Mathematics - 2022-04-13

4:01 AM

I think you should think critically about it.

BAYMAX

4:21 AM

yes will try

That's all you can do.

🤔

Dr. @leslietownes have you considered putting munchkin in a school for like minded "gifted" munchkins?

💝🎁💝

👨‍🏫👩‍🏫

4:37 AM

@leslietownes that is pretty intuitive of her to realize the faux pas...

user, she's weird enough as it is :)

👼

at bedtime she flipped a shirt over her head and wore it like a hat, with the neck of the shirt as a kind of headband. she kept shouting "I'M THE PIZZA MAN" because she associates wearing a shirt this way with a chef's hat, which she has seen depicted on a man on a pizza box

there's a lot going on, is all i'm saying

They say during these developmental years there is the most going on cognitively.

sometimes when she goes to bed i can sense that. her head is just buzzing with questions. we have weird conversations.

4:49 AM

Do you read to her a lot?

:^)

not a huge amount, but a little every day.

her mom reads to her a lot. i spend more of my time on drawing and making up games to play

i used to give my daughter little puzzles like mazes, and graph paper featured a lot. anything that involved precise colouring appealed to her (not her brother, though)

reading is pretty high order

True; deleted.

she used to like trains and similar more until she discovered pink & frilly

sad day when ho scale meant earring size and not model train gauge.

she can recognize her name when it's written out, and write letters on command if you tell her which ones to write, but it's too much like work for her. she tires of it.

she knows the six-digit code we use to unlock the device that we sometimes let her watch videos on.

thankfully she can't navigate menus yet, so if she gets in, she can't do anything. although if the thing she wanted is already open she can start it up.

5:04 AM

i am sure i mentioned the time my daughter 'helped' me by dragging every icon into the trash.

we were in the middle of series b and i had many versions on the desktop

haha yes.

that was back when i had a life

😳 ouch

quite by accident i had a video call with both offspring today, rather unexpected and nice

Coolio.

Video conferencing is taking over.

5:08 AM

i prefer the phone if i must communicate remotely

but their mom prefers video.

I still prefer old fashioned email.

when i grew up phone calls were expensive (relatively) and everyone could listen so there was a 'get off the phone quickly' imperative

when i say expensive, i mean expensive relative to when my parents were growing up since that seems to be the measure of cost that parents apply

i@leslietownes i presume she is inundated with lego and the like?

Speaking of emails, did you all know that the newest version of Yahoo mail requires iOS 13?

i just use telnet

Not to mention the grand opening of Apple Park.

5:16 AM

the space ship?

copper: yes. and a lot of toy trucks and tools. her fine motor control is pretty good.

Head office:

$5, 000, 000, 000

corporate excess

On a royal scale, considering Buckingham palace has been assessed at the same price.

i wrote lots of notes & problems for my kids on microcrap onenote and now i cannot access them.

5:34 AM

If we did these kind of things the old fashioned way, on paper, we wouldn't have these problems.

a few decades ago i decided to try going paperless. did not work out.

the onenote was handy at the time.

we worked with a consultant who had a ton of stuff in some kind of proprietary program. an absolute ton of stuff. he stopped consulting unexpectedly due to family issues and left us with this gargantuan database that nobody could open. all of his notes were in there as well as the materials he was citing.

we did get it open but it wasn't funny for a few days in there.

5:52 AM

I wonder if they've considered digitalizing the Library of Congress.

one of my projects was to write them up in latex, but i guess that will take second place to actually opening the stupid files

this is why steve balmer-monkey called linux a cancer. it doesn't leave you in the lurch like the good microsoft does

the copyright office was still requiring paper copies for a lot of deposits for registrations at the beginning of the pandemic. they wanted the authority to allow all-electronic, and couldn't get it. so literal warehouses were filling up with printed material and nobody was there to go through it. i think they still have a huge backlog.

i'm sure the librarians at the library of congress are all for digitization and it is more a question of getting the money and authority to do it.

its so annoying that i can't access the stupid files!

what's the deal? is onenote not supported anymore? do you need a cdrom from 1997 to unlock the right version?

i remember compatibility between different versions of MS office being a surprisingly huge thing for a really long time. it seems to be better now although maybe that is because docx is less crazy of a frankenstein format than legacy doc files were.

the files are from onenote 2003 and they have gone through a few incompatible versions.

