Mathematics - 2020-12-05 (page 2 of 2)

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10:00 PM

Yeah

I think every closed $3$-manifold is a boundary of a $4$-manifold, modulo orientability hypothesis I might be missing. Every closed $3$-manifold (adjective?) is a surgery on a link in $S^3$, $S^3 = \partial D^4$ so I think you can do clever handlebody stuff at the boundary.

Being nullcobordant is I think a surgery invariant

Yes, every closed $3$-manifold is orientable anyhow :P

This is stuff I knew in grad school but never taught or used, so ... Hippa's meme.

What about $\Bbb{RP}^2 \times S^1$?

Hmm, what am I missing?

Why did you say it's true? I think I get this wrong everytime as well

Because it's a famous fact !! LOL ...

Obviously we don't need simply connected, because every simply connected guy in every dimension is orientable.

10:04 PM

I get this wrong because I keep RP^n as canonical nonorientable guys in mind, but RP^3 is of course very orientable as you'd say.

Yeah, only even $n$ for that one. WTH are we thinking of, darn it?

Maybe it's time for me to retire for good :P

We're probably thinking every orientable $3$-manifold is parallelizable, or something like this

Yeah, that's it.

Whew

Haha yeah no this is confusing

Not really, but glad you sorted it out :)

I'm not sure I ever knew that proof, either. We get a trivial line sub-bundle for free.

There's got to be some obstruction that vanishes because of dimension.

10:07 PM

Yeah, I don't know a proof either! But maybe it follows from our discussion about cobordism?

every orientable 7-manifold is a boundary, apparently

If something is oriented nullcobordant it's parallelizable... by Thom's theorem haha

@TedShifrin I think this is the morally correct idea but I don't know enough

@Thor: It's true in every dimension not divisible by $4$, I believe.

Assuming orientability.

Actually, Balarka, this proof I did know/understand at one point.

crazy

There's an argument based on spin structure, but I thought I remembered something easier.

10:11 PM

hmmm

Ah, PVAL (I miss him!) gave an argument here.

every 3-fold is spinnable?

yeah, orientable, still.

whoops right

oh ok

all SW classes vanish somehow

Odd. I would have started with a trivial line bundle (because $\chi = 0$).

10:14 PM

that definitely means you can take three lin indep sections by the zero locus interpretation

And then I just want to frame the orthogonal complement.

Well, how do you get from mod 2 to $\Bbb Z$?

oh fair enough

i don't know

you're right, lifting the frame bit by bit must be the correct argument

I think I must have read a proof in Steenrod years ago. I wish I'd kept that book.

On a more geometric note, @Balarka, what do you think of this discussion I had with love-sodam on his post?

His book seems to have a number of false questions. It's discouraging him; I think it's sort of funny. It makes me prouder of mine :D

Wow

I was thrown off by the question until I saw your answer saying it's false

You probably did my exercises about its being true if the geodesics are at right angles (or any constant angle).

I'm giggling, because this reminds me of the angry emails I got from the guy who insisted that Fraleigh had a question saying that any number of degree 4/$\Bbb Q$ was constructible by ruler and compass, so stuff in my book had to be wrong.

10:22 PM

@TedShifrin Yeah I know this one I think

I think there's a tendency for students to trust all books. Mistaken tendency.

That's ridiculous!

The pretty argument is with local Gauss Bonnet. Look at holonomy around a geodesic parallelogram. (Usual contradiction proof if $K\ne 0$ somewhere.)

Right

Of course, I love the moving frames proof :P

I think love_sodam gets particularly riled up, though, based on what I've seen here.

10:23 PM

@TedShifrin I found few typos in your book. Though perfect.

Wrong exercises are the worst.

@Karim: I corrected a bunch in a corrected printing. And I do keep a page of errata. But there are no blatantly false exercises. I learned after my first book to write a solutions manual before sending the book to the publisher.

stumbled on this exercise - and I have an answer - but I wonder if there's a more analytic solution. Like idk something about the 20th derivative of $f' - Kf$ being zero or something; If $f$ is differentiable on $[0, 1],\ f(0) = 0,\ |f'(x)| \le K|f(x)|$, then $f = 0$ on $[0, 1]$.

