Mathematics - 2019-01-16 (page 1 of 2)

4:30 AM

So, a question: what kind of mathematical sttucture would be best to analyze a game where every action changes the actions which are available? Something like chess, for example.

Leaky Nun

@Rithaniel markov chains?

Dair

4:50 AM

@LeakyNun Markov chains for a game with no chance? Interesting.

Leaky Nun

hmm nvm then

Dair

@Rithaniel You can traverse a game tree in the case of chess

@Rithaniel a lot of abstract two player strategy games have a fairly straightforward api (more programming, less math, but it can probably be adapted). initial_position, generate_move(position), do_move(pos, mov) and primitive(pos). primitive(pos) should return whether the game is trivially over or not.

you can then do backward induction to solve the game.

Rithaniel

Also, this specific game is apparently turing complete. It's a video game called Opus Magnum, and I don't think a game tree would necessarily work, because the option space for any particular step is colossal.

Dair

Probably not, but it should in theory work for chess (however, chess is too large to actually be solved)

Rithaniel

Yeah, I recall that detail about Chess, but I did know there was effort put into try and mathematically analyze the game.

A game tree might work for Opus Magnum, but I'm still dubious because of the sheer size of the solution space.

Though, what I'm most interested in is proving whether particular scores are impossible. A lot of the community around the game is centered around optimization of different metrics.

Dair

4:59 AM

game looks complicated and I'm too tired to figure out the rules, it may not fall into the category of a two-player abstract strategy game, in which case the game tree will not necessarily work.

Rithaniel

Yeah, more a single player puzzle game. (Also, I'm pretty sleepy, too.)

Dair

(btw it should be generate_moves(pos) not generate_move(pos). Rip.)

Albas

6:32 AM

I noted something interesting yesterday. Take the field Q(rational numbers), if I take the product of all possible non equivalent norms on Q(norm being a function from Q to the non negative real numbers), then the product is unity.
Is this some general property of a field?

Piyush Divyanakar

6:48 AM

do partial differential equations always have a singular solution ?

Leaky Nun

7:05 AM

@Albas of global fields

Akiva Weinberger

8:54 AM

Does $\int_0^{2\pi}\sqrt n\cos^n(x)dx$ approach a limit?

famesyasd

10:25 AM

is there such a function f : E -> R, E subset R such that it is continuous and vanishes on some dense set S of E but is not equal to zero otherwise? (obviously f(x) = 0 forall x in E is not allowed)

Leaky Nun

@famesyasd no

hint: R is T1

famesyasd

T1?

what is T1

Leaky Nun

Separation axiom

In topology and related fields of mathematics, there are several restrictions that one often makes on the kinds of topological spaces that one wishes to consider. Some of these restrictions are given by the separation axioms. These are sometimes called Tychonoff separation axioms, after Andrey Tychonoff. The separation axioms are axioms only in the sense that, when defining the notion of topological space, one could add these conditions as extra axioms to get a more restricted notion of what a topological space is. The modern approach is to fix once and for all the axiomatization of topological...

Alessandro Codenotti

Don't you need $T_2$ here?

Leaky Nun

I don't think so

Alessandro Codenotti

10:31 AM

No I agree you don't in this case, I was thinking of a more general result

Leaky Nun

$I_{n+2} = \int_{0}^{2\pi} \cos^{n+2}(x) \ \mathrm dx = \int_0^{2\pi} \cos^{n+1}(x) \ \mathrm d\sin(x) = -\int_0^{2\pi} \sin(x) \ \mathrm d\cos^{n+1}(x) \\ = (n+1) \int_0^{2\pi} \sin^2(x) \cos^n(x) \ \mathrm dx = (n+1)(I_n - I_{n+2})$, so $I_{n+2} = \frac{n+1}{n+2} I_n$

famesyasd

so we just pick any irrational point x0, pick any rational sequence xn that tends to it and then by continuity |f(xn)-f(x0)| < epsilon => |f(x0)| < epsilon forall epsilon > 0 => f(x0) = 0 but I don't see how T1 applies or whatever