6:03 AM

oh i believe it

I think the pandemic has, in a lot of ways, prematurely awakened the techlogical Frankenstein.

By that^ I mean the medium is the monster.

*technological

unbelieveable, they discontinued windows journal too. pos

6:23 AM

Is there not a relevant stackexchange that you could ask on?

I will try.

I found a legacy windows journal version, but I am annoyed at losing the 2003 onenote stuff. Seems like it violates something, especially when one pays through the nose for it

good night!

Q: Degree of the splitting field of $x^6+5x^3 - 2$ over $\Bbb Q$.

cya

Good luck!🤞

☘️

🍀

love_sodam

7:27 AM

Is there anyone who knows Clifford algebra?

0

Compute the degree of the splitting field of $x^6+5x^3-2$ over $\Bbb Q$. From this answer, I first let $y := x^3$ then the given polynomial transformed to $y^2+5y-2$ so roots of $y$ is ${-5\pm\sqrt{33}\over 2}$. letting $\alpha := \sqrt[3]{-5+\sqrt{33}\over 2}$, I can conclude that $[\Bbb Q(\al...

11:04 AM

Hi guys, can someone explain to me how can I find a closed formula for this series?:

$\sum_{n=1}^{\infty} \frac{1}{n*2^n}$

My solution:

I tried to split this sum of product in a product of sums. It was like this:

$\sum_{n=1}^{\infty} \frac{1}{n} \cdot \sum_{n=1}^{\infty} \frac{1}{2^n}$

But if I do this, I know I'm wrong because this is equal to $2 \cdot \sum_{n=1}^{\infty} \frac{1}{n} = \infty$ I know I'm wrong but I don't know what mistake I'm doing. Please help.

12:04 PM

About Warsaw circle not being the image of unit interval again

I found an elementary proof

If $X$ is the Warsaw circle and $Y$ is the partof it homeomorphic to the unit interval which corresponds to the interval obtained from closure of graph of $sin(1/x) $ plus a little more, then we can find a neighbourhood $U$ of $f^{-1}(Y)$ such that $f(U)$ is entirely contained in $Y\cup V$ where $V$ is open (path-connectedness is used here)

V means a small interval addition to Y so that their union is a space homeomorphic to [0, 1)

Leaky Nun

@MatheusSousa replace the numerator with x^n and differentiate

@MatheusSousa the mistake is that ab+cd is not equal to (a+c)(b+d)

Then $f^{-1}(Z^)$ is open, so we obtain a map from compact $f^{-1}(Z^c)$ onto an interval homeomorphic to $[0, 1)$, but this is impossible

12:33 PM

Quotient of locally connected spaces are locally connected, and if $\colon [0,1]\to X$ is a continuous surjection, then it is in fact a quotient map, since it is closed

In other words continuous images of $[0,1]$ are locally connected

This is (part of) the easy direction of the Hahn-Mazurkiewicz theorem

12:52 PM

Oh, I didn't know quotients of loc connected spaces are loc connected

It's nice that it can be shown in a less direct way

love_sodam

@Jakobian you know what? It's actually one of the exercise problem in Munkres

I haven't read Munkres

1:10 PM

@Jakobian yes it's a nice result

Very false for general continuous images of locally connected spaces though

Q: What is the probability of the given event?

3:18 PM

@AlessandroCodenotti does the concept of one-dimensional continua agree across all definitions of dimension?

Samyak Marathe

Hi

I need help in probability

If someone can help me in this regard

0

Consider 2 Boxes. One has 99 white balls and 1 red ball. Another box has 99 red balls and 1 white ball. Now one has to chose a box at random and take out a red ball. It is given that the person has taken out the red ball. Now what is the probability that the ball has been chosen from box 1 which ...

probability

3:44 PM

@Jakobian small/large inductive dimension and lebesgue covering dim agree for separable metrizable spaces

To me continua are compact metrizable, but for some people they are only compact Hausdorff, so it depends on your definitions

4:02 PM

Ah, thanks. I think this is what people refer to when talking about dimension then

4:33 PM

Yes usually they mean lebesgue covering dim when talking about continua because it's the easiest to work with

But it's the same as inductive dimension anyway

5:01 PM

I was asking because I have an exercise to prove that Siepińsk universal curve is indeed universal

5:26 PM

I am gradually learning point-set topology (at least partly as a foundation for differential geometry). I see this result that a bijection between topological spaces that is continuous one way is continuous the other way, if the domain space is compact and the range space is Hausdorff.

