Mathematics - 2020-06-30 (page 1 of 4)

12:01 AM

What's the question, @T_01?

Well, ask, do'nt ask to ask. Happy that you are here. I will write down the question

Let $F: \Omega \subseteq \mathbb{R}^3$ with $\text{curl} F = 0$, and $\varphi: \Omega \rightarrow \mathbb{R}^3, x \mapsto \int_0^1 <F(tx),x> dt$. I have to show that then $F = \nabla \varphi$

We did not have any advanced integration theorems, I only know that I can put the partial derivatives into the integral because F is continuos differentiable (forgot to mention that)

Ted Shifrin

Yes, this has been answered (including by me) several times on main.

T_01

Okay, I'm sorry. Then i will search for that question!

Ted Shifrin

You will differentiate and use the curl 0 condition to change so that it works out perfectly with integration by parts.

Where are you stuck?

So you write down $\partial\phi/\partial x_1$. What do you get?

T_01

It takes so long to type the latex.

Ted Shifrin

12:11 AM

You should get the sum of two integrals.

T_01

I will take an image of one of my ideas

I had something like this a few sheets of papers ago

Ted Shifrin

Blue pen on dark checked paper is horrible.

Let me try.

T_01

I'm sorry :D

Well I get a sum of three integrals

and the derivative of that, which I put into the integrals...

Ted Shifrin

You have a confusion with indices. Let's just be explicit and do $x_1$.

T_01

Okay, I try that again.

Ted Shifrin

12:14 AM

So one term will come from differentiating $x_1$, and other will come from differentiating $F_k(tx)$ with respect to $x_1$.

So, yeah, you get several terms like that.

T_01

So I get something like this
deriv_{x_1} $\int_0^1 F_1(tx) * x_1 dt + \int_0^1 F_2(tx) * x_2 dt + \int_0^1 F_3(tx) * x_3 dt$

Ted Shifrin

No *s please

Yes. So the first term gives two terms, and the others give one each.

T_01

The second and third gives
$\int_0^1 \partial_{x_1} F_1(tx)x_i dt$

Ted Shifrin

No, $\partial F_i/\partial x_1$.

Now you use the curl condition.

T_01

Oh. Indeed, I am messing things up

So because of zero curl this is equal to $(\partial F_1(tx) / \partial x_i) x_i$

Ted Shifrin

12:20 AM

OK. So you have three terms like that plus one extra term from the first integral.

T_01

Okay give me a minute, I write it down. Maybe I am now able to do it...

Ted Shifrin

Cool.

T_01

I have really no Idea how I can spend hours on such problems and don't manage to get even the easiest steps

[\int_0^1] oh, no big latex here?

Ted Shifrin

I think you got lost in indices. My suggestion is to do it in $\Bbb R^2$ and just write out the terms carefully.

use dollar signs

single dollar sign for in-line, double dollar sign for display

T_01

Oh, okay. Thanks

So this should be correct, is it?
$$\int_0^1 F_1(tx) dt + \Sigma_{x=1}^3 \int_0^1 (\partial F_1(tx) / \partial x_i ) x_i dt$$

Ted Shifrin

12:28 AM

Yes, this is correct.

Now switch the integral and the sum on the right, and recognize what that is.

T_01

i might be completely on the false track, but isn't it $< \nabla F(tx), x>$ in the integral?

Ted Shifrin

You mean $F_1$. Yes. But you're integrating with respect to $t$, so that suggests something you might be looking for.

T_01

Ah, yes, $F_1$.

wait a second this is a part of $\varphi$

no it is $\varphi$

Ted Shifrin

No.

T_01

at least a part of it?

Ted Shifrin

12:33 AM

No, you've got a gradient of $F_1$ dotted with $x$. Nothing to do with $\phi$.

