Mathematics - 2019-12-14 (page 1 of 4)

anakhro

12:10 AM

I see you have made some progress in your understanding of algebra, @Hawk!

Ultradark

good job hawk

I think my new year's resolution is to be more clear and rigorous mathematically

anakhro

A good resolution.

Ultradark

starting january first

actually I'll start now.

Ultradark

12:33 AM

So I am reading Dummit and Foote abstract algebra. The title of the page is Automorphisms

Consider the automorphism, $f:T \to Z.$

where $T$ and $Z$ are sets

Hawk

I probably should stop using this book, but this explanation confuses me. If charF = 0, ka_k = 0 only if k = 0, so how does that imply a_k = 0 for all k?

loch

Ka_k=0 implies k= 0 or a_k = 0! So if k >=1 you deduce a_k = 0

Ultradark

So $f$ is a unary operator

How does one express a map from a function to itself?

I've seen it written as $F:X\to X$

but I want $X$ not to be a set, but rather a function

I don't think I can write $F: f(x)\to f(x)$

anakhro

1:02 AM

@Ultradark automorphisms are not maps between functions, but maps from an algebraic object into itself which is also an isomorphism.

But in the case you do want a function between "functions", you will have to turn your functions into sets somehow. The usual way is looking at it from the perspective of relations. A relation between two sets A and B is a subset R of AxB. A function is a particular type of relation, one which is "total" (for each a in A, there is a b in B such that (a,b) is in R) and "single-valued" (if (a,b) and (a,b') are in R, then b=b').

Relations between A and B which are total and single-valued are functions A-->B.

In this way, you can define a function between functions. You would just be doing functions from relations to relations, which is the same as doing functions between subsets of Cartesian products.

But again, not very related to automorphisms. :P

I'm itching to write a little math pdf on something. But I don't know what.

Ultradark

1:19 AM

oh okay cool! @anakhro

anakhro

The name "auto"morphism is to distinguish it as an isomorphism among the other "endo"morphisms.

Endomorphism is just any old map from the object to itself. No need for isomorphism.

Ultradark

cool, I think I'm looking at endofunctions

anakhro

@Ultradark do not hesitate to ask if you have any questions.

Ultradark

okay

I have a question

anakhro

Go for it!

Ultradark

1:32 AM

@anakhro say you have a endofunction, specifically an involution, however the involution has two disjoint parts, separated by asymptotes. I'm trying to understand a map between these two disjoint parts

I have a relation between points separated by the asymptotes

anakhro

What set is this a function on?

R?

Ultradark

yes $\Bbb R$

anakhro

So what do you mean by "has two disjoint parts"?

Ultradark

Here are the maps that I've found

The two disjoint parts are circled in blue

anakhro

How is this a function on R? It's not defined for many x?

Ultradark

1:37 AM

ohh

my bad

Here's it's domain and range

basically it's defined in the first quadrant, for $x,y \ne 1$

so there's a relation between points in the two blue regions

anakhro

So what do you mean by "map between these two disjoint parts"?

Ultradark

see the rays?

anakhro

Do you mean a function on the graph {(x,f(x)) | x != 1}?

Ultradark

also $f(x) \ne 1$^^

anakhro

Well that's a restriction imposed by f itself.

Ultradark

1:43 AM

the purple ray connects exactly two points on the function

and I want to relate these two points

anakhro

Yes, it's a projection from one curve to the other.

Ultradark

okay

I wasn't sure if it was indeed a projection or not

anakhro

Well you know it is a projection because it satisfies $P^2 = P$, where $P$ is your map.

Ultradark

So for rays $y=mx$ s.t. $m>1$ we get $(x,\phi(x)) \mapsto \big(\frac{1}{\phi(x)},\frac{1}{x}\big),~~x\in(e,\infty)$

Is this related to inversive geometry do you think?

anakhro

No.

Not in any obvious way. I can't see anything, anyway.