My question: is this an important result or just a some factoid?

It is quite, quite useful! It saves tedious proofs from scratch that the map is a homeomorphism.

In differential geometry in particular or in multiple fields?

Any field in which one works with manifolds. I have no idea what you think differential geometry is; most people use the term for things having little to do with differential geometry.

I have a mediocre grasp of point-set topology and I know a couple of facts about manifolds, so my idea of differential geometry is "whatever's in the standard grad textbooks".

5:42 PM

What standard grad textbooks?

Differentiable manifolds? Differential geometry introduces more structure — connections on vector bundles, either Riemannian geometry, complex manifolds, or general connections.

@Novice this is useful in many contextes, it tells you that if you have a continuous injection from a compact space, then it is aa topological embedding

Common cases for the compact spaces are the interval $[0,1]$, so that you are looking at paths in some space, or compact manifolds

BTW, there is also undergraduate diff geo, which is largely curves and surfaces in $\Bbb R^3$, without so much fancy stuff. But what is essential at all levels is strong multivariable calculus/analysis and linear algebra.

@TedShifrin The Smooth Manifolds book by John Lee (admittedly in his preface he says his book is really about manifolds more broadly), and Jeffrey Lee's AMS GSM book immediately come to mind. (I know there are many other books in this genre.)

I picked up Morita's book Geometry of Differential Forms from my school library because some people on the Web were saying it's well written.

I'm hoping to find time to start reading Fecko's Differential Geometry and Lie Groups for Physicsists, and Naber's two books Topology, Geometry and Gauge Fields. (I like when physics is mixed in.)

(I'm not a math student, so I have lots of other things to tend to.)

Lee’s book is definitely not what I consider diff geo. It’s diff manifolds. For physics connections will be very important; that’s what gauge theory is about. Physicists are not fond of differential forms, but they are very powerful.

Okay, thank you. I may pop in once in a while and ask questions. Regarding your videos, I am up to the one where you start proving properties of the exterior derivative.

5:55 PM

You need multivariable analysis (derivative as linear map, inverse and implicit function theorems) and a little bit of point set topology. Definitely need compactness, connectedness, a bit more.

This multivariable analysis stuff is covered in my preceding videos.

I'm 95% sure I saw the inverse and implicit function theorems in Lang's book but they didn't make an impression on me. I will watch your videos on those topics. Thanks.

6:09 PM

@LeakyNun, I put the $x^n$ into the numerator and differentiated the sum. I found the formula for power series:

$$\sum_{n=1}^{\infty} (\frac{x}{2})^{n-1} = \frac{1}{1 - \frac{x}{2}}$$

If I integrate this I'll have:

$$\sum_{n=1}^{\infty} \frac{1}{n} \cdot (\frac{x}{2})^{n} = ln(1 - \frac{x}{2})$$

To equal $\sum_{n=1}^{\infty} \frac{1}{n} \cdot (\frac{x}{2})^{n} = \sum_{n=1}^{\infty} \frac{1}{n \cdot 2^n}$, x must be equal to 1.

So, $$\sum_{n=1}^{\infty} \frac{1}{n \cdot 2^n} = - ln(2) $$

I really don't know what is wrong in this approach, but the result doesn't makes sense to me.

there's a sign error when you integrated 1/(1 - x/2). it's -ln(1 - x/2) (plus a constant, which turns out to be 0 because you want the antiderivative that is 0 when x = 0)

fix that and it's fine

i pointed out the hidden role of the potential +C here because that's another thing that can go wrong, although it doesn't here

Ah ok man, sorry

I made a mistake in integration by parts

Now I found $\sum_{n=1}^{\infty} \frac{1}{n \cdot 2^n} = ln(2) $

Thanks

i also goofed, it's -2 ln(1 - x/2), you might chase through the rest of your work to see how that ends up giving you the same result

factors of 2 were never my strong suit

I was about to scream.

just in time, i guess

hint: the missing factor of 2 was forgotten on both sides of the above

6:20 PM

Holas chat

and heeeeere comes some algebraic geometry

hahah, actually, I'm reading about algebraic topology this week!

there's more to life than algebraic things

Leaky Nun

@TedShifrin dont hold back!

goodness, i'm repeating myself. chat.stackexchange.com/transcript/36?m=58163665#58163665