Alessandro Codenotti

T1 implies {0} is closed and so is its preimage through a continuous function

Leaky Nun

anyway we know that $I_{2k+1} = 0$ and we might as well treat $I_0$ as a constant and not compute it

Alessandro Codenotti

But there aren't many dense and closed sets

Leaky Nun

10:38 AM

$I_4 = \frac34\frac12 I_0$

$I_{2n} = \frac{(2n)!}{4^n (n!)^2} I_0$

So we want to know $\displaystyle \lim_{n\to\infty} n^k \frac1{4^n} \binom{2n}n$, and particularly $k=0.5$ @AkivaWeinberger

Akiva Weinberger

Oh OK

And then we can just Stirling it

(And $I_0=2\pi$)

Leaky Nun

$\dfrac{(2n)!}{(n!)^2} \le \dfrac{e(2n)^{2n+0.5}e^{-2n}}{2\pi n^{2n+1} e^{-2n}} = C\dfrac{4^n}{\sqrt{n}}$

Akiva Weinberger

So I think you end up with $2\sqrt\pi$?

$2\pi\sqrt n\cdot\dfrac1{\sqrt{\pi n}}$

Leaky Nun

oh no

I can't squeeze it to zero

I can only bound it by a constant

your choice of $\sqrt{n}$ was very planned

Akiva Weinberger

$3.54491$

Leaky Nun

10:47 AM

wolframalpha says it is 1/sqrt(pi)

so you have two subsequences, one converging to 1/sqrt(pi), and one converging to 0

Akiva Weinberger

Right and then multiply by $2\pi$

Oh right yeah if it's an odd exponent it's 0

Didn't realize

Leaky Nun

anyhow

Akiva Weinberger

This is like the odds of flipping $n$ coins and having it be half heads half tails

Leaky Nun

fun question

Akiva Weinberger

$\binom{2n}n/2^{2n}$

Or $\binom n{n/2}/2^n$ actually

for $\frac1{2\pi}I_n$

@LeakyNun It's strange how, if you flip $2n$ coins, the odds of exactly $n$ being heads and $n$ being tails involves $\pi$

(approximately)

$1/\sqrt{\pi n}$

Leaky Nun

10:59 AM

I see

Akiva Weinberger

I dunno if $\pi$ shows up unexpectedly elsewhere in probability

Well I guess $\int_{-\infty}^\infty e^{-x^2}dx=1/\sqrt\pi$ counts as probability

in a sense

The two are related probably

Secret

2:23 PM

Last night dream, saw a visual snow with a very complicated cellular structure:

oily water placed on the side for reference of impression

I wonder if there is an algorithm to generate these patterns, especially the way the large and small ovals are herded together

Abhas Kumar Sinha

2:35 PM

How to find the derivative using the first principle? $$x^2 + y^2 = c^2$$, where c is some constant

rschwieb

2:55 PM

Quick poll: how much of mathematics with the term "fuzzy" is actually useful? I've heard of fuzzy set theory and fuzzy logic, and in spirit the idea of de-discretizing boolean values and having a sort of continuum of truth is appealing. But in my experience the adjective "fuzzy" has been used by researchers writing papers of very low quality, often inventing ideas which are clones of existing theory so that they can re-prove all the known results for their new thing.

Overall, from my own experience and a few of my mentors, we have a very dim view of the adjective.

In short, how much of it is just bandwaggon-hopping-on injection of the adjective "fuzzy" and how much of it is serious mathematics?

Mike Miller

You're right

rschwieb

3:09 PM

@MikeMiller Are you agreeing with the part that "fuzzy set theory and logic are actual things"?

Poline Sandra

Dear all how are you

Mike Miller

I am agreeing with the experience of you and your mentors.

rschwieb

That's the insidious part of this whole sordid business is that the subject matter has hooks that let bogus researchers camouflage their ideas with.