I was so upset when I found errata (and typesetting flubs) in my algebra book (first). But then I learned that we all make mistakes and I just accept it. But this new geometry book has some whoppers in it.

Yeah haven't found any false exercises. I have solved half of the book I might keep my solution manual when I teach this class at some point.

I won't post it online

10:25 PM

There actually is one for teachers available from the publisher.

No, please do NOT post solutions online anywhere!

I finally started putting my exam solutions behind a secret password for my students.

There are a few typos in some exercises. Check the errata sheet, @Karim.

Oh, Taylor's remainder formula, @JoeShmo.

Spivak has a few exercises of this flavor, too.

ah ha! I knew it

Thanks

yeah will check it I will definitely use your book when I get chance to teach this class

I did MVT and creeped to $0$

Well, to be honest, @Karim, very few places have students that can handle this sort of course.

Yeah I agree

10:28 PM

And very few places have an integrated linear algebra/multivariable course, let alone a proof-oriented one.

My undergrad didn't have that

It is a shame

Places like Harvard, Yale, Vanderbilt, UGA (for now) have it, but not very many places.

The condition says that the logarithmic derivative of $f$ is bounded, but $\log f$ tends to $-\infty$ at $0$, so that cannot be

Chicago, of course. But they go off the deep end typically and do graduate functional analysis and no multivariable.

ok, that's not a rigorous argument, but hey

10:30 PM

Oh, @JoeShmo, I take it back. I thought your $f$ solved some ODE.

Thor is right. The right proof is based on a famous thing that shows up in differential equations. Gronwall's inequality.

It's a wonderful theorem (and I did add that to Spivak's 4th edition as an exercise).

@TedShifrin I'm guessing it doesn't go any further than this? lol ($f \colon x \mapsto xx^{\top}$) $$ \begin{aligned} \text{d}(\text{tr} \circ f)(x) v & = \text{d} \text{tr}(f(x)) \circ \text{d} f(x) v \\ & = \text{d} \text{tr}(x x^{\top}) (xv^{\top} + vx^{\top}) \\ & = \text{tr} \left( (xv^{\top} + vx^{\top})(xv^{\top} + vx^{\top})^{\top} \right) \\ & = (xv^{\top} + vx^{\top})^{2} \end{aligned} $$

Whoaaa.

It has to be linear. What's all this stuff?

You're misinterpreting basic stuff.

oh no I have forgotten how to do maths >_<

Did someone already tell Thorgott everything he needed to know

Trace is linear, so you take trace of $df(x)v$.

Oh, magical @Mike is here. Yes, but then Balarka and I got stuck on things.

10:34 PM

Stuck on anything non-trivial, though?

;)

Magic Mike lol

Ah, here's how to fix my idea. If $f$ is not the constant $0$ zero function, there is some interval $(a,b)\subset[0,1]$ such that $f(a)=0$, but $f(x)\neq0$ for all $x\in(a,b)$. Then $\log f$ is well-defined and differentiable on $(a,b)$ and has bounded derivative. But bounded derivative on bounded interval implies bounded by FTC, yet $\log f(x)\rightarrow-\infty$ as $x\rightarrow a$.

@Khallil: What does it mean to say $T$ is a linear map? What is $dT(b)$ for any vector $b$?

I'm just playing

Actually, @Thor, if you've never seen Gronwall, look it up. It's beautiful, and it's based exactly on your differential inequality thing.

@MikeM: The "best" proof that any orientable 3-manifold is parallelizable.

10:36 PM

nice, I'll look it up

There is no way to avoid Wu's formulas I don't think

Once you know that an orientable 3-manifold is spin you're basically finished

What if you use Thom's theorem instead

I've not seen it, this is me being ad hoc

How are you getting that M is spin?