6:27 PM

I actually do have a question about CW complexes + cohomology + cup product, but it's not too pressing

unless someone is interested ofc

if nothing is too ugly, cup product is dual to intersection of submanifolds, just do that instead

it's a cheat code

hm, I haven't seen something like that yet. We will treat Poincaré duality soon tho

well, let me post it anyway

I don't know how standard the notation is that is used here

in any case: I don't see why we have that homology isomorphism

$S^{\mathcal O}(U\cup V)$ is the simplicial set corresponding to the $\mathcal O$-small singular simplices

So $\sigma\in S^{\mathcal O}(U\cup V)_n$ is a continuous map from the standard $n$-simplex $\Delta^n\to U\cap V$ with image in $U$ or $V$

and the $C_*$ stand for the singular homology complexes

Shiranai

Hello, I'm reading this answer https://math.stackexchange.com/questions/1406162/do-there-exist-general-conditions-under-which-we-can-conclude-that-continuity-on
and I'm not understanding why R-generated is equivalent to "a subset of X is closed iff its preimage is closed under every continuous map R→X is closed"

could anyone please elaborate? I can see how the quoted proposition implies X is R-generated, but not the converse

6:46 PM

@AlessandroCodenotti do we have any results like embeddability into $\mathbb{R}^n$ for finite-dimensional separable metric spaces (covering dimension)?

I'm just wondering if we can get something similar to $n$-dimensional manifold embedding into $\mathbb{R}^{2n}$

6:58 PM

Guys, can someone give me some hint to start this exercise?

Calculate the sum below:

$$\sum_{n=1}^{\infty} \frac{n^3 \cdot x^n}{2^n}$$

My approach:

I tried to integrate 3 times but it results in something that I don't know your closed formula:

$$\sum_{n=1}^{\infty} \frac{n^3 \cdot (\frac{x}{2})^{n+3}}{(n+1)(n+2)(n+3)} + Cn^2 + Dn + F $$ where C, D and F are constants added by improper integral.

Q: Bounded open set $S$ in $\Bbb R^2$ with no Jordan measure?

differentiate the geometric series sum (x/2)^n once and multiply the result by x to get a series for sum n (x/2)^n (mind some factors of 2). having that, differentiate it and multiply by x to get a series for sum n^2 (x/2)^n (again mind factors of 2). do that a third time.

Spring

Hello! Does anyone have any suggestions on generalizing this example of an open bounded set without a Jordan measure to higher dimensions using somewhat elementary methods? I included in my post what examples I've heard of (as of today) in $\Bbb R,$ but I'm in my second year of undergraduate studies, so I'm not yet familiar with the first answer in the thread linked in the comment

0

Is there any open bounded set $S\subset\Bbb R^2$ which doesn't have a Jordan measure, that is, the characteristic function $\chi_S:\Bbb R^2\to\Bbb R$ isn't integrable? I first thought there isn't any as I was thinking locally, how $\chi_S$ is always continuous around an open neighbourhood of ea...

real-analysis integration intuition

7:35 PM

Apparently any $n$-dimensional continua can be embedded into $\mathbb{R}^{2n+1}$

So this is a little weaker than Whitney's embedding theorem, but still neat

It's a theorem of Nöbeling

@Jakobian 2n+1

You can also require that at most n coordinates are rational

This is written in details in Engelking's dimension theory

7:51 PM

Hello everyone, could somone pls help me with some hints with a exercise concerning algebraically open sets

8:53 PM

@leslietownes sorry man, I tried this multiple times with an calculator. This doesn't work.

I tried to start from $sum_{n=1}^{\infty} (\frac{x}{2})^2 = \frac{1}{1-\frac{x}{2}}$ and differentiate 3 times. In each time I multiply each side by $\frac{x}{2}$ to maintain $(\frac{x}{2})^n$ in the sum.

But the answer shows 26 as a result

I'm getting 13/4.

I don't know why.

the fact that you're off by a factor of 8 = 2^3 is encouraging. double check the factors of 2.

@Spring The example given generalises immediately to $\mathbb{R}^n$, you just need an enumeration of the rationals in $(0,1)^n$.

9:40 PM

I am learning basic topology and manifold theory. While pondering the definition of a fiber bundle recently, I found it impossible to ignore the similarity with manifolds: fiber bundles and manifolds both locally look like one thing, but globally they are different.

