Mathematics - 2020-11-20 (page 1 of 2)

Question: If a compact, self-adjoint operator on a Hilbert space $\mathcal{H}$ exists, does that mean that an orthonormal basis of $\mathcal{H}$ consisting of real valued eigenvectors has to exist, via the spectral theorem on Hilbert spaces?

AttractorNotStrangeAtAll

1:40 AM

Hello everyone.

I have $p(x)=a_n x^n + a_{n-1} + \cdots + a_1 x + a_0$, which is a polynomial with real coefficients such that $n$ is even and $a_n$ is positive. I need to show that this polynomial reach its minimum in some real value, that is, that there exists a $x_0$ such that $p(x_0)\leq p(x)$ for all $x\in\mathbb{R}$.

Any hints on how to proceed?

I feel that I'll be using IVT but I'm not sure where

Thorgott

1. Can you show that local minima exist? 2. Can you show that $p(x)$ diverges to $+\infty$ as $x\rightarrow\pm\infty$?

AttractorNotStrangeAtAll

The only tools I have so far are continuity, sequences of real numbers and some topological notions in $\mathbb{R}$

Continuity defined with $\varepsilon-\delta$, not with $\lim{p(x)}_{x\to a}$

Akiva Weinberger

Schools closed in NYC today

City hit 3% infection rate, that was the threshold

Rithaniel

I think the nation might be hitting the second wave

Akiva Weinberger

The nation's in the middle of a third wave

But NYC's just beginning a second wave

user2103480

1:51 AM

wait, the first wave ended?

Rithaniel

2103480 has a point

user2103480

to bring some more opinion diversity on that

Rithaniel

There were people pointing to a "second wave," but I always interpreted it as just aftershocks from the first wave

Akiva Weinberger

The US definitely had two humps in its cases graph, though the valleys weren't particularly low

Thankfully NYS had a long stretch of <1% infection rate after the apocalypse in April

In any case. No school, starting today

Rithaniel, you're in NJ, right? Or am I misremembering

Rithaniel

SC

Clemson University

Akiva Weinberger

1:58 AM

Oh

Rithaniel

Not the highest place of math education, but also not the worst

AttractorNotStrangeAtAll

What is the feeling in the US for a new lockdown?

Rithaniel

I say we should never have left the first lockdown, but people just needed those haircuts, apparently

And the government was being stupid about funding people

AttractorNotStrangeAtAll

Lol. I need a haircut also. My gf is cutting my hair and it's horrible. But I'm really afraid of going to the hair salon

Akiva Weinberger

I've heard horror stories coming from the Dakotas

Surprised it's managed to penetrate a region so sparse

I haven't cut my hair since Feb. btw

I think I like it long lol

Occasionally trimmed my beard

Rithaniel

2:08 AM

I shave my head, but let my beard grow. It got really long before I just took some scissors to it.
Also, yeah, if it's hitting low density areas, you know it's contagious

Akiva Weinberger

I think SD is at a rate higher than NY was at its peak (7-8 months ago)

according to Johns Hopkins numbers

Dunno how the hell that happened

Here in NYC you can predict the infection rate of a neighborhood by how they vote

AttractorNotStrangeAtAll

How sad. What scary me the most is what comes next all of this, all that I read are bad news, social and economic crisis

theatlantic.com/magazine/archive/2020/12/… an interesting article if you guys have the time

3

Rithaniel

Oh that's delightful, when political affiliation results in disregard for personal safety

AttractorNotStrangeAtAll

(I don't know how The Atlantic is perceived in the US, so...)

Rithaniel

I mean, why did one party decide to downplay the disease? What is the motivation in that? What do they gain?

AttractorNotStrangeAtAll

2:26 AM

Concerning the problem I posted, I think I have a solution.

Because $n$ is even and $a_n$ is positive, $p(x)\geq 0$ for all $x\in\mathbb{R}$. Well, that means that $p[\mathbb{R}]$ has a lower bound, $0$. As $p[\mathbb{R}]\subset \mathbb{R}$, there exists $\inf{(p[\mathbb{R}])}\in\mathbb{R}$, which we'll call $M=\inf{(p[\mathbb{R}])}$. From now I think I might use TVI to prove that there exists a $x_0\in\mathbb{R}$ such that $p(x_0)=M$, and because $M$ is infimum, then $p(x_0)\leq p(x)$ for all $x\in\mathbb{R}$ as desired.

