Mathematics - 2023-06-01

1:07 AM

@onepotatotwopotato but is a maximal ideal $\mathfrak{m}$ unique?

existenece is quite obvious

shintuku

1:22 AM

Suppose $k$ is an algebraically closed field, $\mathfrak m$ is a maximal ideal of $k[x_1,\dots,x_n]$, and $\phi:k \to k[x_1,\dots,x_n]/\mathfrak m$ is an isomorphism. How come $x_i \equiv \phi^{-1}(x_i \mod \mathfrak m) \mod \mathfrak m$?

Ted Shifrin

So, as always, what’s an example?

shintuku

i'll try working over $\mathbb C$ and see what I can get

Ted Shifrin

Sorta irrelephant, but sure.

one potato two potato

ah wait, $k[x]/(x-a)\simeq k$ for any $a$ and $(x-a)\neq (x-b)$ for $a\neq b$.

uniqueness is false

oh $k$ is not $k[x]$-module...

Ted Shifrin

1:46 AM

Why not?

one potato two potato

hmm... you mean evaluation.

Ted Shifrin

No, I mean letting $x$ act trivially.

one potato two potato

that's evaluating at $1$ isn't it?

Ted Shifrin

No? At $0$?

one potato two potato

trivially you mean $x\cdot a = a$ or $ = 0$?

Ted Shifrin

1:52 AM

$0$

I guess you could do yours, too.

one potato two potato

but anyway my example doesn't work.

once I fix evaluation $0$ then $\Bbb C[x]\to \Bbb C$ by $f(x)\mapsto f(a)$ is not $\Bbb C[x]$-linear map

Because $\mathfrak{m}\neq\mathfrak{n}$ does not imply $R/\mathfrak{m}\neq R/\mathfrak{n}$, I doubt such $\mathfrak{m}$ is unique (also nobody in the internet says uniqueness). there should be a counterexample...

leslie townes

2:19 AM

ted: duck crash landed in the pool today while we were in it. i think we're being stalked.

shintuku

i think i'm getting closer. i'm trying to study the specific case of $\phi: \mathbb C \to \mathbb C/\langle x-7,y-i\rangle$, and i'm trying to see whether, if $\phi$ is an isomorphism, it should map $x$ to $7$ and $y$ to $i$

shintuku

2:32 AM

edit: flash of insight

Ted Shifrin

@leslietownes And appropriately so!

Ted Shifrin

2:48 AM

@shintuku Does this make sense?

shintuku

no it doesn't, that statement has the elements in the wrong domain/codomain

for the specific case of an isomorphism $\phi:\mathbb C \to \mathbb C /\langle x-7, y-i\rangle$, we have $x - \phi^{-1}(x+\mathfrak m) = x - \phi^{-1}(7+\mathfrak m)$, i think if i properly understand what this implies i'll get the general case

shintuku

3:12 AM

i get it: the isomorphism $\phi$ must be the composition of the two canonical maps. so $\phi:k \to k[x_1,\dots, x_n] \to k[x_1, \dots, x_n]/\mathfrak m$. now since $\phi$ is surjective, there is an element $c_1 \in k$ such that $\phi(c_1) = x_1 + \mathfrak m$. But furthermore, by our definition, we have that $\phi(c_1) = c_1 + \mathfrak m$. therefore, $\phi(c_1) = x_1 + \mathfrak m = c_1 + \mathfrak m$, so $x_1 - c_1 \in \mathfrak m$, that is, $x_1 - \phi^{-1}(x_1+\mathfrak m) \in \mathfrak m$

thus establishing the nullstellensatz 😎😎😎😎😎😎😎😎😎😎😎😎😎😎😎

shintuku

4:12 AM

i have just read a piece of bourbaki. i am now a participant in the Mathematical Canon

shintuku

4:37 AM

i'm not sure that gives a surjection after some thought

it gives a surjection if you presuppose the nullstellensatz but that's what i'm trying to prove

at ted: any thoughts?

one potato two potato

6:51 AM

interesting: $R$-module category and $M_n(R)$-module category are equivalent

Sourav Ghosh

Is Lebesgue sigma algebra generated by any topology on $\Bbb{R}$?

