Mathematics - 2020-10-20 (page 1 of 3)

user193319

00:45

If $M$ is an $R$-module, does $End_{\Bbb{Z}}(M)$ just denote group homomorphisms of $M$?

Lucas Henrique

No

user193319

Well, then what does it denote?

Lucas Henrique

Oh, sorry. Just saw the group homomorfisms

Yeah, every group endomorphism is also $\mathbb{Z}$-linear

Thorgott

yeah, $\mathbb{Z}$-homomorphisms, which is the same as group homomorphisms (since $\mathbb{Z}$-modules are the same things as groups)

orientablesurface

01:02

0

Q: tangent plane and tangent space coincide

Let $f:\mathbb{R}^3 \rightarrow \mathbb{R}$ be a smooth function. Suppose $f^{-1}(0)$ is a regular level set of $\mathbb{R}^3$. If $X_p\in T_p\mathbb{R}^3$ such that $X_pf=0$ show that for any $h\in C^{\infty}(\mathbb{R}^3)$ such that $h|_{f^{-1}(0)}=0$ we have $X_ph=0$. I was able to prove the...

user193319

01:17

0

Q: Complex Analysis and Bernoulli Numbers from $\frac{z}{2} \cot (\frac{z}{2})$

Define the Bernoulli numbers $B_n$ by $\frac{z}{2} \cot (z/2) = 1 - B_1 \frac{z^2}{2!} - B_2 \frac{z^4}{4!} - B_{3} \frac{z^6}{6!} - ...$ Explain why there are no odd terms in this series. What is the radius of convergence of the series? Find the first five Bernoulli numbers. I understand why t...

complex-analysis power-series analyticity bernoulli-numbers

Lucas Henrique

01:40

@TedShifrin so the exercise is correct?

Ted Shifrin

@Lucas ... depends on the defn. I am not following the room.

orientablesurface

Hiya @TedShifrin ..math.stackexchange.com/questions/3872942/… is this even true?

Ted Shifrin

03:05

@orientablesurface I don't know what you mean by your final sentence/question. What is the "tangent plane about $p$"? ... But, yes, the result is correct.

Clarinetist

@TedShifrin 'Evening, Ted - I have a question for you. Consider a measure space $(X, \mathcal{B}, \mu)$. Let's say I have a nonatomic measure (i.e., for each set $A$ with positive measure, there is a subset $B \subset A$ with $0 < \mu(B) < \mu(A)$) and I have some set $E$ with measure $\mu(E) < 1/3$, and another set with measure $\mu(F) > 2/3 - \mu(E)$. Is there any way I can construct a set using just $X$, $E$, and $F$ which is guaranteed to have measure between 1/3 and 2/3 inclusive?

The intuition behind this intermediate step is getting me more than anything

Oh, and assume $\mu(X) = 1$

At its crux, it's a probability problem

At least I think it is

If it is helpful, $\mu(E) = \sup\{\mu(B): B \in \mathcal{B}, \mu(B) < 1/3 \}$.

I thought about considering the set $E \cup (X \setminus (E \cup F))$. But in this case, I'm only able to demonstrate that it has measure between $\mu(E)$ inclusive and $2/3$. Close, but not enough.

Ted Shifrin

03:22

Clarinet: My days of proficiency with measure theory are 40+ years in the past. I like integration theory, but I'm useless for this. Take some geometry/topology. :)

Clarinetist

Lol, some day, I will

user480696

03:39

Some psychologists have done studies of math skills that range over 40 years.

user480696

iirc, they compared it to remembering how to speak a foreign language

Adam L

06:03

Lol yeah I'm sure there are some that have studied math for 40 years, they still chose the tertiary studies equivalent of the short bus to major in

Sayan Chattopadhyay

Maybe communications in algebra is becoming less about miscommunication. They retracted the Division algebra article

Adam L

I mean if QAnon was the best false flag they could come up with to take heat off the Epstein problem I don't know if psychological warfare is having a good year

Balarka Sen

07:31

@Thorgott lmao

lEarnIng oN mAnIfoLd

What is the proverbial description of a topologist who went from broke to millionaire after they switched to data science?

Ans: Earning on manifolds

northerner

07:59

I would like to clear up some notation convention that wasn't explained well in school. Often I would see something like $y=f(x)=x+1$. So what this says is the set $y$ is mapped to the set $x$ by the function, specifically the function named f, that is x+1

Is this right?

Also I've seen $y(x)$...what's that supposed to mean?

