Mathematics - 2025-01-16

05:03

@HomesickIguana Hmm I'd heard a (generalized) Gauss map using Grassmann bundle, although I don't know any usage of it. It's a "natural generalization" but it's not natural to me because it looks like the purpose of that generalization is really just a generalization (of course this is because I'm not convinced of the real purpose of it).

The example I said is natural to me because it shows its usefulness.

I think I shouldn't ask this question it's a subjective matter.

one potato two potato

08:11

Interesting fact I found: Let $A\in M_2(\Bbb R)$ and $P\in\mathrm{SO}(2)$. If $\phi:M_2(\Bbb R)\to\Bbb R$ by $\begin{pmatrix} a & b\\ c& d\end{pmatrix}\mapsto b-c$, then $\phi(A) = \phi(PAP^{-1})$.

psie

08:54

Let $p,q$ be conjugate exponents, $\nu$ a $\sigma$-finite measure. I'm reading about the theorem that if $\Phi: L^p(E,\mathcal A,\nu)\to\mathbb R$ is a continuous linear form, there exists a unique $g\in L^q(E,\mathcal A,\nu)$ such that for every $f\in L^p(E,\mathcal A,\nu)$, $\Phi(f)=\int fg\,\mathrm{d}\nu$. Moreover, the operator norm of $\Phi$ satisfies $\|\Phi\|=\|g\|_q$.

If we use the notation $\Phi_g$ instead of $\Phi$, they say that the map $g\mapsto\Phi_g$ identifies $L^q(\nu)$ with the topological dual of $L^p(\nu)$ and is necessarily one-to-one. I wonder, how can one verify it's one-to-one? Here's my attempt. If $\Phi_g=\Phi_h$, then does this mean that $\int fg=\int fh$, i.e. that they agree at $f$? If so, it is straightforward to verify injectivity, but this I'm doubting.

Nasser

09:54

I must be overlooking something. The book says x=0 is solution to this ode

4 x (y')^2 = (3 x -a)^2

Where a is constant. But if we say take a=6, and plugin x=0, we get 0=36, so why does the book says x=0 is solution to the ode? Below is screen shot

The book uses p for y'(x) so it is saying the y-axis is solution. Should not then x=0 satisfy the ode at least?

1010011010

11:03

Hi there, I don't have sufficient information or knowledge to articulate a proper question on the main page. Hopefully someone can help me out.

I am trying to back calculate a fraction of moving averages.

By imposing physical constraints on the individual values that make up the averages.

I have coded a sort of rudimentary program that will keep raising individual values of a seed such that the moving average comes closer to the desired moving average.

Vuk Stojiljkovic

11:24

I understand that we write T(x,y) as T(x,0)+T(0,y) and then we project operator T as such H+H\to H (x_{1},x_{2})\to x_{i} but then from T(x,0) we get that T(x,0)=projection_{1}T(x,0)+projection_{2}T(x,0) and since T(0)=0 we obtain that
T(x,0)=proj_{1}T(x,0)=T(x) if I understood correctly, that is we obtain operator A from the matrix A B C D, in a similar manner we proceed with T(0,y) and we get a mapping on T(y) which I believe would be a mapping D. but how do we obtain the b,c operators, sorry for asking a lot, want to understand this

SoG

12:51

(a_n) is a sequence of real numbers and f_n : [0, 1] \to R is a sequence of continuous functions. Does there exist any Borel measure s such that a_n=\int_{0^1}f_n ds?

n=1, 2,...,m

Jakobian

14:27

@SoG If $f_1 = ... = f_m = 0$ then we need to have $a_1 = ... = a_m = 0$

If all $f_n$ are non-zero, pick pairwise disjoint open sets $U_n$ such that $f_n(x)\neq 0$ for $x\in U_n$ and define $s$ on $U_n$ individually

$\int_0^1 f_n ds = \sum_{k=1}^m \int_{U_k} f_n ds = \sum_{k=1}^m a_{nk} = a_n$

well, heuristically at least, $\int_{U_k} f_n\approx f_n(x_k)s(U_k)$ for $x_k\in U_k$

so I think we can phrase it as solving a system of linear equations $\sum_{k=1}^m f_n(x_k)s(U_k) = a_n$

we want to probably choose $x_k$ so that $(f_n(x_k))$ is an invertible matrix

If $f_n$ are not constant, I believe you can do that from continuity

this should solve it in the case that $f_n$ are not constant

If $f_n$ are constant, then $\int_0^1 f_n ds = f_n(0)s(0, 1) = a_n$ so we need for any such two $f_n$ to have $a_n/f_n(0)$ is a constant value

but I think having one $f_n$ to be constant is still okay

Thorgott

16:10

@psie the statement you quoted already has "unique" in it

psie

yes 👍

but one needs to prove it is unique, and I think it follows from my reasoning above, namely that $\int fg=\int fh\implies fg=fh \ \nu\text{-a.e.}\implies g=h \ \nu\text{-a.e.}$

