Mathematics - 2017-11-30 (page 4 of 8)

@TedShifrin I asked when you started because I've developed a fascination with Reconstruction and its effects recently. And I knew there was no way you started at UGA when it was still segregated, but I thought there may be a chance you overlapped with some people who did

But it was 30 years between that and when you started

Ted Shifrin

6:08 AM

That was in the 60s, Kevin.

Daminark

Is it even true that you can get a continuous extension?

Ted Shifrin

One of my long-ago colleagues was involved and told stories about when Charlayne Hunter and ?? (forgot first name) Holmes were first there.

Eric Silva

@Daminark I'm pretty sure any biholomorphism is gonna end up just being scaling

Ted Shifrin

Yeah, Demonark, I think you actually do.

Kevin Driscoll

Oh? All the way into the 60s? I imagine it was a long, continuous process of sorts. The University claims the official date as in 1951

Ted Shifrin

6:12 AM

I don't believe that, Kevin.

orbit-stabilizer

Oh... I think i have somehting

Ted Shifrin

@Kevin: This. Although I think it's right that they were the first to graduate. I never heard about the first to matriculate.

Kevin Driscoll

Oh no, you're right I totally misread the page. They do say 61, not 51

Ted Shifrin

Yeah, they entered in 61, I think.

So your 30 years should have been 20 years, which is what I thought.

Kevin Driscoll

Indeed. That 10 yaers makes a big difference because now Im not surprised there were still some folks around who remembered

Eric Silva

6:16 AM

@Ted I think I know how to do it

Ted Shifrin

Cool :)

Eric Silva

I guess I shouldn't spoil for @Daminark

Daminark

So I've got no particular insight with this and am just experimenting, so please don't stab me if I'm being an idiot, but would it be worthwhile to send this through $e^{rz}$ where $R_1 < r < R_2$? The idea being that it'd be periodic on one but not the other might allow us to work with it

Or maybe something to that effect

Oh no nevermind that was stupid

Ted Shifrin

Annuli aren't good for periodicity.

Vertical strips are.

Eric Silva

I guess that's a spoiler

Ted Shifrin

6:20 AM

Hmm ... I don't think it's that easy.

Eric Silva

I made a harmonic function out of the biholomorphism @Ted

Ted Shifrin

Yeah, so I think I've thought of a Harnack inequality type thing before. But just max principle?

Eric Silva

yeah, i used the biholomorphism to make an auxiliary harmonic function and used max ppl to show it vanished identically

Ted Shifrin

OK, totally different from the approach I assigned.

Eric Silva

that let me explicitly calculate the ratio $\frac{\log(r_{2})}{\log(r_{1})}$

I happen to be doing my PDE pset at this very moment so I guess making a harmonic function was the most natural thing in my current headspace lol

Ted Shifrin

6:23 AM

Granted.

Eric Silva

cool exercise

I miss complex, can't wait to do it again in spring

orbit-stabilizer

Okay, okay I think I have something!

Ted Shifrin

You're supposed to miss diff geo, Eric :)

orbit-stabilizer

$\frac{(-1)^{\floor(sin(\frac{n}{log(n)})}{n}$

Daminark

Okay so I can sorta see why $|z_n|\to 1 \implies |f(z_n)|\to 1$, maybe after you do $R_2/f$

orbit-stabilizer

6:25 AM

darnit

Eric Silva

I miss diff geo more ofc @Ted :)

everything else is a side hustle

Ted Shifrin

@orbit: May I respectfully suggest you write down actual numbers?

Eric Silva

diff geo is the main hustle

Daminark

So assume $f:A(1,R_1)\to A(1,R_2)$, then take a circle in $A(1,R_2)$ and look at its pre-image, that'll be bounded away from $|z| = 1$ and $|z| = R_1$

So if you take some connected region near the $|z| = 1$ circle in the annulus (that is disjoint from the pre-image of the circle), that'll have to map to a connected set in the image which is disjoint from the circle and it has to all lie on one side

Which should basically do it

orbit-stabilizer

Here you go @TedShifrin

desmos.com/calculator/k6kzxjtm5x

I'm looking at the numbers as I increase the limit

Eric Silva

6:31 AM

that's way too complicated

Ted Shifrin

@orbit: You'll learn a lot more by writing down exact numbers and thinking through a pattern explicitly.