T_01

well I meant the integral over that. But yes, we have this gradient thing. Somehow, I still do'nt see how does this help me with integrating with respect to $t$.
I already see the moment coming where I will bang my head on the desk

Konformist Liberal

Hi! After having Eilenberg-Maclane spaces, can I just have product of them to build a space for any homotopy type?

Ted Shifrin

What would you love to have in order to integrate $dt$?

Finitely many, you mean, Konform?

T_01

maybe the derivative of something with respect to t :D or at least something "concrete" where I can grab the $t$

Ted Shifrin

Yes, the derivative of something with respect to $t$.

Konformist Liberal

12:37 AM

@TedShifrin Hmm so infinite product doesn't preserve the homotopy groups? Is there a systematic way to create a space for any homotopy type using Eilenberg-Maclane spaces?

Ted Shifrin

I haven't thought about this in decades, Konform. Talk to the topology folks.

T_01

It feels embarassing, but I still do'nt see how to integrate this. Should I quit mathematics? :(

Ted Shifrin

You aren't recognizing a fundamental formula from derivatives.

T_01

This
$$ \int_0^1 <\nabla F_1(tx), x> dt $$
is what we are looking at, right?

wait can I apply the chain rule backwards or something

Ted Shifrin

LOL, yes, it's the chain rule.

Mike Miller

12:54 AM

@KonformistLiberal I mean, it's called a product because it satisfies the universal property of a product. Maps into a product are the same as a choice of map to each factor. Apply this to pointed maps from X (or from X x I) to see that [X, prod_j Z_j] = prod_j [X, Z_j] for any index set J. In particular the homotopy groups of a product are a product of homotopy groups.

T_01

does $<\nabla F_1(tx), x> = \frac{d}{dt} F_1(tx)$ hold?

Ted Shifrin

Oh, but you did mess up earlier, and I didn't pay attention, @T_01.

Yes, @T_01, but there was something wrong when you differentiated. You forgot the chain rule there.

T_01

where did I differentiate?

Ted Shifrin

At the very beginning. You have $\partial/\partial x_1 \,F_i(tx)$.

T_01

you mean without the $x_i$ factor?

Ted Shifrin

12:56 AM

Yes, the $x_i$ is there, too. But what is this partial derivative?

T_01

I forgot to multiply with $t$ i guess

Ted Shifrin

Yup. Now it will all work perfectly.

T_01

would you believe that universities in germany are giving people like me degrees?

Ted Shifrin

Stop whining :P

T_01

Thank you so much. Somehow I struggle more with analysis than I should

Do you have any general advice on how to survive the third of these courses, too? :D

Ted Shifrin

12:58 AM

I don't know if you've tried, but some of my YouTube lectures might help you.

T_01

Did not know that you do are creating youtube content. this sounds great

Ted Shifrin

112 lectures on linear algebra and multivariable calculus/analysis ...

T_01

+1 sub

Ted Shifrin

LOL, yes.

The video messed up.

T_01

What do you mean?

Calvin Khor

1:50 AM

@Knight good morning, sorry I had to leave. I tracked down your book and I found out why the authors gave you this proof. This is because in the definition of L(f,P) and U(f,P), the infs and sups are over the closed intervals. This means that if you use the partition P from the step function f, $L(f,P) \neq U(f,P)$. You should verify this is the case with e.g. $$ f:[0,1]\to\mathbb R, \quad f(x) = \begin{cases} 1 & x < 1/2 \\ -15 & x=1/2 \\ 0 & x > 1/2 \end{cases}$$ which is a step function

i strongly suspect using the upper/lower sums with sups/infs over open intervals will give the same result for many sufficiently nice functions but I havent given it too much thought

beta_me me_beta

3:13 AM

0

Q: self-complimentary graph and cut vertex

I have difficulty understanding this 'let v be a cut-vertex in a self-complementary graph G . The graph G' − v has a spanning biclique, meaning a complete bipartite subgraph that contains all its vertices.' I narrowed down my argument to following : v be a cut vertex in the self-complimentary gra...