Jakobian

16:26

@psie "but one needs to prove it is unique". The theorem you have quoted has "unique" in it, so the proof will be in the proof of that theorem

it's not that we don't need to prove that, but rather where we chose to prove that

and if you already have it proven somewhere else then repeating argument is redundant

psie

well, they only state "it is necessarily one-to-one, which gives the uniqueness of $g$ in the theorem"

Jakobian

@psie they do that for this theorem?

psie

so there is no proof of $g\mapsto \Phi_g$ being one-to-one

@Jakobian yes

Jakobian

so are those two things actually one

psie

yes, I think so. If $g\mapsto\Phi_g$ is a bijection (it is onto the dual of $L^p(\nu)$, since in the proof one started out with an arbitrary continuous linear form $\Phi$ and showed $\Phi=\Phi_g$), then for any continuous linear form on $L^p(\nu)$, we know it is of the form $\Phi_g$, for some unique $g\in L^q(\nu)$, because $g\mapsto\Phi_g$ is bijective.

Jakobian

16:49

@psie can I see the source

psie

@Jakobian I would love to show it to you, but it's a long proof. I can show you the paragraph where they state $g\mapsto\Phi_g$ is a linear isometry from $L^q(\nu)$ onto the dual of $L^p(\nu)$, and that it is necessarily one-to-one, which gives the uniqueness of $g$. By the way, this is just a more general version of Riesz representation theorem I guess, which I'm sure you know.

the source is this book, theorem 6.7 in the book

chapter 6, section 6.3

Jakobian

yeah I'll just grab the book

they are saying that the proof of uniqueness is the same as when $\nu(E)<\infty$

@psie what is your question again

seems like your problem is to verify that an isometry is necessarily one-to-one

psie

17:10

ah ok, so an isometry is necessarily one-to-one, I have to check that out

Jakobian

yes, always

if $T:X\to Y$ is an isometry from one metric space to another, then $d(Tx, Ty) = d(x, y)$. So if $Tx = Ty$ then $d(x, y) = 0$ i.e. $x = y$

psie

ah, so that's what they meant 👍 it all makes sense now

Jakobian

yeah, sure. Are you going to be following this book?

psie

yes, I think so. I also have a more basic book for probability, which is the course book at the local university for a course in basic probability

Jakobian

ok

Mats Granvik

17:17

Has the Möbius function been accelerated before?

psie

@Jakobian you should check out chapter 5, where they construct product measures without Caratheodory's extension theorem. I think it's very elegant :)

theorem 5.2

Jakobian

@psie I have other things to do

psie

Ok. For the bucket list then :)

Jakobian

17:31

there'll be always something from general topology for me to do that will be more important than reading this particular curiosity

Jakobian

17:41

first on my list is to read the additional stuff that David Gao wrote here. Second is to read the paper about when $X\times Y$ is an $F$-space. And third would be to find all the spaces with only extremally disconnected compactifications. And fourth would be to continue on reading the book about normal spaces

psie

cool

Soham Saha

18:06

Can anyone open the mathforum.org link in the links section of A058897? It shows connection time-out to me.

Xander Henderson

@SohamSaha That is an OEIS link, not a mathforum link?

Sine of the Time

chat is quiet lately

Xander Henderson

18:53

@SineoftheTime Shhh!

Sine of the Time

19:57

JonathanZ

Would people think this question is worth me re-asking. It was posted and then deleted within an hour, after receiving almost zero attention, positive or negative

math.stackexchange.com/questions/5024017/…

I found a fun little counterexample that I'd like to post, but, as is, it's probably a "not enough context", just-dropped-their-homework question.

The answer on this policy that I found on Meta indicated "if you address the criticisms that lead to it being deleted, then re-post away", but I can't tell why it was deleted, and I can't really say anything of why it's a good question, other than "I found a fun answer"

What do people think?