Ultradark

1:51 AM

so let me be specific. I'm mapping points from the larger curve to the smaller curve (the mapping above)

anakhro

Or both curves to both curves, but the image is just the smaller curve.

Ultradark

yeah

since it's an involution we get $(x,\phi(x)) \mapsto \big(\frac{1}{x},\frac{1}{\phi(x)}\big)~~ x=e$

where $x=\phi(x)$

anakhro

It's not an involution.

Ultradark

for the line $y=x$

well the function is

anakhro

The function you have graphed?

Ultradark

1:54 AM

yeah

also above I should have written $m<1$

anakhro

Sure.

Ultradark

and the last map for $y=mx$ s.t. $m>1$ is $(x,\phi(x)) \mapsto \big(\frac{1}{\phi(x)},\frac{1}{x}\big), ~~x\in(-\infty,e)$

So do I write the maps as I have written them, or is there some better way to write them and describe them

anakhro

I can't verify them because I don't know what the original function is actually defined as.

Ultradark

$\phi(x)=f(g(x)).$

$f(x)=e^x$

$g(x)=\frac{1}{\ln(x)}.$

wait...

what??

Soham Chowdhury

2:09 AM

someone's bugging me with an algebra question

can you solve ${(5x+3)}^{1/4} = -2$ over $\mathbf{R}$?

apparently there was no "no solution" option on the test lmao

so sanity check

Leaky Nun

@SohamChowdhury surely you can't

Soham Chowdhury

but maybe if you do something stupid like separate into two square roots, take a positive square root, then a negative square root, you can force it to work ...?

idk how hs math problems work

by taking $x = 13/5$ that is

doesn't sound legal but maybe the teacher thought it'd be a fun way to spice up the problem

anakhro

Well the only way the left hand side is defined is if x>=-3/5, but then it has to be non-negative.

Soham Chowdhury

yeah

anakhro

So either you wrote down the question wrong, or your teacher made an error.

user193319

2:18 AM

0

Q: Weyl-von Neumann Berg-Theorem and Density of the Eigenvalues

I am reading this expository work Numerical Range of Operators by Paul Skoufranis (see this). Here is the theorem and proof I'm working through: Theorem 2.11 Let $N \in B(\mathcal{H})$ be a normal operator. Then $\overline{W(N)} = conv(\sigma(N))$ Proof: By the Weyl-von Neumann Berg-Theo...

Secret

3:00 AM

So, playing with $\pi - e$ again, and found something very strange

The Great Duck

0

Q: Algorithm to get the list of triangles resulting from clipping a triangle against a convex polygon.

We have a 2D triangle. It has 3 points. We have a 2D convex polygon. It has N points. Normally when clipping, one wants to produce the polygon or list of triangles inside the convex polygon. I want to do the opposite. I want to get the list of triangles that do not intersect aside from borderin...

geometry algorithms

Secret

It seems integer multiples of $\pi-e$ follows some strange sequence where their pairwise differences is 3,8,3,8,3,8,3,8,... etc. for any integer multiples of $\pi -e$ that is only $\pm 0.1$ away from its nearest integer

desmos.com/calculator/ycqwiaaits

The Great Duck

@Secret do you have any thoughts?

I'm really stuck on this.

urg

Secret

Rough thoughts: Maybe get the list of all triangles in the region, and then subtract those the clipped to get those outside the clipping region?

so basically, do the usual way of clipping and then take its complement wrt the clipping object

The Great Duck

3:23 AM

@Secret im not sure how I would go about doing that though...

Secret

You have a region with all the triangles, which are a list of tuples. This list is S. Now in S, there is T which are all the triangles inside a given polygon that does the clipping. You knew S already, and you knew the algorithm to get T, so what you need to do is to get S-T

The Great Duck

@Secret technically still don't have T. That's a problem I've also been banging me head into my table regarding. The difference is that I can find enough google results to maybe not go insane.