Is this the way?

Thorgott

I disagree with $p(x)\ge0$ for all $x\in\mathbb{R}$

think of some quadratics

but you can show that there is a lower bound and that will indeed be a way to arrive at the conclusion

AttractorNotStrangeAtAll

@Thorgott Ok, thanks!

Thorgott

3:32 AM

I realized I don't fully buy this. I believe your argument is "if $E/\overline{\mathbb{Q}}$ is algebraic, then $E/\mathbb{Q}$ is algebraic, so $E\subseteq\overline{\mathbb{Q}}$ since $\overline{\mathbb{Q}}$ contains all numbers algebraic over $\mathbb{Q}$ by definition", but that only works if $E\subseteq\mathbb{C}$ to begin with.
Why can this be assumed WLOG? Of course, we can appeal to the FTA and Steinitz and then it's immediate, but that just shifts the work to proving an arguably even harder theorem.

Though, on the other hand, I can't think of a proof of $\overline{\mathbb{Q}}$ being algebraically closed that doesn't implicitly rely on $\mathbb{C}$ being algebraically closed, so perhaps I'm just being unreasonable

I mean, $\overline{\mathbb{Q}}$ is defined within $\mathbb{C}$, so seems hard to say too much about the former without appealing to the latter

Leaky Nun

3:51 AM

@Thorgott we do need $\Bbb C$ to be algebraically closed

epic_math

4:39 AM

Hello

Edward Evans

6:26 AM

What do I get if I complete a polynomial ring? rofl e.g. $\widehat{K[t, t^{-1}]}$

Sayan Chattopadhyay

Is resultants the only way to do intersection numbers? It seems so tedious

Edward Evans

Still on AlgGeo?

Sayan Chattopadhyay

Yeah man, I have the endterms next week, and I do not understand Intersection numbers

Edward Evans

:56209252 $K((t))$

love_sodam

What is the algebraic closure of a finite field F_q?

Edward Evans

6:37 AM

$\widehat{\Bbb Z}$

err

that's the Galois group lol

love_sodam

Professor said that it's the union of all finite subfields

Why is that?

Sayan Chattopadhyay

Well over $\Bbb{Z}_p$ you have polynomials of the form $x^{p^n} - x$, and the splitting field of theses polynomials give you $F_{p^n}$, so all these guys have to be there in the closure. So the union of these guys is there in your closure. I do not know how to show this union is the algebraic closure

Edward Evans

I think you take $\varinjlim \Bbb F_{p^n}$ with the $\Bbb F_{p^n}$ ordered by divisibility

Sayan Chattopadhyay

Yeah that ^

Edward Evans

I can't remember the deets

love_sodam

6:47 AM

I don't know varinjlim

haven't learn that

I google this problem and many says that it's a union of all finite field

But professor said it's a union of all finite subfield

I'm confused

Edward Evans

Every finite field of characteristic $p$ is $\Bbb F_{p^n}$ for some $n \in \Bbb N$

love_sodam

Yea

Algebraic closure of F_p is F_p?

Edward Evans

No it's $\overline{\Bbb F_p} = \bigcup_{n \in \Bbb N} \Bbb F_{p^n}$ as your prof said

love_sodam

That's what professor said? he said the union of all finite 'subfield'

isn't that mean it's a subfield of F_p?

Edward Evans

$\Bbb F_p$ is the prime field of characteristic $p$, there aren't really any subfields of $\Bbb F_p$

love_sodam

7:02 AM

Ok I what is the meaning of 'the union of all finite subfield'? I don't know why this is equivalent to \overline{\Bbb F_p} = \bigcup_{n \in \Bbb N} \Bbb F_{p^n} for the case F_p

Edward Evans

can you use $ signs around your latex

your prof probably just means the union of all finite char p fields, that's literally what the algebraic closure is

love_sodam

Ok. Then the algebraic closure of $\Bbb F_{p^n}$ is $\bigcup_{n\in\Bbb N}\Bbb F_{p^n}$?

same as $\Bbb F_p$?

This is equivalent to saying that $\bigcup_{n\in\mathbb{N}} \mathbb{F}_{p^{n!}}$ right?

