@EE18 Well, Bėzout's identity implies Euclid's lemma, which leads to the Fundamental Theorem of Arithmetic (that every natural number has a unique prime factorisation).

@PM2Ring I think this is what I was thinking about!

I think my book went with Euclidean division algorithm implies FTA, and then got Euclid's Lemma as a corollary

So I See Thorgott's point about order of all this being arbitrary

is any one good at differential equations here?

if $b_t(s)$ satisfies a pde where t is time and s is space

and you add up discrete space slices to obtain a function strictly of time and look at what happens at $t \to \infty$

I wonder if that has a differential equation intepretation

How important is it to know elementary number theory for analysis?

I feel like there are a lot of facts out there in the ether which I don't know

Bezout, Chinese Remainder, Extended Euclid, and I feel like so many more

I figured out how to post 2 questions simultaneously without waiting the 40minutes

I conjecture that it works for $N$ simultaneous questions

implying that I could fill the entire front page

with all my questions

from my calculations there are 100 questions shown at any given time on the front page

Well I guess I will start writing

will I get banned?

I conjecture that 'no' I won't get banned

as it's just a loophole in the system

@EE18 for analysis, none of these matter

OK, will not get sidetracked then

Will I guess leave them for some other study at another time

BTW Thorgott, what do you think the authors have in mind with the below. Exercise 6.3 was just that there are $n \choose k$ $k-element$ subsets of an $n$-element set, and Theorem 8.4 is the binomial theorem. Obviously putting the two together gives me $2^n = \sum_{k=0}^n n \choose k$ but I don't see how that gets me where they're suggesting. I was going to just start by induction but that doesn't seem like what they're suggesting?

so if I'm constructing a map from a quotient group to another quotient group, I'm mapping congruence classes to other congruence classes right

e.g. $f:\mathbb{Z}_{10}\to\mathbb{Z}_5$

defined by $f([x]_{10}) = [x]_{5}$

the kernel is $\{[0]_{10},[5]_{10}\}$ I think

how would one describe $SL_2(\mathbb{Q})$? Like $$\left\{A=\begin{bmatrix}a&b\\c&d\end{bmatrix}\bigg| a,b,c,d\in \mathbb{Q}, ad-bc=1 \text{ and }A^{-1}\in GL_2(\mathbb{Q})\right\}$$?

I guess I don't have to really write that since it's implied in the definition

w/e

idunno

how bad does $A\in B \ni C$ look :D

when I'm too lazy to erase and rewrite $A,C \in B$

@XanderHenderson might smite me for that

@Obliv looks correct.

@Obliv A^{-1} has rational entries if A does. Moreover, since A A^{-1}= I, you also have det ( A^{-1})= 1 so actually A^{-1} is in SL_2.

oh ok

how do inverses work in $SL_2(\mathbb{Q}$? Like, if I wanna show $SL_2(\mathbb{Q})\subset GL_2(\mathbb{Q}$ I should show for $A,B \in SL_2(\mathbb{Q})$ that $AB^{-1} \in SL_2(\mathbb{Q})$ but it's matrix multiplication and addition in $\mathbb{Q}$

oh wait maybe this is a ring not a group?

anyway, this seems so trivial but I have to show it :\

i don't mean to be annoying, but the details of this depend on how you have defined SL_2(Q). i would not assume that because there is a standard notation for that object, there is a single standard definition of it

isn't it just the set of 2x2 matrices with entries in $\mathbb{Q}$ that have det = 1 and invertible

i.e. your question is likely entirely about how to get it out of your definitions, and not just something you can ask someone who knows what SL_2(Q) is, because they don't know what you know, or are "allowed to use," about that object

obliv: that's one definition. is it your definition?

what do you know about determinants?

