Mathematics - 2024-08-27

Debug

05:40

Just wanted to let everyone know that @Shaun is our new moderator! 🤓 See here:
https://math.meta.stackexchange.com/questions/37708/2024-election-results-congratulations-to-our-new-moderator?cb=1

Congrats to Shaun and to everyone who voted for him.

✌

Sourav Ghosh

06:04

@leslietownes Could you recommend any books, journals, or papers on dilation theory in finite dimensions?

Ryder Rude

06:23

@SoumikMukherjee this is what intuitively seems to be the answer. but upon inspection, i think both choices are identical except that less people are dying in the other choice

in both choices, ur decision is responsible for the death of 'm' people and the life of 'n' people. so, without further information, the choices with n>m must be the rational one

but intuition says that doing nothing is what we should do. but i think intuition can flip if we change the scenario to 1 million vs 1 person

Shaun

06:46

@Debug Thank you :)

Bml

07:11

Hi everyone, I repost my question. Could someone help me solving the inequality $(a+b+c-x)^2 x \leq 4abc$, with $a, b, c > 0$ and $0 < x < a + b + c$? Is there a method to solve this type of irreducibile cubics?

Binky McSquigglebottom

09:24

Hi

Soumik Mukherjee

Hi

Bml

@BinkyMcSquigglebottom Hi, did you manage to solve the inequality we talked about yesterday?

Binky McSquigglebottom

09:51

No :⁠-⁠(

Pizza

Hi

Sine of the Time

hi

Sine of the Time

10:06

@Bml add more context and then ask it on MSE

Ryder Rude

10:21

hi

Binky McSquigglebottom

@Bml What you can do is maybe take a function approach with a derivation to do

Don’t really know if it works or not. If someone have inputs on that I am taking them.

Pizza

10:47

I was doing this exercise: Determine the relative and absolute extrema of the function $f(x, y) = 2x^2 + y^2-x$ in the set $D = \{(x, y) : x^2 + y^2 ≤ 1\}$

I found a relative minimum point $f\left(\frac{1}{4},0\right) = -\frac{1}{8}$

Now I have considered the edge $x^2 + y^2 = 1$ , so $x = \cos(\theta) , y = \sin(\theta) , \theta \in [0,2\pi]$

So the function becomes: $f(\theta) = 2\cos^2(\theta) + \sin^2(\theta) - \cos(\theta)$ , $f'(\theta) = -\sin(2\theta) + \sin(\theta)$

so i have to find when $\sin(\theta) = \sin(2\theta)$

I'm stuck here :(

Sine of the Time

$\sin(2x)=2\sin x \cos x$, $\sin x-\sin(2x)=0 \implies \sin x(1-2\cos x)=0$

Pizza

Oh yes right

Soumik Mukherjee

@RyderRude that's the main point, how do you quantify human lives?

Pizza

11:16

@SineoftheTime So after calculating the function at these points that I found, I also have to manually check at the vertices, right?

Sine of the Time

you have to check the end points of $[0,2\pi]$

Pizza

11:33

@SineoftheTime I found that $f(0) = 1, f(\pi) = 3, f(\frac{\pi}{3}) = \frac{3}{4}, f(\frac{5\pi}{3}) = \frac{3}{4}, f(\frac{1}{4},0) = -\frac{1}{8}$

Sine of the Time

your notation is confusing

it's not a good idea to use $f$ for the parametrization

it's better to use $g(\theta)$

Pizza

$g(0) = 1 \\ g(\pi) = 3 \\ g(\frac{\pi}{3}) = \frac{3}{4} \\ g(\frac{5\pi}{3}) = \frac{3}{4} \\ f(\frac{1}{4},0) = -\frac{1}{8}$

$g(2\pi) = 1$

Sine of the Time

I did not check the computation

but this is the general strategy

Pizza

11:56

@SineoftheTime Sorry I was doing something ... Anyway I checked on wolfram and I should have done it right

Only that: Maximize[{2x^2 + y^2 - x, x^2 + y^2 <= 1}, {x, y

I find that point 3 is an absolute maximum

Which would be at π so at the point (-1,0)

So I should have done well

Sine of the Time

$g(\pi)=f(-1,0)$

Pizza

@SineoftheTime But in that case, it would also be right to use Lagrange multipliers, right?

