Mathematics - 2019-02-04 (page 2 of 2)

user193319

21:07

1

Let $G$ be some group, and let $S_1$ and $S_2$ be some subgroups. Then $S_1$ and $S_2$ are said to be commensurable iff $|S_1 : S_1 \cap S_2|$ and $|S_2 : S_1 \cap S_2|$ are finite. I am trying to show that this is an equivalence relation on the collection of all subgroups of $G$. Reflexivity an...

This problem has been giving me too much trouble...

1010011010

All right guys, either I'm about to point out a glitch in the matrix or I'm extremely dumb

If I perform partial integration on $\operatorname{Li}_s(x)/(x+\alpha)$ for example, and I pick the polylog as the function I'm integrating in both terms, I get on the RHS the following

Mike Miller

@Shaun Perhaps you got the point already, but I don't like either answer, since they don't seem to tell you what's really going on.

1010011010

$$
\int \frac{\operatorname{Li}_s(x)}{x+\alpha} dx = \frac1{x+c_1}(-1)^s \left[x\sum_{i=2}^{s} (-1)^{i}\operatorname{Li_{i}(x) - (1-x)\log(1-x) -x\right] + \sum_{i=2}^s (-1)^{s+i} \int dx \frac{x}{\left[x+c_1\right]^2} \operatorname{Li_{i}(x) + (-1)^{s+1} \int \frac{1-x}{\left[x+c_1\right]^2}\log(1-x)dx +(-1)^{s+1} \int \frac{x}{\left[x+c_1\right]^2} dx
$$

Umm it's not formatting?!

Mike Miller

An element of S_n may be written by partitioning {1, 2, ..., n} into smaller sets, and applying a cyclic permutation on each of them. This is usually written like (134)(256)(7), say.

1010011010

The content compiles just fine in my tex editor:(

Mike Miller

21:14

A cyclic permutation on k letters has order k, so we should understand what that says about products of cyclic permutations (where the sets they act on are disjoint).

@1010011010 see LaTeX in chat on the sidebar.

1010011010

\begin{equation}
\int \frac{\operatorname{Li}_s(x)}{x+c_1} = \frac1{x+c_1}(-1)^s \left[x\sum_{i=2}^{s} (-1)^{i}\operatorname{Li}_{i}(x) - (1-x)\log(1-x) -x\right] + (-1)^s\int dx \frac1{\left[x+c_1\right]^2} \left[x\sum_{i=2}^{s} (-1)^{i}\operatorname{Li}_{i}(x) - (1-x)\log(1-x) -x\right]
\end{equation}

@MikeMiller Yeah thanks I know about that, there was just a bug in the code somewhere, I guess the other comment can be deleted

Mike Miller

Gotcha

@Shaun Let me write some notation to simplify what I want ro say.

1010011010

The above appears to have the same integral on both the RHS and LHS (just take the $i=s$ term of the summation under the integral sign on the RHS)

Does this automatically mean I cannot take the polylogarithm as the function to be integrated?

Mike Miller

@Shaun Let $\sigma_1, \sigma_2, \cdots, \sigma_k$ be cyclic permutations on some subset of $\{1, \cdots, n\}$. We say that $\text{Supp}(\sigma_i)$ is the set of elements so that $\sigma_i(x) \neq x$. We demanded above that $\text{Supp}(\sigma_i) \cap \text{Supp}(\sigma_j) = \varnothing$ for all $i \neq j$.

(I am introducing notation instead of specifying the subsets because the latter would make the notation way more hellish.)

Now, let's say $|\text{Supp}(\sigma_i)| = s_i$. Then $\sigma_i$ has order $s_i$, and $\sum_{i=1}^k s_i \leq n$. We have equality iff none of the $s_i$ are zero.

Now, taking $(\sigma_1 \sigma_2 \cdots \sigma_k)^j = \sigma_1^j \sigma_2^j \cdots \sigma_k^j$. This is because the domains of these permutations are disjoint (so they all commute with one another).