It's the definition on wiki and I just assume that's the only one I guess

(i also have bad news re: "it depends on how you have defined the thing" and determinants, but maybe we don't need to go there)

well probably, but isn't it typical to show $H \subset G$ via $ab^{-1} \in H$ for some $a,b \in H$?

to show H is a subgroup of G

So if (V,+,.) is an n dimensional vector space over R, where . is scalar multiplication, then the vector space (V,+, p), where p is another scalar multiplication need not have dimension n right?

OK. it might help to preface a question contextualized like that with "i'm just reading stuff randomly on wikipedia and don't have any particular set of definitions in mind, or any particular limit on what i am allowed to use. if SL_2(Q) is defined as in [web page], [question]"

to answer your question, if you have something in SL_2(Q), it is (among other things) an invertible 2x2 matrix, and its inverse in the sense of the usual algebra of 2x2 matrices is also going to be in SL_2(Q)

you also seem to be maybe asking if SL_2(Q) is a subgroup of GL_2(Q), but i can't be sure

GL_2(Q) is pretty much SL_2(Q) but with the invertible matrices that don't just have det=1

so it's pretty obvious (at least to me) that it'll be a subgroup. I just have to show that for any A,B in SL_2(Q) the det=1 property is preserved under matrix mult.

I was going to type out all the matrix operations algebraically

but the message was too long lol

let A,B be in SL_2(Q) with AB=(a_1,a_2,a_3,a_4)(b_1,b_2,b_3,b_4) arranged s.t. a_1,a_2 are the top row entries and a_3,a_4 are the bottom (same for B), then AB=(a_1b_1+a_2b_3,a_1b_2+a_2b_4,a_3b_1+a_4b_3,a_3b_2+a_4b_4) and since det=1 for A,B separately, we know a_1a_4-a_3a_2 = 1 and b_1b_4-b_3b_2 = 1 so for AB we have (a_1b_1+a_2b_3)(a_3b_2+a_4b_4)-(a_3b_1+a_4b_3)(a_1b_2+a_2b_4) multiply it out to get (a_1b_1+a_2b_3)(a_3b_2)+(a_1b_1+a_2b_3)(a_4b_4)-(a_3b_1)(a_1b_2+a_2b_4)-

(a_4b_3)(a_1b_2+a_2b_4)

then manipulate that stuff and hopefully det=1

obliv: this is where not starting with some set of definitions might hurt you. it isn't particularly clear from an "expanded out" formula for det as an express polynomial in the matrix entries that det(AB) = det(A) det(B), which is basically what you're wanting to use here.

it helps to add at least one layer of abstraction to prove that property of the determinant. although there's no reason why the idea of the argument you sketch couldn't go through.

I dont know of any other definitions since my linear algebra course was not abstract at all lol

But I'm more than happy to learn of some if it'll help with proofs

so you wouldn't even know a definition of the determinant for, say, 5x5 matrices? or nxn matrices?

maybe they gave you something

but yeah, if they didn't, i'd grab a book that does give you a definition. det(ab) = det(a) det(b) is something that you will see proved pretty soon after any definition.

which isn't to say that the proof is always nice to look at

Hello!

@Koro this is an interesting question. offhand i would guess that the dimensions do have to be the same, but also that it matters that your field is R and not some arbitrary field

i'm thinking, given some more general field k, if there is a field endomorphism f: k to k that is not surjective, then {1} is a basis for (k,+,*), but if you give k the funny scalar multiplication k.x = f(k)*x, then {1} is not a basis for (k,+,.). which is not all by itself a proof that (k,+,.) isn't one-dimensional, is suggestive that maybe it doesn't have to be

e.g. if k is something like Q(t) and f is the automorphism sending p(t) to p(t^2), it feels like maybe {1,t} could be a basis for (k,+,.)

but R doesn't have any endomorphisms that aren't automorphisms

i guess maybe actually the fact that {v} isn't a basis for a nonzero vector v in V is a proof that V isn't one dimensional. it's just, this fact all by itself wouldn't tell us what the "not one dimensional" dimension is

just thinking out loud there

hi EM4

how are you doing?

pretty good, i had a light weekend. yourself?

great to hear, I am coming back to math.