Can someone look over thsi question

*this

Knight

3:39 AM

@CalvinKhor Oh Kay! I have understood that now.

(Given your sleep time I want to know if you’re an Asian?)

Calvin Khor

yup lol

Knight

Would you mind telling me your country?

As a friend, you can easily put my request to Trash :-), it’s just a curiosity

Calvin Khor

Malaysia

where you at?

Knight

US

Calvin Khor

cool :)

Knight

3:49 AM

So, can you explain how the author came up with that inequality?

Calvin Khor

so you mean the rest of the proof?

Knight

Yes, I agree with your explanation that we cannot use partition $P^{\ast}$ for calculating the upper and lower sums because of closed and open interval issues

So, how do we go on for proving the integrability of step functions?

Calvin Khor

looking at an arbitrary partition, it should be clear that there are two cases to consider when trying to estimate each term in the sum that makes up U(f,P)-L(f,P)

Knight

First of all tell me why the author has never said explicitly that $f$ will not take infinite values (or undetermined value) at end points $z_0, z_1 , \cdots z_k$?

Calvin Khor

if the function is constant on $[x_{i-1},x_i]$, then this summand is the same for U and for L, so this summand is actually 0

Knight

3:56 AM

Yes

lemme check the book

You downloaded it?

yup lol

:)

they did say it cant take infinite values because they said its a function $[a,b]\to{\Huge\mathbb R}$

the values at the endpoints can be whatever, as long as it fits the definition I thought a textbook would stress this possible point of confusion (my lecturer did) but oh well

Knight

4:00 AM

Okay

Calvin Khor

I can't reply to my own message lol. but anyway you have those where the summands are 0, and you have some others where they aren't. Here you instead use that the gap of the paritition is small, so that this summand in U(f,P) - L(f,P) is small

Knight

How can be derive that inequality? I couldn’t understand what crossing index was

Calvin Khor

do you get what I said above? I was trying to draw a picture with words lol not so easy

Knight

Lol

Calvin Khor

4:15 AM

This is a graph of a step function in black, with some random partition $P$ which I drew in yellow

its more or less the same step function I defined earlier, I just changed -15 to something easier to graph

when you compute U(f,P) or L(f,P), you will add up a bunch of terms. Most of these terms are the same for both U and L. In this case, with this exact choice of partition, there's one summand that is different

that term is the term that causes U(f,P) ≠ L(f,P), and it is the reason for defining this "crossing index"

@Knight can you follow?

Knight

Yes

I can see that not for some closed intervals $M_i\neq m_i$

Calvin Khor

ok great, then the rest of the proof is just working out the details about how to track these kinds of terms and make sure that the terms -> 0 fast enough

the crossing index is the observation that, for this term where $M_i \neq m_i$, you must have had one of the points in $P*$ (not $P$) in the closed interval $[x_{i-1},x_i]$

maybe there's more than one point of $P^*$, but at least one

Knight

4:31 AM

I couldn’t understand the definition of crossing index and that line “each partition point of $P^{\ast}$ can belong to at most two partition intervals of $P$”

Can you please explain these two things?

Calvin Khor

the above is hopefully explaining the definition of crossing index

in the picture $P^*$ would consist of only 3 points, 0 ,1/2, and 1

Knight

Okay, $P^{\ast}$ consists of only three points then?

Calvin Khor

yeah

well you could pick other refinements of P* but lets stick with three points

so in the notation of the book $k=2$. For this yellow partition, there are 5 partition points, so $n=4$

Arjun

@Knight,@CalvinKhor, hi!