Mike Miller

I think mostly it's more irrelevant than sordid.

Poline Sandra

can someone help me on this : math.stackexchange.com/questions/3075840/…

rschwieb

3:11 PM

@MikeMiller OK... but they really did not have opinions about fuzzy logic or set theory. Just the "clone" fuzzy theories of our own fields that seemed to have nothing tangible to add.

I think this sort of semi-fraud does enough harm to be sordid :)

Although... maybe it is good we have this junk around to keep us on our toes.

@MikeMiller THanks for your input, though, really.

Good to have a sanity check when dealing with this sort of thing

Poline Sandra

who knows the connected compenent in topology ???

rschwieb

@PolineSandra What about connected components?

Poline Sandra

this question please: math.stackexchange.com/questions/3075840/…

rschwieb

@PolineSandra YOu really need to add more context to the question, or it's going to be closed. You've got a start, right? you should edit it in. If you've already expressed a space as a disjoint union of nonempty closed connected subsets, isn't it clear that each of those connected subsets is maximal-connected?

Poline Sandra

3:28 PM

no i donàt know from where i must begin

rschwieb

@PolineSandra Well, that would be a good place to begin. Show that each of the $F_j$ is not properly contained in another connected subset.

I guess that's your definition of "connected component"?

Rithaniel

3:48 PM

Suppose you have $\mathbb{Z}_n$ and you define $U(\mathbb{Z}_n=\{[a]|\text{gcd}(a,n)=1\}$. It's true that if $[a]\in\mathbb{Z}_n$ and $x\in [a]$, then $\text{gcd}(x,n)=\text{gcd}(a,n)$, correct? Also, that if $[a],[b]\in U(\mathbb{Z}_n)$, then $[ab]\in U(\mathbb{Z}_n)$, correct?

Alessandro Codenotti

What do you think?

Rithaniel

I'm a little dubious, because the next question is "show that if $[a]\in U(\mathbb{Z}_n)$, then show that $[a^m]\in U(\mathbb{Z}_n)$ for some $m\in\mathbb{Z}$," and, by just restating the second statement, that should be true for every $m\in\mathbb{Z}$.

Yet it's weighted as a 5 point question when the previous statement is weighted as a 3 point question, which feels like it should be a more difficult question, and so I'm suspicious that I'm missing something.

(As for what I think, I think that both of those initial statements are true.)

(I'm mostly looking for assurance.)

Ah, well, $m\in\mathbb{Z}$ must be non-negative, but that's actually a side effect of me copying the problem down incorrectly. Should be $\mathbb{N}$, not $\mathbb{Z}$.

rschwieb

4:06 PM

@Rithaniel Why would it have to be $\mathbb N$? Where are you getting that idea?

Rithaniel

Well, $\mathbb{Z}_n$ is integers mod n. An integer to a negative power is not an integer, so that would violate the structure being a group.

rschwieb

I would say that if $[a]$ is a unit of $\mathbb Z_n$, then $[a]^m$ is a unit for ALL $m\in\mathbb Z$. Even the negative ones. Because $[a]^{-1}$ is a unit as well.

Thorgott

Both of these statements seem true to me. The first is a reformulation of $\gcd(a+kb,b)=\gcd(a,b)$ for integer $k$ and the second is a reformulation of $\gcd(a,n)=\gcd(b.n)=1\Rightarrow\gcd(ab,n)=1$. Both of these statements can be derived from Bezout's Lemma. Though I am also confused by that deceptively easy looking second question.

rschwieb

An integer to a negative power is not an integer true, mostly, but $[a]$ is not an integer.

Rithaniel

Ah, it's an equivalence class, good point, but we're talking about $[a^m]$, and not $[a]^m$.

rschwieb

4:09 PM

@Rithaniel But $[a^m]=[a]^m$

OK well, i've got off to a bad start explaining this

you're right, of course, that $a^m$ isn't an integer when $m$ is an arbitrary integer.