Leaky Nun

08:12

It means the variable y is the same as the function f applied to the variable x, which is equal to x+1

y(x) means y is a variable that depends on the variable x

northerner

so y(x) means variable y depends on variable x, BUT f(x) doesn't mean the function x depends on x...that wouldn't make sense, the the function is more of an artificial construct which doesn't depend on anything. It does take the argument x.

am I looking at it right?

TL;DR would it be incorrect to say given f(x), the function f depends on x?

Leaky Nun

08:32

yeah basically

Buddhini Angelika

09:02

If $T$ is a linear transformation from $\mathbb{Z}_p \times \mathbb{Z}_p$ where $p$ is a prime >3, and $T$ satisfies $T^3=I$, then can I say that the minimal polynomial of the linear transformation divides $x^3-1$?

Edit:
If $T$ is a linear transformation from $\mathbb{Z}_p \times \mathbb{Z}_p$ to $\mathbb{Z}_p \times \mathbb{Z}_p$ where $p$ is a prime >3, and $T$ satisfies $T^3=I$, then can I say that the minimal polynomial of the linear transformation divides $x^3-1$?

Leaky Nun

@BuddhiniAngelika sure

Buddhini Angelika

09:42

Thanks @LeakyNun how can I justify it?

Leaky Nun

what's the definition of minimal polynomial?

Buddhini Angelika

The minimal polynomial m(x) is the monica polynomial of smallest degree such that m(T)=0

Leaky Nun

@BuddhiniAngelika do you know that it divides any such polynomial?

Buddhini Angelika

Yes, I got it now.

Thanks. :) Also, a matrix representing the above T should always be a 2 X 2 matrix isn't it?

T(v)=T.v for any vector $v \in (\mathbb{Z}_p \times \mathbb{Z}_p)$, all v's are 2 X 1 vectors, so then shouldn't T be having dimension 2 X 2?

user193319

10:07

0

Q: Complex Analysis and Bernoulli Numbers from $\frac{z}{2} \cot (\frac{z}{2})$

Define the Bernoulli numbers $B_n$ by $\frac{z}{2} \cot (z/2) = 1 - B_1 \frac{z^2}{2!} - B_2 \frac{z^4}{4!} - B_{3} \frac{z^6}{6!} - ...$ Explain why there are no odd terms in this series. What is the radius of convergence of the series? Find the first five Bernoulli numbers. I understand why t...

complex-analysis power-series analyticity bernoulli-numbers

To calculate the $n$-th Bernoulli numbers, wouldn't I just take the $n$ derivative of $\frac{z}{2} \cot (\frac{z}{2})$, evaluate it at $z=0$, and then divide by $n!$?

strawberry-sunshine

Hi! Two questions are in need of urgent moderator intervention.

Who do I reach out to?

Two users have attempted to ask questions from an ongoing examination at my institution, in order to plagiarize the solution and compromise the sanctity of the examination.

I know the OPs.

I've requested for deletion by flagging one of them

flagged both now

user91500

10:27

@user193319 $\frac z2\cot(\frac z2)=1-\frac{B_2}{2!}z^2+\frac{B_4}{4!}z^4-...$

user193319

@user91500 Yes, I know that. And if $f(z) = \frac{z}{2} \cot (\frac{z}{2})$, that means $\frac{f^{(2)}(0)}{2!} = \frac{B_1}{2!}$, $\frac{f^{(3)}(0)}{3!} = \frac{B_2}{4!}$, etc...right?

which gives me a way of calculating the Bernoulli numbers, right?

Actually, I should've written $\frac{f^{(1)}(0)}{1!} = \frac{B_1}{2!}$, etc....That is, the index of the derivative order should be one less.

In other words, more generally $\frac{f^{(n)}(0)}{n!} = \frac{B_n}{(2n)!}$, right?

Stupid question inc

11:46

does doing drug makes you learn math better?

Astyx

No

Balarka Sen

lol

Stupid question inc

12:05

I have been on weed for a weeks and I found i learn rudin better than ever

Edward Evans

@Stupidquestioninc I smoked like 4g a day for a year and it decimated my memory and it's only just recovering

everything in moderation

also waddup

Astyx

I finally understood what blowing up was

Edward Evans

when you eat a super hot curry

Stupid question inc

umm i don't want to be caught in school for smoking

Edward Evans

... don't do it at school then? looool

Astyx

12:08

I think smoking will only get you bad consequences in the long run

I'm no expert though

Sayan Chattopadhyay

Wasn't there some story about TSA catching math people on an airport who were talking about blowups?