Calvin Khor

@Arjun yo

Arjun

4:42 AM

@Knight, i recommend you to this book,books.google.co.in/books/about/…

about maths

Knight

@CalvinKhor yes

5 partition points

Calvin Khor

writing $P = \{x_0,x_1,x_2,x_3,x_4\}$ where $x_i$ is in increasing order, we see that $0\in[x_0,x_1], 1/2 \in [x_1,x_2],$ and $1\in[x_3,x_4]$

so the set of crossing indices is $\mathcal C = \{1,2,4\}$

Knight

Okay

Calvin Khor

does this example make the crossing index definition clearer

its just a way to track intervals where elements of $P^*$ appear

Knight

Yes

Calvin Khor

4:47 AM

these are important because they are the points where $f$ can change, and if $f$ is not constant, then it contributes to $U\neq L$

Knight

And what about this statement “each partition point of $P^{\ast}$ can belong to at most two partition intervals of P”

Calvin Khor

so in my example, each partition point of $P^*$ belongs to only 1 partition interval of $P$

which I guess partition interval means the closed interval defined by consecutive partition points

Can you find a partition where 1/2 belongs to more than one partition interval

Knight

Yes.

Calvin Khor

and it can never belong to more than 2

so at most 2

Knight

$P=\{ 0, ,1/4, 1/2, 3/4 , 1\}$

Calvin Khor

4:50 AM

yup

Knight

Thank you so much Calvin

Calvin Khor

glad to help!

Knight

You are very friendly and knowledgeable

Calvin Khor

nice to hear :)

gonna make lunch, ping if theres anything

Knight

Have a good lunch :-)

Calvin Khor

4:53 AM

thanks!

Knight

7:33 AM

Lunch done, Calvin?

Calvin Khor

7:46 AM

@Knight yup

shouldnt you be/have been asleep lol @Knight

Knight

8:36 AM

@CalvinKhor I’m back

As you said that step function can take arbitrary large values at end points of sub intervals of $P^{\ast}= \{z_0, z_1, z_2 \cdots z_k\}$ then how can we have a $M \gt 0$ such that $$-M \lt f(x) \lt M$$ ??

Calvin Khor

8:51 AM

@Knight thats OK because the M depends on the step function

each step function can be arbitrarily big, but a step function's image is finite

a finite set has a maximum and a minimum

Knight

9:05 AM

@CalvinKhor Didn’t get that? A step function can be arbitrarily big but it’s image is finite? Doesn’t it’s image at end points gonna be infinite (if we allow that)

Calvin Khor

9:21 AM

no

give me any step function, and I'll give you an M st |f|<M

once you see this M, you can go, aha, now I have a worse step function

but thats allowed, I'll just change the M

like the above explicit example, i could choose M=200

Thorgott

A step function can not be arbitrarily big

You can find arbitrarily big step functions (plural)

Calvin Khor

Is the A in italics cuz its not so prevalent. But yes that is the emphasis

Thorgott

But each single one is bounded

Calvin Khor

mmm i did say each step function can be arbitrarily big. I should have said what you said. Step functions can be arbitrarily big

Knight

I need a proper clarification, I’m not a very smart student like @LeakyNun :-)

(I’m tight)

Calvin Khor

9:26 AM

A step function defined by $N$ partition points can take how many values?

Knight

$N$

Calvin Khor

no

there's N-1 intervals, and then the N partition points, thats 2N-1

Knight

Yeah

No we have N+1 points

Calvin Khor

well you either have the N points which you label up to the $(n-1)$th, or $N+1$ points and you label until the $n$th, since your book's lists begin with 0

but OK, either way, you have at most, 2N+5 points. that's a finite set of real values.

Knight

Yes

MathStudent

9:30 AM

Hi all, quick question: Say, you have an iterative algorithm. So, for example $t = 0, 1,...:$ 1. Set $a_t = f(a_{t-1})$ (where f is some function dependant on $a_{t-1}$) and 2. Set $b_t = g(b_{t-1}, a_t)$ (where g is some function dependant on $b_{t-1}$ and $a_t$. Now, if you want to talk about the {b_t}, what would you call them? I've seen them be called "the iterates {b_t}$ but I think that may be wrong.