Daminark

By the same logic, $|f(z_n)|\to R_2$ as $|z_n|\to R_1$

Ted Shifrin

OK, Demonark. You haven't got to the hard part yet.

Eric Silva

@Daminark that's the easy bit

damn sniped

Daminark

Yeah I'm not exactly sure why this helps yet

Eric Silva

6:35 AM

@Ted im wondering what the different approach you were talking about is

Ted Shifrin

Schwarz reflection.

Eric Silva

ah

Ted Shifrin

Hi demonic Alessandro.

Daminark

Never heard of that

Ted Shifrin

As the president would say, "Saddd."

Daminark

6:36 AM

I was thinking of recomposing $f$ with itself a bajillion times

Eric Silva

you guys didnt do it in bootcamp?

Ted Shifrin

Have you done the Schwarz lemma yet?

Eric Silva

/Nori didnt do it?

depressing tbh

Daminark

Nori didn't talk about conformal maps for long, he just mentioned them for like, 20 minutes at the end of a class and gave a few problems on it

Ted Shifrin

Schwarz reflection and Schwarz lemma are basic deep facts from complex analysis.

Nothing to do with conformal maps.

Eric Silva

6:38 AM

schwarz reflection is an analytic continuation thing

Alessandro Codenotti

Hi chat

Ted Shifrin

And Schwarz lemma gives you structure of maps on a disk.

Daminark

Schwarz lemma about the whole, $f:B(0,1)\to \mathbb{C}$ where $f(0) = 0$ and $|f'(0)| = 1$? We did do that

Ted Shifrin

Hmm, demonic Alessandro yet again says "hi chat" and not to me.

Well, mostly right, Demonark.

Eric Silva

I guess I don't personally really think of the reflection principle as a complex analysis thing really

Ted Shifrin

6:39 AM

Well, it works for harmonic functions, too, of course.

Daminark

Nori never really talked about analytic continuation at all

Eric Silva

or elliptic equations in div form @Ted

Daminark

And in the bootcamp, I think it might've come later on in the complex section when I was at the conference in Notre Dame

Ted Shifrin

Well, ok, but it still belongs in a beginning (graduate) complex course.

Eric Silva

definitely

Daminark

6:41 AM

That or it just didn't happen. We were a good deal behind schedule

Alessandro Codenotti

Hi @Ted, @Dami and @Eric I meant

Ted Shifrin

Of course you did, Alessandro.

Eric Silva

a lot of elliptic stuff I guess really should be thought of trying to see how far complex analysis extends beyond the best case scenario

Ted Shifrin

Well, right, a lot of complex is just ellipticity of $\bar\partial$. But not all.

Eric Silva

yeah

Ted Shifrin

6:44 AM

OK, I'm outta here. I'm still partially on eastern time.

orbit-stabilizer

HAHA it worked! My example worked!

Eric Silva

see ya @Ted

Ted Shifrin

Good luck proving it, orbit.

Daminark

Could we try to consider $f^n$ and see what it does to $A(R_1,R_2)$? Like maybe if we could get that they're equicontinuous and then do some Arzela-Ascoli stuff to it

Ted Shifrin

I have no idea how to prove anything with that example, orbit.

orbit-stabilizer

6:45 AM

@TedShifrin, thanks Ted!

Ted Shifrin

Yes, Demonark, there's a proof that goes that way.

Better have $R_1=R_1'=1$ and $R_2'<R_2$ or something.

I also assigned that approach for homework last time I taught the grad course.

Eric Silva

interesting

Daminark

Well I was thinking of like, if you had $f:A(1,R_1)\to A(1,R_2)$

Eric Silva

i never wouldve thought to iterate f

Ted Shifrin

You need $R_2<R_1$.

Daminark

6:47 AM

Oh yeah true

Ted Shifrin

You'll find something subtle comes up there, too, but that's a great approach.

Daminark

@Eric yeah I mean the idea was something about it thinning out a region or something

Eric Silva

I like my max ppl

Daminark

It seems interesting, though I don't really know anything about harmonic functions so yeah, that wouldn't have worked

Ted Shifrin

Do you know normal families stuff, Demonark? Do you know you get a subsequence uniformly convergent on compacts?