Astyx

LOL

Edward Evans

hahahaha

Sayan Chattopadhyay

Dude the explanation that would proceed after that would be hilarious

Astyx

"Have you tried blowing up the real plane ?"

3

Stupid question inc

12:10

erdos on amphetamine

Sayan Chattopadhyay

@Astyx This reminds of The Dictator helicopter scene lol

Astyx

I haven't seen this movie yet. Is it good ?

Stupid question inc

I think amphetamine helps you do math too but haven't try it yet

Sayan Chattopadhyay

Anything by Sacha Baron Cohen is gold. Though this one is relatively okay compared to his other work

Balarka Sen

i only liked ali g

Sayan Chattopadhyay

12:13

Dude did you see Who is America

Balarka Sen

no man

Edward Evans

The best part of Borat is that it caused Sacha Baron Cohen to be declared persona non grata in Kazakhstan

Sayan Chattopadhyay

Watch it dude

Balarka Sen

ill pass haha

his type of comedy is too retarded for my taste

but also thats exactly what makes it funny

Sayan Chattopadhyay

Lolol, I am a massive sucker for Political Comedy.

Balarka Sen

12:15

why watch political comedy when i can watch US politics live

Edward Evans

@Sayan have you heard of "The Thick of It"

Sayan Chattopadhyay

Nope

Okay on the list

Edward Evans

rofl

Balarka Sen

btw morales' party got elected in bolivia again

Edward Evans

it's quite funny

Astyx

12:16

@BalarkaSen Facts. Trump just said that people shouldn't vote for Biden because Biden would listen to scientists.

Balarka Sen

after they got couped by CIA last year

memes golden

Astyx

"just" meaning "one or two days ago"

Edward Evans

The commons just voted through a bill that essentially establishes a UK Stasi

hopefully the Lords shit on it

Sayan Chattopadhyay

@BalarkaSen Maybe it's another Pinochet waiting to erupt

Rithaniel

12:42

So, I'm told that $\mathcal{H}$ is a Hilbert space, but then asked to show that $\text{Ran}(T)$, where $T:\mathcal{H}\to\mathcal{H}$ is a bounded operator, is a subspace of $\mathcal{H}$ that isn't always closed. Isn't this contradictory because $\mathcal{H}$ is a Hilbert space?

Mike Miller

Why is that contradictory?

Rithaniel

Because any subspace of a Hilbert space is closed, right?

Astyx

What's Ran ?

Mike Miller

Why is that true?

Rithaniel

The range of $T$

Because a Hilbert space is complete. I thought one of the main descriptors of a Hilbert space was "all subspaces of a Hilbert space are closed"

Alessandro Codenotti

12:44

@Rithaniel What's your favourite example of an (infinite dimensional) Hilbert space?

Mike Miller

You're telling me a lot of things about Hilbert spaces but you still haven't given me an argument that all subspaces are closed

Rithaniel

I may have misplaced my memory, because I immediately thought of $l^2(\mathbb{N})$, but then $l_0(\mathbb{N})$ is a non-closed subspace

What is the main "attractive" property of Hilbert spaces, then, @MikeMiller

Alessandro Codenotti

The inner product

Rithaniel

You can have an inner product space that isn't Hilbert, though. Like $C[0,1]$

Mike Miller

Sure, but then you lose completeness, so you might have sequences that want to converge but don't.

I assume you're using the inner product from $L^2[0,1]$.

Every separable (inf-dim Hilbert space) is isomorphic to $\ell^2(\Bbb N)$, aka has a complete countable basis. You can use the basis (and the inner product) in arguments.

Many nice properties are shared between Hilbert and Banach spaces but the existence of a basis makes Hilbert spaces stand out.

Alessandro Codenotti

12:49

Sure but in an Hilbert space the inner product and the topology of the space are related in a good way

Rithaniel

Okay, so completeness/existence of a basis are the the things you get from a space being Hilbert

Mike Miller

(Completeness is part of the definition)

Rithaniel

Yeah, so I should say "what is meant when you say a space is Hilbert"

I seem to recall there was something of importance about closed subspaces of a Hilbert space

Alessandro Codenotti

I'm sure there's more intuitive good properties of closed subspaces of a Hilbert space, but the thing that comes to mind right now is that they are always complemented

Astyx

Sanity check : do continuous functions on a closed interval with $L^1$-norm form a Hilbert space ?

Rithaniel

12:57

Wait, is it that Hilbert spaces themselves are closed?