That's what i noticed

however $d(s)$ is the divisor function which I don't know how to extend to the complexes

I know that it is possible but I would have to learn the construction

@leslietownes Yes, let's consider field R only. I was thinking of defining a scalar multiplication p: R\times V--->V as p(r,v)=0 for all r and v.

But it doesn't work because we require p(1,v)=v for all v.

:)

consider it asked: math.stackexchange.com/q/4911648/460999

Hey! I've tried to solve this problem for quite a few times now and I'm just not getting it. I looked up the solutions online and most of them don't even get the answer 41. Any help would be appreciated!

@Swan Do you know about permutation groups?

@SoumikMukherjee I do

to make a long story short, it isn't any of those answers. 25 maybe comes the "closest" (it counts the number of such f that are not the identity)

@leslietownes Man I had gotten 25 itself in my first attempt (I had forgotten to map the numbers to themselves). It's so weird they have not made any clarification about if this question was given bonus or something. Such a wastage of time and effort

oof, yeah. at least it kept you off the street and (one presumes) out of trouble?

@Swan What exam is this from?

@leslietownes 😭😭

@SoumikMukherjee IISER APTITUDE TEST 2021. This one gets you into the 7 IIISERs and IISc (from last year)

for MSc?

As in IISc added this exam as one of their modes of admission from last year. That's the reason the exam is getting popular

@SoumikMukherjee No, BS (Research) for IISc and Integrated BS+MS for IISERs

IISER making such a mistake is totally unexpected

@SoumikMukherjee There are more errors (or I guess ambiguity) in the biology section. (You have to study PCMB to get the top ranks). And they don't even have an official answer key on their site.

Ohhh

What in the world in CSA Lie (H)?

@leslietownes huh?

Oh. Scalar multiplication

Not scalar product

@EE18 to show two numbers are equal, show their difference is $0$

@Obliv I think this is perfectly fine

@AlessandroCodenotti any progress on that question about $A\subseteq \mathbb{R}^3$ consisting of points $(x_1, x_2, x_3)$ with exactly one $x_i$ rational being connected?

I didn't think about it, but I have 10 minutes now

Let's see, what happens in $\Bbb R^2$?

Its totally disconnected there

thats because $(r_1\pm x, r_2\pm x)$ for $r_i$ rational and $x$ arbitrary lies in the complement

but for $\mathbb{R}^3$, the same trick doesn't work, it makes up some kind of net instead of a $2$-dimensional surface you could enclose points with

one thing for certain is that $A^c$ is connected

sure

that's all I came up with myself

Hmm yeah it's not clear

I have a really terrible question

say your warping functions in some warped product metric satisfy some differential equation. What conditions allow one to pass the differential equation onto the geometry described by the metric?

nevermind

@JohnZimmerman that's terrible

just joking I don't even know what this means

I realized

that this question needs revisions

yeah I'm not insulting your intelligence or anything

explaining that this is a joke is just in case someone thinks I'm being rude

@JohnZimmerman This looks like word salad to me.

yeah it definitely looks like a word salad, but maybe ODE and differential geometry people understand it

there's definitely a lot of definitions in differential geometry that I have no idea about

Lie algebras aren't one of them

@XanderHenderson Well you maybe correct. The warping function $g_t(x)$ depend. on some time parameter $t$ changes the geometry of the manifold by changing $t$ itself

so I don't have a good grasp about what occurs when $g_t(x)$ is also depend. on a partial differential equation

This still looks like word salad to me.

sufficiently nice one of course

What is a "warping function"?

What is "the geometry" of a space?

How does a function "depend on" a DE?

$$g_{\phi} = \frac{1}{\phi_t(u)^{2}}\, du^{2} + \phi_t(u)\, dv^{2}.$$

for example

What is a "warped product metric"?