Daminark

6:50 AM

I have not heard of that stuff just yet

Abcd

Hello everyone.

orbit-stabilizer

Well, maybe it doesn't work. Sigh idk. Goodnight all.

Eric Silva

I used nothing about harmonic functions except that their max is on the boundary

Ted Shifrin

You'll need that.

Night, @orbit.

OK, I'm gone.

Daminark

Yeah our complex class was a bit weak. In hindsight Danny's was better, they got pretty far and apparently have been talking about stuff like elliptic curves and all, but it seems both spent a bit too much time reviewing 203

See you @Ted!

Eric Silva

6:51 AM

Danny's great

Daminark

Yeah. Like I still definitely like Nori, his lectures are quality and he's definitely fun, but I think he was a bit too light. He took way too long on Cauchy's integral formula since he was reviewing a lot of basic facts about convergence, and I think he started off first week literally reviewing stuff like complex numbers and norms

Eric Silva

7:06 AM

just made some dope ass dan bing

Daminark

Dan bing?

Eric Silva

food52.com/recipes/27369-dan-bing-taiwanese-egg-crepe

Daminark

Sick!

Eric Silva

do people who arent math majors take complex

Daminark

Not too many, I don't think

It does require you get through the whole year of analysis at which point you're almost done with the math major

Eric Silva

7:14 AM

cuz charlie also reviewed analysis for too long

Daminark

(ODEs can be done just with "Math methods in physics" class so that actually is often taken by people who aren't math majors)

Eric Silva

i was talking about his complex analysis class.

the one i took

Daminark

Yeah I wasn't responding to you, I was just like, I think ODEs is the one that non-math majors take

Eric Silva

ah ok

Daminark

But yeah if the prereq is 205 they should just hit the ground running and then get to later topics

Eric Silva

7:16 AM

i was just thinking, it seems like people spend so much time reviewing analysis in complex, but it's a prereq, like what's the point

unless a bunch of people who arent math majors are walking in

maybe that used to be the case, and the curriculum hasnt caught up w changes yet

Daminark

I guess? I dunno, I wish I took it with Schlag though, he seemed to actually do a good amount

Eric Silva

yeah but that's schlag

Daminark

I mean I'm not saying he's like, the model for how all classes should be taught, I could see a case being made that he teaches an honors complex analysis class and then everyone else does something intermediate

Eric Silva

im saying he generally like, just is harder than other people

Daminark

Or better yet, Soug complex analysis, that'd be a time

Eric Silva

7:19 AM

iirc i heard from someone that it used to be a thing a lot of other stem people took because of it's relevance for PDEs, which a lot of other stem people take

but i think structural changes in other departments made this stop happening

i wasnt just wildly speculating about curriculum changes

Daminark

I see, maybe

Eric Silva

i think i remember schlag taught a math methods class at some point

Daminark

Oh snap

Wonder how that went

Eric Silva

probably hard but rewarding

Daminark

True

Eric Silva

7:27 AM

i would love a reading course w him

Daminark

I did ask Schlag about his harmonic analysis class and he did say he thinks I should have enough background to do it

Eric Silva

i wonder what the content will be

Daminark

Now, I may just audit it since I already have a bunch of things floating around that quarter

Eric Silva

im almost definitely taking it

Daminark

He said he's still not sure what he's gonna teach

Eric Silva

7:30 AM

I think im doing physics in spring so idk if ill have room for more stuff than that and complex

Daminark

This spring I'm sorta between a bunch of stuff

Galois theory with Emerton, complex with Marianna, descriptive set theory with Marianna, combinatorics with Laci, formal languages with Kurtz, harmonic with Schlag

Santana Afton

Not to butt in, but you're taking six math classes in a semester?

Daminark

Nah, I'll be choosing 4 of the 6

But yeah I mean, combinatorics is 100% happening

The only case I could see not doing Galois (and this probably a shit idea) is if I don't get into algorithms and then try to do a reading course on the stuff right now

Complex with Marianna is also def happening

Alessandro Codenotti

8:09 AM

I'd do descriptive set theory and formal languages but I don't think that's your kind of thing

Daminark

As of now I'm inclined toward formal languages, but I am curious, why don't you think it'd be my thing?