Therefore, the order of $(\sigma_1 \cdots \sigma_k)$ is the least $j$ so that all of the $\sigma_i^j = 1$.

That is, the order of $\sigma_1 \cdots \sigma_k$ is the greatest common denominator of the orders of the $\sigma_i$.

The question "what are the orders of elements of $S_n$", then, is the same as "if $p = \{s_1, \cdots, s_k\}$ is some sequence of numbers with $1 < s_i$ and $\sum s_i \leq n$, what is $\gcd(s_1, \cdots, s_k)$? If we allow $p$ to vary over all partitions of numbers at most $n$, what are the possible numbers this produces?"

Then in those questions, people carried out this number theoretic calculation.

Mostly it is a matter of brute computation. The fact that $1 < s_i$ reduces the amount of computation somewhat.

In particular, if you want an element of order exactly 2n, the discussion in one of those posts shows that this cannot be any prime power n. This reduces us to 6, 10, 12 as our first possible examples.

It is not true of n=6: there the possible partitions as above are {2,2}, {2,3}, {2,4}, {2,2,2}. The gcds in each case are 2, 6, 4, 2: never 12.

For n=10, take the partition {5, 4}.

I do not see a good reason to star all of these. I'm going to remove every star but the first one (which I guess can serve as a marker to find this conversation later), unless somebody objects.

Shaun

21:31

Sorry, just keeping track - I'll stop. Thank you for answering

Mike Miller

No worries. BTW, there is a feature that lets you save conversations to check for later. But I am having trouble figuring out how to use it.

Shaun

I was thinking about this problem over the last couple of hours (in the back of my mind). Partitions did occur to me. I don't think I could make my thoughts on it any more precise than yours, so, yeah, thank you.

Mike Miller

I was pretty disappointed in those answers for not just saying this explicitly.

Alessandro Codenotti

The annoying think with that feature is that they're not saved privately

Mike Miller

Do you know how to use it? I can't find the button

Anonymous

21:35

FWIW, you can bookmark conversations from the transcript. You need to select the starting and ending messages. (Click on "bookmark a conversation".)

Alessandro Codenotti

I used to know, let me see if I can figure it out again

Mike Miller

I unfortunately only see the permalink button...

permalink | history

Alessandro Codenotti

It's on the right, a blue button "bookmark a conversation"

Shaun

See the Group Theory chatroom . . .

in Group Theory, 1 min ago, by Shaun

in Mathematics, 18 mins ago, by Mike Miller

@Shaun Let me write some notation to simplify what I want ro say.

Mike Miller

Ahhhh

Anonymous

21:36

Alessandro Codenotti

And all bookmarked conversations can be seen in this tab‌, which is what I don't like about this feature: it's public

Mike Miller

OK, @Shaun, if you want you can go to the transcript (using Blue's link above), click a button on the right that says "bookmark conversation", and then choose the sequence of messages to be saved. As Alessandro said this will save it publicly for anyone to check back on if they want.

Anonymous

@AlessandroCodenotti Yeah, that's unfortunate. :/

Mike Miller

I will leave up the first star from this conversation so that you can click that to return to the thread, if you want.

Alessandro Codenotti

I don't mind people seeing conversations I bookmarked, but it makes them way harder for me to find than if I could have a tab on my profile or somewhere listing all the stuff I bookmarked

Mike Miller

21:38

Yes, I don't really care to look through other people's bookmarks!

Shaun

@MikeMiller That sounds like a great idea. I hope I can access it on the mobile site, though, as that's pretty much all I use; it's okay if not.

Mike Miller

I mostly use mobile, and for a lot of featueres you need to quickly change to the full site to reach them. It's a hassle.

You can access it, but usually through a little bit of pain.

Anonymous

@AlessandroCodenotti The individual bookmarks have permalinks. So could save them in a local file (as a workaround).