Sine of the Time

yes, you can

Pizza

In case I forget for example the values of the various angles

I can use this other method

Sine of the Time

I don't know if it's simpler though

Pizza

12:10

@SineoftheTime Maybe it depends on the type of domain that the exercise gives us

Sine of the Time

yes

Bml

@SineoftheTime OK.

@BinkyMcSquigglebottom Thank you for your attempt.

Ryder Rude

13:14

@SoumikMukherjee people die regardless of what one chooses here. if we dont quantify, we would have to choose randomly. if we assume a quantification doesnt exist, then perhaps random choice becomes the rational choice

but i think some weak quantification exists which can guide peoples' morals irl here. wiki says, in studies, majority tend to choose to kill one person. but not by a high margin. like 68% of people

psie

Consider the polar decomposition of a function $f:X\to\mathbb C$. We have $$f=\mathrm{sgn}(f)|f|,\quad \mathrm{sgn}(z)=\begin{cases} z/|z|& z\neq 0\\ 0 &z=0.\end{cases}$$ I am trying to verify that $\mathrm{sgn}(z)$ is Borel measurable, i.e. that $\mathrm{sgn}^{-1}(U)$ is a set in the Borel sigma-algebra, for $U$ open.

As $\mathrm{sgn}(z)$ maps points to the unit circle, I have figured out that the preimage of a set that does not contain the origin and any points on the unit circle is empty, hence measurable. Now, the preimage of a set that contains nonzero points on the unit circle must be a union of rays from the origin (but not including it). My question is; are the rays open sets in $\mathbb C$?

Ryder Rude

but at the same time, in the variant problem, people would never choose to make the other option legal for doctors

this reveals some inconsistency in peoples' morals

Thorgott

do rays look open to you?

Ryder Rude

but mostly, it just reveals that the variant problem has a different context

Ben Steffan

@psie stop for a second and think of what characterizations of open subsets of $\mathbb{C}$ (or $\mathbb{R}^2$) you know

Also, it might be worth noting that $\operatorname{sgn}$ is continuous on $X \setminus \{0\}$ :)

psie

13:26

ok, I stopped for a second and came to my senses. We couldn't possibly fit a ball into a line. But this complicates matters for me, I think. If $U$ is open and contains points on the unit circle (and not $0$), its preimage will be a union of rays (right?). And since the rays are not open, what is its preimage (open/closed/something else?)

Ben Steffan

What can you say about preimages of open sets under continuous functions?

psie

Alright, ok, they're open indeed

Ben Steffan

good, so the only thing you need to be careful around is if $0 \in U$

psie

right, so if $0\notin U$, the preimage is open, and if $0\in U$, then we get an open set $V$ union with $\{0\}$ I guess

Jakobian

13:48

@psie correct

psie

@Jakobian roger :) (thanks)

zeta space

14:10

How many smooth compact submanifolds of dim. $2$ does the $3$ sphere embedded in dim. $4$ admit?

Just wondering if this is part of a known problem/theorem

Sine of the Time

@zetaspace $\pi$

Ben Steffan

The embedding in dim. 4 is irrelevant, no?

All compact orientable surfaces embed in $\mathbb{R}^3$ and therefore in $S^3$

zeta space

yes you're right about the embedding in dim. 4 being irrelevant

smooth?

Ben Steffan

...on the other hand no embedding of a closed surface into $S^3$ is surjective (obviously) and therefore they all factor through $\mathbb{R}^3$, so your question reduces to what surfaces embed into $\mathbb{R}^3$.

All surfaces are smooth.

Pizza

Compute the following double integral: $$\int_D \frac{x}{y} dxdy$$ where $D = \{(x, y) : x + (y − 2)^2 ≤ 4, y ≤ 2 − x , x ≥ 0\}$ , initially I observe that $0 \leq r \leq 2$, I switched to polar coordinates $\begin{cases} x = r \cos(\theta) \\ y = r \sin(\theta) \\ x^2 + y^2 = r^2 \\ dxdy = r dr d\theta \end{cases}$, then for $x \geq 0 \Rightarrow r \geq 0$ for $y \leq 2 - x \Rightarrow r \leq \frac{2}{\cos(\theta) + \sin(\theta)$

Ben Steffan

14:20

@Pizza you're missing a closing } at the end

Pizza

I'm on the phone and I didn't notice :(

It doesn't let me edit anymore

Ben Steffan

ah, well

it's still pretty readable :)

Pizza

* then for $x \geq 0 \Rightarrow r \geq 0$ for $y \leq 2 - x \Rightarrow r \leq \frac{2}{\cos(\theta) + \sin(\theta)}$

For $x^2 + (y-2)^2 \leq 4 \Rightarrow r(r-4\sin(\theta)) \leq 0$ then $r \leq 0$ and $r \leq 4 \sin(\theta)$, but $r$ cannot be $\leq 0$ so I should find the minimum between $4\sin(\theta)$ and

$\frac{2}{\cos(\theta) + \sin(\theta)}$?