Alessandro Codenotti

I was referring more to descriptive set theory than formal languages

Formal languages would make sense if you're leaning toward compsci

Daminark

Ah

Secret

Set theory (music)

Musical set theory provides concepts for categorizing musical objects and describing their relationships. Many of the notions were first elaborated by Howard Hanson (1960) in connection with tonal music, and then mostly developed in connection with atonal music by theorists such as Allen Forte (1973), drawing on the work in twelve-tone theory of Milton Babbitt. The concepts of set theory are very general and can be applied to tonal and atonal styles in any equally tempered tuning system, and to some extent more generally than that. One branch of musical set theory deals with collections (sets and...

This is lolololol

Alessandro Codenotti

wtf

Daminark

8:23 AM

Oh ffs

Vrouvrou

please I have a function $f:\mathbb{R}\to [0,+\infty)$ such that

$(i)$ $f$ is convex and continuous

$(ii)$ $f(t)=0$ if and only if $t=0$

$(iii)$ $\lim_{t\to0}\dfrac{f(t)}{t}=0$ and $\lim_{t\to+\infty}\dfrac{f(t)}{t}=+\infty$

$(iv)$ $f$ is even

Can we say that $f$ is bijective on R_+?

Thank you

someone here ?

Leaky Nun

@Vrouvrou "$f$ is even"

??

nvm

?

8:35 AM

I think (i) and (ii) are enough

Vrouvrou

it is bijective ?

Leaky Nun

I think so

Vrouvrou

but how to prove?

if i take $f(t)\frac12 t^2$ it is not bijective

Leaky Nun

well it's continuous and strictly increasing

it is

Vrouvrou

why strictly incrasing please ?

Leaky Nun

8:38 AM

because it's convex?

Vrouvrou

i don't find this property

Abcd

8:59 AM

@LeakyNun I am unable to ping you in the other room...

Leaky Nun

ok

Perturbative

11:04 AM

Hey everyone!

user8469759

0

Q: Approximating the computation of arctan via a boundary value problem

Suppose $$ f(x) = \arctan(x) $$ We can write $$ f^{(2)}(x) = 2x\left( f^{(1)}(x) \right)^2 $$ Subject to some boundary condition $f(a) = c, f(b) = d$. I'm not an expert in this boundary value problems, but I'd like to know what kind of iteration I can use to solve such kind of problem. I've...

Secret

ams.org/publications/journals/notices/201711/rnoti-p1313.pdf

Vrouvrou

or $f$ is increasing $\Longleftrightarrow x\leq y\Longleftrightarrow f(x)\leq f(y)$

Secret

I have very different interpretations of these artworks here, which mostly look like copper plates and lines in a geometric sketch

Vrouvrou

Hello, we say $f$ is increasing $\Longleftrightarrow x\leq y\Rightarrow f(x)\leq f(y)$
or $f$ is increasing $\Longleftrightarrow x\leq y\Longleftrightarrow f(x)\leq f(y)$

Secret

11:25 AM

mathworld.wolfram.com/IncreasingFunction.html

seems an iff

Erik the Outgolfer

@TedShifrin haha, looks like you were pinging me too

use the reply arrow ;-)

Secret

I don't know if it is even possible to have that without the converse hold. For sure, that cannot be drawn on paper

berrygreen

12:06 PM

So in differential geometry, $dx_i(E_j)=\delta_{ij}$ where $dx_i$ is an element of the basis of the dual space of the tangent bundle and $E_j$ is a standard vector field. Is this correct?

So then for $X=a_1E_1+...+a_nE_n$, $dx_i(X)=a_i$, correct?

Tug' Tegin

1:19 PM

Hello! Is anybody here?

Akiva Weinberger

@Secret Definitions tend to use "if" where they really mean "iff"

I'm here kinda

Secret

@AkivaWeinberger that's a really confusing practice, but ok

But since such practice is common, how do they define something that only holds in one direction of implication?

Akiva Weinberger

That wouldn't be a definition

Tug' Tegin

What do you think, learning abstract algebra from older books like Soviet mathematician Kurosh's "A course in higher (abstract) algebra" book published in 1968 good (or is that book on equal status with the modern ones)? Thanks in advance!

Akiva Weinberger

I don't see why that wouldn't be

(good)

Tug' Tegin

1:28 PM

I wondered if there were some, say, "inventions" in math after 60s and they weren't included in older books.