If you happen to know, what they're called in German that'd be even better (I got stuck because I wanted to translate a sentence into German and the word iterates seemed wrong after a google search)

Knight

How $2N+5$ ? 5?

Calvin Khor

I just chose a number big enough so that ≤2N+5 is correct. If you don't like it then you can make it 2N+100. finite is all that matters

MathStudent

In case it helps to clarify what I mean about talking about the {b_t}: In the particular text I was looking at, it talked about how "the AMP iterates $b_t$ converge to the MLE" where AMP is an iterative algorithm

Calvin Khor

this finite list of numbers lets say $L$ has a maximum and a minimum. Take the largest in absolute value, and call this absolute value $M$. then $|v| \le M$ for each $v\in L$. But for each $x$, $f(x)$ always belongs to this list $L$. so $|f(x)|\le M$.

@MathStudent that sounds like something that either doesn't have a catchy name, or would only have context-specific names. iterates? updates? just "the sequence $b_t$"? these sound fine to me...

@Knight If you want to be precise, you should have gotten at most 2N+1 values for N+1 points. But if you take at most 2N+1 points, you also take at most 2N+1000....I like to "give myself some room" if its available, sorry if that's confusing

MathStudent

@CalvinKhor ah yeah, the latter two are simple known names that I couldn't think of, thanks! Btw, with them I also just found a book calling them "the iterates (i.e., the sequence of updates) ".

Calvin Khor

9:49 AM

i think something like that is called updating if you're doing kalman filters @MathStudent

Knight

@CalvinKhor let’s agree that we have $2N+1$ values of $f$. Then my question is what ascertain that $f$ is bounded?

Calvin Khor

if I give you 2N+1 numbers $a_0,\dots,a_{2N}$, can you find a number that's bigger?

Knight

Hmm... but who’s to claim that none of $a_0, a_1, \cdots a_{2N}$ is infinity?

Calvin Khor

the fact that $f:[a,b]\to{\Huge \mathbb R}$

Knight

Okay! Means if ever it is written that $$ f : [a,b] \mapsto {\Huge \color{blue}{\mathbb R}}$$ then $f$ is always bounded?

Konformist Liberal

9:59 AM

Is there a cw-complex for every homotopy type?

Calvin Khor

No, $f$ can have arbitrarily large values, but all of them are Real Numbers, so infinity is not part of the game

Knight

Okay

Calvin Khor

for example $f:[0,1]\to\mathbb R$ defined by $f(x) = \begin{cases} 0 & x=0 \\ 1/x & x\neq 0 \end{cases}$

Knight

Okay

Calvin Khor

in the special case of a step function, being a map $f:[a,b]\to \mathbb R$ automatically means that its bounded, because its image is finite, and a finite set is a bounded set

but the example above shows that this isnt true for all functions

Knight

10:05 AM

How can we write $f(x) = 1/x$ in that mapping form of domain and range ?

Calvin Khor

the domain cannot contain 0

strictly speaking, the definition of the function comes with the domain and range, but one answer to your question could be domain $\mathbb R\setminus \{0\}$ and range $\mathbb R\setminus \{0\}$

Knight

Okay I agree

So, we got an $M\geq 0$ such that $$ -M \leq f(x) \leq M$$

That means for any sub interval $[x_{i-1}, x_i]$ we have $$ | M_i| \leq M \\ |m_i| \leq M$$ right?

Calvin Khor

yyyyup

Knight

How can we obtain $$M_i -m_i \leq 2M$$ ?

Calvin Khor

triangle inequality

Knight

10:17 AM

I don’t know about that

Is there any other way?

Calvin Khor

alternatively/equivalently note that saying $|a|\le b$ means that both $a\le b$ and $-a\le b$

so $M_i \le M$ and $-m_i \le M$

Thorgott

You can't tell me that you don't know the triangle inequality

Calvin Khor

since hopefully you have proven already that ($A\le B$ and $C\le D$ implies $A+C\le B+D$) is true, this implies the result

also yes you should probably learn the triangle inequality lol

Knight

@Thorgott Why ? Is that something very obvious?