Rithaniel

It does? (I'll be honest, I'm not familiar with power operations on equivalence classes)

rschwieb

But I think the spirit of the problem is to show that the equivalence class's powers are all units

Thorgott

That equality holds only for $m$ non-negative

rschwieb

@Thorgott Yes, i agree, I botched the line of explanation.

I shouldn't have disagreed earlier :)

@Rithaniel I'm not familiar with power operations on equivalence classes . You make it sound so complicated! it's just repeated multiplication...

Rithaniel

Alright, so we would define $[a]^m$ as equal to $[a^m]$, then? I got an impression that $[a]^m$ would be some operation on the equivalence class.

Thorgott

4:14 PM

If $[a]\in U(\mathbb{Z}_n)$, then $[a^m]\in U(\mathbb{Z}_n)$ for all non-negative $m$. If the modular multiplicative inverse $[a]^{-1}$ exists, then you should even have $[a]^m\in U(\mathbb{Z}_n)$ for all integer $m$, but of course $[a^m]$ doesn't exist for the negative values of $m$:

Rithaniel

Which, effectively, $[a^m]$ already is.

Thorgott

$[a^m]=[a\cdot...\cdot a]=[a]\cdot...\cdot[a]=[a]^m$. This can be made rigorous by induction as a consequence of $[ab]=[a][b]$ being well-defined.

Rithaniel

Now, this next one seems much more impressive: "Show that if $n=p_1^{a_1}p_2^{a_2}p_3^{a_3}···p_m^{a_m}$ with each $p_i$ prime and each $a_i\in\mathbb{N}$, then then $|U(\mathbb{Z}_n)|=\Pi_{i=1}^m (p_i^{a_i −1}(p_i−1))$"

Also, I went ahead and decided to go with mathematical induction to show that the assertion holds form every $m\in\mathbb{N}$, too. @Thorgott

rschwieb

@Rithaniel . I got an impression that $[a]^m$ would be some operation on the equivalence class. It is exactly that. I'm not sure why you'd think of it any other way.

It happens to have the property that $[a^m]=[a]^m$ whenever $a^m$ is defined

Thorgott

4:33 PM

Hint: First consider the case where $n=p^a$ is a single prime power, then generalize.

Rithaniel

Hmmm, two passes of induction?

One for adding new primes and one for increasing powers? (Though, increasing powers is just adding new primes, effectively, so maybe just one inductive pass)

Thorgott

I don't think you even need induction. Start with the simplest case; when does $\gcd(a,p)=1$?

Rithaniel

When $a<p$, so you have $p-1$ such $a$.

Well, in $\mathbb{Z}_p$, it might be better to say $a\neq 0$, right?

Thorgott

Yes, since $\gcd(0,p)=p$ (generally $\gcd(0,n)=n$), so its exactly the classes $[1],...,[p-1]$. We have $\gcd(a,n)=\gcd(b,n)=1\Rightarrow\gcd(ab,n)=1$, so when does $\gcd(a,p^2)=1$?

Rithaniel

When $\text{gcd}(a,p)=1$, so we just multiply the number of instances of elements in $U(\mathbb{Z}_n)$ by $p$?

Thorgott

4:53 PM

That's the idea. Each $a\in U(\mathbb{Z}_n)$ gives rise to $p$ elements in $U(\mathbb{Z}_{p^2})$ (namely $a,a+p,...,a+(p-1)p$). You just need to check that there is no other $a$ satisfying $\gcd(a,p^2)=1$. This can be done by noting that all the elements we haven't yet considered are integer multiples of $p$ and thus share the divisor $p$ with $p^2$ (and the case $a=0$).

user330477

5:46 PM

Is it true that if $f$ is continuous on $[a,b]$ and $f=0$ on $(a,b)$, the $f=0$ on $[a,b]$? It is not trivial just by definition of continuity.

Mike Miller

It is, you just haven't applied it to the right open set.

user330477

@MikeMiller What is wrong with what I said?