What does it mean to "pass a DE onto the geometry described by a metric"?

Warped product is a generalization of the (semi-)Riemannian metric used on a product manifold

As @Jakobian said, it is very possible that an expert in whatever part of mathematics you are studying might know what you are talking about, but to me, it is nonsense.

Xander these are amazing questions!

But the “pass a DE onto the geometry” makes no sense to me

@JohnZimmerman No, they are not "amazing" questions. These are the basic questions that you should be able to answer before you even pose the question that you are trying to ask.

You need to know the answers to these questions, and you need to be able to explain those answers to others. Otherwise, you are jumping into the deep end, and you need to go back and revise.

why can't pose them retroactively assuming i can do so in <k days where k is a very low constant?

For example, one of the motivating questions of my phd work could probably be summarized by the sentence "Under what conditions does the geometric zeta function associated to a metric space which carries a measure provide meaningful information about the geometry of that space?"

To an outsider who is not immersed in the very niche field I work in, I expect that this is word salad.

because we're too lazy to find a meaningful way in which something makes sense unless its very basic stuff

If I were to ask someone else a question about that work, I would need to be able to explain what I mean by a "geometric zeta function", what aspects of the "geometry" of a space are "meaningful", and what kinds of conditions might be reasonable to expect for a metric space carrying a measure (e.g. in what ways are the metric and measure compatible or not?).

I thought we were talking about metrics in geometry and not about metric spaces

@Jakobian I was giving a completely different example.

oh, right

it was so similar to what @JohnZimmerman is trying to do in recent days that it blended in

@XanderHenderson I meant that your questions were good

@JohnZimmerman I know what you meant. But my questions are only good in the sense that they are the basic, most fundamental questions that you should be able to answer while posing the question that you are asking.

These are not insightful or deep questions.

It is the basic kind of thing you need to be able to do for yourself.

@Thorgott OK, I will think on this. I am not sure what you and the author have in mind in terms of using still, but maybe it will come to me while I bike to work today...

I figured it out: it's just taking a collection of surfaces (say spatial slices) and stitching them together over time prescribed by the diff eq. to generate some new space which may or may not be a manifold in general

@EE18 it's a matter of looking at it the right way that I don't wanna spoil, but you truly don't need anything but the binomial theorem

BTW, I am trying to figure out how to argue that if $E$ is the embedding of $R$ into $C$ then $|E(x)|_C = E(|x|_R)$. Because $E(x)^* = E(x)$ this reduces to showing that $\sqrt{E(x)^2} = E(\sqrt{x^2})$, but I am stuck on showing how $E$ a field homomorphism (or I guess just ring homomorphism gets me here in terms of passing it through square roots (the rest is obvious).

I have this vague sense that I'm meant to separate the sum on the RHS of the binomial theorem into one sum over even subsets and one sum over odd subsets. But I can't see how to formalize that. But will keep thinking and maybe ping you at EOD if I still can't get it

To be clear, $E(x) = (x,0)$, the standard embedding (I'm not sure, maybe there are other pathological ones, though given the nature of these theorems I'd guess it's unique?)

that $E$ is a field homomorphism should follow from howeveryou have defined the multiplication on $\mathbb{C}$

Oh, now that I think about it, the argument might be something as simple $E(\sqrt{a})E(\sqrt{a}) = E(\sqrt{a}\sqrt{a}) = E(a)$, whence $E(\sqrt{a}) = \sqrt{E(a)}$ by uniqueness of square roots in $\Bbb R$ (I guess I would need to use the field homomorphism properties to prove that existence and uniqueness of square roots is true in the image of the embedding too, but won't get sidetracked)

Have you ever heard a statement like two plane fields tangent to something are homotopic? @Thorgott I'm wondering the definition of homotopic plane fields

Hi, guys! Does anybody there know how to prove $f(x) = \frac{\sinh(x)}{x}$ with $f(0):=1$ is a smooth function? (Actually, it seems to be analytical, but smooth would be fine for me)