Hawk

@loch if charF = k = 0, then a_k isn't necessary 0.

loch

3:41 AM

i'm not sure which line are you confused by in the picture you attached ---
it's saying that f' = 0 <=> ka_k = 0 for all k>= 1 (clear)

if char F = 0, then ka_k = 0 implies a_k = 0 since you can divide by k

k is not char F !

Hawk

oh ok

hang on why can't a_k \neq 0?

i mean the equation ka_k = 0 forces the charF = k = 0 right?

The Great Duck

no clue

loch

char F and k are different things - don't write char F = k = 0!

our assumption right now is f' = 0
f' = 0 implies that ka_k = 0 for all k >= 1,

now k>= 1 , so you can divide by k [ here we are using the fact that char F = 0, so you don't have k=p=0]

if you want just pretend you're working over rational numbers

Hawk

hang on so what does charF = 0 then have to do with this?

that if charF \neq 0, we can't divide?

The Great Duck

just wait and let him get to it

loch

3:50 AM

i mean - technically you can always divide by non-zero elements (because you're in a field)

the point is that p = 0

The Great Duck

you're not ready to consider that yet

loch

say you're working over char p

The Great Duck

;-)

loch

take f(x) = x^p

then f'(x) = px^{p-1}

but p = 0

so f'(x) = 0

The Great Duck

and therefore f(x) is constant

loch

3:51 AM

no because f(x) = x^p

The Great Duck

but p is 0

so f(x) = x^0

which is constant

loch

so i guess the upshot is try think through what your text is saying in the case 1) F = \F_p and 2) F = \Q

Hawk

i mean i get what you trying to do but I don't think that's what they want me to take away.

loch

no technically if you want to be pedantic there is some abused notation here
x^p is x*x*...*x
px^{p-1} is x^{p-1} + x^{p-1} + ... + x^{p-1} = 0

Hawk

is the idea that if charF \neq 0, say p, then we might arrive at pa_p = 0 for k = p

then it might not be determinable that a_k = 0 for all k

but f charF = 0

The Great Duck

3:54 AM

it has something to do with polynomals and which coefficients are 0 when the derivative is 0

it's that simple

Hawk

ka_k =0 are distinct

its just that one line that confuses me

The Great Duck

If the derivative of a polynomial is 0 for all x, then certain coefficients must be 0

that theorem tells you which ones are 0

loch

@Hawk then i'm not sure what you're confused by since you seem to understand what's happening in char p. are you confused if the underlying field is \Q or \R or \C?

Hawk

No

The Great Duck

it is an extension of the rule of thumb in regular calculus that the derivative of a linear function is a constant and that the derivative of a constant is 0. What that theorem is telling you is that there are fields with a characteristic such that nonlinear functions and nonconstant functions have a derivative of 0 everywhere.

Hawk

3:57 AM

i know you are trying to identify Fields with charF = 0 with (say) Q and argue it from there. But I think there is a different argument the book is sayin

The Great Duck

basically that (x^2)' could be 0 everywhere

loch

ka_k = (1+1+...+1)*a_k

1 is an element of F
so is 1+...+1 (k times)
this is non-zero (since F has char 0), therefore it is invertible
multiply its inverse on both sides => a_k = 0

The Great Duck

for some field with a characteristic value compatible with the conditions in that theorem

@loch we're looking at the case when it's not 0.

Hawk

@loch ah ok

The Great Duck

im heading out

good luck

Secret

4:06 AM

yesterday, by vzn

Mathematician Proves Huge Result on ‘Dangerous’ Problem (Tao on Collatz!) / quanta mag

Collatz is not the problem that really grip me (because it is too abstract)

The problem that grip me are $\pi -e$ and $e^e$

Any two irrationals can be represented by a convergence sum of the form $\sum_{n=1}^{\infty} \frac{a_n}{b^n}$

where $b$ is the base

Thus, given two irrationals $s,t$ such that $s-t$ is rational, it means the terms $\frac{a_n}{b^n}$ has to be cancelled out for all $n > M$

Since the sums are convergent, we can without lost of generality, write any partial sums (including cases where the terms are nonconsecutive) as some rational number that is the result of said partial sum. Then, a permutation of these terms will ensure all the matching rationals (if any) in the two irrationals to match up in descending order