Obviously such topology (if exists) would be strictly finer than the usual topology.

D Left Adjoint to U

@shintuku some more twin prime crackpottery by me:
https://math.stackexchange.com/questions/4710485/how-to-project-larger-prime-gaps-onto-strictly-smaller-nonzero-prime-gaps-does

one potato two potato

7:37 AM

How can I get an isomorphic $Q\otimes_{M_n(R)}P\simeq R$ as an $(R,R)$-bimodule where $Q$ is a set of row vectors and $P$ is a set of column vectors. It should be $a\otimes b\to ab$ but I can't find the inverse of this map. $R$ is a unital ring.

one potato two potato

8:01 AM

Ah it's doable.

PNDas

I asked my friend for some measure theory questions. He couldn't remember it exactly but the question has something to do with $\mu<<\lambda$ and $\lambda<<\mu$ then something something happens.

Have you seen any such questions?

one potato two potato

it's a notion of absolute continuity

flowian

The definition of *hyperbolic* metric space is:
The geodesic metric space $X$ is hyperbolic if there exists $\delta \geqslant 0$ so that for any geodesic triangle $[x,y]\cup [x,z]\cup[z,y]$ and any $p \in[x,y]$ there exists some $q\in [x,z]\cup[z,y]$ with $d(p,q)\leqslant \delta$. Such $\delta$ is called hyperbolicity constant of $X$.

I want to prove that $\Bbb R^2$ is not hyperbolic. For this I negated the quantifiers in definition. Is this correct? I have:

$\Bbb R^2$ is not hyperbolic if for any $\delta \geq 0$ there exists geod. triangle $[x,y]\cup [x,z]\cup[z,y]$ or there exists $p\i…

PNDas

@onepotatotwopotato I know. But I am asking if anyone remember any questions along those lines.

one potato two potato

I recently read something about the Hausdorff measure and its dimension. It's surprisingly interesting despite it's a measure theory.

PNDas

8:07 AM

I guess we have to take $\mu,\lambda$ positive. By RNT, $d\mu=hd\lambda$ and $d\lambda=gd\mu$. So $h,g$ have to be non negative. and $d\mu=hgd\mu$. So $\int_E hg-1d\mu=0$ for all $E$.

So $hg=1$ a.e. .

But can we conclude something about $\mu,\lambda$ here?

Got it. I think he was talking about this: math.stackexchange.com/questions/3260420/…

Dannyu NDos

8:21 AM

This is just a silly question emerged from ignorance, but can double sharps and double flats appear in musical isomorphisms?

s.harp

@DannyuNDos What are the domains and codomains of these functions?

Koro

If X is a metrie space and A closed in X, show that a map $f:A\to S^n$ can
always be extended over a neighbourhood of A.

This means that we have to show that there exists an open set U in X containing A such that f can be extended (continuously) to U.
Is my understanding correct? Thanks.

Or, do we have to show for every neighborhood of A?

Dannyu NDos

@s.harp I mean, do the symbols 𝄪 and 𝄫 appear in any occasions?

@Koro "extended over a neighborhood of A" – that marks existence.

s.harp

@DannyuNDos Morally: No, the domain of sharp differs from its codomain, so you cant apply it twice. Same for flat. Technically: It could, if you have two different sharps you might as well compose them. Practically: Never.

Dannyu NDos

@s.harp Thx!

Sourav Ghosh

8:50 AM

@Koro See here

Dannyu NDos

Dang it, ninja'd!

I knew I gotta make use of Tiezte, but got confused with the retraction part.

Koro

@DannyuNDos thanks :-).

@DannyuNDos I think you are right that only existence has been asked. I just wanted to confirm it anyways.

With existence, $U=\tilde f^{-1}(R^{n+1}-\{0\})$ works.