I mentioned this to a math acquaintance who told me that fiber bundles *are* manifolds. I've been searching textbooks for definitions of fiber bundles just to see whether I can verify this for myself. Long story short, I checked books by John Lee, Jeffrey Lee, Nicolaescu, Walschap and Tu, and they seem to have definitions that …

Now, my understanding of the (general) definition of a manifold is that it is a topological space that is locally homeomorphic to some reference space, be it Euclidean, Banach, or whatever.

So, my question is: are fiber bundles manifolds?

(Meta question: does this question even matter? Is it enough to simply see that they're similar?)

9:53 PM

Yes, fiber bundles are manifolds with extra structure. You need to concentrate on examples, rather than general abstract theory.

i think i've seen at least one book define fiber bundles at a level of generality where they might not be automatically manifolds. e.g. simply because you can make the fiber look like anything with a trivial bundle [base] x F, where F is any topological space you like. but i can't think of any interesting examples or applications of the more general definition.

Cylinders and the Möbius strip are simple examples. So is the Hopf bundle $S^3\to S^2$.

so if you don't bake manifoldness into the definition, my guess is you bake it into other stuff fairly quickly.

Leslie: You’re thinking of fibration rather than fiber bundle. Of course you can do bundles over non-manifolds, but I doubt that’s relevant here .

well i'm thinking about stuff like the definition here en.wikipedia.org/wiki/Fiber_bundle, which doesn't seem to require manifoldness, although different fibers are required to be the same thing.

9:59 PM

No, it’s not required.

although every example they give is a manifold, except maybe that stuff with quotient spaces, where obviously a lot of stuff can break.

@copper.hat I like your comment here. My first thought was "Look at $f(x) - x$ and apply the intermediate value theorem", but the proof-without-words approach is nice.

his comment... is entirely words.

[kidding.]

:P

Hey guys can i get some hints on this problem pls. Prove that if a convex set is contained in the union of a finite family of halfspaces in $\mathbb{R}^{d}$ then it is contained in the union of some $d+1$ (or fewer) halfspaces from the family (covered by some $d+1$ subspaces).

I assume i have to apply Helly theorem but it seems like some reverse Helly or something

10:01 PM

Except copper should say curve, not line!

copper, someone is asking about convex sets

@TedShifrin No, line is right. He is comparing $y=f(x)$ to $y=x$. The "line" being drawn is the graph of $y=x$.

He called the graph a line.

Oh, there are two uses of line.

Nevermind.

Yeah, the second line should be a curve.

is copper doing lines in public again? i thought we told him to stop that

10:03 PM

Heh.

Helly’s Thm is something I’ve never studied.

@AlekMurt see page 13 of whitman.edu/documents/Academics/Mathematics/parish.pdf

Whitman, eh? My sister went there for her BA.

seems like i dont know how to write a contrapositive :(

@robjohn Our dubiously competent acquaintance is back at work.

@leslie That looks nice, except the author doesn't understand the difference between "less" and "fewer." Sigh.

10:18 PM

@TedShifrin To be fair, neither to most native English speakers.

AMDG

For some reason, dictionaries seem to fail to describe the relationship between adjectives and whether one may be stronger, weaker, or equivalent in strength.

(Especially grocery store managers.)

@TedShifrin Thanks. My understanding is that the cylinder is trivial because it's just a product space, but the Möbius strip is not trivial, intuitively because of the twist. I need to think more about this. (I'm also aware that tangent bundles and vector bundles are fiber bundles.)

Yup.

Here’s another example. The space of (oriented, if you want) lines in the plane is a bundle over the circle.

AMDG

Imagine getting 13 views per 5 hours on your question. SO is still quite active these days, right?

Arief

10:34 PM

Hello there, just want to share my recent video on IMO 2014 shortlist problem that involves Tetrominos. Thanks.

https://www.youtube.com/watch?v=TOuQhFAjKhw&list=PL48gowPA25i6o8sy2dlr8WSXATzJ1CKdn&index=10

ted: the difference between what and what, now?

10:48 PM

What?

@TedShifrin Is this related to real projective spaces? I've never taken the time to understand what they are, but I probably ought to.

With unoriented lines, that’s what they are — lines through the origin in one higher dimension. Mine was all lines in the plane. You can make them oriented or not. Oriented lines double cover unoriented ones, since a line has two possible orientations.

reading wiki about coverings it says "Hence the cardinality of a subcover of an open cover can be as small as that of any topological basis", this implies if for example i'm in d dimension space can i say that i can find a subcover of d open sets for a set already covered by finite opens?