Thus, the problem of irrationality reduces to finding the smallest M such that their tail matches

because if the tail cancels out, it will be a rational

That is, at least for subsets of $s-t$, their difference is rational if their tail cancel out in at least one numeral system X

The case however, is tricker for $s-t$ which has the following expansion:

$s = \sum_{n=1}^M \frac{a_n}{b_n} + \sum_{n=M+1}^{\infty} \frac{a_n}{b_n}$
$t= \sum_{n=1}^M \frac{c_n}{d_n} + \sum_{n=M+1}^{\infty} \frac{c_n}{d_n}$

where $\sum_{n=M+1}^{\infty} \frac{a_n}{b_n} = \sum_{n=M+1}^{\infty} \frac{c_n}{d_n}$

That is tricker since the only requirement for the tail is that the two sums converges to the same limit, and this limit can be any real number

In other words, we are dealing with this case: $s-t=u+x - (v+x)$ where $x$ is real and both $u,v$ are rational

using this, we have:

$s-t = (u-v) + (x-x)$

$K(s-t) = K(u-v) + K(x-x)$

Since $u,v$ is rational, $K(u-v)$ is an integer

Thus, we want:

$|K (s-t) - K| < \epsilon$

or $|s-t-1| < \frac{\epsilon}{|K|}$

Hawk

4:28 AM

@loch idk if u are still here. but if charF = p, why does $p \neq | k \implies a_k = 0$ i tried applying division algorithm to this k = pm + r and so $0 = ka_k = pma_k + ra_k = 0 + ra_k$ but we have a contradiction?

loch

you have 0<r<p

Leaky Nun

@Hawk literally divide by $k$

loch

in particular r!=0

Hawk

@loch right so doesn't that mean $ka_k = 0$ only if p divides into k?

loch

r!= 0 so ra_k = 0 => you can divide by r to deduce a_k = 0

Hawk

4:31 AM

oh u are saying $r \neq 0$

Secret

Thus, $s-t$ is transcendental if

$|K(s-t) - L(K)| > \epsilon$

where $L(K)$ is an integer

Hawk

I thought $r$ MUST be $0$ since we have exhibited a number $r<charF = p$ such that $ra_k = 0$

loch

you're given ka_k =0 and trying to deduce that a_k = 0

Hawk

so that isn' a contradiction?

Secret

Every irrational can be written like this:

loch

4:37 AM

i dont know what contradiction are you referring to lol
here's the claim: ka_k = 0 and k \ne 0 mod p implies a_k = 0
proof: in short - divide by k => a_k = 0

in slightly more words: divisoin algorithm => k = pm + r for some 0<r<p, since k\ne 0 mod p

therefore ka_k = (pm+r)a_k = ra_k

so we have ra_k=0
r is non-zero because 0<r<p. divide by r on both sides, get a_k = 0

Secret

$s = u+x, t = v+y$
$s-t = (u-v) + (x-y)$
where $u,v \in \Bbb{Q}, x,y \in \Bbb{I}$ and $x-y \in \Bbb{I}$

Multiply by a suitable integer $K$ we have:

Hawk

no i mean charF = p should be the SMALLEST integer such that pa_k = 0, but we arrived at ra_k = 0, so r has to be 0?

Secret

$K(s-t) = K(u-v) + K(x-y)$

$K(s-t) - K(u-v) = K(x-y)$

where $K(u-v) = L(K) \in \Bbb{Z}$

loch

yes p is the smallest integer such that 1+...+1 = 0 (p times)
0< r < p
*therefore* you deduce that r is not zero, because it is smaller than p
so becaues ra_k = 0, this *implies* that a_k = 0

*if* for some reason you know that a_k is not zero, *then* you get a contradiction (since you're working in a field there are no zero divisors)

Secret

Thus we have for irrationality:
$|K(s-t) - L(K)| = |K(x-y)| > \epsilon$

Hawk

4:43 AM

oh wow ok

so basically

its because we know hat ra_k = 0 doesn't make sense in charF = p, UNLESS a_k = 0.

haha

loch

yes!