I'll add it to my reading list

10:37 PM

Let's clarify this in the next 2 minutes so I can get back to commenting on projects

oh I dunno spin, I was thinking you can prove by hand 3-folds are orientable nullcob, so parallelizable

I thought there was a straight-forward obstruction theory argument (e.g., in Steenrod).

Coming from having a trivial line bundle and then being left with a $2$-plane bundle.

here's a proof based on the MVT - suppose WLOG $f(b) > 0$, then on $f(b) = f(b) - f(0) = f'(c) (b-0) = f'(c) \cdot b \le Kb \cdot f(c)$. You could iterate $c \rightarrow 0$, whence by continuity of $f$ you have $f(b) \le 0$

Why does that 2-plane bundle have w_2 = 0? This is not clear to me.

@BalarkaSen What is your argument?

There are of course plenty of guys which are orientable null-cob but not parallelizable

oh er i guess

i dont have argument sorry

10:41 PM

I don't even know how to use that to show that w_2 = 0. All of the SW numbers have w_1 as a factor (or are w_3 = e mod 2 = 0)

@TedShifrin $T(ax+by) = aT(x) + bT(y)$ for scalars $a,b$ and $\text{d}T(b)v = T(v)$

yeah good point

IF someone can show that w_2 = 0 you get that the tangent bundle is classified by a map $M \to B\text{Spin}(3) = \Bbb{HP}^\infty$. Cellular approximation (AKA obstruction theory) implies this is null-homotopic.

Oh, right $w_1(L)=0$, so no info.

Showing that w_2 = 0 is necessary in any argument (obstruction theoretic or otherwise).

10:42 PM

@Khallil: Right, $dT(b) = T$. Look what you wrote above.

Ted's approach (TM = R + P) gives us that w_2(TM) = w_2(P) but what then?

Maybe there is something but I don't see it :)

Time for me to go

Bubye. Thanks.

Oh my goodness

Never do maths after midnight >_<

Thank you, @TedShifrin

its like 4 AM and im about to start my assignment

Balarka: You're a bad influence on everyone here.

You break everyone's sleeping.

10:45 PM

i should put up a CONTENT WARNING

CONTENT WARNING: The author does not endorse severe insomnia and sleep deprivation as a lifestyle in any way, shape or form

6

Nevertheless, some readers may find ...

lol

This PhD situation is really getting me down, @TedShifrin. I know that I'm in a very privileged position & all, but, well, that kind of makes me feel worse, like I've wasted the opportunity.

What makes you want to be an academic, and at what level of academe?

I'm getting old

The 4am lifestyle is no longer viable

Alessandro Codenotti

10:51 PM

@BalarkaSen I thought you were supposed to be the author of your assignments, but apparently not

Ted, is there a geometric "interpretation" to the logarithmic and exponential functions?

LOL @Khallil. Define "old."

@AlessandroCodenotti Lmfao

early to bed, early to rise?

Hmm, area under curve @JoeShmo for log? I don't know what you mean. I can parametrize a hyperbola by $\cosh$ and $\sinh$, analogous to the circle.

10:52 PM

@Alessandro A track

Alessandro Codenotti

Oh Porcupine Tree, noice

Back when Wilson didn't do pop

yeah man

wtf is his new material

To be skilled enough to teach at a university level and get paid to do research seems to me a noble goal. I think I'd get a lot of satisfaction from inspiring students, hopefully, the way some lecturers have done with me. I get a lot of pleasure out of studying and I would be honoured to do it for a living.

Alessandro Codenotti

@BalarkaSen I haven't really been keeping up with his stuff after Hand. Cannot. Erase.

good lol

he's doing EDM now

wait let me find a meme

10:54 PM

What country are you in, Shaun?

England.

he spelled programme earlier

@Ted I just had a look at the Grönwall inequality, very neat and natural result. It's also precisely what you get when you generalize my logarithmic derivative argument from above (though you'd have to pay a bit more attention to signs and zeros in general when taking that approach).

So I don't know the variety of opportunities there. In the US there are different levels of colleges, and a number of teaching-oriented people end up in community (or 2-year) schools and find it fulfilling to make a difference to students there.