Xander Henderson

@JonathanZ It was deleted by the author. Likely because they figured out a counter example. Personally, I don't think that is an interesting enough question to bother with reposting.

psie

20:14

In Rudin's PMA, he shows $C(X)$ is a complete metric space when the metric is the sup-norm (according to Rudin, $C(X)$ is the space of all complex-valued bounded continuous functions on the metric space X). I wonder, what extra steps/assumptions do I have to take/make to show $C(X)$ is a Banach space?

Xander Henderson

(But note that this is just an opinion---you can try reposting it, and see how the community responds.)

Thorgott

it's a Banach space by definition now, no?

Xander Henderson

What is your definition of a Banach space?

psie

@Thorgott doesn't $X$ have to be compact?

Xander Henderson

@psie Why?

You have a set $C(X)$. As a set, this is the collection of all bounded, complex valued functions $u : X \to \mathbb{C}$, yes?

psie

20:18

yes, indeed

Xander Henderson

What additional structure on that set do you need in order to check whether or not it is a Banach space?

psie

@XanderHenderson a norm?

Xander Henderson

@psie Among other things, yes.

Is there a norm?

psie

@XanderHenderson yes, the Euclidean norm would work. I suppose there would be no difference if $X$ is a locally compact metric space (in the statement that $C(X)$ is a Banach space), because the reason I was asking in the first place is because it's claimed that if $X$ is a locally compact metric space, then $C_0(X)$ is a Banach space, and somehow I think one needs to use that $C(X)$ is a Banach space.

$C_0(X)$ are the real continuous functions that tend to $0$ at infinity.

I think I'll figure it out.

@XanderHenderson sorry, not the Euclidean norm, but the sup-norm.

Jakobian

@XanderHenderson If $f$ is unbounded, then is the map $a\mapsto af$ from $\mathbb{R}$ to $C(\mathbb{R})$ continuous?

@psie bounded?

JonathanZ

20:34

@XanderHenderson Thanks for the feedback

Jakobian

@psie there is no place in the world for which someone would denote $C(X)$ as the set of all bounded continuous functions

psie

Rudin does :D I'll send you a screenshot in a second

Jakobian

just tell me the page

Sine of the Time

Rudin is a menace

psie

Definition 7.14.

Xander Henderson

20:39

@psie What do you mean by "the Euclidean norm would work"? The problem either gives you a norm, or it does not.

What norm lives on the space?

Jakobian

well he doesn't write $C(X)$, he writes $\mathscr{C}(X)$

Xander Henderson

@Jakobian That was a pedagogical question, meant to get @psie to think about the theorem they are/were quoting.

Jakobian

@psie $\mathscr{C}(X)$ to me a little more acceptable for bounded continuous function than $C(X)$, so I wouldn't call Rudin a complete weirdo

I'd still avoid using $C(X)$ to denote that

psie

ok

Xander Henderson

@psie You still haven't really answered my original questions: you defined $C(X)$ to be the space of all bounded, continuous functions $u : X \to \mathbb{C}$. You claim that this space is complete with respect to the sup-norm, then ask what extra assumptions are necessary in order to conclude that it is a Banach space. (1) What is your definition of a "Banach space"? (2) What properties or structures do you need to check for with respect to $C(X)$?

psie

20:47

ok, I will answer

Xander Henderson

Note, also, @Thorgott's comment:

31 mins ago, by Thorgott

it's a Banach space by definition now, no?

You might think about why they said that.

psie

@XanderHenderson (1) it is a complete normed linear space (2) we need to check that $C(X)$ is a normed linear space and that every Cauchy sequence in $C(X)$ converges (with respect to the sup-norm)

Xander Henderson

Regarding your answer to (2), you asserted that $C(X)$ was complete, so you are claiming that the condition about Cauchy sequences has already been satisfied, no?

You also provide a norm form the space, in the form of the sup-norm.

psie

@XanderHenderson yes, the condition about Cauchy sequences has already been satisfied

to check it's a normed linear space, well, ... let me get out my linear algebra book

If $V$ is a vector space over $\mathbb R$ or $\mathbb C$, then $V^S$ is a vector space, where $S$ is nonempty

So $C(X)$ is definitely a vector space.

Xander Henderson

@psie Sure. It is also pretty simple to verify directly that $C(X)$ is a vector space, without invoking theorems.

In any event, it seems that you don't need to make any additional assumptions. All the applicable properties are verified, n'est-ce pas?