Calvin Khor

i will say it wasnt obvious the first time i saw it. but its used over and over and over again, so its worth burning into your memory

Knight

10:20 AM

Oh okay! I didn’t know it’s name. I always say “sum of two sides of triangle is greater than the third side”

I almost never used the term “triangle inequality”, and I learned this fact in 7th Grade

Oh gosh! I made myself a joker

Calvin Khor

lol turns out im the one with the dunce cap, being shown up by 7th graders

i spent days trying to prove the triangle inequality lmao

Thorgott

I'd say when put in terms of triangles, it actually is obvious

Intuitively, I mean

You definitely ought to know this inequality, because it is used in proving much more fundamental facts in real analysis than anything in integration theory

For example you use it to prove that if $\lim a_n=a$ and $\lim b_n=b$, then $\lim a_n+b_n=a+b$

Calvin Khor

oh its blindingly obvious...but it was hard for at least one uninitiated person to translate it into symbols (past me)

Knight

So, how can we get $M_i -m_i \leq 2M$ from triangle inequality?

Calvin Khor

u wanna give it a go or shall i tell you

builds character

Knight

10:24 AM

@Thorgott That’s a good advice for future, I shall surely follow it.

@CalvinKhor Let me try

Calvin Khor

:P oki

Knight

$$ -M \leq M_i \leq M \\ -M \leq m_i \leq M$$

Am I correct this far?

Calvin Khor

yes

though im not sure this leads to an answer via triangle inequality, this can lead to the answer

FYI triangle inequality in the form most useful for you right now: If $a,b$ are real numbers, then $|a-b|\le |a| + |b|$

Thorgott

Yeah, translating anything into symbols is hard when you're uninitiated

Knight

Considering only this $$M_i \leq M \\ -M_m_i$$ and now using the practical rule we have $$ M_i -m_i \leq 2M$$

Thorgott

10:28 AM

It's kind of hard everytime anew when you learn new symbols

Knight

Considering only this $$M_i \leq M \\ -M\leq m_ii$$ and now using the practical rule we have $$ M_i -m_i \leq 2M$$

Thorgott

Which is why it's good to keep intuition in mind to not lose track of what one is doing

Calvin Khor

yup, thats it

Knight

Thanks Calvin

Calvin Khor

Now that you have one proof, I'll give you the triangle inequality proof...$$ M_i - m_i = |M_i - m_i| \overset{\star}\le |M_i| + |m_i| \le 2M$$

and the inequality step marked with a $\star$ is of course where the triangle inequality in the above form is used

Knight

10:31 AM

Wow

Calvin Khor

@Thorgott I've only recently started to step away from "everything needs to be formalised before I call it a proof". this is something I 'knew', but now its something I 'believe', if that makes any sense

@Knight glad you like it, as practice, you should prove the triangle inequality if you haven't. Its very very similar to your proof

im off, see you guys!

Knight

@CalvinKhor Bye Calvin

Thorgott

cya!

Thorgott

10:58 AM

@Balarka Kreck just told me the story about the punctured torus' fundamental group. If you blow up the puncture, which even is a homeo, you get what looks like two crossing strips (which of course retract to two wedged circles). The loop around the puncture in the punctured torus corresponds to the loop traversing the entire boundary of this figure, which, in terms of the loops around the two circles in this figure, which are the generators of the fundamental group of the torus, is their commutator. On the fundamental square, puncturing can be thought of as removing all 4 corners and this l…

I think I'm starting to grok

Balarka Sen

Yes, the punctured torus is just a small neighborhood of the wedge of a meridian and a longitude inside the torus

Thorgott

This is also the 3-chart atlas you described to me then

Balarka Sen

Well, my chart was the complement of the wedge

That's an R^2

$T^2 \setminus \{p\}$ however is not homeomorphic to $\Bbb C \setminus \{1, -1\}$, even though they are homotopy equivalent (to $S^1 \vee S^1$)

Thorgott

Yeah, I'm just saying the charts fit into the same picture

Balarka Sen

We briefly discussed that $(T^2 \setminus \{p\}) \times \Bbb R \cong (\Bbb C \setminus \{1, -1\})\times \Bbb R$ before, right?