@Derso sure. Simplest way to do this would be to write $\sinh(x)$ as a power series

@Derso here's the analogous proof for sin(x)/x: math.stackexchange.com/questions/2238427/…

Oh, now I got what a guy said on some forum... I was trying to look at the power series of $\frac{\sinh (x)}{x}$

Sure. Thank you guys!

it does depend a bit on what your definition of sinh is

e.g., if sinh is defined as a series then it's pretty much immediate

$\sinh(x) = \frac{1}{2}(e^x-e^{-x}) = \sum_{n=0}^\infty \frac{1+(-1)^{n+1}}{2n!}x^n$

Just $\frac{\exp(x) - \exp(-x)}{2}$

yeah. which pushes the question back to how e^x is defined

a series definition of e^x is again immediate

lol. It's defined as being the function such that $(e^x)'=e^x$ :P

pfff

now notice how for $n = 0$ the term is $0$

i'm fond of the classic (1+x/n)^n as n->infinity myself

so $\frac{\sinh(x)}{x} = \sum_{n=0}^\infty \frac{1+(-1)^n}{2(n+1)!}x^n$ for $x\neq 0$

and for $x = 0$ this power series is equal to $\frac{1+(-1)^0}{2\cdot 1!} = 1$

which i guess would make the problem: Let $f_n(x)=\frac{1}{2x}\left[(1+x/n)^n-(1-x/n)^n\right]$. Show that $f_n(x)$ converges to analytic function as $n\to\infty$

so your function is given by a power series $\sum_{n=0}^\infty \frac{1+(-1)^n}{2(n+1)!}x^n$ so its analytic

(and $f_n(0)=1$)

I don't think it matters that much which definition of $e^x$ we use as long as one can show $e^x = \sum_{n=0}^\infty \frac{1}{n!}x^n$

yeah, it's just a matter of how much labor you feel obliged to do

as in, for the purposes of this problem it doesn't matter

well, the funny thing about the $f_n(x)$ formulation is that $f_n(x)$ is just a polynomial for integer $n$

but i would guess that $f_n(x)$ being polynomial (and therefore trivially analytic) for finite $n$ is not by itself sufficient to establish that the limit is also analytic

@Jakobian Sure, but you don't actually have to do anything with the power series at all if you don't want to. The question was about smoothness, not about whether or not the function is analytic. One approach is to show that the function is analytic (and, therefore, smooth), but if the $u' = u$ definition of the exponential, a direct proof that all derivatives exist might be just as simple.

this is true

E.g. $\sinh(x)/x$ is smooth away from zero, so we only really need to look at derivatives near zero.

and has a nice fact associated to it, namely that sinh(x)/x is the 0th modified spherical bessel function

I'm not saying that this is easier, but it is another possible approach to the problem.

hence can be defined as the solution to $y''(x)+(2/x) y'(x)=y(x)$ with $y(0)=1,y'(0)=0$

@Semiclassical That's cute.

(of course, if we're talking bessel functions then there's almost certainly a useful integral representation but i cba to look it up)

idk, this looks overengineered

@Jakobian The phrase you are looking for is "swatting a gnat with a nuke."

@onepotatotwopotato no clue

Any hints on a closed form?

@onepotatotwopotato It means what you think it should mean: there's a continuous family of plane fields interpolating the two.

This can be phrased as a homotopy of sections of the appropriate Grassmann bundle.

I like the part where the root 3 is on the RHS

In Folland's book he proves that, given two $\sigma$-finite measure spaces $(X,\mathcal M,\mu), (Y,\mathcal N,\nu)$ and $E\in\mathcal M\otimes\mathcal N$, then the functions $$x\mapsto\nu(E_x),\quad y\mapsto\mu(E^y)\tag1$$ are measurable on $X$ and $Y$ respectively (page 66).