But a non-Hilbert infinite dimensional vector space isn't necessarily closed, as with $C[a,b]$

Astyx

Whatever the topology on $X$ is, $X$ is closed, so no

Sayan Chattopadhyay

Not with $L^1$ I think, maybe parallelogram law doesn't hold. You have to check though to be sure.

Rithaniel

Every finite dimensional subspace of a Hilbert space is closed, but this is true in any vector space, right?

Sayan Chattopadhyay

@Astyx Maybe take $x$ and $1-x$ on $[0,1]$

Astyx

I just need to show the set of continuous functions is a closed subset of $L^1([0,1])$, which is true because continuous functions on [0,1] are uniformly continuous (?)

Mike Miller

13:14

@Astyx No dude, approximate the indicator of [1/2, 1] by piecewise linear functions

Astyx

Oh yeah

I'm dumb

Sayan Chattopadhyay

But isn't it also enough to say that the norm is not coming from an inner product?

Just asking

Astyx

I think it does come from an inner product because it is induced by the norm on $L^1([0,1])$ which is a Banach space

Sayan Chattopadhyay

@Astyx Why so? A banach space norm need not come from an inner product

Astyx

Ignore me, I'm saying nonsense

Rithaniel

13:28

(Currently trying to come up with a proof that $\text{Ran}(T)$ might not be necessarily closed)

(I am thinking about constructing an example bounded operator $T$ in $l^2(\mathbb{N})$ for which $\text{Ran}(T)=l_0(\mathbb{N})$, but I have a sneaking suspicion that there is an easier way)

Clarinetist

13:49

Measure theory question for you all:

Consider a measure space $(X, \mathcal{B}, \mu)$. Let's say I have a nonatomic measure (i.e., for each set $A$ with positive measure, there is a subset $B \subset A$ with $0 < \mu(B) < \mu(A)$) and I have some set $E$ with measure $\mu(E) < 1/3$, and another set with measure $\mu(F) > 2/3 - \mu(E)$. Is there any way I can construct a set using just $X$, $E$, and $F$ which is guaranteed to have measure between 1/3 and 2/3 inclusive, assuming $\mu(X) = 1$?
The intuition behind this intermediate step is getting me more than anything.

user193319

With respect to the natural action of $M_n(k)$ on $k^n$, where $k$ is a field, I am asked to "describe" $Hom_{M_n(k)}(k^n,k^n)$.

What exactly do you think is meant by describe? Completely characterize?

Since the natural action of left multiplication by matrices includes scalar multiplication, it's not hard to see that any $f \in Hom_{M_n(k)}(k^n,k^n)$ is also a linear map.

Sayan Chattopadhyay

@Rithaniel Maybe this works, take $T : l^{\infty} \to l^{\infty}$, $T(\{x_n \}) = \{\frac{x_n}{n} \}$. Then consider $x_n = (1 , \sqrt{2}, \sqrt{3}, \cdots , \sqrt{n}, 0,0, \cdots )$ then, $T(x_n) = T((1 , \sqrt{2}, \sqrt{3}, \cdots , \sqrt{n}, 0,0, \cdots )) = (1, 1/\sqrt{2}, 1/\sqrt{3}, \cdots, 1/\sqrt{n}, 0, \cdots, 0, \cdots)$. Then $\{T(x_n) \}_n$ converges to $x = \{1/\sqrt{n} \}_{n}$. But there is no element in $l^{\infty}$ which maps to $x$.

user193319

Is there anything else that can be said about $Hom_{M_n(k)}(k^n,k^n)$?

Sayan Chattopadhyay

Hence the $\operatorname{Ran}(T)$ is not closed.

Rithaniel

Hmmm, okay, I think that works. All I need to do then is to show that $T$ is bounded

Astyx

14:00

@user193319 For every $y$, there's an $M\in M_n(k)$ such that $y=Me_1$ (where $e_1=(1,0,\dots,0)$)

user193319

@Astyx Yes, I used that to show that fact to show $k^n$ is a simple $M_n(k)$-module.

But are you saying that fact can also be used to describe $Hom_{M_n(k)}(k^n,k^n)$?

Oh, wait...a module homomorphism between two simple modules is either trivial or an isomorphism, right?

Astyx

This means that any $f\in Hom_{M_n(k)}(k^n, k^n)$ is completely determined by $f(e_1)$

Sayan Chattopadhyay

Which it does right, if $||x||_{\infty} \leq 1$, then for all i $sup(|x_i|) \leq 1$, but then $|x_i|/i \leq |x_i|$, so you have that it is bounded

user193319

Oh, so $Hom_{M_n(k)}(k^n,k^n)$ is cyclic?

in some sense?