Thorgott

11:04 AM

And yeah, I was also told about this, they don't have the same ends

Yeah, that was your example for non-cancellation of $\mathbb{R}$

Balarka Sen

Cool.

Thorgott

11:15 AM

Horrible picture, but this shows how the two wedged circles lie in each of these spaces. The boundary of the punctured torus here has just one piece, which is the single end, whereas the boundary of the doubly punctured plane has three pieces, which are its three ends respectively. So they are homotopy-equivalent and not homeomorphic.

Balarka Sen

Yup

Thorgott

How exactly does this difference get killed when multiplying with R?

Balarka Sen

Since R has nontrivial end, the end of X x R changes substantially from that of X.

Precisely, End(X x R) is End(X) x R glued to X x End(R) = X x S^0 along End(X) x End(R) = End(X) x S^0

So it's two copies of X with a tube End(X) x R glued along the ends of X

Like a connected sum, but at infinity

So End((T^2 \ p) x R) is two copies of T^2 \ D^2 tubed together along the boundary circle, i.e., T^2 # T^2

Similarly, End((S^2 \ {3 points}) x R) is two copies of S^2 \ {3 disks} tubed along the boundary circles, i.e., T^2 # T^2 again

Thorgott

11:37 AM

Hmm, I don't quite see why gluing two triply punctured spheres together along the boundary circles gives the connected sum of two tori

Balarka Sen

You get genus 2, no?

Let me draw a picture

The red and blue guys are triply punctured spheres

This is a genus 2 surface, aka T^2 # T^2

Thorgott

12:05 PM

oh, yuh, makes perfect sense

can I see geometrically how the twisting is undone by multiplying with R?

Balarka Sen

12:16 PM

@Thorgott Yeah. If $M$ is a manifold with boundary, what is the boundary of $M \times [0, 1]$?

Working in topological category, no funny corner business

Balarka Sen

12:31 PM

Oh when you say geometrically you mean for that specific situation

Observe that both of those surfaces $\times \Bbb R$ is a neighborhood of $S^1 \vee S^1$ in $\Bbb R^3$

So a genus 2 handlebody

Thorgott

$M\times\{0\}\cup M\times\{1\}\cup\partial M\times[0,1]$?

Balarka Sen

Yeah, also known as the double of $M$.

Gluing two copies of $M$ along $\partial M$

So that's the end of interior of $M \times [0, 1]$ i.e., $\text{int}(M) \times \Bbb R$

Thorgott

Yeah, it makes sense that both are neighborhoods of $S^1\lor S^1$, but why is the punctured torus one not "twisted" anymore? Somehow times R gives this enough space to un-twist itself (I think), but I'm not sure how exactly that takes place.

Balarka Sen

It's like how a neighborhood of the Mobius strip in R^3 is $S^1 \times D^2$

Thorgott

that does sound analogous

I'll have to convince myself the latter works real quick

Balarka Sen

12:39 PM

It's in essence because if $M$ is the Mobius strip bundle over S^1, $M \oplus M \cong \varepsilon^2$ is trivial of rank 2

The normal bundle to the Mobius strip, when restricted along the center circle, is another Mobius strip

I guess the most fundamental reason is a reflection can be realized by a rotation one dimension up. $O(n)$ embeds in $SO(n+1)$.

Thorgott

12:59 PM

Actually, isn't the statement about the bundle better? It says that if you have these two twists, you can essentially undo them independently. The neighborhood statement, I'm trying to picture this atm, just says that if you thicken up the strip it looks, well, like a doughnut.

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