Edward Evans

yes, for every $n \in \Bbb N$ there is a finite field with $p^n$ elements, these are all the splitting fields of polynomials over finite fields, so just stuff them all together I guess

love_sodam

I think the above expression makes easier to prove

Edward Evans

what why

love_sodam

7:17 AM

maybe inclusion relation between them?

But I just notice that

Doesn't change that much

Edward Evans

$\Bbb F_{p^3}$ isn't in that union

love_sodam

But contained in $\Bbb F_{p^3!}$

Edward Evans

lol yeah you're right

i gotta go, lectures call, just look up the proof, it's essentially stuffing all the splitting fields of polynomials over finite fields together

assets.press.princeton.edu/chapters/s9103.pdf

love_sodam

in the proof, why the consider $f\in U[x]$?

Only need to show $f\in \Bbb F_q[x]$ splits in $U$

Leaky Nun

@EdwardEvans complete wrt what

love_sodam

7:31 AM

@EdwardEvans I don't understand why $U\subset\Bbb F_q$ since $\Bbb F_{q^n}$ is a subset of $\overline{\Bbb F_p}$.

I mean the red part

I think the red part is not necessary

Leaky Nun

I think you misread

love_sodam

Why is that?

Leaky Nun

it says U is a subset of F_q-bar

Edward Evans

@Leaky I'm not really sure tbh, $K$ is a complete valued field, maybe a topology inherited from $K$?

Leaky Nun

7:47 AM

I think if you take the topology that has $t^n K[t]$ as a neighbourhood basis of $0$ then you do get $K((t))$

love_sodam

Yes but I don't know why the red part is necessary. What does it show?

Edward Evans

I'm playing with $\bigoplus_{i \in \Bbb Z}\Bbb C_K(i)$ where $K$ is a p-adic field and $\Bbb C_K(i)$ is $\Bbb C_K \otimes_{\Bbb Z_p} \Bbb Z_p(i)$ and $\Bbb Z_p(i)$ is some twist of $\Bbb Z_p$ by the cyclotomic character, and this dude can be identified with $\Bbb C_K[t, t^{-1}]$ rofl

and I'm trying to prove that the $\operatorname{Gal}(\overline{K}/K)$ invariants of that guy and it's field of fractions are both $K$ and the tip is to embed it into the field of formal Laurent series over $\Bbb C_K$, but idk why I can do that

$\operatorname{Quot}(\Bbb C_K[t, t^{-1}]) = \Bbb C_K(t)$ and then complete that wrt the topology you mentioned up there I guess

love_sodam

8:13 AM

How can I prove $|(\Bbb F_q)^2| = \frac{|\Bbb F_q|+1}{2}$ where $(\Bbb F_q)^2 = \{x^2:x\in \Bbb f_q\}$

Leaky Nun

use the fact that squaring is a 2-1 map on the units

love_sodam

Oh that's amazing

Does it hold for any characteristic field?

I mean for any prime p?

epic_math

Hello

love_sodam

I think for only odd p

epic_math

you all know about Goldbach's conjecture

for every even integer $n$ greater than 2, there exists an ordered pair $(p_x,p_y)$ where $p_x$ and $p_y$ are odd primes such that $p_x+p_y=n$

I want to share some of my ideas. Can I?

First consider all pairs $(x,y)$ such that $x$ and $y$ are odd integers such that $x+y=n$

And now consider pairs $(u,v)$ where $u$ and $v$ are odd integers divisible by an integer $>2$ and $u+v=n$

if for all odd integer $n$ we have $$o(x,y)>o(u,v)$$ where $o(x,y)$ is the number of pairs $(x,y)$ and $o(u,v)$ is the number of pairs $(u,v)$ then Goldbach's conjecture is true.

The next step is to count all pairs (u, v) where u + v = n and either u is not prime or v is not prime. To do this, we must find all pairs where the x or y coordinate is divisible by 3 but $x \neq 3$ and $y \neq 3$. Then find all pairs where x or y is divisible by 5, then 7, then 11 etc. until we reach a prime number p, such that there are no pairs divisible by p.

Then we sum up all the non-prime pairs. If for all even integers n, the total of non-prime pairs is less than the total number of pairs, then we have proven Goldbach’s conjecture

you people following what I am saying?

now we can analyze the cases but lol I don't know what it leads to

I just shared my ideas :)

love_sodam

9:07 AM

Is there anyone know the latex code of this kind of thing?