Mike Miller

@Blue There is a question in the associated chat you made, that you would have a quicker answer to than me. I didn't find the right thread in a half minute of googling.

Anonymous

I think someone had written a script to save individual bookmarks on the site itself (visible in a separate page). I'll let you know if I find it.

Alessandro Codenotti

21:39

Sure, but I'm saving the permalink to the message starting the conversation I'm interested in in a local file directly instead

Mike Miller

Nevermind!

Eran

Is there an element of order $80$ in $S_{20}$?

I'd like to have a hint please

Leaky Nun

@Eran what is the order of $(1~2~3~4~5~6)(7~8~9~10)(11~12~13~14~15~16~17~18~19~20)$? (don't worry it isn't 80)

Mike Miller

Seems like you should read the above discussion.

Eran

lcm of it's cycle lengths

Leaky Nun

21:48

there you go

Mike Miller

@Shaun the most recent comment makes me realize I mistakenly said "greatest common divisor" instead of "least common multiple" up above :P

Leaky Nun

Permutation on finite set

35 mins ago, 16 minutes total – 20 messages, 2 users, 0 stars

Bookmarked 8 secs ago by Leaky Nun

@Eran ^

Eran

So I should consider all possible partitions of 20 and then check it's lcm sounds like too much work

thanks

Leaky Nun

@Eran I'm sure you can come up with a better method

Mike Miller

you probably don't need to think about literally every partition.

Leaky Nun

21:52

you don't need to think about any partition at all

Eran

Only partitions to primes maybe, this way ill minimize the gcd and maximize the lcm ?

idk

Leaky Nun

you're closer

also @Mike hi!

Is it true that $H^n_{dR}(M) = \Bbb R$ for smooth connected compact orientable $M$ of dimension $n$ just by Poincare duality?

without boundary

Mike Miller

'just'

Leaky Nun

I mean...

"as a trivial corollary of"

Mike Miller

yes of course

Eran

21:55

@LeakyNun Closer primes or closer gcd haha

?

Leaky Nun

@MikeMiller and then as a corollary that every top form that integrates to zero is exact?

Mike Miller

that's not a corollary, that's the definition of what you just said

well, i guess you don't start with the integration functional at hand.

Leaky Nun

you still have to do a tad bit of work to show that $\int : H^n_{dR}(M) \to \Bbb R$ is surjective hence injective

Mike Miller

but yes.

Leaky Nun

what's the dimension of the empty manifold? is it even a smooth manifold? and why on earth am I thinking about edge cases again?

Mike Miller

21:58

the third question is the most important

Leaky Nun

also can this observation be generalized? "every vector field on R^2 that integrates to zero along every closed curve is the grad of some function"

Mike Miller

the definition that makes dimension additive under product is $\dim \varnothing = - \infty$, the definition that lets you make sense of null-cobordisms as a cobordism from one manifold to another says $\dim \varnothing$ is whatever you want it to be today

@LeakyNun exercise

Leaky Nun

hmm

Mike Miller

think about surfaces and de rham's theorem

Leaky Nun

I don't know de Rham's theorem

Mike Miller

22:00

mild irritation in sufficiently large dimensions

meh

Leaky Nun

ok I googled

how do you find Whitney approximation? @MikeMiller

Mike Miller

what does that mean?

Leaky Nun

Whitney approximation theorem: any continuous map between smooth manifolds is homotopic to a smooth map (or something like that)

Mike Miller

do it in charts or use convolution

to do it in charts without using convolution appeal to stone-weierstrass

Leaky Nun

@loch!

Shaun

22:06

@MikeMiller It happens t' best of us :)

Leaky Nun

Suppose we have a vector field $f : \Bbb R^3 \to \Bbb R^3$ that integrates to zero along the surface of every sphere ($S^2_{r,c} = r S^2 + c$). Let $g = \nabla \cdot f$. Divergence theorem says $\displaystyle \int_{B(c,r)} g = 0$, so $g = 0$, so $f$ is the curl of some function... @MikeMiller does this sound right?