By the word inventions i meant new theorems, subjects...

Silent

Which answer should I accept if I have multiple answers, and all are equally good?

Leaky Nun

@Silent the one which helped you the most

Silent

@LeakyNun, :) what if all were saying same thing?

Leaky Nun

plagiarism?

Akiva Weinberger

@Tug'Tegin Hm, maybe. I don't know, then

Leaky Nun

1:32 PM

well you need to make a decision

MatheinBoulomenos

@Daminark not doing Galois would be huge mistake

Leaky Nun

it's up to you

you can also accept none of the answers

Silent

yup

that is good

Leaky Nun

lol

rather than make a choice you would make no choice?

MatheinBoulomenos

not everyone accepts choice in mathematics

BAYMAX

1:35 PM

Hi

I have an expression

a bit complicated

Akiva Weinberger

@Silent Battle to the death

Either that or random number generator

BAYMAX

$\chi = \frac{2(1 - k^2 sin^{2}(\frac{\beta h}{2})) + \sqrt{-4k^2 - k^4 sin^{2}\beta h - 2k^2 sin^{2} \frac{\beta h}{2} + 4k^4 sin^{4}\frac{\beta h}{2}}}{2(1 + k^2cos^{2} \frac{\beta h}{2})}$

Is $\chi < 1 \forall k ?$

MatheinBoulomenos

what kind of abomination is that?

BAYMAX

$k > 0$

Haha Hulk vs Abomination :)

yes if we remove the square root part

then for any $k >0$ value of $\chi <1$

but the square root makes it complicated!

If I have done right then $\chi$ is one of the roots of $\chi^2 (1 + k^2 cos^2(\frac{\beta h}{2}) - \chi (2 - 2k^2sin^{2}\frac{\beta h}{2})) + (1 + k^2 sin^2(\frac{\beta h}{2})) = 0$

Akiva Weinberger

Try graphing

berrygreen

1:44 PM

With regards to notation of permutations how does on distinguish one-line notation from cyclic permutations? Like $(1, 3, 2, 4)$ could mean $1\to1$ or $1\to 3$ depending on whether it's cyclic or not

Leaky Nun

@MatheinBoulomenos how dare you defy the axiom of choice

Akiva Weinberger

@berrygreen I think cyclic is much more common

so if it's not explicitly stated I would assume it's cyclic

berrygreen

@AkivaWeinberger I'm writing a little something that uses permutations. I guess I will just state what notation I'm using. Thanks

BAYMAX

@AkivaWeinberger perhaps we can something from product and sum of roots?

Secret

[Random or challenge(?)] (Its short, I promise)

We know what a cantor set is. Now consider trying to do a cantor set like construction on the "unit rational interval" $\Bbb{Q} \cap [0,1]$ instead, what will happen

Akiva Weinberger

1:56 PM

What, like it's pseudoboundary?

Alessandro Codenotti

Isn't that the same as intersecting $\Bbb Q$ with the Cantor set?

Akiva Weinberger

I don't think pseudoboundary is a real word but you know what I mean hopefully

The endpoints of the intervals that make up its complement

Secret

Ah right, because each step in the construction only open intervals with rational endpoints were deleted, we still have rationals left behind

Akiva Weinberger

@AlessandroCodenotti Not quite, for the standard Cantor set at least, 'cause $\frac14$ is in it but isn't on the pseudoboundary

($0.020202\dots_3$ in base 3)

Secret

hmm... I think the pseudoboundary is dense

Since the pseudoboundary is countable, it has measure zero

I wonder what happens if we modify the construction of the cantor set such that we remove closed middle thirds intervals instead of open intervals. Perhaps we also end up with a countable set that is dense since the left and rightmost end for each interval in the construction is left untouched

in particular, the points 0 and 1 should be left behind among countably many others

Secret

2:14 PM

Unrelated question: What are the applications of $\omega_1$ in other fields of mathematics outside of set theory. So far I am only aware of measure theory and the long line?

MatheinBoulomenos

@Secret if you give $\omega_1$ the order topology, you get a pretty weird topological space, which can serve as a basis for counterexamples to several statements in topology. (e.g. you can construct a sequentially continuous function that is not continuous)

Secret

I see. I think I can comprehend such function as based on the fact that no sequences of ordinals can converge to $\omega_1$ without having $\omega_1$ as one of its elements

I am more interested on its appplications to theorems analysis and geometry, besides constructing counterexamples

do you aware of any of those?