Learner

Hey everyone! Is someone familiar with the Stolz-Cesàro theorem? I have the following question, and I've made an attempt, but I would really appreciate some feedback:

1

Q: Stolz-Cesàro $0/0$ case: is $\limsup \frac{a_n}{b_n}\le \limsup\frac{a_{n+1}-a_n}{b_{n+1}-b_n}$?

The general form of Stolz-Cesaro $\infty/\infty$ case states that any two real two sequences $a_n$ and $b_n$, with the latter being monotone and unbounded, satisfy $$\liminf\frac{a_{n+1}-a_n}{b_{n+1}-b_n}\le\liminf\frac{a_n}{b_n}\le\limsup \frac{a_n}{b_n}\le \limsup\frac{a_{n+1}-a_n}{b_{n+1}-b_n...

Thorgott

Take a sequence $(x_n)_n$ in $(a,b)$, s.t. $x_n\rightarrow a$. You have $f(x_n)=0$ for all $n$ and $f(x_n)\rightarrow f(a)$ by continuity, hence $f(a)=0$. The same works for $b$.

user330477

@Thorgott Thank You.

Thorgott

6:00 PM

np

rationalpi

Let K be a subfield of C not contained in R then if Char(K)=0 then K is dense is easy to see as it will have Q as a subfield. Is this true when char(K)=!0?

Rithaniel

Okay, I've reached a point where my proof needs one more element and I'm unsure how to prove this part. Given $\text{gcd}(m,n)=1$, then I want to say that $|U(\mathbb{Z}_{mn})|=|U(\mathbb{Z}_{m})||U(\mathbb{Z}_{n})|$

Thorgott

Can you use the CRT?

Learner

@Thorgott Sir, may I paste here to you the proof I wrote under that question?

Rithaniel

What does CRT stand for?

Ah, chinese remainder theorem?

That's actually the next problem after this one. So I can prove it there and then refer to it in this problem. How do I utilize it, though?

Thorgott

6:15 PM

The CRT gives you a bijection (a ring isomorphism even, though that's not the point here) between $\mathbb{Z}_{mn}$ and $\mathbb{Z}_m\times\mathbb{Z}_n$. You can restrict this to a bijection between $U(\mathbb{Z}_{nm})$ and $U(\mathbb{Z}_n)\times U(\mathbb{Z}_m)$ upon noting that $\gcd(a,mn)=1$ is equivalent to $\gcd(a,m)=\gcd(a,n)=1$. Sets in bijective correspondence have equal cardinality and $|A\times B|=|A|\cdot|B|$, so that's what you want.

Rithaniel

Would it be correct to just say that because there is a bijection between $\mathbb{Z}_{mn}$ and $\mathbb{Z}_m\times\mathbb{Z}_n$, that there is also a bijection between $U(\mathbb{Z}_{mn})$ and $U(\mathbb{Z}_m)\times U(\mathbb{Z}_n)$? Or would that warrant an argument?

I see how that argument would go, but I wanna make sure that it's worth the time investment.

Thorgott

You need to a) prove that this new mapping is well-defined, i.e. that the image of some $a\in U(\mathbb{Z}_{mn})$ under the bijection is in $U(\mathbb{Z}_m)\times U(\mathbb{Z}_n)$ and b) that the new mapping is a bijection (injectivity is clear since it's a restriction of an injection, so surjectivity remains)

@Learner I'm not too familiar with Stolz-Cesaro and only skimmed over your proof, but from my first impression, I don't see why you can assume that $(b_n)_n$ is monotonic. Also, you start with an $\alpha>\limsup\frac{a_{n+1}-a_n}{b_{n+1}-b_n}$ and then show $\alpha>\limsup\frac{a_n}{b_n}$, but, unless I'm missing something, this does not generally imply $\limsup\frac{a_{n+1}-a_n}{b_{n+1}-b_n}>\limsup\frac{a_n}{b_n}$.