$\tilde f$ is a (Tietze) extension of f.

flowian

Can anyone explain why $\Bbb R^2$ is not hyperbolic?

Koro

specifically, $\tilde f$ is obtained by Tietze on every $f_j$ (jth coordinate map of f).

Sourav Ghosh

Yes. We can apply Titze's extension theorem for component functions

Koro

9:03 AM

I suspect that it is not necessary for f to continuously extend to all of X.

of course, it is not necessary.

Silent

9:36 AM

@SouravGhosh I am extremely thankful for this insight, Sir! :-)

porridgemathematics

9:49 AM

@Koro Tietze extension would be overkill here, and in any case, it would extend it to all of X

whereas the question is only asking about a neighbourhood

so its clear that they want you to use specific properties unique to metric spaces (not normal spaces) to furnish the extension

you can use tietze, but then youre kind of missing the point of the exercise imo

Sourav Ghosh

Is Lebesgue sigma algebra generated by any topology on $\Bbb{R}$?
Obviously such topology (if exists) would be strictly finer than the usual topology.

porridgemathematics

@SouravGhosh you should be able to construct such a topology, based on the fact that a lebesgue measurable set looks like $B \cup N$, where $B$ is a borel set and $N$ has lebesgue measure zero

so replace $B$ with open subsets (in the euclidean topology), clearly the result is a topology on $\mathbb{R}$

and clearly this generates the lebesgue measurable sets

but this is definitely finer than the euclidean topology

also, you can play the same game with any completions of borel sigma algebras

this is not unique to $\mathbb{R}$

as for the properties of this topology, idk, but its at least hausdorff because it contains all euclidean opens

Koro

10:11 AM

@porridgemathematics it won't, in case of S^n in codomain.

@porridgemathematics in my case, Tietze's been proven for metric spaces using the fact that two disjoint closed sets can be separated by a continuous map.

but the later statement is otherwise known as Urysohn's lemma... For metric spaces though its proof doesn't use the full generality of normal spaces.

porridgemathematics

oh okay, i missed the S^n part

yeah indeed you wont be able to then, and if tietze was only proven for metric spaces then that proof is essentially what I thought they wanted you to reproduce to show this for a neighbourhood

you could extend it to all of $X$ if you had partitions of unity, but for metric spaces you wont necessarily always have those

using basically the whitney theorem

(well the proof of the whitney theorem)

Jakobian

10:27 AM

@porridgemathematics this is not closed under arbitrary unions

porridgemathematics

@Jakobian less than a measure zero set then, so $U \setminus N$, where $U$ is euclidean open

this should still generate the lebesgue measurable sets but wont suffer from the previous issue

oh wait

ok maybe this doesnt work either

hmm

i thought subtracting would fix the issue, because I was erroniously just taking arbitrary unions of sets of the form $U \setminus N_{\alpha}$

where $U$ stays fixed..

but im not sure how to handle the general case

(so I just get $U \setminus \cap_{\alpha} N_{\alpha}$ and im subtracting another null set)

oh wait this could work but i retract my previous statement about it being generalizable to arbitrary completions

but it will work for all second countable topological spaces

basically we can just reduce to the case of $U$ being the same with different null set subtractions, because we only ever take countable unions when our underlying topology is second countable

so in $\bigcup_{\alpha} U_{\alpha} \setminus N_{\alpha}$, we let $O_{i}$ , $i=1,2,3,...$ be a countable basis, so that our union looks like $\bigcup_{j \in J} O_{i} \setminus N_{i}$ from the previous step, where $J \subset \mathbb{N}$ is some subset

so what I mean by that is rewrite each $U_{\alpha}$ , $\alpha \in A$, as a countable union over the $O_{i}$, then for every $i$, we are taking an arbitrary union $\bigcup_{\alpha \in K_{i}} O_{i} \setminus N_{\alpha}$, for some subset $K_{i} \subset A$

and this arbitrary union is of the form $O_{i} \setminus N_{i}$, where $N_{i} = \bigcap_{\alpha \in K_{i}} N_{\alpha}$

so this works for every second countable topological space, i.e. the completion of the borel sigma algebra for such a space is realizable as the borel sigma algebra w.r.t a finer topology

and yeah my previous effort was nonsense, because it would just generate the discrete topology..