So, again, think about @Thorgott's comment:

43 mins ago, by Thorgott

it's a Banach space by definition now, no?

psie

20:59

ok 👍

Jakobian

21:10

please stop using $C(X)$ to denote all bounded continuous functions on $X$

if you want to talk about the latter, denote is, say, by $C_b(X)$

$C(X)$ usually denotes all continuous functions on $X$, which is not even a topological vector space

psie

yeah, $C_b(X)$ is better, noted

Sine of the Time

or $BC(X)$

Xander Henderson

21:49

@Jakobian in an informal chat, when the mission has already been introduced and used, why does it matter? Why do you care? Yes, it is not the standard notation... And? You weren't even a party too the discussion---if you don't like it, you are free to not engage.

Jakobian

@XanderHenderson that's not a good stance for an educator

why does it matter? There is plenty of reasons why it matters and they are quite obvious

why do I care? That doesn't matter

I weren't part of the discussion? That doesn't matter

You are also as free to not respond as I am free to not engage

maybe you should worry about that

Thorgott

Hey @BenSteffan, how do you figure $G$ must be divisible in your question?

Shaun

22:15

I wish more people knew about the internal logics of topoi. Some lack the law of excluded middle. They're concrete examples of how the laws of logic are not absolute. I guess it's also a blessing, as there'd be loads of woo surrounding them.

Sometimes, I'm afraid I might be a crank, taking them seriously, but they make sense to me. I want to study them properly. Maybe I could do a Master's in topos theory after my PhD . . .

I'm not sophisticated enough philosophically to be an intuitionist. I guess I have to be one though, naïvely.

I'm aware of at least one topos theorist who is not an intuitionist.

I forgot his name though.

The metalogic is classical after all.

I lament the fact I have few people to discuss these ideas with.

Ben Steffan

22:35

@Thorgott Let $p$ be a prime and consider the short exact sequence $0 \to G \xrightarrow{\cdot p} G \to \operatorname{coker} G =: G' \to 0$. Applying $K({{-}}, 2)$ yields a fibre sequence $K(G, 2) \to K(G, 2) \to K(G', 2)$. This gives rise to a homological Serre specseq. which you can use to show that $G'$ must be 0: Since $M$ must also be a $M(G, 2)$, it suffices to show that if $G' \neq 0$ then it has homology in an inconvenient degree. The easiest possible degree you could hope for is 4.

Now observe that $G'$ is an $\mathbb{F}_p$-vector space. In particular, it suffices to figure out whether $H_4(K(\mathbb{F}_p, 2); \mathbb{Z}) \neq 0$. This I leave as an exercise for the reader :)

Thorgott

hmm, I'll have to think this step over

but if we take this for granted, I think the argument is almost complete

every divisible group is a direct sum of copies of $\mathbb{Q}$ and $\mathbb{Z}[p^{\infty}]$s

Ben Steffan

But I do need somebody to give me a sanity check: Both kernel and cokernel of $G \xrightarrow{\cdot p} G$ are $\mathbb{F}_p$-vector spaces, right? The kernel because every non-zero element of it has order $p$, and the cokernel because you can write it as $G \otimes \mathbb{Z} \xrightarrow{\mathrm{id}_G \otimes (\cdot p)} G \otimes \mathbb{Z} \to \operatorname{coker} \mathrm{id}_G \otimes (\cdot p) \cong G \otimes \mathbb{Z} / p$.

@Thorgott yeah, if I have made no mistake I should be able to finish tonight

Thorgott

$K(\mathbb{Q},n)$ for $n\ge2$ has infinite (co)homological dimension, so it remains to calculate $K(\mathbb{Z}[p^{\infty}],n)$, but $\mathbb{Z}[p^{\infty}]$ is a direct limit of $\mathbb{Z}/p^n\mathbb{Z}$, so surely this calculation can be carried out with more mental energy than I have right now

Ben Steffan

That claim about $K(\mathbb{Q}, n)$ is false.

Thorgott

oh true

@BenSteffan either is $p$-torsion, so sanity check granted

Ben Steffan

22:42

$H^*(K(\mathbb{Q}, n))$ is concentrated in a single dimension when $n$ is even

@Thorgott thanks a bunch :)

@Thorgott I think I can finish with another spectral sequence. I'd rather not have to think about what $H_*(K(\mathbb{Z} / p^k, 2))$ is, or try the literature again.