@Astyx Does this mean that $f \mapsto f(e_1)$ is an isomorphism from $Hom_{M_n(k)}(k^n,k^n)$ to $k^n$?

Module isomorphism?

We can view $Hom_{M_n(k)}(k^n,k^n)$ as an $M_n(k)$-module, right?

Tobias Kildetoft

What structure do you have on the Hom?

How?

user193319

14:09

Nevermind...$Hom_{M_n(k)}(k^n,k^n)$ has a ring structure, not a module structure.

So, every $f \in Hom_{M_n(k)}(k^n,k^n)$ is completely determined by $f(e_1)$...is there anything further I can say?

Tobias Kildetoft

Well, you were right about some of the above things

Thorgott

it has both a ring and a module structure

user193319

hmm...so is $Hom_{M_n(k)}(k^n,k^n)$ isomorphic to $k^n$ in any sense?

Tobias Kildetoft

Yes, in the sense you would most commonly expect given the notation

user193319

As modules?

Thorgott seems to suggest that $Hom_{M_n(k)}(k^n,k^n)$ has a module structure.

Azmuth

14:13

C'mon

I need new haters, old ones became my fans

Tobias Kildetoft

Well, a k-module structure

user193319

Oh, so isomorphic as vector spaces.

Tobias Kildetoft

right

user193319

Ah, okay. Thank you! Let me see if I can write this up.

Thorgott

I don't believe this

Maybe I'm missing something, but the only elements of $\operatorname{Hom}_{M_n(k)}(k^n,k^n)$ should be homotheties, no?

Tobias Kildetoft

14:21

should be what?

Thorgott

scaling transformations

multiples of the identity

Tobias Kildetoft

why?

Thorgott

Cause an $f\in\operatorname{Hom}_{M_n(k)}(k^n,k^n)$ commutes with scalars from $M_n(k)$, so is an element of $Z(M_n(k))\cong k$

Tobias Kildetoft

Ahh, right

So what is preventing us from just sending $e_1$ to $e_2$ and extending this to a map on the entire space by the above mentioned uniqueness?

user193319

I don't understand...Are you guys saying that $f$ just acts by scalar multiplication?

Thorgott

14:29

uniqueness doesn't give existence

Tobias Kildetoft

right

Thorgott

aha, turns out I was wrong about something, tho

$\operatorname{Hom}_{M_n(k)}(k^n,k^n)$ is not a $M_n(k)$-module

apparently non-commutative rings are that ugly

user193319

So elements of $Hom_{M_n(k)}(k^n,k^n)$ are clearly linear operators and therefore have matrix representations. So are you saying that a matrix representation commutes with everything in $M_n(k)$?

Tobias Kildetoft

You generally need a Hopf-algebra to get a proper module structure on Hom spaces

Thorgott

but we have $\operatorname{Hom}_{M_n(k)}(k^n,k^n)\cong k$ as rings

user193319

14:33

Wait...so I'm confused now...

Tobias Kildetoft

it will probably help if you ignore everything I have said so far

(that holds in general, not just for this conversation)

user193319

hmm...okay...so back to square one then...What can we describe $Hom_{M_n(k)}(k^n,k^n)$ as?

Astyx

@user193319 If $F$ is the matrix of $f$, then for every $M, x$, we have $f(Mx) = FMx = Mf(x) = MFx$, leading to $FM=MF$ so F commutes with every matrix

user193319

Okay, so $f$ lies in the center of all linear operators, which means that $f$ is a scalar mutliple of the identity.

Thorgott

ye

Astyx

14:42

Can you do this if you don't impose $f(x+y) = f(x)+f(y)$ ?

Oh but that's induced by the $M_n(k)$ action

user91500

14:58

@user193319 The series I've written is different. Check your question once again. Also note that $B_{2n+1}=0$

Thorgott

@Astyx yes, I believe you don't need additivity

Mike Miller

15:33

@Astyx A normed space which is completely metrizable is in fact complete to begin with

I don't remember the proof but you can google it

It follows that the $L^1$ topology on $C[0,1]$ is not even homeomorphic to the standard topology

Astyx

That makes sense, cheers

Thorgott

16:10

Is there an easy/standard example showing that Diff doesn't have equalizers? I think I have one, but it's not particularly pretty.