Topological basis? What? And what do you mean “in d dimensional space”?

Cover a circle by overlapping tiny intervals.

11:03 PM

@TedShifrin I have commented on this low quality answer.

Is the guy trying to get incompetently promoted with horrible rep on here? @robjohn

He seems worse than most good undergraduates, truly.

I have not seen a good answer from them, not that I've been anxiously reading their answers.

Nope. Only trumped-up (ugh) pretend questions.

But he seems truly to have a doctorate. Minor google research. Just shows to go.

@TedShifrin If you believe that that is really them, then that is just horrible.

The picture at the university agrees with the picture here.

Not a true math position, at least.

11:14 PM

I've been trying to see what is needed to be a "Research Professor"

Depends whether it’s Princeton or Timbuktu.

@TedShifrin which university are you looking at?

I think I lack the background to figure out how the "space of (oriented, if you want) lines in the plane is a bundle over the circle", but I'll keep thinking about it. Thanks.

at Leslie U you can endow a professorship of research and populate it with a person of your choosing by converting $5 million US to lesliecoin.

@leslietownes such a deal.

11:22 PM

yeah, $5 million US is basically worthless anyway.

@robjohn I looked a week ago. I remember not.

Ok, I was snooping around and didn't find anything that looked as if it were them.

@Novice You can think about how to parametrize the space easily.

Well, unless I'm badly misunderstanding, for any number from $[0, \pi)$ you can get an unoriented line in the plane, and for any number from $[0, 2\pi)$ you can get an oriented line in the plane. But those are only lines through the origin, so I think I'm missing something.

@robjohn Here's some snooping.

@Novice: Good start. So how can you quantify parallel lines in a certain direction?

11:36 PM

@TedShifrin It's been a long time since I logged into LinkedIn. My computer started going slow after I did. Something was going on there.

I left and now my browser seems calm again.

the profile is public and can be viewed without login. i wore a copper hat while clicking just in case.

I don’t log in. I do not belong.

It asked me to log in to see the page. hmm

that's a shame. you're missing out on a lot of inspirational posts about boosting productivity from third-degree connections.

maybe it's recognizing a cookie and trying to lure you back. opens fine in private browsing for me.

@leslietownes I feel so ashamed...

11:38 PM

the answers seem consistent with someone of that background. very mechanical, do this, do that, not a lot of theory behind it, and also not entirely wrong, just not mathy.

A doctorate from Wash U, really?

a doctorate in something other than math? absolutely.

when computing derivatives defined by the implicit function theorem and using the chain rule, care needs to be taken that wasn't.

Pathetic, regardless.

it made me do a captcha which asked which image of a buffalo was "the correct way up." who am i to decide that for the animal?

11:41 PM

Who made you?

the linkedin website.

I didn’t do nothing.

Better cleanse your browser.

no, it did. it maybe didn't like my IP or the fact that i was opening it in a "private browsing" session.

Today’s Wordle I wasted a move on a stupid guess. One guess better in French, again.

@TedShifrin every time I use that link, I get asked to log in.

11:43 PM

the worst captchas ask you to click all the photos including something that might reasonably appear in traffic, and some of the images have a tiny version of what it's asking for in the background. like a traffic light, or a motorcycle. you ask yourself if the machine knows that it's there.

Right. Or is that a palm tree in disguise?

TIAA-CREF had a good one that is basically a childrens game. it asked you to click on a cartoon starfish that's hiding behind a skyline.

i had my daughter do it.

She’s ready to retire.

I tried using my home IP and the UCLA VPN. Both hit the authwall and asked me to log in.

I get a security thing when I click on the link on my ipad.

@robjohn Supposedly, Vel Tech Univ. Does it exist?

11:50 PM

@TedShifrin Maybe if you kept track of a vector that indicated the translation of a line that originally passed through the origin. But that doesn't quite work, because many of those translations will produce the same line. So I guess I need some kind of equivalence relation.

@Novice Here's a helpful hint. Translate orthogonal to the line. :)

@TedShifrin I could get there with a private window. There must be a cookie that LinkedIn put in my browser a long time ago.

@TedShifrin Seems to be

I need to restart my browser. Things are still slow.

BBL

@TedShifrin If you make that assumption then all you need is to keep track of the signed length of the translating vector, so then I guess you just need a second real number. So just $[0, 2\pi) \times \mathbb R$.