Secret

i.e. $|K(s-t) - \lfloor K(s-t) \rfloor | > \epsilon$

or more succinctly: $|\{K(s-t)\}| > \epsilon$

$e^{|\{K(s-t)\}|} > 1$

Hawk

ah actually I think i know why

they identified a_k \to a_j^p and factored it out.

Secret

$e^{x - \lfloor x \rfloor} > 1$

$e^x > e^{\lfloor x \rfloor}$

actually that does not work

e.g. $e^{0.8} > e^{\lfloor 0.8\rfloor}$

but 0.8 is rational

Leaky Nun

5:14 AM

@loch

loch

@LeakyNun what's up

Secret

$e^{K|s-t|-K|u-v|} > 1$
$e^{K|s-t|} > e^{K|u-v|}$

Leaky Nun

@loch is there any connection between a.g. and d.g.?

loch

alggeom and diff geom?

Leaky Nun

yeah

loch

5:23 AM

hmm what kind of connections are you looking for
the trivial answer is of course - complex smooth varieties are manifolds

Leaky Nun

I was reading about maximum principle

so maybe you wanna solve PDEs on manifolds

but do you solve PDEs on complex sooth varieties?

what do you do to varieties actually

compute cohomology?

loch

here's a cute one

you know faltings theorem tells you that curves of genus g>=2 have only finitely many rational points

g=0 you have infinite (assuming you have 1 rational point), and g=1 you have mordell weil

so somehow there are three 'types' of curves
g=0, g=1 and g>=2

it turns out also that g=0 has positive curvature
g=1 flat
g>=2 negative curvature

@LeakyNun yes a lot of people care about cohomology too

Leaky Nun

cool

why the curvatures?

loch

hm
actually i forgot how this story goes let me find a reference..

@LeakyNun 99% sure some people do

@LeakyNun so eg some people care about computing cohomology of moduli space of curves

Leaky Nun

i see

loch

5:32 AM

one way to motivate that is that as you saw earlier today cup product really has something to do with intersections

and computing intersections on a moduli space allows you answer enumerative questions

*to

Leaky Nun

lol were you here?

loch

i said something about bezout and lurked

if you learn more AG you'll see that there are these things called chow groups which is similar to cohomology

but cohomology is easier to compute so a lot of times people just work with cohomology

but yeah some people care about complex geom which sits in the intersection of ag and dg and people use things from both sides to do stuff with it

then as you know there are also a lot of things in ag motivated from dg

Leaky Nun

I guess

loch

smoothness vector bundles etc.

but maybe a more interesting one is hodge theory --- if you've heard of recent buzzwords one of them is p-adic hodge theory

i know nothing about that but from the name you can guess it's probably kind of motivated from hodge theory

:p

or eg if you know weil's conjectures
part of the proof involves figuring out what l-adic cohomology is and use that it has a lefschetz fixed point theorem (though there's much more) --- this is motivated from the corresponding theorems in singular cohomology for manifolds (or for more general spaces i guess)

Secret

Let {s} be fractional part of s

Then s is irrational if for all integers K, |K{s}|>0

Secret

6:05 AM

Actually, slightly more is true. Given q rational, if s is irrational then |q{s}|>0 for all q

That means, irrationality results when no rational can nuke the fractional part of a real number

Balarka Sen

8:43 AM

@LeakyNun Given a compact complex manifold $X$, producing holomorphic embeddings in $\Bbb{CP}^n$ (for some $n$) is an important question. In dimension 2, degree-genus gives strong restrictions on when you can embed in $\Bbb{CP}^2$, for example: a Riemann surface of genus $2$ does not embed in $\Bbb{CP}^2$ as $4$ is not a product of consecutive integers. Every compact Riemann surface holomorphically embeds in $\Bbb{CP}^3$.