Mike Miller

sounds plausible

good argument

Leaky Nun

ok so I need to generalize this using de Rham's theorem

Mike Miller

NOPE drop that last thing I said

more subtle in general as you observe

Leaky Nun

I mean this argument works for at most $n-1$ forms

the most straightforward generalization is a criterion for exactness for $n-1$ forms in $\Bbb R^n$

or actually more generally for closedness

I only used the cohomology being 0 for closed => exact

I drop that step and so I can generalize it

however de Rham's theorem talks about closed forms

Mike Miller

22:18

@LeakyNun disagree, your argument will show that if $\omega$ is a $k$-form on $M$ that integrates to zero on every closed $k$-submanifold, then $d\omega = 0$

Leaky Nun

ok more generally it's a identity theorem

ok you sniped me

Mike Miller

it's much the same except instead of determining that it's zero at every point you see that the value is zero on every k-plane

Leaky Nun

hmm

yeah I realized Stokes theorem isn't the crucial (i.e. new) step

it's the identity theorem bit

@MatheinBoulomenos!

Mike Miller

i do not know what the identity theorem bit means

Leaky Nun

i.e. a criterion for a function to be zero

MatheinBoulomenos

22:20

hi @LeakyNun

Leaky Nun

I... copied the name from the "identity theorem" from complex analysis

@MatheinBoulomenos we're discussing (i.e. I'm thinking about) how to generalize the observation that any 2D vector field that integrates to zero along any closed curve is a grad of some function

Mike Miller

i really disagree with that choice of notation

your deleted comment was on the right track, in some sense - this is a two-part problem

and what you described is part 1

Leaky Nun

wait what did I say?

ah I said it needs to be a boundary

Mike Miller

that was when you were talking about integrals over bounded manifolds

that gives you $d\omega = 0$

also, i can see every deleted comment now

:O

Leaky Nun

oh right...

Mike Miller

22:24

the raw power courses through my veins

Leaky Nun

@MikeMiller and is there a name to the thing we found?

Mike Miller

what thing did we find?

Leaky Nun

well... if something integrates to zero along any closed submanifold then it's zero

Mike Miller

you have not proved that yet

you have proved that if something integrates to zero along any closed submanifold, then $d\omega = 0$

this is probably best known as "exercise left to the reader", whose proof you supplied above

Eran

Could you please master tell me an example of a PID which is not Euclidean domain besides $\mathbb{Z}\ [\frac{1+\sqrt{-19}}{2}]$?

Mike Miller

22:32

it would be pretty silly if that was the only PID which wasn't a euclidan domain, wouldn't it

the full statement is best known as 'the de rham theorem', but actually proving it requires extra technology I will point out once you write a proof with a hidden subtlety

until I see a near-proof I will stay quiet

Eran

@Mi

@MikeMiller

Leaky Nun

by near-proof you mean near-proof of de Rham's theorem?

Mike Miller

near proof of the statement "if $\omega$ integrates to zero on every closed oriented submanifold of appropriate dimension, then $\omega$ is exact"

Leaky Nun

ok let's see...

MatheinBoulomenos

@Eran $\Bbb{Z}[\frac{1+\sqrt{-44}}{2}]$, $\Bbb{Z}[\frac{1+\sqrt{-67}}{2}]$, $\Bbb{Z}[\frac{1+\sqrt{-163}}{2}]$

Leaky Nun

22:36

@MatheinBoulomenos did you google that?

Eran

@MatheinBoulomenos what's with this form?

MatheinBoulomenos

@LeakyNun I knew that there were more among ring of integers of imaginary quadratic fields, but I had to look up the exact numbers, yeah

Mike Miller

rings of integers of the field $\Bbb Q(\sqrt{n})$, where $n \in \Bbb Z$, are always of the form $\Bbb Z[\sqrt{n}]$ or $\Bbb Z[\frac{1+\sqrt{n}}{2}]$ depending on parity

-163 is a famous one though

Leaky Nun

@MatheinBoulomenos then you should have told him to google

depending on whether $n \equiv 1 \pmod 4$ @Mike

Eran

Why didn't you tell me to gooogle yourself licky noun?