@BenSteffan :)

Sine of the Time

14:41

why polar coordinates centered in $(0,0)$?

leslie townes

gotta center 'em somewhere

Sine of the Time

you have a point

@BinkyMcSquigglebottom this was asked here @Pizza

I have to learn how to reply to my own message

Pizza

Are you on PC?

@SineoftheTime I'll look now, thanks

Sine of the Time

yes

Pizza

But do you mean to do this

59 secs ago, by Pizza

Are you on PC?

Sine of the Time

14:50

yes

Pizza

You have to go to the message, then where the grey square with the arrow is, it should be written permalink, you have to press the right button and copy the link address

2 mins ago, by Sine of the Time

yes

Maybe you just have to click on the grey square and copy the link address

Binky McSquigglebottom

Hi

Thorgott

I suppose the more interesting question is how many isotopy classes of embeddings of surfaces into $S^3$ there are

Binky McSquigglebottom

18 secs ago, by Binky McSquigglebottom

Hi

zeta space

hi

Sine of the Time

14:54

2 mins ago, by Pizza

You have to go to the message, then where the grey square with the arrow is, it should be written permalink, you have to press the right button and copy the link address

cool, thank you

isn't there also a method to reply to my own message?

@SineoftheTime maybe something like this

It worked

Binky McSquigglebottom

🤔

Pizza

How did you do it?

I'm on phone now

Binky McSquigglebottom

Copy the link from your message

Pizza

@BinkyMcSquigglebottom Where did you get that exercise from?

Sine of the Time

you click on the blue triangle, then click on permalink. At the end of the link, there's a number (that is the number of the message). You put :[number]

Binky McSquigglebottom

14:58

@Pizza Wdym

Sine of the Time

@SineoftheTime here is the exercise Pizza

Sorry for the ping BInky

Pizza

@BinkyMcSquigglebottom That exercise is the same as mine

@SineoftheTime 👍

?

Binky McSquigglebottom

😭

Bye ...

Im back

@Pizza yes

Binky McSquigglebottom

15:20

have you seen the Brawl Stars collaboration with Spongebob?

Shaun

I have about a week to answer this:

-1

Q: What is the equivalent of a normal closure in an inverse semigroup?

Quick question: What is the terminology for the analogue of the normal closure of a set in a group, applied to an inverse semigroup? I do believe such a notion exists; I just have forgotten the name for it and, naturally with these things, I can't find it anywhere, not even Lawson's, "Inverse S...

Sine of the Time

man the downvotes to your Q/A are crazy

Shaun

The question hasn't been received favourably so far so I'm bringing it to people's attention here now.

@SineoftheTime Whose?

Sine of the Time

In general, when you ask a question it's immediately downvoted by someone

Shaun

@SineoftheTime Haha, yeah; I must have made some enemies having closed a lot of quesitons.

Ben Steffan

15:32

I'm a little surprised that this isn't caught by the "suspicious voting" mechanisms

but then again I also have no insight into how those work

Jakobian

@BenSteffan You're assuming that all of the votes are unjustified

Ben Steffan

No. I'm assuming that it is patterned voting

Jakobian

And in particular that it's unjustified

Pizza

@SineoftheTime I found that $-2 \leq r \leq 2$ so $r \leq 2$ , only that for the second condition that is $y \leq 2 - x$ I find $2 + r \sin(\theta) \leq 2 - r \cos(\theta)$ , so

$r(\sin(\theta) + \cos(\theta)) \leq 0$

I don't know how to proceed here

Sine of the Time

If you draw a picture, you don't even need the inequalities

Jakobian

15:37

There was a lot of critique of Shaun's posts in the past, for example by @Xander that critiqued how they are structured

Sine of the Time

@Pizza are you using polar coordinates centered in $(0,0)$?