The way to go about producing such embeddings is producing ample line bundles $\ell$ on $X$. I forget the story but it's something like this: Take $n+1$-sections $s_0, s_1, \cdots, s_n$ of $\ell$, and then the candidate embedding is $[s_0 : s_1 : \cdots : s_n] : X \to \Bbb P^n$.

This has to be well-defined, so eg they don't simultaneously all vanish. And then you need it to be an "embedding", yadda yadda. There's some algebraic conditions $\ell$ needs to satisfy to admit such sections

A theorem of Kodaira says that if $X$ is compact Kahler, $\ell$ is ample if and only if $\ell$ satisfies some purely geometric condition called positivity: if $\nabla$ is the curvature $2$-form of $\ell$ then $i\nabla$ is a $(1, 1)$-form.

I don't know this very well but Ted surely does, we should ask him. But I think this is one of the instances of surprising connections between AG and DG I have seen

@BalarkaSen *positive definite $(1, 1)$-form, which is the whole point :p

ÍgjøgnumMeg

9:17 AM

Moooornin'

Balarka Sen

yo

ÍgjøgnumMeg

wazzup

Balarka Sen

learning how to draw engames lol

its slightly better than losing

ÍgjøgnumMeg

rofl ncie

nice*

I was on the chess team in secondary school and I'm absolutely aids at chess

they just picked people who were likely to say yes to being on the team hahaha

Balarka Sen

lmao

i shoulda joined some club but found math more interesting

ÍgjøgnumMeg

9:20 AM

I was just in the team because I got a day off school every time we went to go play

Balarka Sen

it was going to cut away my math time

damn nice

i bunked school anyway so

ÍgjøgnumMeg

fair one

I should probably start working on my seminar talk for topology

it's on the 20th of January and I've done 0 for it rofl

Balarka Sen

what are u talking about

ÍgjøgnumMeg

covering spaces

Balarka Sen

Oh nice

ÍgjøgnumMeg

9:26 AM

yeah, I don't even really know the definition yet ahahaha

Balarka Sen

thats why its a good idea to give a talk on it, no?

ÍgjøgnumMeg

well yeah it'll force me to learn the shit out of it

Balarka Sen

if u should know something but u dont want to learn it just commit to a talk

ÍgjøgnumMeg

aye

Balarka Sen

its good fire on your backside to get things rolling

ÍgjøgnumMeg

9:28 AM

truths

tbf it won't be very demanding to learn it

all the other talks have been essentially verbatim out of Munkres

Balarka Sen

ya you can learn it in a week if u want to

learn from hatcher

ÍgjøgnumMeg

Inderd

well everything is going from Munkres so they want us to learn from Munkres

Balarka Sen

ah ok

i need to decide what i want to talk about in the student seminar as well

something i should know but dont feel like learning... whacoulditbe

Soham Chowdhury

it's going to be algebra isn't it

Balarka Sen

:(

Soham Chowdhury

9:30 AM

lmao

i may have been drawing domains in the complex plane all morning but i will stand up for algebra if nobody else does

ÍgjøgnumMeg

there was a seminar course on Riemann surfaces this semester which I wish I'd attended, but which I didn't really have the background for i guess

Balarka Sen

oh u know what

i could learn and talk about the proof of Riemann mapping theorem (and uniformization thereof)

ÍgjøgnumMeg

do you have free reign over topics?

Soham Chowdhury

heyy you could probably explain it to me to test out ur talk :) :) :)

Balarka Sen

@Soham hell we could even learn it togather

@ÍgjøgnumMeg no, I am not going to learn Iwasawa theory

if that was your intention

ÍgjøgnumMeg

9:32 AM

hahaha

Soham Chowdhury

:(

ÍgjøgnumMeg

No I merely meant to ask if you can do whatever you want for the seminar

Balarka Sen

yeah basically

Soham Chowdhury

@BalarkaSen sounds like it could be fun, when is ur seminar

Balarka Sen

it should be accessible to undergrads sort of

ÍgjøgnumMeg

9:33 AM

(although Iwasawa theory is cool af so)

Mats Granvik

The Rahm cohomology can sometimes be recovered from the cardinalities of ergodic flow patterns. Taking the vector magnitude of the topological L2 norm of the spectral density of the accumulated angles of depression of the apothem, then gives us the upper bound needed on the contrapositive congruence of sinusoids. This in turn means that we have a variable segment of the tautochronic Schur measure of the symmetric angle decomposition of the parallellogram expansion.