Edward Evans

\begin{cases} \end{cases}

$$A = \begin{cases}1\ \text{if}\ n\ \text{odd}\\ 0\ \text{if}\ n\ \text{even} \end{cases}$$

love_sodam

I have a problem

$\Bbb F_q = \{0,1,\alpha,\alpha^2,...,\alpha^{p^n-1}\}$ where $\alpha $is a generator of $(\Bbb F_q)^\times$. Hence the sum of all elements of a finite field $\Bbb F_q$ is $1+\Sigma_{m=1}^{p^n-1}\alpha^m = 1+\frac{\alpha(\alpha^{p^n-1}-1)}{\alpha-1} = 1$ as $\alpha^{p^n-1}= 1$

epic_math

yes tell it

love_sodam

Assuming $q\neq p$

so $\alpha\neq 1$

But for $\Bbb F_4$, the sum of all element (0,1,\alpha,\alpha+1) is 0

My conclusion above is 1 for such thing

What's the problem?

Edward Evans

@love_sodam you have too many elements here

love_sodam

9:19 AM

Too many?

Edward Evans

count the number of elements in the $\Bbb F_q$ you wrote down above

love_sodam

OH

1 is already there

0

Q: Computation of $[\mathbb{Q}(\sqrt[p_1]{q_1}+\sqrt[p_2]{q_2}+\sqrt[p_3]{q_3}):\mathbb{Q}]$

If $p_i,q_j$ for $1\leq i,j\leq 3$ are all distinct primes and $\alpha = \sqrt[p_1]{q_1}+\sqrt[p_2]{q_2}+\sqrt[p_3]{q_3}$, then what is a degree of $\mathbb{Q}(\alpha)$ over $\mathbb{Q}$? I think I need to use some Kummer extension or radical extension to solve this but I don't know how to appr...

Is there anyone can do this?

TheReal__Mike

9:50 AM

Hi guys

and gals

123

Hi guys. Pls explain in complex number how these two sides of equality is different?

|z_1| - |z_2| <= | |z_1| - |z_2| |

Alessandro Codenotti

10:10 AM

@EdwardEvans what does completion mean here

Edward Evans

10:26 AM

@Alessandro I mean, we're completing fields wrt metric topologies and then defining rings related to these fields and then "completing them". They're not $I$-adic completions or something, so I'm just trying to work out what topology we're working with tbh

robjohn

10:57 AM

@123 If $|z_1|\lt|z_2|$, then the left side is negative

love_sodam

11:16 AM

If $\varphi(n)\mid\varphi(m)$ then is there any relationship between $n,m$?

$\varphi$ is an euler-phi fucntion

yh05

Hello all, I have a question about accuracy of converting decimal to binary floating-point number, and to decimal. Please help if possible, thank you.

https://math.stackexchange.com/questions/3914836/proof-of-accuracy-of-converting-decimal-to-binary-floating-point-number-to-decim

user2103480

12:11 PM

@EdwardEvans get rekt

Leaky Nun

@EdwardEvans you're stepping into perfectoid space territory

Edward Evans

well, we're doing p-adic Galois reps and going from Schneider's book, which replaces Fontaine's original methods with perfectoid methods from sChOlZe

love_sodam

I think $\Bbb Q(\sqrt[p_1]{q_1}+\sqrt[p_2]{q_2}+\sqrt[p_3]{q_3}) = \Bbb Q(\sqrt[p_1]{q_1},\sqrt[p_2]{q_2},\sqrt[p_3]{q_3})$

geocalc33

12:33 PM

Edward, can you analytically continue $\big\{ k\zeta(s): k\in \Bbb N \big\}$ (so multiple copies, one for each $k$)

feynhat

@BalarkaSen iNtErGaLaCtIc

love_sodam

Oh I think I solved that

RewCie

Great!

love_sodam

12:51 PM

Oh

AttractorNotStrangeAtAll

Hello everyone. Here https://math.stackexchange.com/a/693531/838368 Schunang says that

For any $\varepsilon>0$, there is a $N>0$ such that for all $\left|x\right|>N$ we have $\left| \sum_{k=0}^{2n} a_k x^{k-(2n+1)} \right|<\varepsilon$.