Sorry, @JoeShmo — I'm talking here and answering on main simultaneously.

user image

@Alessandro

Alessandro Codenotti

10:56 PM

Just noticed that Vildhjarta got 25 days left to deliver their promised 2020 album, why do I fall for it every time

lol

Salut @Astyx.

Hello

damn they keep delaying

Hi

@Thor: Yes, it is logarithmic derivative, with a limiting argument at $0$.

Very powerful tool, though.

10:57 PM

That sounds like the equivalent of a sixth form college, @TedShifrin, where students get their A Levels. I'm much more oriented towards conducting research than I am, teaching.

Alessandro Codenotti

@BalarkaSen their modus operandi seems to be "release a 90s long teaser and then wait for a few more years for the next one" (it sounds great I must say though)

thats hilarious

Academic life is full of lots of responsibilities other than research, unless you are a research super-star. And let me just say, from my personal experience, that research is itself far more pressure-filled than teaching.

Alessandro Codenotti

woops I meant to link a specific one

the comments below those videos

yeah im listening to the one you linked

10:59 PM

which book do I have to read to acquire the sacred knowledge that oriented 7-manifolds are null-cobordant

is this in Milnor-Stasheff

why do you want this exactly

I think so, @Thor. I no longer have it on my shelf to check.

is this for the exotic sphere

Alessandro Codenotti

At least the album will be good when it comes out, Architects keep releasing new singles from the upcoming album and they are each worse than the previous one

yes

it's literally the second sentence

11:01 PM

You need to know the Wu formulas, identities for Steenrod operations, and the fact that a manifold is null-cobordant iff all of its Stiefel-Whitney numbers vanish

So yes that book

@TedShifrin pressure from the pubiish or perish community?

I was hoping, failing a career in academia, to work in the R&D department of some scientific company, so that I may have the opportunity to make a significant contribution knowledge.

And, in the US certainly, from university administrations to bring in grant money and publish.

I feel like I'll have to read at least half that book as preliminary to this paper

I'm scared

what book?

11:02 PM

You don't need a PhD to work in industry, Shaun.

But you probably do need computation skills.

@Thorgott that paper is too hard

i dont have the patience to learn the algebra to understand it

characteristic classes are algebra at the end of the day

It's also important understand that, say, 10% of PhDs at top schools ultimately land tenure-track jobs; this is not because 90% are poorly skilled but rather because there are comparatively few open positions (and none are opening much); plenty of incredibly intelligent and incredibly diligent people do not get tenure-track jobs, and not for lack of trying, and this causes a lot of distress

algebra isn't scary

but the topology is scary

This is not to say you shouldn't pursue a PhD, which I'm not commenting on here; it's information one should be equipped with

I agree with you @Thorgott.

11:05 PM

The system is a bit different in England, though, @MikeM. However, now that Brexit has happened, they're separated from the rest of the EU and that has a definite impact on education and mobility of both faculty and students.

characteristic classes are the perfect topology an algebraist like you'd read thorgott

its just manipulating with laurent polynomials

Fair enough, I'm commenting specifically on the US here --- sorry

hopefully

I'll have to figure all this stuff out within the next, uh, less than 3 months

He's worried about proving that SW numbers 0 => (unoriented) boundary

Well, I'm not sure how universal it actually is. I know that France has for years had a separate research environment that the best young people went into, and then they blended into teaching. I'm not sure what that status is now.

11:07 PM

I've often compared wanting to be an academic to wanting to be a professional football player, except it's less glamorous and more demanding. I'm aware the odds are against me. It looks like dropping down to the Master's is my first step towards an alternative career. But a PhD at least seemed attainable. My friends & family know I'm after a PhD. I don't know what to do. It's embarrassing.

You giving a lecture on it, @Thor?

What do people who don't get tenure track jobs do?

a seminar talk

Ambitious for anyone, certainly an undergrad. Do the best you can, and black-box some of the worst.

@Anixx I don't really want to do this conversation right now, probably should not have made my original comment (edited to clarify who this is directed to)

Yes you should black-box this fact

11:08 PM

@Shaun: They're not judging you (at least they shouldn't judge you), so you need to stop judging yourself.