In some lecture notes based on Folland's that I'm reading, there is a theorem that claims that if $f:X\times Y\to\mathbb C$ is a measurable function, then the functions $g:X\to\mathbb C$ and $h:Y\to\mathbb C$ defined by $$g(x)=\int f_x\,d\nu,\quad h(y)=\int f^y\,d\mu,\tag2$$ are measurable. It seems to me $(2)$ is a bit more general than $(1)$ and I wonder, is it possible to prove $(2)$ from $(1)$?

plane field is just a global section of a rank 2 vector bundle over a manifold right?

No

A plane field is a rank 2 subbundle of TM

@BalarkaSen Hmm if it's a bundle then why call it a field?

@psie yeah, that seems like it should be a proof by standard methods

@Derso :) even this is good because the ODE allows you to recursively show that all the derivatives at the origin are $1$, and so Taylor’s formula with remainder allows you to immediately conclude that the Taylor series equals the function. But the other way of proving smoothness of $\sinh(x)/x$ would of course be L’Hopital’s rule and induction (on the form of the derivative)… but this is unnecessarily tedious

@onepotatotwopotato it's like a vector field but instead of vectors at every point you have planes at every point

@peek-a-boo This "unnecessarily tedious" way was my first attempt! So tedious that I came here to ask lol

@psie btw Folland does prove (2); it’s inside the proof of his Fubini-Tonelli.

@BalarkaSen I see. But it seems to me that one can always homotope one plane field to another because I can simply move one plane to another No?

@peek-a-boo indeed! :) thanks, I wasn't aware

@onepotatotwopotato Wrong. S^3 has infinitely many homotopy classes of 2-plane fields

Well locally?

B^3 has infinitely many homotopy classes of 2-plane fields relative to delB^3.

@Derso haha yea but sometimes this tediousness cannot be avoided (e.g for the non-analytic smooth function $f(x)=e^{-1/x}$ for $x>0$ and $f(x)=0$ for $x\leq 0$)

Non-relatively what you said is right, but that's silly

A 2-plane field on an oriented three manifold is essentially the same as a line field: choose a Riemannian metric and take orthocomplement. A line field on an oriented three-manifold is a map M -> RP^2. Homotopy classes of 2-plane fields on M is thus in 1-1 correspondence with [M, RP^2]

it's not a section of a vector bundle, so you can't straight-line homotope or anything

If you restrict to oriented plane fields & homotopy through those, this is [M, S^2] = [M, CP^infty] = H^2(M, Z). The classifying element is the Euler class of the bundle

@BalarkaSen Oh interesting. Where can I learn such plane fields on a manifold?

Geiges, introduction to contact topology

Oh I didn't know it's from contact topology

What is the most effiecient way to find an $n$-digit number , only containing zeros and ones in the decimal expansion with the maximum possible value $\omega(n)$ ? In other words , the number of distinct prime factors should be maximized.

oeis.org/A077807 is related

1, 10, 110, 1110, 10010, 101010 seem to be the first few terms

1, 10, 110, 1110, 10010, 101010, 1111110

it seems like the $n$-digit number $x$ with the most amount of $\omega(x)$ is usually of the form $1111...10$

in all of those 7 terms it also was unique

next one is 11111100, breaking the pattern a little

for the $9$th digit numbers we no longer have uniqueness

there is 6 such numbers

the smallest is 100110010

the largest is of the form 111111000

it seems to me that for large $n$, the largest $x$ which will maximize $\omega(x)$ is of the form $11...1100...00$ where there is certain amount of $0$'s and $1$'s

@Peter

thats my hypothesis at least

so those will be numbers of the form $10^a\cdot \frac{10^b-1}{9}$ for some $a, b$

or maybe it is that since $\frac{10^b-1}{9}$ maximizes the previous one, and the way those numbers increase seems to be slower than $n$, sometimes we can get away with multiplication by $10$ as in, getting the previous number which worked... hmm

Every day some kitten scratches me.

also, they keep meowing when they are near me and if I don't pay attention to them, they bite me in my feet.

