shifrin

What's the pointwise limit?

GFauxPas

Let $\{e_n\}_{n=0}^\infty$ be an orthonormal basis. then for $x \in H$, $||\langle x,e_n\rangle e_n - \langle x,e_m\rangle e_m || = 1$ for $m \ne n$

Ted Shifrin

That's not true for all $x$.

Try $x=0$.

GFauxPas

17:37

the sequence is $T_n x = x + \langle x, e_n \rangle e_n$

Ted Shifrin

Try $x=e_1+e_2$.

GFauxPas

oh oops this isn't a pointwise limit I'm sorry it's saying that

Ted Shifrin

I don't know where you got that norm equality.

GFauxPas

$||T_n - T_m|| = 1$ in the operator norm

Ted Shifrin

You understand the difference between that and what you said?

GFauxPas

17:38

yes yes definitely

$\sup_{h \in H, ||h|| = 1} ||(T_n-T_m)h||$

Ted Shifrin

Right.

I guess it's not obvious to me instantaneously that that difference has operator norm $1$. At least $1$ is clear.

GFauxPas

okay well let's start from the beginning of the problem if you're willing to work on this together with me Ted

I didn't have to do every problem on the final but I want to know how to do this one

Ted Shifrin

So you're trying to show that $T_n$ can't converge pointwise?

GFauxPas

show $T_n \to I$ pointwise

hint: pythagorean theorem

Alessandro Codenotti

@GFauxPas That converges pointwise to $x$, right?

GFauxPas

17:41

that's the first part of the exercise

the latter term is is the projection of $x$ onto an orthogonal basis so

Ted Shifrin

OK, so that operator norm condition certainly shows you can't have uniform convergence. Uniform convergence would imply uniform Cauchy-ness.

Alessandro Codenotti

$||\langle x,e_n\rangle||$ must go to $0$ just because $e_n$ is a basis

Eric Silva

sup nerdoids

Ted Shifrin

and hi @Alessandro

heya nerd Eric.

Alessandro Codenotti

Hi @Eric @Ted @GFauxPas

GFauxPas

17:43

$||\sum \langle x,e_n \rangle e_n ||^2 = \sum ||x_n||^2$

Ted Shifrin

Too many bars flying around, @GFauxPas.

GFauxPas

wait no

Ted Shifrin

And no.

GFauxPas

too many - yeah lol

Eric Silva

i like to write $||| \cdot |||$

Ted Shifrin

17:45

fires Eric

Eric Silva

for the sake of transparency i am lying

Ted Shifrin

I don't believe that operator norm 1 thing.

You're not even translucent, Eric.

GFauxPas

I didn't get up to it yet!

mayhbe i didnt write out all hypotheses

Ted Shifrin

@GFauxPas: You haven't told us what $x_n$ is. You mean $\langle x,e_n\rangle = x_n$? That's a scalar, so no double bars.

GFauxPas

let's start from the beginning because I lost track of whats a scalar

let $H$ be a hilbert space, infinite dimensional, with orthonormal basis $\{e_n\}$

countable

Ted Shifrin

17:48

Sure.

GFauxPas

$T_nx := x + \langle x,e_n\rangle e_n$

Ted Shifrin

$T_n(x)$.

GFauxPas

claim: $T_n \to I$ pointwise. Hint: pythageorean theorem

Ted Shifrin

Or Parseval's identity or ...

GFauxPas

or CS

?

Mary Star

17:50

A DIN A4 sheet is divided into thirds. A rectangle is the root of the tree, the other two rectangles are each divided into thirds again. Two rectangles form the branches - one to the left, one to the right - the others are again divided into thirds and so on.

Ted Shifrin

CS?

GFauxPas

Cauchy-Scwarz

Mary Star

Could you give me a hint how we could calculate the area of that tree?

Secret

(some notes)
GFauxPas and philmcole is not *dies not exist*. Do not include them in the tally

Eric Silva

which schwarz is cauchy-schwarz

Ted Shifrin

17:52

I don't see how it's Cauchy-Schwarz, @GFauxPas, but maybe I'm missing something obvious.