Learner

@Thorgott The monotonicity of $b_n$ is a hypothesis of Stolz-Cesaro

As for the reasoning about $\alpha$, I think that's normal. It depends on the fact that we we worked with any $alpha$

Thorgott

6:45 PM

Ah, I see. I missed that $\alpha$ was arbitrary. I'll give it a closer look now.

Learner

Thanks!

user10478

7:25 PM

youtu.be/vr0sTKbV7lI?t=145 Is the infinitesimal volume in the video an actual cube (so that $dx$, $dy$, and $dz$ are some arbitrarily small but equal $h$), or is it a miniature version of the larger volume (so that, i.e., the ratio of $dx$ and $dz$ equals the ratio of the $x$ and $z$ in the larger volume)?

Ted Shifrin

7:37 PM

@user10478 The guy says "cube" but the dimensions should be arbitrary (and "small"). You're never going to fill up the big rectangular solid with cubes if the dimensions are different.

Learner

Hi @TedShifrin!

Ted Shifrin

Hi @Learner

user10478

@TedShifrin Why not? Is it required to use the same number of small volumes in each dimension?

Learner

@TedShifrin Are you familiar with the Stolz-Cesàro theorem?

Ted Shifrin

I don't know what you mean by that. If you use cubes, then no matter how many you use, you're going to end up with the same net change in each dimension.

I don't know the name, @Learner.

@user10478: There's nothing that says you have to divide the region evenly, just as you don't have to in one-dimensional integrals. I don't know if you know the actual definition of the integral. Most books don't do a good job.

user10478

7:43 PM

Okay, yeah I think I understand the 1-dimensional case pretty well, I just wasn't sure how the differentials of the independent variables were supposed to relate to one another if at all.

Ted Shifrin

No, they're totally independent.

Learner

@TedShifrin It's about real sequences, though it comes with different flavours. One version is that any two real two sequences $a_n$ and $b_n$, with the latter being monotone and unbounded, satisfy

$$\liminf\frac{a_{n+1}-a_n}{b_{n+1}-b_n}\le\liminf\frac{a_n}{b_n}\le\limsup \frac{a_n}{b_n}\le \limsup\frac{a_{n+1}-a_n}{b_{n+1}-b_n}.$$

user10478

Okie, thank you

Learner

In the following:

1

Q: Stolz-Cesàro $0/0$ case: is $\limsup \frac{a_n}{b_n}\le \limsup\frac{a_{n+1}-a_n}{b_{n+1}-b_n}$?

The general form of Stolz-Cesaro $\infty/\infty$ case states that any two real two sequences $a_n$ and $b_n$, with the latter being monotone and unbounded, satisfy $$\liminf\frac{a_{n+1}-a_n}{b_{n+1}-b_n}\le\liminf\frac{a_n}{b_n}\le\limsup \frac{a_n}{b_n}\le \limsup\frac{a_{n+1}-a_n}{b_{n+1}-b_n...

real-analysis calculus sequences-and-series proof-verification limsup-and-liminf

Ted Shifrin

@Learner: Nah, I don't know it. I've never come across that.

Learner

7:45 PM

@TedShifrin Got it, thanks for hearing me out

Ted Shifrin

Too bad I no longer have my Pólya-Szegö.

Thorgott

@Learner I read the proof and I think it's a nice one. I only had some minor (and probably irrelevant) remarks, but they were too long for this chat, so I posted them as answer to your question. Hope you don't mind.

Learner

@Thorgott Absolutely not, thank you so much! And I think you make good points, upvoted and accepted!

Thorgott

Thanks and no problem :)

CaptainAmerica16

8:33 PM

@Ted Induction can only be used within $\Bbb N$?

Thorgott

You should be able to use induction on any set equipped with a well-order

CaptainAmerica16

I can't think of any other sets of numbers that have a least member.

Fargle

@CaptainAmerica16 The natural numbers starting at 2?

CaptainAmerica16

Hm... :P

Fargle

Seems like a joke answer, but it is important to note that you can also apply induction to "shifts" of the natural numbers---sometimes a statement is only true for $n \geq 5$.