Thorgott

@BenSteffan right of course

then it seems I don't actually understand Moishe's point

Ben Steffan

uh-oh

Thorgott

what exactly rules out $G=\mathbb{Q}$?

Ben Steffan

well I am sitting here now wondering why I believed Moshe's comment

this is troubling

Thorgott

oh no...

Ben Steffan

22:51

this might really get me stuck now

there's no more obvious homological or homotopical information to exploit for that case

maybe I need to break into the 4-manifolds literature

On a sidenote I'm very surprised how little I've found on this sort of question, say, "for what value of $G$ and $n$ does $K(G, n)$ admit a manifold model?"

somehow I would have expected this to be something someone at some point would have studied

Thorgott

to be honest I would expect a negative answer for all $n\ge2$, but the problem is (as just evidenced again) very different probably

Ben Steffan

in particular since intuition reasonably suggests that the answer is "no" in general when $n > 1$

yeah

Thorgott

I'm not too surprised the references are scarce cause people typically restrict to f.g. $G$/compact manifolds (and those are ruled out, of course)

Ben Steffan

yes

Thorgott

there's a MO question (that you've most likely already seen) about modeling $K(G,n)$ as a colimit of manifolds, but that's much more flexible still

Ben Steffan

22:54

yes, I've come across that

also the literature on EML spaces is quite frankly pretty terrible

"oh yeah this was done in the seminaire h. cartan"
are you sure? have you read the transcripts? all ~20 of them that are relevant for this? I doubt it

case in point people always point at the seminaire abstractly but never cite anything directly or use any of the techniques

Jakobian

@Shaun being a crank is not about studying something, but doing so with the attitude resembling that of ignorance. All it takes, is to not make outrageous claims, and be careful about your mathematics. Which is what every mathematician already has the skills for

Thorgott

step 1. run the Serre spectral sequence
step 2. repeat

but I don't really have familiar with these calculations either

Ben Steffan

@Thorgott I tried to be lazy for the spectral sequence calculations I did and went to look up the results I needed about low degree homology in Eilenberg & Mac Lane's papers on the topic

in the end I decided to do the calculations myself instead because even understanding the notation in that paper would have taken me longer

but it's beautiful: the calculations paper is from '53, before the authors apparently learned to appreciate Serre's thesis, and there is not a spectral sequence in sight

instead the authors are, in their own words, "forced to treat a number of new and sometimes quite bizarre functors of $\Pi$" (where $\Pi$ is an abelian group)

Shaun

@Jakobian Indeed. Thank you :)

Thorgott

@BenSteffan oh I didn't know that

sounds meesy tho

Ben Steffan

23:09

if you look for a while you come across pretty strange things

here's a paper by somebody called barcus on the stable homotopy groups of EML spaces that develops its own suspension spectral sequence maths.ed.ac.uk/~v1ranick/papers/barcus.pdf

Thorgott

@Ben by the way, I just remembered this answer: math.stackexchange.com/questions/2015592/…

might be helpful

Ben Steffan

I've also learned about baues book, which is pretty mindblowing

Thorgott

the one on the homotopy category of simply connected $4$-manifolds?

Ben Steffan

homotopy category

@Thorgott I think we've both just embarassed ourselves terribly: $H^*(K(\mathbb{Q}, n))$ is polynomial, not exterior

when $n$ is even

it obviously has to be like this, because it can't be polynomial when $n$ is odd

so Moishe's comment makes perfect sense, of course

Thorgott

I'm confused now

probably cause it's been a while since I last thought about this stuff

isn't $K(\mathbb{Q},n)$ the rationalization of $K(\mathbb{Z},n)$?

cause I thought $S^n\rightarrow K(\mathbb{Z},n)$ is a rational homotopy equivalence for even $n$

Ben Steffan

23:17

@Thorgott it is

@Thorgott this should be odd $n$ I believe (?)

Thorgott

bruh is this a sign error

mixed up even and odd...

of course, Hopf invariants vanish in odd degrees, not even

Ben Steffan

@Thorgott yeah

as I said, it's a bit embarrassing

Thorgott

but ok, at least that rules out torsion-free divisible

@Thorgott unfortunately, this answer suggests $K(\mathbb{Z}[p^{\infty}],2)$ is a Moore space

Ben Steffan

I think I have a full argument now.

Let me try to write it up, though it will take a second.

Thorgott

oh wow, I am still clueless about torsion

Transcript for

$Mathematics$ Mathematics