Edward Evans

Hey @Tobias, do you have any idea about smooth representations?

if you're here lol

Tobias Kildetoft

Not much

Now, algebraic ones I know about

Edward Evans

tbh I don't think you need to know much to tell me I'm an idiot with the question I'm about to ask

Tobias Kildetoft

Let's find out, shall we?

Edward Evans

Suppose I have a rep $(\pi, V)$ of a (locally profinite but who cares) group G and define $V^\infty = \bigcup_{K\ \text{compact open subgroup}} V^K$ (the spaces fixed by the action of $K$). Is it obvious that $V^\infty$ is $G$-stable? Since every $v \in V^\infty$ is fixed by some compact open $K$, can't I just write $G = \bigsqcup gK$, so every $g^\prime \in G$ is $g^\prime = gk$ and then $\pi(g^{-1}g^\prime)$ fixes $v$

and then just.. do that for all the K's

(This $V^\infty$ is probably going to be the maximal smooth subrep of $V$)

Tobias Kildetoft

16:23

But your $g^{-1}g'$ is in $K$

Edward Evans

durr

Tobias Kildetoft

So all you concluded was that the elements of any such $K$ fix that subspace

(which is certainly true)

Edward Evans

(but not what I wanted)

yeah thanks lol

Tobias Kildetoft

On the other hand, you don't want it to be fixed, you just want it to be stabilized

Edward Evans

oh I read "G-stable" as "fixed by G"

Tobias Kildetoft

16:25

So you need to show that if $v$ is fixed by all elements of some open compact subgroup, then so is $gv$ for any $g$ (but not necessarily the same open compact subgroup)

Edward Evans

that sounds easier

and/or possible

Tobias Kildetoft

I assume the point here is that compact open subgroups will only have trivial smooth reps?

Edward Evans

I'm not sure, I think the point is just to construct a smooth rep from a not necessarily smooth rep.

Thorgott

I'll regret this, but what's a smooth rep

Edward Evans

Just a representation such that every $v$ is fixed by a compact open subgroup

Thorgott

16:31

why's that called smooth

Edward Evans

idfk I think it's equivalent to a continuity condition

like $G \times V \to V : (g, v) \mapsto \pi(g)v$ is continuous

where $\pi, G, V$ are the obvious notations

there's also "admissibility" which is that the spaces fixed by compact open subgroups are finite-dimensional, but I don't think I'd have guessed that by the word admissible

it's all just made up to confuse me

Thorgott

I mean, people call everything admissible when it's nice

but I don't see the connectino between this and classical smoothness

Edward Evans

shrugs locally profinitely

Thorgott

like, uh, if $G$ is Lie and $G\rightarrow\operatorname{Aut}(V)$ is smooth, then is that also smooth in your sense?

Alessandro Codenotti

@Thorgott Lie groups have very few compact open subgroups

Edward Evans

16:37

I think a lot of examples of locally profinite groups are p-adic Lie groups

so

Thorgott

"p-adic Lie group"

Edward Evans

don't ask me I just say what the book says man

Alessandro Codenotti

@Thorgott What else could you have possibly meant when you said Lie group

Edward Evans

p-adic Lie group in the sense that the groups have the structure of a p-adic analytic manifold obviously

P-adic quantum mechanics

p-adic quantum mechanics is a collection of related research efforts in quantum physics that replace real numbers with p-adic numbers. Historically, this research was inspired by the discovery that the Veneziano amplitude of the open bosonic string, which is calculated using an integral over the real numbers, can be generalized to the p-adic numbers. This observation initiated the study of p-adic string theory. Another approach considers particles in a p-adic potential well, with the goal of finding solutions with smoothly varying complex-valued wave functions. Alternatively, one can consid...

Thorgott

who even considers groups that aren't given by the classifying etale topos of a p-adic scheme

Astyx

16:54

A primary ideal Q of a commutative ring Q is an ideal such that if $xy\in Q$, then $x\in Q$ or $y^n\in Q$ for some $n>0$.
Why is the definition of a primary ideal not symmetric in x and y ?

Thorgott

it is , is it not

since the ring is commutative

Edward Evans

I always get the urge to play Skyrim when I hear the music, and as soon as I start playing Skyrim I realise that I find it kinda boring and what I really wanted was to literally be there

and then I cry in the foetal position

Astyx

Oh ok I get it

I can't play skyrim cause it gives me motion sickness

Balarka Sen

17:12

Primary ideals are amazing

Thorgott

Is there an easy proof of invariance of domain for smooth functions?