Secret

(hopefully SE chat is smart enough to pick out "dies not exist" when I finally search for it some time later)

Alessandro Codenotti

The functional analyst is here!

Ted Shifrin

@MaryStar: I don't understand enough to say anything.

GFauxPas

no it's Parseval you're right, I got them confused

Alessandro Codenotti

Hi @0celo7

Ted Shifrin

17:53

Yup, 0celo renders me redundant. Yea!

looks for another R

GFauxPas

$||T_nx -Ix|| = ||x + \langle x, e_n \rangle e_n - x|| = ||\langle x, e_n\rangle e_n||$

Daminark

Henlo

GFauxPas

that's not 0, I did something wrong

Ted Shifrin

Hi Demonark.

It is $0$ in the limit, @GFauxPas. That's what Parseval tells you.

GFauxPas

ooh otherwise it wouldnt be square summable

Ted Shifrin

17:59

Right.

GFauxPas

Okay ^^ so $T_n \to I$ pointwise

thanks

Ted Shifrin

Right, I'm skeptical. I say it's at least $\sqrt2$.

You mean $n\ne m$.

GFauxPas

yes I do

$\sup_{||x||=1}||T_n x - T_m x|| = \sup_{||x||=1}||\langle x,e_n \rangle e_n - \langle x, e_m \rangle e_m||$

Ted Shifrin

Oh, never mind. I say it's at least $1$, not at least $\sqrt2$.

How do we know it can't be bigger?

GFauxPas

it could be a typo, let's see

Alessandro Codenotti

18:06

I think it's correct that is $1$

Eric Silva

@TedShifrin immediate, no? or are you prodding GFauxPas

Alessandro Codenotti

$||(T_n-T_m)(x)||$ is maximized when $x$ has components only along $e_n$ and $e_m$, right @GFauxPas?

Eric Silva

that's kinda just giving it away I think

Ted Shifrin

twiddles thumbs

GFauxPas

okay I did it using parallelogram law

which I'm allowed to do because Hilbert

hides from Ted for not using geometry

Ted Shifrin

18:09

LOL

Eric Silva

sounds like overkill to me but ok

GFauxPas

::rocket launcher::

what's an easier way

dinking

Daminark

Oh that's a new one

Eric Silva

$x_{n}^{2} + x_{m}^{2} \leq \sum_{i = 1}^{\infty} x_{i}^{2} = \|x\|^{2}$

Ted Shifrin

But $|x_n|+|x_m|$ isn't the sum of squares :P

Do we get to use CS now?

GFauxPas

18:12

So $T_n \to I$ pointwise but $T_n \not \to I$ uniformly

Eric Silva

the norm of $T_{n}(x) - T_{m}(x)$ is $x_{n}^{2} + x_{m}^{2}$ @Ted

GFauxPas

last part: assume now that$H$ is finite dimensional. Prove $T_n \to I$ uniformly

Ted Shifrin

Um, really?

Eric Silva

yes? Am I doing something stupid

Ted Shifrin

Yes.

GFauxPas

18:13

but that's equal to $2$ and I have it's $1$

lol

Eric Silva

or square root of that i mean

Ted Shifrin

I don't see that, either. I see an inequality.

Eric Silva

$T_{n}(x) - T_{m}(x) = (0, \cdots, x_{n}, \cdots, x_{m}, 0, \cdots)$?

if we're using the fact that ONB gives iso to $\ell^{2}$

GFauxPas

whoa acronyms

Ted Shifrin

OK, so I had $\|T_n(x)-T_m(x)\|\le |x_n|+|x_m|$.

GFauxPas

18:15

orthnormal basis i smart

Eric Silva

all $0$s except $n$-th coordinate and $m$-th coordinate

norm of that guy is definitely the root of $x_{n}^{2} + x_{m}^{2}$.

Ted Shifrin

I can get to yours by CS.

We're all ending up at the same place.

Well, @GFauxPas, you should get to that easily.

Eric Silva

sure, but im like 99% sure im not talking crazy tho

Ted Shifrin

I only said you were crazy when you said norm was norm squared.