Sourav Ghosh

11:10 AM

Finally I have managed to show that $\int_{0}^{\infty}\frac{\sin x}{x} dx=\frac{\pi}{2}$

Jakobian

11:20 AM

@porridgemathematics there is a paper by S. Scheinberg, "Topologies which generate a complete measure algebra" which discusses different topologies on $\mathbb{R}$ which generate the Lebesgue measurable sets as Borel sigma-algebra. This is one of them, in the paper it's called $T'$-topology

porridgemathematics

@Jakobian thanks for the reference, ill take a look!

@SouravGhosh how did you do it? i can think of one way via a clever contour integral trick (that i just remember and probably wouldnt be able to reproduce without memory of it), and another via a fourier transform

Sourav Ghosh

Contour integral.

porridgemathematics

thats using e^{iz}/z and then a semi-circular contour with a small bump at zero, right?

plus the residue theorem

Sourav Ghosh

Yes.

porridgemathematics

ah okay, yeah i guess its not necessarily a trick, theres another way via contour integration that avoids the bump, but requires modifying the numerator to something else

Sourav Ghosh

11:34 AM

Yes.

porridgemathematics

oh yeah, subtract one

so there is a removable sing at the origin

and then just use the residue theorem and take imaginary parts

Sourav Ghosh

$f(x)=\frac{\sin x}{x}$ is not Lebesgue integrable on (0, \infty) $

one potato two potato

11:55 AM

modular representation theory gave me a headache

user 858770

2:10 PM

Good morning Professor @TedShifrin

jay

4:28 PM

anyone see how this inductive argument works ?

s.harp

Do it by hand for $k=1$ and $k=2$

porridgemathematics

4:44 PM

@jay you can perform the swap (y_0,...,y_{k-1}) -> (y_{k-1},...,y_0) by the inductive hypothesis

then use tonellis theorem to move the \mu(dy_{k-1}) to sit next to the p_{t_{k} - t_{k-1}}(y_{k-1},dy_k), and use the k=1 case to transpose (y_{k-1},y_k) -> (y_k,y_{k-1})

you are given the k=1 case because that is the same as saying p_{t}(x,dy) \mu(dx) is symmetric in (x,y)

finally use tonelli again to move the \mu(dy_{k}) back to the right so its integrated against last

noballpointpen

5:23 PM

Hello, Ted! What is the standard proof of boundness of the image of uniformly continuous real function on bounded set $E\subseteq R$?

I have constructed two proofs so far, first was with Heine-Borel and continuous extension, but recently after your notes on my writing I found it messy (and I didn't bother to prove continuity), so at first instead of doing it, I tried to construct a cleaner argument, so the second deals with looking at the preimage and using Heine-Borel, too. Much cleaner and shorter.

shintuku

i use uniform continuous implies continuous implies bounded

Semiclassical

trying to figure out a pithy way to say and generalize the following

noballpointpen

shin, but continuity does not imply boundness. 1/x is continuous.

shintuku

continuous over interval my bad

noballpointpen

But we can't guess if the interval is closed.

shintuku

5:27 PM

theorem is only true for closed interval, no?

noballpointpen

Theorem I mentioned? No, it is true for every bounded set, which might be open interval (a,b).

shintuku

oic

Semiclassical

suppose i have two vectors $a,b$ in R^3 which i project onto the xy-plane. then the scalar product of these vectors before/after projection are $a_x b_x+a_y b_y+a_z b_z$ and $a_x b_x+a_y b_y$, i.e., they differ only by the product $a_z b_z$

it's basically just splitting up $\sum_k a_k b_k$ in some way, so it's not very interesting. but i feel like i'm forgetting a name for that fact

leslie townes

i dunno if it has a name beyond, like, orthogonality? if a hilbert space is an orthogonal sum of two subspaces, the inner product is the sum of the inner products in the subspaces. it sometimes 'looks more interesting' if they aren't coordinate subspaces but i don't know if it has a name.

maybe your people have a name for that :)

Sourav Ghosh

@noballpointpen Let $E\subset \Bbb R$ bounded then cl(E) is compact. Since $f$ continuous $f(cl(E)) $ is also compact. Hence $f(cl(E)) $ is bounded. Implies $f(E) $ is bounded.