CaptainAmerica16

8:41 PM

Well, that is true.

Thanks

Thorgott

You can easily define a well-order on the integers and use induction on it. It's important to note though that in that case the induction step is not of them form $A(n)\Rightarrow A(n+1)$.

Ted Shifrin

@Thorgott: CaptainAmerica is just starting out in math. No need to go overboard.

Fargle

Hey @Ted

Ted Shifrin

Heya @Fargle

CaptainAmerica16

So, I guess just any specifically defined set that has a least member.

Even if it isn't in the natural numbers.

Ted Shifrin

8:43 PM

No. What would you do with all the nonnegative real numbers?

Fargle

Not necessarily. Good luck inducting on [0,1]

Ted Shifrin

wonders if @Leaky has discovered the meaning of "is" yet

CaptainAmerica16

son of a gun

Leaky Nun

nope

Thorgott

Ah, I wasn't aware. I didn't mean to go overboard.

CaptainAmerica16

8:45 PM

Well, I get the gist. I'm not going to overthink what I know.

Ted Shifrin

LOL, I didn't mean to sound harsh. But it's hard enough keeping CaptainAmerica on track :P

Leaky Nun

@TedShifrin I don't understand how you induce an orientation on the boundary of an orientable $n$-manifold in the case of $n=1$

Fargle

You don't only need a least element, you need every element to have a clearly-defined "next" element too.

Ted Shifrin

You assign a sign to each point, @Leaky.

CaptainAmerica16

Yeah, I get that part. @Fargle

Leaky Nun

8:45 PM

in what way?

Fargle

Gotcha. Just running my mouth :P

Ted Shifrin

It's a $+$ sign if the oriented basis for your $1$-manifold points "out" and a $-$ sign if it points "in."

Leaky Nun

and more formally?

Ted Shifrin

That's consist with the rule that you look at $n,v_1,\dots,v_{n-1}$.

CaptainAmerica16

I'm on part b of question 3 in ch.2

Thorgott

8:47 PM

You have a least element. Remove that least element from the set. The least element of this new set is the next element. @Fargle

Ted Shifrin

@Leaky: That is formal. What's the definition of an outward-pointing normal in general?

Fargle

@Thorgott That's only valid if we already have an inductive set in the first place: I cite again the example of [0,1]

Ted Shifrin

@Thorgott had well-ordering.

You guys aren't on the same page.

Fargle

Ah, understood

Mike Miller

@LeakyNun Let's chase definitions (this is what you're asking in the end). If $M$ is oriented, how do you define the orientation on $\partial M$?

Never write down a fixed dimension.

Leaky Nun

8:48 PM

@TedShifrin I want to define it as a section of the $\{-1,1\}$-bundle

should we go over to the other room?

Mike Miller

Whatever.

Ted Shifrin

Huh, @Leaky. We need to do what Mike just said and what I said before him.

You need to define the boundary orientation in general.

So you need to know what an outward-pointing normal vector is at the boundary.

Thorgott

Yeah, I meant on a well-ordered set and then it's valid

Mike Miller

He wants to say it in terms of orientation double covers, but if we write $\Lambda(E)$ to mean the orientation double cover corresponding to some vector bundle, you ultimately need an isomorphism $\Lambda(TM\big|_{\partial M}) \cong \Lambda(T \partial M)$. And that, ultimately, amounts to what we're asking.

No way to avoid this in the end.

Fargle

Yeah, I agree in that case. Sorry for the misunderstanding @Thorgott

Thorgott

8:51 PM

All good

Ted Shifrin

Yup, what Mike said. It's a special case of the adjunction formula for the canonical bundle of a divisor in a complex manifold :D Tee hee hee.

Mike Miller

fancy words for a man who's just splitting an exact sequence :p

Leaky Nun

I don't even know what an orientation is; online sources say that I only need to give an oriented chart for the interior