Eric Silva

oh ya that was just misspeaking

mistyping

Ted Shifrin

18:17

And I actually never called you crazy. :P

Eric Silva

the inequality i first wrote was still right and gave us the keys to the kingdom

i called myself crazy

GFauxPas

trying to find the point in my work where I assumed infinite dimensionality

Ted Shifrin

Actually, I don't understand the question. There are only finitely many operators if the dimension is finite. So WTF?

trashes the examiner

GFauxPas

oh it's asking generically

prove pointwise convergence in a finite dim Hilbert space is uniform convergence

pointwise convergence of a sequence of operators

well you can just use a covering argument, can't you

Ted Shifrin

Oh ...

Covering? Huh?

GFauxPas

18:25

wait nope that didnt make sense

Ted Shifrin

You mean compactness of the unit ball?

GFauxPas

yeah but that proves it the other way

Ted Shifrin

I'm not following.

GFauxPas

that's how you can prove that uniform convergence implies pointwise convergence on compact sets

but I want the harder implication, that pointwise implies uniform

Ted Shifrin

Uniform convergence always implies pointwise convergence. What you talking about?

You're confuzled.

GFauxPas

18:27

I'm saying I was confused as to which implication direction I was doing

Ted Shifrin

I still say it's what I said.

GFauxPas

anyway I have to go on to my take home assignment

thanks guys, especially Ted

Ted Shifrin

OK, bubye.

GFauxPas

:) I'll still be around, I just have to do more myself now because this isn't review it's actually something I have to do

Adeek

18:46

hi @TedShifrin

I missed talking to you

0celo7

@AlessandroCodenotti lmao I just noticed the proof also involves 3.1 and 3.9

Adeek

@TedShifrin I just came to update you with news :) I won a 15 k scholarship. My Thesis also won best thesis award.

I have been doing nothing but math lately. Today since I have been working hard. I will invite my wife out for a dinner outside and thank her for all the support.

GFauxPas

Awesome Adeek

congratsw

make sure to bring your thesis to the dinner so you can talk about it with your wife

Adeek

yeah.

I will cya guys later.

Alessandro Codenotti

19:02

@0celo7 they're just messing with you at this point, are you sure it's not a prank from your collegues :P?

0celo7

@AlessandroCodenotti the claim is actually so elementary that it's embarrassing that I'm having so much trouble with it

0celo7

19:14

@AlessandroCodenotti the guy is working on the ball of radius 6...maybe I should go to the ball of radius 10 so I can work with all integers

Semiclassical

Welp

Alessandro Codenotti

That's why people usually say "pick a big enough ball" and skip the numbers

Balarka Sen

Usually people don't say that twice though

Daminark

Rekt

Alessandro Codenotti

People with an awful internet connection do

Hi @Balarka

0celo7

19:22

now he randomly wrote a $C^3$ estimate

Balarka Sen

hi

0celo7

even though everything is second order

I think this is an April Fool's paper

@BalarkaSen dude are your exams finally over

Balarka Sen

ye

0celo7

I need you to read some dumbass topology I wrote

Balarka Sen

im physically and mentally way too exhausted to look at math at this moment

Alessandro Codenotti

19:23

Say I have a surface (smooth, in $\Bbb R^3$) that contains a line. I see that this line must be a geodesic. Must the gaussian curvature on points of the line be 0? I think so because the gauss map is constant in the direction of the line @Balarka

Balarka Sen

Both statements are false

0celo7

Why do you think it is a geodesic

Balarka Sen

Hyperboloid of one sheet, a line in it's standard pencillization or whatever its called

Alessandro Codenotti

Because the covariant derivative of the tangent field looks zero to me

Balarka Sen

One of those lines

Neither geodesic nor K = 0 along them

0celo7

19:25

Weird, I asked this question like two weeks ago

Balarka Sen

Yeah

0celo7

What you need is the angle between the surface and some plane to be constant along the line

then K=0

Balarka Sen

Ahhh that was an exercise in Ted's book

Alessandro Codenotti

Hmmm, I'm missing something, how can it not be a geodesic?

0celo7

it's also a claim in a Yau paper, Balarka

But I doubt Yau read ted's book

Balarka Sen

19:27

Both are Chern students ;)

Alessandro Codenotti

If $\gamma$ is a straight line don't I have $\gamma''=0$ and so the covariant derivative of $\gamma'$ is also zero?