CATastrophe

koro: nature is telling you to adopt one of these cats

@leslietownes I thought those are his cats

It's finally raining here 🥹

Consider proposition 1.15 in Folland. I'm getting hung up on a small detail. We want to show $\mu_0$ is a premeasure on $\mathcal A$, the algebra of all finite disjoint unions of h-intervals. We need to show countable additivity, and the author writes we need to check this for $\{I_j\}_1^\infty$ being a sequence of disjoint h-intervals with their union in $\mathcal A$.

"... a sequence of disjoint h-intervals...", does this mean every $I_j$ is an h-interval? Each $I_j$ must also be in $\mathcal A$, right? So each $I_j$ is in fact a finite disjoint union of h-intervals. Is a finite disjoint union of h-intervals an h-interval?

I'm reading the section on product measures actually, but a similar premeasure is there established, so I'm reviewing this proposition.

i would interpret that language to mean that each (a,_j, b_j] is an h-interval, and that these sets are disjoint from one another

please bear in mind that the term "h-interval" is, as far as i know, peculiar to folland, or at least used only in some textbooks and not others, and in particular, i don't know what an "h-interval" is

but whatever an "h-interval" is, that language tells me that the author intends each (a_j, b_j] to be one in the hypothesis

An h-interval is a set of the form $(a,b],(a,\infty)$ or $\varnothing$, where $-\infty\leq a<b<\infty$.

okay, so by that definition, many finite disjoint unions of h-intervals will not be h-intervals

hmm, but for countable additivity we need the $I_j\in\mathcal A$, right?

i'm not sure that i understand the relationship between that question and what has come before it

if A is defined to be the set of finite disjoint unions of h-intervals, then A will, among other things, contain every h-interval

so if I_j = (a_j, b_j] is an h-interval, you will indeed have (a_j, b_j] in A

you're also saying "for countable additivity we need . . ." and i'm guessing this has something to do with some discussion somewhere else, as 1.15 appears to involve only mu_0 as applied to finite unions of h-intervals, and not (for example but more generally) mu_0 as applied to countable unions of elements of A

you're also introducing this notation I_j, and i'm not sure if that is just defined to be shorthand for (a_j, b_j] or if it is part of some other discussion. the gifs don't make reference to anything called I_j, and you don't define it, except to discuss them in a way similar to the way the gifs are discussing (a_j, b_j]

is it possible to reduce your question to a sequence of yes-or-no questions about the algebra A or the properties of the function mu_0

so there's {A_j}, a sequence of disjoint sets in A (which contains more than h-intervals), which might be different from your I_j, which sounded like h-intervals

@leslietownes I will try, let me think

if A_1 is in A, you can find disjoint h-intervals I_1, ..., I_k with A_1 their disjoint union. similarly if A_2 is in A, you could find disjoint h-intervals J_1, ..., J_m with A_2 their disjoint union. if A_1 and A_2 are assumed disjoint, then I_i intersect J_j will also be empty, for any 1 <= i <= k and 1 <= j <= m.

that's just a thought i notice, that connects disjointness of elements of A to disjointness of collections of h-intervals

most elements of A will be representable in more than one way as disjoint unions of h-intervals, for roughly the same reason that (0,2] and (0,1] union (1,2] are the same set. you might be able to get uniqueness of a representation if you require more out of the representation than just that its union be the given element of A

but unless the results you are trying to prove are crying out for that, i wouldn't wade into that

just tossing some random thoughts out there on the off chance that they will help

a lot of measure theory results are formulated so they are assuming just enough about what the sets they are referring to "look like," but not any more

even if when thinking about those sets you might want to know more

as a rough analogy, any open subset of R is a union of open intervals, and can even be uniquely expressed as a disjoint union of open intervals (as long as you allow "infinite" intervals), but this kind of representation is not always the most helpful way of viewing or proving things about open sets