We need only continuity of $f$ .

porridgemathematics

5:46 PM

@SouravGhosh you need uniform continuity to be able to extend to cl(E), otherwise continuity will get you things like tan on (0,pi/2), which does not have bounded image

noballpointpen

@SouravGhosh the problem is that $f$ can miss limit points of $E$, so that your $f(cl(E))$ is meaningless.

More precisely, $E$ can miss its limit points, so that $f$ is not defined on them. We cannot just use $f(cl(E))$.

porridgemathematics

@noballpointpen I dont think you need anything about continuous extensions: Consider $f(E)$, and suppose there are points $f(x_n) \in f(E)$ such that $|f(x_n) - f(x_{n-1})| \geq n$. Let $x_{n_k}$ be a subsequence converging to $x \in \text{cl}(E)$, since $f$ is UC, $f(x_{n_k})$ is cauchy , and thus convergent. Contradiction.

all I used was heine-borel

and uniform continuity

and it was basically two sentences

shintuku

anyone know where i can find exercises on k-algebras

porridgemathematics

just remember UC -> cauchy subseq for image sequences

then you dont need to worry about extending anything

noballpointpen

Why such sequence would exist?

porridgemathematics

5:54 PM

huh?

the whole point is $f(x_n)$ as I have defined doesnt exist..

if they did exist, $f(E)$ would be unbounded

since they dont, $f(E)$ is bounded.

thats why I end the proof with the phrase contradiction

noballpointpen

Now I see.

Didn't comprehend such a definition deals with assuming unboundness.

But it still seems suspicious. You define it by a growing gap between terms.

Sourav Ghosh

@porridgemathematics For my proof $f:\Bbb R\to \Bbb R$

noballpointpen

At first glance it seems to me that we could very well have a sequence in $f$ without a growing gap but which still shows our assumption about unboundness.

So it is not at all clear to me.

Am I right that you defined it basing on assumption about growing gap? Maybe I missed your idea.

Here is how I've done it in my second proof

Let $X$ be any infinite subset of $\overline{f(E)}$. Look at the preimage of $X\cap f(E)$. It is an infinite subset of $E$, which is bounded, so it has a limit point $e$ in $\overline{E}$. Applying $f$ to every Cauchy sequence in the preimage converging to $e$ we have it converging to a point in $\overline{f(E)}$. So $X$ has a limit point in $\overline{f(E)}$, so the set is compact, and since $f(E)\subseteq\overline{f(E)}$, we are done.

Overline is closure in Rudin's notation, btw.

Sourav Ghosh

@noballpointpen What is your question? What is the domain and codomain of $f$?

noballpointpen

Initially I asked Ted what is a standard approach. I am looking for a cleaner and accepted way. Just to look how it could be done by expert.

Just scroll up. My question is not so far ahead.

porridgemathematics

6:07 PM

@noballpointpen what does not bounded mean?

noballpointpen

For any given $M>0$, existence of $x$ for which $|f(x)| > M$.

Sourav Ghosh

Let $E\subset \Bbb R$ bounded and $f:E\to\Bbb{R}$ is uniformly continuous. Then $f(E) $ is bounded.

noballpointpen

I am not looking for help proving it, notice.

Yes, this is the statement, Sourav.

porridgemathematics

Good, now let M vary over the integers. Fix a point z in your unbounded set, and define the following sequence: x_1 is z, x_2 is a point outside the disk of radius 2 center 0, x_3 is a point outside the disk of radius 3+ |x_2| , and so on. For the converse, assume your set had a sequence with a growing gap but was bounded. Then the growing cap sequence has a convergent subsequence, which is absurd because such a subsequence is not cauchy.