0celo7

Yau doesn't even read his own books

@AlessandroCodenotti $\gamma''=0$ in R^3, but not relative to the surface.

Balarka Sen

You need to project $\gamma'$ to the tangent plane of the surface and then take it's derivative as you move

That's nontrivial because of curvy stuff my man

0celo7

@Semiclassical So I found the ultimate procrastination tool

@Semiclassical it's glorious i.gyazo.com/5e08db09e7e83ae87f6493cd900a54f8.png

Alessandro Codenotti

Hmmm I'm confused

You shouldn't nest \text{} and dollar signs or so they say

0celo7

19:36

who

are you talking about me

Alessandro Codenotti

Yes

0celo7

why are you hating

it works perfectly

Balarka Sen

I use \mathsf{Theorem}: instead of a proper theorem environment

does that trigger you

0celo7

you're a maniac

Alessandro Codenotti

A lot of stuff which people knowledgable about latex suggest avoid work perfectly in most cases

Daminark

19:37

I nest \text and dollar signs lmao

0celo7

my preamble could probably kill a small child because it's so messy

ÍgjøgnumMeg

^

Alessandro Codenotti

@Balarka I don't understand. Say I have a vector field $X$ along a curve $\gamma$ contained in a surface $S$, with $X(t)\in T_{\gamma(t)}S$ for all $t$. Then the covariant derivative of $X$ along $\gamma$ is the projection of $X'(t)$ on $T_{\gamma(t)}S$, right?

Balarka Sen

Yes

The projection in the case of a lines in the hyperboloid would not be zero. Think, eg, about what happens when you project the direction vector of the line at a point on the waist of the hyperboloid (where the tangent plane lies vertically)

Alessandro Codenotti

For a geodesic in particular I use $X=\gamma'$

Daminark

19:44

@0celo7 In my general quest to be the anti-0celo7

Alessandro Codenotti

Isn't $gamma''$ zero before projecting? And so zero afterward as well?

Balarka Sen

Oh maybe you're right

Daminark

Like that's it, after that it's just my name, \maketitle, and then it starts the document

Balarka Sen

I've just forgotten everything. All the more reason to not ask me math rn

On that note I should go to sleep and rest

0celo7

According to MSE, the line is a geodesic

But that does not mean it's a line of curvature

That was a conjecture of mine, sad to see it's wrong

and then more

Alessandro Codenotti

19:47

I'll summon @Ted

@0celo7 that's disappointing

Daminark

Well, for a paper I wrote it was a bit longer

0celo7

here it is for a note I wrote to my advisor yesterday

Alessandro Codenotti

Good night @Balarka

0celo7

Daminark

God

Alessandro Codenotti

19:50

You guys are aware that you can just put everything into a preamble.tex file and then include it in every document and forget about the horror you created?

0celo7

wot

ÍgjøgnumMeg

w h a t

0celo7

oh but I have different settings for notes, homeworks, and my thesis

keeping track of that would be hard

Daminark

I use an online compiler and also my stuff is usually very short but I've heard of this

0celo7

my thesis was supposed to be short

it didn't work out that way

ÍgjøgnumMeg

19:52

Daminark

Well, my stuff = my preamble

Alessandro Codenotti

I even put chapter and stuff into different .tex files

ÍgjøgnumMeg

too organised

Daminark

It's literally the default plus like, at most 5 packages

ÍgjøgnumMeg

hahaha

Alessandro Codenotti

19:53

My .tex documents are the only well organised thing in my life

My phone really likes sending messages twice tonight

0celo7

@ÍgjøgnumMeg that's so algebraic it's not even funny

at least I have Hom though...I'm not a total n00b

Alessandro Codenotti

@ÍgjøgnumMeg why \roi instead of \O?

0celo7

then again I probably only use that for the Hom bundle...

ÍgjøgnumMeg

@Alessandro I can't remember, I think I tried that and it didn't work for some reason

that, or I'm just a moron

@0celo7 I think there was a hom somewhere in there for a dual basis

but I only used it once so I didn't put it in the preamble

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$Mathematics$ Mathematics