MatheinBoulomenos

Yeah the fact no sequence of ordinals converges to $\omega_1$ is the main point

titusAdam

Hi, just a quick question: Suppose we have two sets of edges I and J, such that |I|<|J|. Furthermore, we know that if we match these sets to a set of nodes called V, the resulting graphs (V,I) and (V,J) both will have at most one circuit. Is there a way to prove that there always exists an element e in J\I, such that (V,I+e) has at most one circuit?

MatheinBoulomenos

I don't know a lot about analysis and geometry, but somehow I don't think that $\omega_1$ is relevant there

Secret

2:24 PM

hmm I see...

One thing that puzzles me about ZF and ZFC is what motivates us to have a hierarchy of infinite well orderings that has larger cardinality than the previous. I mean, I can understand why we need different infinite cardinalities because of powersets and cantors theorem, but I see no motivation on why we need well orderings beyond countable well orderings

The fact that no (countable) sequences can converge to $\omega_1$ basically means $\omega_1$ is in some sense unreachable from below, thus its existence depends entirely on hartog's number ( or axiom of choice if using ZFC). But what is a rationale to justify defining them?

@AlessandroCodenotti @AkivaWeinberger may be interested

Put it in the context of hartog numbers, why are we interested in having ordinals that don't inject to countable ordinals (sure we can easily prove them, but why we want to define such mapping in the first place)?

I always felt the answer to this might lie in some theorems in real and functional analysis, I might need to dig deeper...

Alessandro Codenotti

@AkivaWeinberger sure but if you do the middle thirds construction on $\Bbb Q\cap[0,1]$ you'll still get 1/4 in the set in the end, no?

MatheinBoulomenos

@Secret I don't know about Hartogs number, but from a topological point of view, I interpreted the weird properties of $\omega_1$ as an example that for a topological spaces, sequences alone are insufficient to describe the topology, unlike for metric spaces. Note that in real and functional analysis, a lot of the spaces that people care about are metrizable or pseudometrizable, so this pathologies don't arise.

But there are spaces in functinoal analysis that are not first-countable, which means that their topology is not characterized by sequences, for example dual spaces of infinite dimensional topological vector spaces with the weak topology

This is not directly related to $\omega_1$, though

But it means that there are spaces in functional analysis which behave like $\omega_1$

in the regard that the toplogical properties can't be deduced from the convergence of sequences alone

Balarka Sen

2:46 PM

@AkivaWeinberger Your dad knows what's up. I agree with him.

Secret

I see, I might need to read more on that later, but a preliminary check shows these dual spaces are quite important in defining linear operators in some hilbert spaces as well distributions

MatheinBoulomenos

dual spaces are really important in functional analysis, that's true

Secret

though at my current knowledge, I am not sure if uncountable well orderings in such space is useful though, I think I can only answer that after a trip to functional analysis

but for many topologies, we can do just fine without thinking much about well orderings in general, I think...

We only need linear order in order to define nets (generalisation of sequences)

MatheinBoulomenos

there's the Hahn-Banach theorem which is also about dual spaces which you can prove using the well-ordering theoerm

if you want Hahn-Banach in full generality, countable choice or determinacy is not enough

Alessandro Codenotti

A lot of stuff in functional analysis breaks down without AC

Balarka Sen

2:56 PM

axiom of choice is obviously true

so it doesn't matter

Secret

I am not terribly worried about the usefulness of axiom of choice (because we know how important it is in many functional analysis theorems) What I am puzzled about is the utility of $\omega_1$ and ordinals that represent higher cardinality well ordering

MatheinBoulomenos

the existence of well-orderings is equivalent to AoC

I don't think that specific ordinals like $\omega_1$ are relevant in functional analysis

Balarka Sen

the well-ordering theorem is obviously false

Secret

Yes, but do we actually use uncountable well orderings directly?
(ok sniped, you just answered)

hmm.........

Semiclassical

3:24 PM

hi chat

MatheinBoulomenos

hi

Antonios-Alexandros Robotis

hey

Balarka Sen

I accidentally stumbled upon a configuration which could lead to the resolution of the paper cube problem

The configuration I have is two adjacent cubes, one of them missing a face and the other missing two of them

Secret

what is the paper cube problem?