Pizza

No, I'm using $x = r\cos(\theta), y = 2 + r\sin(\theta)$

Sine of the Time

Are you having trouble solving $\sin \theta +\cos \theta \le 0$ ?

Pizza

I found $\tan(\theta) \leq -1$ , so $\theta \leq -\pi/4$

But shouldn't it be $0 \leq \theta \leq 2\pi?$

So I don't have to consider this theta that I found

Sine of the Time

$3\pi/2\le \theta \le 7\pi/4$

Pizza

15:53

but these steps I did should be correct, I don't understand how to continue

How did you find those theta values?

Sine of the Time

the conclusion you draw from $\tan x\le-1$ are not correct

@Pizza a picture of the domain does the job

Pizza

@SineoftheTime You mean it can't be solved like this

@SineoftheTime mm ok

Sine of the Time

it can, but you have to be carefull

start with the condition $x\ge 0$

Pizza

r ≥ 0

Sine of the Time

this is always true

Pizza

15:59

Yes

Sine of the Time

why did you simplifiy $\cos \theta$?

Pizza

ah, you meant this: $\cos(\theta) \geq 0$

Sine of the Time

how do you solve $2x\ge 0$ ?

I don't want to be harsh, but you must not do these silly mistakes

@Pizza yes

Pizza

@SineoftheTime I divide both sides by 2

Sine of the Time

same thing with $r\cos \theta$, since $r\ge 0$ you divide by $r$

now solve for $\theta$, keep in mind that you are considering $\theta \in [0,2\pi]$

Pizza

16:05

I got $\theta \geq \frac{\pi}{2}$

Sine of the Time

wrong

Pizza

$-\pi/2 ≤ \theta ≤ \pi/2$

$0 ≤ \theta ≤ \pi/2$ U $3\pi/2 ≤ \theta ≤ 2\pi$

Sine of the Time

ok

to solve $\sin \theta +\cos \theta\le 0$ you can use different methods

for example you can note that $\sqrt 2(\frac1{\sqrt 2}\sin \theta+\frac1{\sqrt 2} \cos \theta )=\sqrt 2\sin (\theta+\pi/4)$

Pizza

16:25

Yes

Sine of the Time

now solve $\sin t\le 0$ and then let $t=\theta +\pi/4$

Pizza

$\sin(\theta + \pi/4) \leq 0$

@SineoftheTime ok

The sin in negative in $t \in [\pi, 2\pi]$ so $\pi \leq \theta + \pi/4 \leq 2\pi$

Sine of the Time

go on

Pizza

$3\pi/4 \leq \theta \leq 7\pi/4$

Sine of the Time

now find the range of theta

using also the restriction you found before

Pizza

16:36

$\theta \in \left[\frac{3\pi}{2}, \frac{7\pi}{4}\right]$

Sine of the Time

ok

did you find the range of $r$?

Pizza

0 ≤ r ≤ 2

Sine of the Time

ok

now you're ready to compute the integral

Pizza

Yes, thank you very much for your help

Sine of the Time

@Pizza it's clear that you're studying a lot, but still you're struggling with basic algebra

this might be a problem, expecially because you have 2h for the exam

Gian's Pizzeria

16:50

+pressure

Sine of the Time

having Pizza and Pizzeria in the same chat is funny

Pizza

17:12

@SineoftheTime yes indeed I know... I hope to fix this thing

@Gian'sPizzeria Nice name

The Empty String Photographer

19:00

Be careful: if you put pineapple on a pizza and someone else notices, there’s a 50% chance that they’ll get angry and beat you on the head.

Sine of the Time

19:24

If you are in Italy, the chance increases to $95$%

zeta space

19:35

How many isotopy classes of embeddings of surfaces into $S^3$ there are?

I got $7$

For $S^2$ I got $1$

Thorgott

19:51

@zetaspace there are infinitely many surfaces, so that doesn't make sense

@zetaspace this is correct

zeta space

20:09

I wonder how to construct a nontrivial example of a smooth $n$-manifold, $M^n$ whose isotopy classes of embeddings of codim. $1$ hypersurfaces into $M^n$ are finite for all $n$

Nishant Jani

Hi folks, learning about covariance matrix and eigen values for PCA. If I have a dataset of two variables x1 and x2 that drive an outcome y, when creating covariance matrix should we consider x1, x2 and y?

leslie townes

@SouravGhosh dilation of linear operators? i don't know of any references that would be specific to the finite dimensional context, or any constructive aspects. vern paulsen's book on completely bounded maps has a good survey of the standard results in the area. the original papers are very readable

Sine of the Time

@Pizza did you manage to evaluate the integral?