Mike Miller

22:38

I like licky noun

Daminark

Lmao

Eran

everyone loves licky noun

Mike Miller

thanks, @LickyNoun, I remembered it was something mod 4 but not the precise condition (so parity was the wrong word to use! I should have said depending on a congruence condition)

Eran

hahahah awesome, licky noun you're the smartest guy here

even smarter than the owner of the room

Leaky Nun

Let $\eta = \mathrm d\omega$. Suppose $\eta_p \ne 0$. Let $x : U \ni p \xrightarrow\sim \tilde U \subseteq \Bbb R^n$ be a chart, and then WLOG suppose $\eta_p > 0$ (otherwise look at $-\eta$). Pick $\varepsilon$ such that $\eta|_{B(x(p),\varepsilon)} \ge \varepsilon/2 > 0$. Integrate $\omega$ along the boundary of a ball with appropriate dimension to get a contradiction @Mike

Mike Miller

22:42

Sorry, you just keep showing that $\omega$ is closed

You want to show that it is exact too

Leaky Nun

oh wait what

Mike Miller

that was your goal

Leaky Nun

that's... interesting

user330477

For determining half-lives of radioactive isotopes, it is important to know what the background radiation is in a given detector over a specific period. The following data were taken in a γ -ray detection experiment over 98 ten-second intervals:

               58 50 57 58 64 63 54 64 59 41 43 56 60 50 46 59 54 60 59 60 67 52 65 63 55 61 68 58 63 36 42 54 58 54 40 60 64 56 61 51 48 50 60 42 62 67 58 49 66 58 57 59 52 54 53 53 57 43 73 65 45 43 57 55 73 62 68 55 51 55 53 68 58 53 51 73 44 50 53 62 58 47 63 59 59 56 60 59 50 52 62 51 66 51 56 53 59 57

Assume that these data are observations of a Poisson random variable with mean λ.
(a) Find the values of x and s2.

Mike Miller

it's the verbatim generalization of your original claim!

user330477

22:44

Is there a way to find $x$ bar and $s^2$ without using a calculator?

Leaky Nun

well I would need some global lemmas... since every closed form is locally exact

Mike Miller

I have said the name of the relevant theorem a few times

Leaky Nun

oh sorry

Mike Miller

I am not asking you to prove that theorem - I am asking you to see that what we want is (almost) a quick corollary of it

Leaky Nun

I'm a bit slow

Mike Miller

22:45

Nothing to be sorry about

If I was frustrated I probably would have stopped talking a while ago :D

Leaky Nun

I do need Whitney approximation though don't I

I mean, I need it to even state the de Rham theorem

whatever

Mike Miller

@LeakyNun do you?

Leaky Nun

so we know that $\omega$ is closed. Consider the corresponding element $W \in \operatorname{Hom}(H_p(M;\Bbb R), \Bbb R)$ that sends $[c]$ to $\displaystyle \int_c \omega$. By hypothesis that integral is zero, so $W = 0$, so $W$ is exact, so $\omega$ is exact

@MikeMiller because otherwise $\int_c \omega$ makes no sense

Daminark

Pullback on charts + partitions of unity, no?

Mike Miller

he's saying that $c$ is a continuous simplex

but usually the de Rham theorem is stated for the chain complex of smooth simplices

and it is a lemma that the inclusion of this into the chain complex of continuous simplices is a homology iso

(indeed, a Whitney-type result)

Leaky Nun

22:51

how's my near-proof?