Balarka Sen

@Soham ill probably give one on second week of January

Soham Chowdhury

is that from the math generator

nvm it's too internally coherent

Balarka Sen

yeah haha

thats what i thought

Soham Chowdhury

someone wrote that by hand

Balarka Sen

9:35 AM

also "The Rahm cohomology"

smells like "de Rham cohomology"

dhoram cohomology

thats a good name

Soham Chowdhury

also known as an extraordinary Chicago cohomology theory

ÍgjøgnumMeg

oh my god I hadn't heard of the math generator

Soham Chowdhury

there's also a conty phil generator

and a CS generator

Balarka Sen

cont phil is A+

Soham Chowdhury

just like the generated stuff

Alessandro Codenotti

9:37 AM

@ÍgjøgnumMeg it's just a stack of pancakes

Balarka Sen

there should be some general cohomology theory which solves all the major problems in mathematics in one shot

ÍgjøgnumMeg

A homomorphism $U$ is null if $t^\prime$ is quasi-invertible, real, Pappus and Fibonacci

@Alessandro this, I know

Balarka Sen

we could call it the boom kaput cohomology theory

ÍgjøgnumMeg

rofl

Balarka Sen

sorry very bad jok

Soham Chowdhury

9:38 AM

you know i was reading about branches and was like "why not just move up along some other axi- oh right"

Alessandro Codenotti

Bring an actual stack of pancakes to the seminar talk and serve them to people, they'll forget you actually never talked about math

ÍgjøgnumMeg

hahaha

a fine plan

Soham Chowdhury

people should really teach riemann surfaces before defining complex log /s

Balarka Sen

stack of records

DSBM records

Soham Chowdhury

it all comes crashing down

because sad

Balarka Sen

9:39 AM

@SohamChowdhury honestly, that actually helps

i never understood log before understanding covering spaces

ÍgjøgnumMeg

I'm so glad I know about this math generator now, I'm amazed by it

Balarka Sen

every nuclear space is pseudo-Grothendieck, quasi-Euclidean and finite

ÍgjøgnumMeg

I'm gonna start churning out papers and posting them on facebook to see if anyone notices

Balarka Sen

It is a classical result which every student is aware of that $\Theta = -1$

It was a groundbreaking result of X. Pythagoras that monoids are anabelian

ok ill stop

ÍgjøgnumMeg

Let us suppose we are given a compactly reversible group acting canonically on a finitely Minkowski, subgeometric, bounded topological space $Y$. Then there exists a compact null, coprime, real subring.

imagine if this thing just accidentally generates a proof of RH

Soham Chowdhury

9:44 AM

@ÍgjøgnumMeg

news.mit.edu/2015/…

> The program was crude, but it did the trick: In April of 2005 the team’s submission, “Rooter: A Methodology for the Typical Unification of Access Points and Redundancy,” was accepted as a non-reviewed paper to the World Multiconference on Systemics, Cybernetics and Informatics (WMSCI), a conference that Krohn says is known for “being spammy and having loose standards.”

Balarka Sen

Rooter lol

Ultradark

Do the isotopic subgroups of all abelian varieties commute if and only if the dual category of representable functors project into locally ringed spaces

Soham Chowdhury

i had to parse that unusually carefully

afraid i was being had

ÍgjøgnumMeg

rofl

Ultradark

I actually posted this to math stack exchange

And someone said it was gibberish lol

Soham Chowdhury

9:47 AM

it's algebraic geometry, that's part of the charm

Ultradark