Semiclassical

16 hours ago, by Ted Shifrin

Balarka Sen

3:33 PM

16 hours ago, by Ted Shifrin

So ... fold that into a cube, folding only along the obvious given creases.

Semiclassical

16 hours ago, by Ted Shifrin

So ... fold that into a cube, folding only along the obvious given creases.

Balarka Sen

Sniped

Semiclassical

retro-sniped on my part

Secret

@BalarkaSen but how does that resolves it, that net does not seems to allow having two separate fragments of a cube?

Balarka Sen

Read what I wrote again. I didn't say I solved the problem.

Just that it could lead to a solution.

Ah OK solved it

Beautiful problem. You're not actually supposed to get it quickly; the creases need to battered enough to obtain the "key move".

Secret

3:43 PM

How can one continue to fold that net if any fold along the 1x3 squares will block any folding adjacent to it?

One thing that puzzles me is that any valid fold in the 8 square region will create a 1x3 wall which then blocks any further folds adjacent to it

and I don't see any way to fold away the corner squares in the 8 square region without turning them into triangles by introducing a new crease

Antonios-Alexandros Robotis

I feel like I should take a stab at this.

This seems fun. I haven't any other work to do. /s

Balarka Sen

lol

Try it, it's really fun

I'd say it's the best paper rage game I have seen

Antonios-Alexandros Robotis

yeah I will haha. I'll use it as a break before returning to my suffering.

Balarka Sen

Dare I say this is the Dark Souls of origami?

Semiclassical

@BalarkaSen I forget. Did I blather on about positive semidefinite matrices yesterday?

Balarka Sen

3:56 PM

I saw that you found that your variety is determinantal

Semiclassical

yeah

I can sharpen that further: Each point on $f(x)\geq 0$ in [0,1]^3 corresponds to a positive semidefinite matrix.

Balarka Sen

cool

Semiclassical

yeah

that makes it easy to argue that it's a convex set as well

Balarka Sen

Sure

Hm. Is the volume bounded by a closed, positively curved surface in $\Bbb R^3$ always a convex set?

Semiclassical

probably?

i dunno.

Balarka Sen

3:59 PM

Let's see. I know that $K > 0$ implies that the surface curves away from it's tangent space

Semiclassical

right

Balarka Sen

I.e., that locally it's bounded on one side by the tangent space.

It's a proof-worthy fact, but it can be done

Semiclassical

so it's definitely 'locally' convex at each point on the surface

Balarka Sen

Hm, right.

Semiclassical

(it's locally the boundary of a convex set, that is)

Balarka Sen

4:01 PM

That sounds right

Semiclassical

and if it's locally the boundary of a convex set, I have a hard time seeing how it could fail to actually be a convex set in R^3

Balarka Sen

4:20 PM

Hi @Daminark

Leaky Nun

@BalarkaSen @MatheinBoulomenos must the kernel of a monomorphism be isomorphic to the zero object?

MatheinBoulomenos

I assume you're working an abelian category

Yes

Leaky Nun

is it false in non-abelian category?

MatheinBoulomenos

Well, you need zero objects and kernels to exist

Balarka Sen

@LeakyNun i am of no help to you

MatheinBoulomenos

4:29 PM

just check that the zero objects satisfies the UMP of the kernel

Leaky Nun

@MatheinBoulomenos let's say they exist

fair enough

but then it's circular: I need to prove that everything that satisfies the equalizing property factors through the zero object

MatheinBoulomenos

there's nothing circular about it. Let $f: A \to B$ be a monomorphism. we claim that the unique map $0 \to A$ satisfies the UMP of the $\operatorname{ker}(f)$. Let $g: X \to A$ such that $f \circ g= 0 = f \circ 0$, then as $f$ is a monomorphism, we get $g=0$, so $g$ factors through the zero object

you need to prove that $0 = f \circ 0$ for every morphism $f$

Leaky Nun

say what

MatheinBoulomenos

but that's just by definition of initial object

Leaky Nun

that's interesting

MatheinBoulomenos

4:36 PM

I think you need Abelian for the other direction $\operatorname{ker}(f) = 0 \Rightarrow$ $f$ is monomorphism iirc

Leaky Nun

I see