Mike Miller

@LeakyNun what is $c$, precisely? I am skeptical of the phrase "By the hypothesis, that integral is zero"

Leaky Nun

ok

Mike Miller

It's the right idea but will need a little nudging

Leaky Nun

$W$ sends $[\sum a_i \Delta_i]$ to $\sum a_i \int_{\Delta_i} \omega$

I hope $\Delta_i$ is orientable

Mike Miller

:p

Leaky Nun

22:53

... is it a theorem that only taking injective simplices gives you the same homology groups?

Mike Miller

doubt it

iirc akiva asked about that once

Leaky Nun

this is not good

Mike Miller

don't worry about injectivity right now too much

Leaky Nun

pick an atlas, do barycentric subdivisions, do everything locally, boom

Mike Miller

you can replace the statement with "if for every smooth map $f: N^k \to M$ from a closed oriented $k$-manifold to $M$, $\int_N f^*\omega = 0$

where is the submanifold, in the end? why don't these simplices meet (say) orthogonally at the boundary?

(you are seeing the point! the injectivity comment was good)

Leaky Nun

22:56

I mean, if I do barycentric subdivisions enough times, all of my smaller simplices fall inside one of the charts

Mike Miller

why does that help?

Leaky Nun

so I can do everything locally

can I use your strengthened hypothesis?

Mike Miller

sure, but I don't see how this helps you find a closed manifold with boundary

Leaky Nun

I don't actually know how you define $\int_{\Delta_i} \omega$ to start with

Mike Miller

smooth simplices, as above

(can you prove the desired result for surfaces?)

Leaky Nun

22:58

so we're pulling it back?

Mike Miller

yeah

which is the same as your barycentric subdivision idea

since restriction to a chart == pullback

Leaky Nun

I don't know if this is what you're looking for, but $\Delta_i$ is the closed manifold (without boundary)?

Mike Miller

what is the boundary of a simplex?

Leaky Nun

more simplices?

Mike Miller

of a triangle? of an interval? :p

yeah, it's not closed

sorry, is the issue in my terminology 'closed'? that is topologist language for compact without boundary

Leaky Nun

22:59

oh that's what you mean

homotope the simplex to a sphere locally

Mike Miller

you can't do that for one simplex at a time - you need all the boundaries to cancel out

or your $c = \sum a_i \Delta_i$ is not a cycle

(If you get frustrated, let me know and I will say the point - I am enjoying the nudging)

Leaky Nun

ok I need some notations to tell apart the simplex and the image of the simplex

let's say $c = \sum a_i f_i$ where $f_i : \Delta^k \to M$

Mike Miller

really strongly suggest thinking about oriented surfaces to start

that case you can prove (I think, I am not sure what calculations you know) and you will see the issue in general

Leaky Nun

I need to show that $\displaystyle \sum a_i \int_{\Delta^k} f_i^\ast \omega = 0$ right?

Mike Miller

yeah, what is your idea/approach?

Leaky Nun

23:03

so my idea is that WLOG (barycentric subdivision) suppose that $f_i(\Delta^k) \subset U_i$

where $x_i : U_i \xrightarrow\sim \tilde U_i \subseteq \Bbb R^n$ is a chart

user330477

For determining half-lives of radioactive isotopes, it is important to know what the background radiation is in a given detector over a specific period. The following data were taken in a γ -ray detection experiment over 98 ten-second intervals:
58 50 57 58 64 63 54 64 59 41 43 56 60 50 46 59 54 60 59 60 67 52 65 63 55 61 68 58 63 36 42 54 58 54 40 60 64 56 61 51 48 50 60 42 62 67 58 49 66 58 57 59 52 54 53 53 57 43 73 65 45 43 57 55 73 62 68 55 51 55 53 68 58 53 51 73 44 50 53 62 58 47 63 59 59 56 60 59 50 52 62 51 66 51 56 53 59 57

Mike Miller

shoot, i have 10 minutes left - tell me your idea and then I will sketch what I have been thinking

user330477

@MikeMiller Hello. Long time no see.