That proves unboundedness coincides with “has a sequence with a growing gap”

So now you should be comfortable with the growing gap sequence I defined as a standin for being unbounded

noballpointpen

I think I got it how we could have such a sequence. Thanks for clarification.

porridgemathematics

6:13 PM

Yeah but you wanted to know about the converse

why having such a sequence implies unboundedness

if you werent then i have no idea what you mean by its not clear to you

noballpointpen

I already told you that now I appreciated how we could construct such a sequence in assumption of unboundness.

Anyway, I am going to leave right now. Thank you, porridge and Sourav, for the talk! :)

porridgemathematics

sure but that wont be enough for the validity of my proof is all, since im actually assuming the existence of such a sequence implies unboundedness…

alright

also what I gave you is a “standard approach” justf fyi

noballpointpen

@porridgemathematics Good. Then it is all done with. I am just saying you that such a sequence is tied with unboundness.

porridgemathematics

great, honestly the converse implication isnt super obvious on paper

its obvious by virtue of heine borel or visualizing a growing gap sequence

so you were right to question it

anyway night

PM 2Ring

6:44 PM

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Jakobian

@AlessandroCodenotti I found a cute proof that discrete space of size continuum is realcompact, you might be interested

biject it with $[0, 1]$, and using that any ultrafilter on $[0, 1]$ converges, take a local base $U_n$ of its limit $x$. If the ultrafilter has countable intersection property, then $\{x\}$ belongs to it, so it's principal.

So you just use that $[0, 1]$ is compact and first countable

Semiclassical

@leslietownes yeah. it's just one of those features that makes orthogonality such a powerful tool

someone on main was asking for an explanation of $E[(X-EX)(Y-EY)]=E[XY]-E[X]\cdot E[Y]$. if you think of random variables as vectors with inner product $E[XY]$, then $E[XY]=E[(X-EX)(Y-EY)]+E[X]E[Y]$ is that principle in action

Jakobian

all first countable compact Hausdorff spaces have at most size continuum, so we can't prove anything about bigger discrete spaces this way

s.harp

7:12 PM

@Jakobian The wiki page says something is realcompact if and only if its homeomorphic to some closed subset of $R^X$ for some $X$. If you have a discrete space $D$ isn't it hoemeomorphic to the "axis vectors" $\{ \delta_k \mid k \in D\}\subset R^D$?

Jakobian

7:43 PM

why are you saying this

it's not like I am looking for a proof of this, or using that definition

besides it's kind of not entirely true, since you need to add the zero vector

otherwise it won't be closed

noballpointpen

8:05 PM

To porridge: actually I thought about having only a sequence which grows with a constant gap, not thinking that we could just use its subsequence to arrive at the sequence in question :) Everything else is ok and understandable.

s.harp

@Jakobian cause I've never heard of this "realcompact" and looked it up

noballpointpen

8:44 PM

@porridge, also I am not so sure why you said "wont be enough for the validity". We have a bidirectional relation: such a sequence exists iff the set is unbounded. We can either assume $f(E)$ is not bounded and have such a sequence or just assume existence of such a sequence, as you did, which is less wordy, of course; both lead to the same contradiction.

noballpointpen

9:22 PM

Oh, it's just your wording "assuming". You could add "which is of course true", then, to make your sentence less confusing. Actually in your proof it is the implication: unboundness $\implies$ exists such sequence. Since we arrive at the condition of non-existence, this implies boundness, by contrapositive.

Petya Balabanov

9:34 PM

Hello everyone. Am I right that \ddot{x} = t^3/x^3 is equivalent to the system (d/dt)(x_1, x_2) = (x_2, t^3/x_1)? My advisor says that something is wrong in this

Ted Shifrin

10:12 PM

Did you forget $x_1^3$?

But otherwise this is standard. Please use dollar signs so your ChatJax will display.

Transcript for

$Mathematics$ Mathematics