I don't get it, where is this from?

love_sodam

no there is a very serious error in my proof

epic_math

hello

@AttractorNotStrangeAtAll you can ask in the comments

AttractorNotStrangeAtAll

Oh, ok! Didn't know if that was alright because this is an answer from 2014

epic_math

he will surely get it in the inbox

so no matter how old it is

can someone evaluate $$\sum_{k=1}^{n}\frac{1}{k!}$$?

love_sodam

12:55 PM

Let $K = \mathbb{Q}(\sqrt[p_1]{q_1}+\sqrt[p_2]{q_2}+\sqrt[p_3]{q_3})$. Then as $K\neq \mathbb{Q}$, $[K:\mathbb{Q}]\neq 1$. Since $K\subset\mathbb{Q}(\sqrt[p_1]{q_1},\sqrt[p_2]{q_2},\sqrt[p_3]{q_3})$, $[K:\mathbb{Q}]\mid p_1p_2p_3$. WLOG, assume $p_1\mid [K:\mathbb{Q}]$.

epic_math

$n$ being a natural number

love_sodam

Then $\sqrt[p_1]{q_1}\in K$ since if not, $[K(\sqrt[p_1]{q_1}):K]\neq 1$ as $x^{p_1}-q_1\in K[x]$ has $\sqrt[p_1]{q_1}$ as a root and $p_1$ is a prime, $[K(\sqrt[p_1]{q_1}):K] = p_1$ which is a contradiction as $K(\sqrt[p_1]{q_1})\subset \mathbb{Q}(\sqrt[p_1]{q_1},\sqrt[p_2]{q_2},\sqrt[p_3]{q_3})$ (Note that $p_1,p_2,p_3$ are distinct).

Now, suppose $\sqrt[p_2]{q_2}\notin K(\sqrt[p_1]{q_1})$. Then, $[K(\sqrt[p_1]{q_1})(\sqrt[p_2]{q_2}):K(\sqrt[p_1]{q_1})]=p_2$ using the same argument before. But then, as $K(\sqrt[p_1]{q_1})(\sqrt[p_2]{q_2}) = \mathbb{Q}(\sqrt[p_1]{q_1},\sqrt[p_2]{q_2},\sqrt[p_3]{q_3})$, $[K(\sqrt[p_1]{q_1}):\mathbb{Q}] = p_1p_3$. Hence, $\sqrt[p_3]{q_3}\in K$ which is a contradiction.

RewCie

Is it some human based language y'all talking?

love_sodam

This was my proof but I think it's wrong

epic_math

the answer is $$\frac{e\Gamma(n+1,1)}{\Gamma(n+1)}-1$$

but I dunno why

@RewCie math is like talking in a code language

like your dot chat

RewCie

12:58 PM

What is $\Gamma (n+1, 1)$? Isn't $\Gamma: \mathbb R \rightarrow \mathbb R$

epic_math

it is the incomplete gamma function

Incomplete gamma function

In mathematics, the upper and lower incomplete gamma functions are types of special functions which arise as solutions to various mathematical problems such as certain integrals. Their respective names stem from their integral definitions, which are defined similarly to the gamma function but with different or "incomplete" integral limits. The gamma function is defined as an integral from zero to infinity. This contrasts with the lower incomplete gamma function, which is defined as an integral from zero to a variable upper limit. Similarly, the upper incomplete gamma function is defined as ...

internet is not working well here

I am not good at evaluating partial sums

love_sodam

1:18 PM

Changed my proof

Could you check this?

I didn't use the fact that q_i's are distinct

RewCie

Do I need to introduce myself here?

Balarka Sen

@feynhat lol

RewCie

Imma computer science engineer. I can resolve any tech issue that you've ever encountered or any that exists in this universe ever. I have jacked into the Matrix all the times. I'm an IT God. Look upon me and despair.

love_sodam

Something must be wrong in there

KeithMadison

2:30 PM

Anyone happen to know of a good comparison series for n^2 / n^4-n-3

epic_math

@RewCie okay cure my computer. It has memz trojan, cryptolocker ransomware, plugX malware and mydoom worm.

@KeithMadison is it $$\frac{n^2}{n^4-n^3}$$ or $$\frac{n^2}{n^4}-n^3$$

Tobias Kildetoft

@RewCie I don't think I have seen anyone use the term "computer science engineer" before.

epic_math

you should put brackets correctly

love_sodam

Is $[\mathbb{Q}(\sqrt[p_1]{q_1},\sqrt[p_2]{q_2},\sqrt[p_3]{q_3}):\mathbb{Q}] = p_1p_2p_3$?