Leaky Nun

@MikeMiller and then I need to prove $\displaystyle \sum a_i \int_{\Delta^k} f_i^\ast \omega = 0$ for the new $i$

Mike Miller

Sure

Leaky Nun

23:05

and then since each $\Delta^k$ falls inside $\Bbb R^n$, I can homotope them to spheres

and the integrals should be the same

Mike Miller

the issue is that when you do that, you move the boundary of each simplex - when you perform your homotopy, you need to do it on all simplices at once, so that at each stage of the homotopy $\partial c_t = 0$

(where $c_t$ means the cycle you get by performing your homotopy along all the simplices up to time $t$)

otherwise $\int c_0 \neq \int c_1$, or at least you can't prove that

Kasmir Khaan

hi yall

Leaky Nun

ok what's your idea then?

Kasmir Khaan

none leak , I just came in

Mike Miller

let me clarify the situation for surfaces first

$H_1(\Sigma_g; \Bbb Z) \cong \Bbb Z^{2g}$, each factor generated by a certain loop

assuming $\int_M \omega = 0$ for every smooth submanifold $M$ of $\Sigma_g$, we see therefore that the map $\int \omega: H_1(\Sigma_g;\Bbb Z) \to \Bbb R$ is identically zero - it vanishes on the generators

and therefore, we may apply de Rham's theorem

now the way one should really think of $H_k(M)$ (imo) is that elements are "bordism classes" of "singular" manifolds, where the singularities arise in codimension 2 - I have written about this a few times before, and this is certainly how Poincare was thinking in the early 1900s

Leaky Nun

23:10

5

A: invariance of integrals for homotopy equivalent spaces

Your expression doesn't make sense. If $\omega$ is a form on $\Sigma$, then it's not a form on $\Sigma'$! You need a map $f:\Sigma'\to \Sigma$ to compare them. So a better version of your claim would be $\int_{\Sigma'} f^*\omega=\int_\Sigma\omega$ when $f$ is a homotopy equivalence. This is now f...

> Theorem If $f,g:\Sigma'\to \Sigma$ are homotopic maps and $\Sigma'$ a smooth $k$-manifold and $\omega$ a closed $k$-form on $\Sigma$, then $\int_{\Sigma'}f^*\omega=\int_{\Sigma'}g^*\omega.$

Mike Miller

to be clear, Kevin is being lazy and means that $\partial \Sigma ' = \partial \Sigma = \varnothing$

now there is a big question - what classes of $H_k(M;\Bbb Z)$ may be represented by submanifolds?

this was a big question in the 50s and 60s, and there are examples where homology classes cannot be represented by submanifolds, or even by smooth maps from a smooth manifold

Kasmir Khaan

@LeakyNun Do you know why {1/n} does not converge in the metric is all the positive reals?

Leaky Nun

that is not a metric

Kasmir Khaan

with d(x,y) = |x-y|

why not

Leaky Nun

it doesn't converge because it would have to converge to 0

but 0 is not a positive number

recall that "converge" means "there is a point that it converges to"

it is an existence proposition

Mike Miller

23:13

so what I have to tell you is that to get the final result you need a big machine ;) Thom proved in the 60s via cobordism theory that every class of $H_k(M;F)$ can be represented by a smooth manifold if $F = \Bbb Z/2$ or if $F = \Bbb Q$

Kasmir Khaan

since when converge to an element is required that the elemnt is in the set?