Balarka Sen

My favorite kind of commutative algebra

What is your version of invariance of domain

Thorgott

smooth injection U->R^n with U open in R^n is open map

Balarka Sen

Ya it's definitely easier than topological invariance of domain

Hm, maybe not.

Not sure, sorry.

Thorgott

how about an easy reason for why there is no smooth injection $M\rightarrow N$ if $\dim M>\dim N$?

Balarka Sen

17:17

Find a regular point by Sard's theorem

The preimage is a $\dim M - \dim N$ dimensional manifold

In particular more than a singleton.

NeverEnoughTime

What's the precise statement you're asking about? Is it

Let U be an open subset of R^n given induced smooth manifold structure, and f:U-->R^n be a smooth injective map, then with induced manifold structure f gives an isomorphism of smooth manifolds between U and f(U)?

Thorgott

@Balarka this sounds incredibly silly, but how do we preclude that $M$ doesn't inject as a measure zero subset

@NeverEnoughTime yes

Balarka Sen

What? I don't understand. Sard's theorem is the statement that there is always a regular value.

Why are you thinking measure zero etc at all.

Thorgott

take the zero map R->R, the differential is everywhere zero, so no point is regular

of course there are regular values, but they have empty preimages, so are useless

Balarka Sen

Oh fine. Yeah I don't know you can argue somehow

Thorgott

17:25

Sard's says that if f:R^n->R^m is smooth enough, the image of the set of critical points has measure zero. For your argument, we want a regular value with non-empty preimage, so the existence of a regular point. To get that from Sard, we need that the image has positive measure.

Balarka Sen

Yeah I don't care haha

You can argue

Thorgott

Sard's works to show that there is no surjection N->M

Balarka Sen

Find some measure theory theorem which says some preimage has positive measure lol

This is a waste of my time

Joe Shmo

does anybody know how to strike through an integral in latex?

$~\int~$

doesn't work

NeverEnoughTime

There's $\fint$?

Oh that uses a package

Joe Shmo

17:35

package.. yeah

NeverEnoughTime

Apparently the esint package

Joe Shmo

but what package?

graze

NeverEnoughTime

It's not a straight strikethrough though

Joe Shmo

I want a flat line

wtf why is that so difficult

NeverEnoughTime

Did you try googling it lol, I find lots of answers doing that in the first results

Joe Shmo

17:37

I did

NeverEnoughTime

11

A: Average integral symbol

The following example defines macro \mint{<symbol>} (\limits|\nolimits|\displaylimits)* _{...} ^{...} The first argument is the symbol that is put in smaller math style in the middle of the integral symbol. Then a limits specification follows, any number and order. The last one is used. Then ...

You saw that?

Joe Shmo

I was hoping for something easier..

whatever. that'll have to do

NeverEnoughTime

If something easier existed, why would it have 11 upvotes lol

Someone would write a comment "Wow, this is retarded, you can do this in 1 line by blah"

You can make a terrible version of it by hand

Joe Shmo

as in, theres gotta be a package out there

NeverEnoughTime

$\int\!\!\!\!\!\!-$

Joe Shmo

17:39

nope

you suck at this

NeverEnoughTime

??

$\int\!\!\!\!\!\!-$

Is that not what you want?

Joe Shmo

lol.. what do you think?

NeverEnoughTime

??

Joe Shmo

the answer on SE worked fine

NeverEnoughTime

$$\int\!\!\!\!\!\!\!-$$

Joe Shmo

17:40

almost

now you screwed it up again

keep trying. I'll check back in in an hour

NeverEnoughTime

Not sure if you're being a dick on purpose

Joe Shmo

no..

you were playing with the integral weren't you..? you even almost had it at some point

user193319

18:10

If $R$ is a commutative ring and $M$ and $R$-module, does the following define an $R$-module structure on $Hom_{R}(M,M)$? $(r \cdot f)(x) := f(rx) = rf(x)$?

Thorgott

yes

$R$ being commutative being integral here

user193319

Cool. Thanks!

Thorgott

@Balarka oh, it's actually just a consequence of the rank theorem

Lucas Henrique

hey

Astyx

hi

Lucas Henrique

18:19

sup

Ted Shifrin

Salut @Astyx. Hi, @Lucas. Did you ever settle the definition in your problem?

Astyx

Salut Ted

Ted Shifrin

@JoeShmo: When I've done "by hand" LaTeX fudges (e.g., making a transversality sign), I often end up using rlap and llap.

Lucas Henrique

hey @Ted. yeah, $\mathrm{Div}(f) := \{p \in \mathcal{P}(\mathbb{F}): \exists q\ (f = pq)\}$

Ted Shifrin

Hmm, so that includes non-proper divisors, so your complaint was right.