KeithMadison

@epic_math the former, and thanks!

epic_math

2:54 PM

so friends, I want to ask a question to all of you

what is the most complicated thing in math you know? (I did not write fields, I wrote thing, so note that)

The most complicated thing I know is a very hard proof of onw of ramanujan's formula

too long to even mention here

Tobias Kildetoft

So by "thing" do you mean a proof, or?

epic_math

anything

like a theorem, formula, proof or anything

(which you understand easily)

(and which is the most complicated you know)

Tobias Kildetoft

Possibly that would be the proof that parabolic projective functors in type A are restrictions of projective functors

Though it might take me a bit if I needed to recall the full proof properly

epic_math

okay

it should be complicated

what about others?

Book Of Flames

3:17 PM

Though it is not complicated, but I think it is a very tricky question: Is there a correct proof for 2 + 2 = 5?

But I think the proof is not much of a "math" kind. Answer: Interchange the symbols that represent the numbers four and five.

Symbols 4 and 5, namely.

Book Of Flames

3:30 PM

And, rest all math objects are the same.

Thorgott

it depends on how you define the numbers

if you define 5 as the thing that we usually call 4, then 2+2=5 is true, if you define 5 as the thing that we usually call 5, then 2+2=5 is false (all assuming 2 is defined as it usually is)

though nothing interesting is happening in either case

Edward Evans

-2

Q: Is $\int 1 dx^{2}$ possible to solve?

I think this integral, $\int 1 dx dx$, should be possible by a substitution I just do not know which to use.

calculus

Thorgott

should say that as $1$-forms on $\mathbb{R}$, $dx\wedge dx=0$, so the integral is $0$

for max confusion

Edward Evans

3:46 PM

lol

geocalc33

If f: R_ -> (0,1) with f(x)=exp(x) then what is the product manifold (0,1)^3? I think I know the answer actually but i just want a second opinion

epic_math

I have a very simple proof of PNT

which I made myself

here it is:

we know that $1+1=2$, which implies $e^{\pi i}+1=0$, from which it easily follows that the nontrivial zeroes of the zeta function have the real part of half, which in turn implies the abc conjecture is true. Now it is obvious that PNT is true from these statements.

can you find any flaw in my proof?

geocalc33

4:01 PM

That’s completely wrong

epic_math

4:22 PM

@geocalc33 you can't follow my reasoning

This is why you're saying it

Tobias Kildetoft

@geocalc33 It is mostly a waste of time to read what epic_math writes here

9

epic_math

Lol

@TobiasKildetoft you are completely right

I will take it as a compliment :)

Tobias Kildetoft

It was not meant as one

epic_math

But I will still take it as one

It is getting stars quickly and one of them is mine

Lol my insult is being loved here

Hey I don't always talk waste

RewCie

4:43 PM

@TobiasKildetoft That's why I use it!

@epic_math Take the plug out!

Mary Star

Is someone of you familiar with polynomial interpolation?

Tobias Kildetoft

Also, never before heard anyone using any of those three words to describe themselves advertise that what they do well is "tech support"

I usually pretend to not be able to do anything remotely similar to tech support (except when the customer calls of course)

user2103480

@TobiasKildetoft since I had to work in tech support, my respect for tech support grew immensely

Mary Star

Here is my question:

0

Q: Give an upper bound for the uniform error

Let $f\in C^5(\mathbb{R})$, the $5$th derivative is bounded uniformly by $B$. Let the interpolation point be $(x_i)=(-2\ 1 \ 3 \ 4 \ 5)$. Give an upper bound for the uniform error of the polynomial interpolation by the unique polynomial of degree $4$ for $f$, for the interval $[-2,5]$. I have don...

numerical-methods interpolation error-function

user2103480

So many quirky stupid details to remember

Tobias Kildetoft

4:47 PM

@user2103480 Ohh, I have plenty of respect for people who do tech support well. I just try not to have to do it due to the terrible questions that tends to get asked

user2103480

@TobiasKildetoft sadly, I was not among the people who did it well

due to the terrible questions

Tobias Kildetoft

So I only do tech support for a system where it is not the primary thing, and where I have full access to all source codes and so on

user2103480

@TobiasKildetoft one time, my bosses assigned me to help someone with a question about a program's professional/admin version. There was neither an FAQ nor had I access to the admin version.

epic_math

Oh I forgot to tell you people that today is Benoit Mandelbrot's birthday.