Jam

20:46

how can i calculate 123454321 mod 36 wihout calculator??

leslie townes

just straight up long division is pretty hand calculable for a number of that size

Jam

really?

damn i must be bad at long division

Pizza

@SineoftheTime actually I got a $\log(0)$ in my calculations

Sine of the Time

that's bad news

Pizza

I saw a solver do this: $\lim_{\delta\to 0}\int\limits_{0}^{2-\delta}{f\left(r\right)}{\;\mathrm{d}r}$

Sine of the Time

20:58

the integral was $\int_0^2\int_{3\pi/2}^{7\pi/4}\frac{r\cos \theta}{2+r\sin\theta}rd\theta dr$ right?

Pizza

Yes

Sine of the Time

how did you procede?

Pizza

With $u = 2 + r \sin(\theta)$

So $r^2 \cdot \int \frac{1}{r u} du$

Sine of the Time

I don't see how you got $\log 0$

Pizza

and I got $r \cdot \log|2+r \sin(\theta)|$

evaluated at the extremes I get:

$r\left(\log(2-\frac{\sqrt{2}}{2}r\right) - \log(2-r))$

Up to this point everything should be fine, now the problems have started

leslie townes

21:04

@Jam calculator.net/…

Pizza

I integrated and I get a result that when I go to evaluate it I get a $\log(0)$

Sine of the Time

when solving $\int_0^2 r\log(2-r)$?

Jam

im gonna sound crazy but apparently i had forgoten how to do long division because i did the whole division in my mind like i would think how many times does X exist in Y and would take guess and do the multiplication and find out .For exable 1992 divided by 13 i would think i got 1300 and 692 more so i got 650 so 42 more i got 39 so 3 is the remainder hahahaha

so i forgot the long division algorithm

Pizza

@SineoftheTime yes

leslie townes

i realize the irony of pointing you to an online calculator for a request about something being done by hand :) but seven steps of the long division algorithm is not something i would look for a clever way around, i would just do those divisions

Sine of the Time

21:09

did you procede integrating by parts?

Jam

@leslietownes totally agree i did it after freshing up the algorithm twas easy hahaha

Pizza

@SineoftheTime yes

Jam

its funny i almost re invented the algorithm i was trying to do the division and i was thinking what i fi divide with just the first digits will it work?

leslie townes

the long division algorithm feels like cheating

haha

Jam

yes

and i felt that must not work

leslie townes

21:11

there's no way this is gonna... oh it does work

Jam

im dividing the first digits whats the relation

Pizza

$\int \ln(2-r) dr = -\ln(2-r) (2-r) - r$

Sine of the Time

you can't apply immediately IBP

first perform the sub $2-x=u$ and then apply IBP

Pizza

Yes that's what I did

Sine of the Time

does it work?

Pizza

21:15

that result above comes out, so when I go to evaluate it with 2 it will come out log(0)

Jam

imaging someone saying ooh to do 123454321 by 36 divide 123 by 36 first hahahaha like bro

Pizza

you can also see on wolfram : int ln(2-x) dx

Sine of the Time

but you're computing $\int_0^2 r\log(2-r)$

Pizza

@SineoftheTime $\frac{\ln\left(2-r\right)\,{r}^{2}}{2}-\frac{{r}^{2}}{4}-r-2\,\ln\left(\left|r-2\right|\right)$

Now I should evaluate it at the extremes

Sine of the Time

if you solve it this way, you have an indeterminate form

Pizza

21:29

@SineoftheTime what should be done?

Sine of the Time

I don't know when you found that antiderivative, but if you want to do it this way you should group the terms with log

$\int_0^2r\log(2-r)dr \overset{2-x=u}{=}\int_0^2 (2-u)\log u du=2\int_0^2\log udu-\int_0^2 u\log u du=\dots$

Pizza

$\int^2_0 \log(u) du = [u \log(u) - u]^2_0$

can this be calculated?