Mike Miller

The latter is enough. While we start with a functional $H_k(M;\Bbb Z) \to \Bbb R$, it clearly factors through $H_k(M;\Bbb Q) = H_k(M;\Bbb Z) \otimes \Bbb Q$

Kasmir Khaan

aint this like limit ?

in fact we use the same notation

Leaky Nun

well in limit you worked in $\Bbb R^n$ presumably

Mike Miller

and since this group does have manifold representatives for every class, we see that the integration functional vanishes here

Leaky Nun

23:14

(or worse, just $\Bbb R$)

where this issue doesn't occur

but more generally your limit point needs to be inside your set

Kasmir Khaan

our sequence is always positive

Mike Miller

(I did not expect you to come up with this, of course; I thought it would be a good idea to struggle with it for a bit so as to see where the many issues come in)

Kasmir Khaan

okay thanks

Leaky Nun

because points outisde the sets do not exist in the point of view of the set itself

Mike Miller

I really liked when you caught the injectivity problem

Leaky Nun

23:14

it doesn't even know that 0 exists

Kasmir Khaan

seems fair as a point

Leaky Nun

@MikeMiller thanks :)

Kasmir Khaan

but we never did it that way

Leaky Nun

because you only did it for $\Bbb R^n$

where this issue doesn't occur

Kasmir Khaan

hmm

Leaky Nun

23:17

and Cauchy noticed that at least in $\Bbb R^n$, an equivalent way of formulating that the statement "$a_n$ converges" without referring to any point of convergence

is that the sequence is, what we today call, Cauchy

however this "Cauchy implies converges" is unfortunately false in general metric spaces

for example in the space of positive reals as you've seen

Kasmir Khaan

cauchy converge is the better criteria

it does not care what happen in the "end"

only at certain concecutive points

cauchy was a genius :D

Leaky Nun

but, the upshot is, that you can complete any metric space to get a new metric space where every Cauchy sequence converges

Kasmir Khaan

cool stuff

Leaky Nun

so the statement is that for any metric space $(X,d)$ there is a metric space $(\overline{X}, \overline d)$ and a continuous injection $f : X \to \overline X$ with dense image and that commutes with the metric, such that every Cauchy sequence in $\overline X$ converges

and for the case of $\Bbb Q$ with the usual metric it gives you $\Bbb R$ with the usual metric

(please, can someome verify)

2

Q: What is the Cauchy completion of a metric space?

I was wondering what the "Cauchy" completion of a metric space is. I can't find any helpful information on Google. Feel free to post any links to sources you find relevant.

real-analysis metric-spaces cauchy-sequences

the answer that is not accepted is surprisingly nice

there's a universal property with the Cauchy completion

in fact it has quite some analogy with the case of algebraic closure

if $i_1 : K \to L_1$ and $i_2 : K \to L_2$ are field homomorphisms such that $i_1$ is algebraic and $L_2$ is algebraically closed, then there is a field homomorphism $L_1 \to L_2$

if $f:(X,d_X) \to (Y,d_Y)$ and $g:(X, d_X) \to (Z,d_Z)$ are continuous isometries such that $f$ has dense image and $(Z,d_Z)$ is complete, then there is a continuous isometry $(Y,d_Y) \to (Z,d_Z)$

and furthermore, given $i : K \to L$ such that $L$ is algebraically closed, we can construct an algebraic closure of $K$ (one that is both algebraic and algebraically closed) as the integral closure of $i$

and furthermore, given $\varphi : (X, d_X) \to (U, d_U)$ such that $(U, d_U)$ is complete, we can construct a completion of $(X, d_X)$ (one that both has dense image and is complete) as the closure of (the image of) $\varphi$

I see I'm being ignored by @KasmirKhaan

Kasmir Khaan

23:36

@LeakyNun No no sorry, was reading my book

@LeakyNun thanks for the answer, I only wanted basic answer for the moment , still not there in the reading material

Leaky Nun

ok

Leaky Nun

23:49

@AkivaWeinberger you might be interested in this: so there's this age-old problem of "how many times in 12 hours do the hour hand and the minute hand overlap?" and it corresponds to counting the number of intersections of the (12,1) knot and the (1,1) knot around the torus: so given coprime pairs (p,q) and (r,s) what is the number of intersections of the (p,q) knot and the (r,s) around the torus?

"the (4,1) knot and the (1,1) knot intersects 3 times"

Permutation on finite set

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