Tobias Kildetoft

@user2103480 ouch

user2103480

4:49 PM

yeh

epic_math

Okay bye

Don't insult me in my absence XD

user2103480

bye, no worries!

RewCie

Anyone alive during COVID

?

user2103480

hopefully most after

RewCie

There's no after

Astyx

4:59 PM

hi chat

RewCie

hi Astyx

user2103480

For motivation: splitting short exact sequences are exactly the tool to show that a group can be decomposed as a sum of two others, right?

Sayan Chattopadhyay

A different kind of a sum @user2103480. A semidirect product of the two others

user2103480

Thanks!

Thorgott

if you work in any abelian category (so, in particular R-modules over a commutative ring R or abelian groups), then a SES is left split iff right split iff middle term is isomorphic to direct sum of outer terms

user2103480

5:09 PM

@Thorgott no thanks

jk good to know

Thorgott

if you just work in the category of groups (which is not abelian), a SES right splits iff the middle term is a semi-direct product, as Sayan says, but i believe left splitting is still equivalent to being a genuine product

the remark about abelian categories is general nonsense, but it's a useful thing to know for modules

gets relevant when you study projective modules, for example

Tobias Kildetoft

@Thorgott The last iff is not quite right

being isomorphic is not enough

user2103480

I only care about what makes me understand my topology course hahaha

Thorgott

I should say the sequences are isomorphic

Tobias Kildetoft

right

Thorgott

5:18 PM

ah right, the thing is that a left splitting allows you to define a nice map $C\rightarrow A\times B$, which makes the diagram commute, but a right splitting doesn't allow you to do the same thing as you would only get a map from the coproduct $A\ast B\rightarrow C$ instead, so it's a consequence of products in Grp not being biproducts

Tobias Kildetoft

Right, the right splitting might not correspond to the inclusion of a normal subgroup

user2103480

I should more often make vague statements about algebra

it seems to generate a lot of discussion

Has there been any more mochizuki drama btw? I love that

Sayan Chattopadhyay

@Thorgott Oh I did not know this

geocalc33

5:49 PM

What does a direct product between O(1,1) and itself look like?

O(2,1)?

Thorgott

6:30 PM

tfw this (aimed at first-semesters) physics lecture spent one week defining vectors, operations on vectors and then differentiation rules and now it's suddenly doing Frenet theory

is this normal progression

user2103480

idk but I still feel like that in current lectures

our spde lecture goes from 0 to "mild solutions" of stochastic navier stokes in 7 weeks, and before that introduces a stochastic integral that integrates stochastic processes with hilbert-schmidt operators as values in one lecture, hille-yosida and semigroup theory applied to pde in the next one, and the week after follows a 5-page proof of existence of solutions defined via semigroup generators and said stochastic integrals

I see WoW's Illidan before me, screaming "YOU ARE NOT PREPARED"

But I am more prepared to that than the computational neuroscience grad students I take another course with

Thorgott

@user2103480 accurate description

on the other hand, a mild case of masochism should be a requirement to sign up for an spde lecture

user2103480

6:46 PM

There the same lecturer introduced in two lectures: fourier/laplace transforms, brownian motion, stochastic integrals, SDEs and their flow properties + the fokker-planck pde satisfied by the densities of some distributions (and the infinitesimal generator + its adjoint), stopping times and boundary value problems.

I really don't know how any of them manages to follow, they only had some maths prep course 2 months beforehand lmao

@Thorgott yeh. Problem is that I was lured in by all the cool general functional analysis generalization of the stochastic integral. It's super elegant

The PDE things, including pages of inequalities.... not so much

Sayan Chattopadhyay

Something similar happened in my mechanics course. The instructor had mentioned that he was going to take a very mathematical route to Classical Mechanics, focusing on symplectic and Poisson Geometry. But being titled classical mechanics, many physics students took the course.