I probably need a break, thanks for the thoughts though :)

if $\hat{A}$ is a Hermitian operator then $\hat{A}(\psi^*) = (\hat{A}\psi)^*$?

or no

allie: it seems possible, but this is very physicsy notation for a math chat :) it might help to drop the hat decoration from A if you don't plan on using it. note that it looks like A is being regarded as acting on at least two spaces here (wherever "psi" lives, and wherever "psi*" lives), and the answer is probably in the definition of how * is defined and how A is defined (presumably via *) to act on the space other than the one it's defined on

i guess i should ask in the physics chat haha

in math notation, you are probably going to get what you want from the equation <A phi, psi> = <phi, A psi>, where here < , > denotes the inner product on a hilbert space that A is defined on and the equality is for all phi and psi in that space. the LHS is a strong candidate for the definition of "A psi*" applied to phi, and the RHS is a strong candidate for (A psi) * applied to phi

and that equation of inner products for all phi, psi is equivalent to the definition of "A is hermitian"

anyone here good with euler characteristic?

specifically good with vanishing euler characteristic

here * denotes the map from the space that A acts on to its dual, by way of sending phi to the functional that takes the inner product of its input with phi (what a mathy person might write as sending phi to the linear functional <-, phi>)

and A is regarded as acting on the dual space by defined to send the 'dual vector' <-, phi> to the functional <A -, phi>

hmmm, i feel like im missing something stupid here

this is some bastard version of bra ket notation

i don't know if these notations or definitions are conventional in physics, but my memory is that something like these is

leslie: After some break, I think I can answer my own question now. Folland, in proving countable additivity of $\mu_0$ in proposition 1.15, introduces $\{I_j\}_1^\infty$ to be a sequence of disjoint h-intervals. However, we are dealing with the algebra of finite disjoint union of h-intervals, not just h-intervals.

So I thought each $I_j$ should be a finite disjoint union of h-intervals, but as a sequence, this just gives us a sequence of h-intervals anyway, where the first $n$ elements of the sequence are the disjoint h-intervals that make up $I_1$, the next $m$ elements are the h-intervals that make up $I_2$, and so on. Capeesh?

yeah, that sounds right

great :D

ugh i feel like an idiot i dont know where im going wrong. i have somthing like $\int \psi A \psi^* - \psi^* A \psi$ and that would be like the inner products <psi*, A psi*> - <psi, A psi>, which given A as hermitian should equal <A psi*, psi*> - <psi, A psi> and so putting it back into the integral form its $\int (A \psi) \psi^* - \psi^* (A \psi)$ which looks like it's 0? what am i doing wrong here im sure im missing something

althought maybe it wasnt this part thats incorrect

I just learned why the bar construction is called bar construction

elucidating moment

Hello, geometers out there! Tell me if I'm wrong: the metric on $\tilde{SL(2,\mathbb{R})}$, one of the eight Thurston's model geometries is twisted, right?

$\tilde{SL(2,\mathbb{R})}$ is the universal cover of $SL(2,\mathbb{R})$, the $2\times 2$ matrices with real coefficients. It acts on the half-plane model of the hyperbolic plane by Möbius transformations (which are isometries).

And using Iwasawa decomposition, we can get a universal covering map from $\mathbb{H}^2\times \mathbb{R}$ onto $SL(2,\mathbb{R})$. Then we can define a riemannian metric on $SL(2,\mathbb{R})$ in the usual way (for Lie Groups) and then induce a metric on the universal covering space.

Thurston's is certainly the twisted one, right?

*I forgot to add "with determinant one"

Let $n\in\mathbb{N}\setminus \{0,1\}$. Is it true that the number of odd integers between $1$ and $n$ is $\lfloor (n+1)/2 \rfloor$ and the number of even integers between $1$ and $n$ is $\lfloor n/2 \rfloor$?