Lucas Henrique

18:21

I was being dumb. The exercise does not imply that every space is cyclic.

Ted Shifrin

Oh, I guess I didn't pay enough attention.

Sal

What does the vertical bar mean in a lot of maths that I see?

It typically looks like this: $f(x)\vert_0^\infty$

Lucas Henrique

Actually, if you have a polynomial $p = f_1 ^{\alpha_1}\cdots f_k^{\alpha_k}$ and an operator primary decomposition $V_{p(T)} = \bigoplus\limits_{j = 1}^k V_{f_j^{\alpha_j}(T)}$, it's not hard to show that, given $v_j$ with $m_{v_j} = f_j^{\alpha_j}$ (also easy to show that exists), $\sum\limits_{j=1}^k v_1$ has min poly $p$

Ted Shifrin

@Sal: $f(x)\big|_a^b$ is shorthand for $f(b)-f(a)$.

Sal

Thanks!

Ted Shifrin

18:26

@Lucas: Oh, $\sum v_j$ ?

Lucas Henrique

@Sal you may also see things like $f\big{|}_A$ which means the restriction of the function $f$ to the domain $A$.

In this context, it's what Ted said

Ted Shifrin

That was pretty clearly a calculus context, @Lucas :)

Lucas Henrique

@TedShifrin yeah, since the minpolys of $v_j$ are each two coprime

Ted Shifrin

I was just checking, Lucas. You had a typo.

Lucas Henrique

oh! lol

this linear algebra class is not fun at all

every theorem looks so damn arbitrary

5000 preliminary lemmas which do not look anyhow related. that's probably what happens in research, though. you try everything

can you confirm, @Ted?

Ted Shifrin

18:30

Well, I'm not a fan of that teaching style.

Confirm what?

Lucas Henrique

those apparently unrelated and arbitrary, not obvious lemmas do show up while in math research?

our professor said that it's a math thing. in physics, things look less arbitrary since reports have all the process to get the results. in math, people do a clean up and the dirty work disappears

Ted Shifrin

No, in research, you do things as you stumble upon them. You don't stack up a pile of preliminary lemmas. That's a style of book-writing and teaching which, as I said, I do not like.

Tobias Kildetoft

Often in research, the lemmas don't really become lemmas until you write up the final thing and realize that some part of the proof is more clear as a lemma on its own

Ted Shifrin

Sometimes in research, people use the "method of wishful thinking" and assume a result (as reasonable) to see if it gets them an interesting result. If it does, then after they come back to try to justify it.

Sha Vuklia

Wews, @TedShifrin, I saw you were in the chat and I just wanted to say hi.

Ted Shifrin

18:35

Hey there, @Sha. Nice to see you!

Tobias Kildetoft

And sometimes, that wishful thinking becomes a conjecture and several papers are written assuming it. And then someone goes and disproves the conjecture :)

Sha Vuklia

Also, olas @Astyx

Astyx

Oh wow, a wild Sha has appeard

How are you ?

Ted Shifrin

@Tobias, yes, there are instances of that :P

Sha Vuklia

XD

Tobias Kildetoft

18:36

(glad I was not the one to write those papers)

Sha Vuklia

I'm doing alright! Am reading in Atiyah & MacDonald atm

Joe Shmo

alright, thanks @TedShifrin I ended up doing it by hand..

Sha Vuklia

How are you's

Ted Shifrin

Wow, @Sha, a far throw from all your physics days.

Sha Vuklia

Ye, I sort of almost quit physics, tbh.

At least this year I'm only taking math courses.

Sayan Chattopadhyay

18:37

@Thorgott Sorry for interrupting but how exactly? The rank theorem needs the jacobian to have constant rank in a neighborhood, how do you guarantee that for a smooth function $f: M \to N$, dim(M)> dim(N) ?

Thorgott

every smooth map has constant rank somewhere

cause the set of points where it has maximal rank is open

Ted Shifrin

Two points to Thor.

@Sha: I remember the young and innocent Sha. :)

Sayan Chattopadhyay

Oh yes nice nice.

Sha Vuklia

@TedShifrin xD

Ted Shifrin

@Astyx: Is this like trying to land on a Möbius strip?

Sayan Chattopadhyay

18:58

Man I get distracted by the chat too much, I had set to do some AG, I ended up doing some functional analysis and smooth manifolds.

Transcript for

$Mathematics$ Mathematics