Initially it was smooth, but then he went full algebro-symplectic mode. And the last week was on Kontseivich's deformation quantisation

AttractorNotStrangeAtAll

lol

Sayan Chattopadhyay

For the rest of it he wants to do, Scalar Field theory and some operad stuff. Let us see how that goes. I mean it is hard for me to fully follow, but he's good with stuff. But I really imagine the plight of the physics majors

user2103480

@SayanChattopadhyay sounds crazy

@Thorgott yeah ... I checked and he covers about half of a book which written as a result of a two semester grad course... in half a semester, before christmas

fml

Thorgott

lmao

user2103480

7:00 PM

I dont even see how any of this can have any physical application ever

Sayan Chattopadhyay

I am really liking this categorical approach to physics, especially Classical Mech that he has taken. I would have never have thought of physics this way. The categorical approach really cleans up a lot of mess. For instance the entire mess of units in physics can be cleared by defining G-Torsors. There's the diffeological spaces stuff that cleans up a lot of infinite dimensional lie algebra stuff that physicists do.

user2103480

From all I've seen, I'm pretty impressed by physicist's way of thinking, which doesn't seem so analytic in essence

Thorgott

nonono, you're not supposed to speak positively of categorical approaches in this chat

user2103480

Hahaha thorgott you know that I took a lecture on HoTT last semester?

love_sodam

Some one just answer the question I sent here long time ago. I think now everything is fine except that why q_i's should distinct primes

user2103480

7:03 PM

Categories with families as models for type theories, kan complexes and such

Sayan Chattopadhyay

@Thorgott Right, the geometry brains will devour me

Thorgott

yeah

starting to realize I'm surrounded by mutants

AttractorNotStrangeAtAll

If I'm finding more difficulty in exercises with $\varepsilon$-$\delta$ or $\varepsilon$-$N$, does that mean I didn't understand very well the definitions and what's going on with limits and continuity?

Because I would say that yes, I understand these concepts, but somehow these kinds of proofs in specific functions are not coming easily, especially when it's something like $\lim \frac{a^n}{n!}$

user2103480

It was cool, though too fast and sketchy oftentimes. I got hung up on details that the lecturer didn't like to spell out, and had to work through too many pages about simplicial set combinatorics that I should've just taken for granted by geometric intuition

also, for the life of me, I just couldn't wrap my head around many things involving pullbacks. I still know no good explanation for what they mean geometrically. I know substitution, equations and fiber bundles are a sort of special case, but in my fields I don't have a lot of applications

non-algebra brain

Thorgott

pullbacks are just fancy intersections, kind of

@Attractor I don't believe there's any reason to involve $\varepsilon-\delta$ to calculate that limit

that just obscures what's going on

user2103480

7:12 PM

yeah, I got so much from simplicial sets, but it wasn't so much help oftentimes

@Thorgott I think it's an exercise

Thorgott

and what's going on is that you always multiply the nominator by the same number, yet the denominator by larger and larger numbers

user2103480

I remember that one from analysis I, iI think it needed some kind of small trick to show that it converges to 0

Thorgott

yeah, I hate exercises that force students to use $\varepsilon-\delta$ in places where there's no need to

user2103480

It's a necessary skill to get fluent in it

construction of lebesgue measure and integral needs enough such approximations

Thorgott

I agree, but there are enough other things one could do where epsilon-delta comes naturally as proof strategy and doesn't just feel contrived

user2103480

7:15 PM

So it's better to get fluent earlier than later

@Thorgott yeah but for example, they don't have logarithms available

I'd assume

Thorgott

no need for logarithms

what I said about the nominator and denominator can be turned into a perfectly clear proof

AttractorNotStrangeAtAll

Indeed, it's an exercise, but still... I just don't want to be discouraged to keep studying math just because I'm finding these proofs painful

user2103480

Yeah true I tried and logarithms just obscure. But I'd say most proofs of this fact are messy if they're in detail

Thorgott

7:36 PM

it doesn't take that much work

so here's perhaps a starting point: if you think about constructing the sequence inductively, you multiply with $a/n$ at the $n$-th step and this term will eventually be less than $1$

user2103480

it should be easiest to do some inequality with powers of two I'd think

user2103480

7:59 PM

Ok, so I have a topological space $X$ and I want to describe homology groups as direct sums of path-components $X_i$. Does anybody want to check what algebraic assumptions I'm implicitly using when calculating around?

Since every singular simplex's image is path connected, it must be contained in exactly one path component, and so a formal sum of simplices can be decomposed as sums of simplices with values in respective path components, i.e. if $C_n, C_n^i$ are the singular $n$-chains in $X, X_i$, then $$C_n \simeq \bigoplus_{i \in I} C_n^i \text{~~~(direct sum)}$$

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$Mathematics$ Mathematics