Sine of the Time

$\lim_{u\to 0^+}u\log u=0$

Pizza

21:45

ah I have to do this thing

But are there other cases where this can be applied? Because this is the first time I've seen this thing

zeta space

does anyone know if a two lines intersect in R^2 then how do you say that you can't non intersect them

Pizza

in the evaluation of integrals

Ben Steffan

@zetaspace what do you mean "can't non intersect them"?

zeta space

I just don't understand at all how to preserve the intersection. Like if I embed them into R^3 then I believe you can easily seperate the lines.

Ben Steffan

"preserve the intersection" under what operation(s)?

zeta space

21:49

This is has been a frustrating sticking point for me for a while

Sine of the Time

@Pizza this is the definition of improper integral

Ben Steffan

two lines in $\mathbb{R}^2$ intersect iff they aren't parallel

zeta space

@BenSteffan okay got it. And in terms of intersecting surfaces in R^3 then we have to use a different notion like transversality?

Ben Steffan

I don't see how transversality fits into this

transversality is a property of intersections

I'd still say that two planes in $\mathbb{R}^3$ intersect iff they are not parallel

(all of this is assuming that you're talking about affine subspaces of course)

zeta space

okay

I was assuming a smooth affine manifold

yeah I guess this is as easy as imposing the condition that the objects intersect

Jam

22:07

@zetaspace if 2 lines intersect in R^2 then every other line must intersect one of these lines since if it didnt u could find a thrid independent vector in R^2 which is impossible

same argument does not hold in R^3

zeta space

Take $X=(0,1)^n$ for $n \ge 3$. Fix all pairs of points $p,q$ s.t. $\text{dist}_n(p,q)=\sqrt{n}.$ Consider for all pairs of points, smooth foliations of $X$ with $(n-1)-$dim. leaves which are mutually diffeomorphic to the class $M_n=(0,\sqrt{n})\times S^{n-2} $ accumulating to $p,q$. I calculated through smooth deformation retractions that for a single choice of $p,q$ the leaves are mutually diffeomorphic to once punctured planes, this is excluding the one degenerate leaf

Assuming we consider all pairs $p,q$ I believe theres not a simple picture in terms of once punctured planes

because if you try to smoothly deformation retract the same way as before, the intersections prevent you from going further

I can't visualise this in terms of once punctured planes actually

lol

psie

I'm reading Theorem 2.10 in Folland's text. It's about the fact that every complex-valued measurable function $f=g+ih$ can be approximated by simple functions (maybe in the top 3 of the most important theorems in measure theory?).

Anyway, my brain protests here against something basic it seems; we have an expression of the form $$\phi_n=(\psi_n^+ - \psi_n^-)+i(\xi_n^+ - \xi_n^-),$$ where $\psi^+$ and $\psi^-$ are positive/negative parts to $g$ and similarly for $\xi_n^+, \xi_n^-$. These all converge uniformly to their respective limit functions, but what's the underlying result behind that $\phi_n$ converges uniformly to $f$?

I know the sum and difference of two uniformly convergent sequences of functions converges uniformly, but is $\phi_n$ a sum of two uniformly convergent sequences of functions? The fact that there's an $i$ in $\phi_n$ confuses me.

EDIT: the sequences of functions converge uniformly on sets where $f$ is bounded, which is what I am considering here.

Ben Steffan

22:29

I think there's two ways about it: Either you know the analogues of the basic convergence properties for $\mathbb{C}$, in which case $i$ is simply a constant and should pose no problem, or you invoke that $\mathbb{C}$ is really just $\mathbb{R}^2$ and treat $\phi_n$ as function of the form $x \mapsto (\psi_n^+(x) - \psi_n^{-}(x), \xi_n^+(x) - \xi_n^-(x))$.

psie

@BenSteffan ok, good idea. So if we view $\mathbb C$ as $\mathbb R^2$, the $+$ in $g+ih$ is not viewed as a summation, right?

Ben Steffan

@psie you could view it as a summation, but you don't have to

the way I wrote it above it's not a summation

but you could write it as $x \mapsto (\ldots, 0) + (0, \ldots)$, using the usual addition on $\mathbb{R}^2$

psie

ah ok 👍

zeta space

22:54

0

Q: Counting isotopy classes in $\mathfrak G_n$

I'm trying to count isotopy classes $\Gamma$, of smooth embeddings of closed codimension $1$ submanifolds into a geometric template, $\mathfrak G_n$. Question: Is my formula $\Gamma=2^{n-1}-1$ correct? My attempt: I'll work in the category $\mathrm{Diff}$ and count only the isotopy classes of s...