That's fine

Honestly, when I was a Ph.D. student at Berkeley, more than 60 or 70% of the students ended up finishing with just a masters or less.

oh wow

that's a shocking statistic

it is impossible to do topology without algebra its so annoying

fuck it ill leave topology

Become a banach space theorist

11:09 PM

I don't know the figure at UGA, but certainly over a third of our students finished with masters or less.

@MikeMiller probability on banach spaces

wow..

So funny, a @Balarka, since Balarka 1.0 and 1.2 were all about algebra.

yeah I spoke to my cousin yesterday (did his doctorate at scripps, right by you)

diagram chasing is fun man

11:10 PM

What did 1.1 do?

I think he skipped that version.

lol

"We prefer not to talk about that version"

wanted to be an academic when he went in for his doctorate, quickly switched to wanting to go into a private lab, and eventually just wanted to get into industry

the maturation process of a doctoral student

what is this $\lambda$ invariant man

why would you define this

WTF milnor

11:12 PM

It's a relative characteristic class it makes sense

It's the same thing as the Chern-Simons functional

Oh really?

where can i read the Chern-Simons approach

ah yes, of course

Was Milnor around the same time as Chern-Simons (early 70s)?

Liquid tensor experiment, by Peter Scholze via Kevin Buzzard

11:17 PM

My friend recently linked me an old post in AoPS by Peter Scholze where he gives a pure algebra proof that $\Bbb Z^n$ admits no nonconstant positive harmonic functions

profinite set lmao

He was like 14 then

You have something discretized on an n-manifold with boundary (in the case of the Chern-Simons functional it is $\int_M tr(F_A^2)$ in $8\pi^2 \Bbb Z$ for A a connection on a bundle over a 4-manifold, in the case of Milnor it is the expression $45\sigma(M) + p_1^2(M) \in \Bbb 7Z$).

Then for (n-1)-manifolds, take an n-manifold bounding it and calculate that same expression. It will depend on the n-manifold, but only up to a discrete ambiguity: take two n-manifolds with the same boundary Y. Glue them together along their boundary. The expressions above are additive under gluing, and they lie in this discrete subspace.

Hence the ambiguity between the various things we are computing lies in this discrete subspace (8pi^2 Z in the first case, 7Z in the second case) while the general value can be anything (in R or Z). Thus there is a well-defined value: cs(A) for A a connection on a bundle over a 3-manifold (this lies in R/8pi^2 Z), or lambda(M) for M a closed 7-manifold (this likes in Z/7Z)

This is also the same as the proof of the Jordan-Brouwer separation theorem.

Wow.

@TedShifrin have you looked through Thomas' 14th edition calculus textbook?

11:19 PM

Hell no.

I hated Thomas from the outset.

It hasn't been Thomas in 4 decades.

Well, it's not over yet. The thing is that I have an assessment on Wednesday and my supervisor & I are both pessimistic. By some miracle, I could scrape by with the minimum necessary for the continuation of my doctorate. It just doesn't seem likely, and this is largely because I've been unwell, not because anyone thinks I'm incapable. I have produced original research; I just have to write it up as a paper.

The writing it up seems to be where lots of people get stuck (I don't mean this in a mean way).

@Thorgott that's a nice one

All I will say is that grad school is tough enough when one is mentally strong.

thanks

11:23 PM

I guess I should really say this is the same as the definition of mod-2 intersection numbers.

If you can find the ambiguity, when you mod out by it, you have something well-defined.

@MikeMiller Explicily, the integrand in CS is $A \wedge dA$ plus some terms?

@Balarka: You should look at the original CS paper.

something like that

Asking because I think $\int A \wedge dA$ for a magnetic potential $A$ computes linking number of field lines or something like this

@TedShifrin Hm ok not sure if I will understand though

oh but the connections here are on principal $SU(2)$-bundles, unlike $U(1)$ in EM?

The original CS paper was written for physicists? Nah.

They actually do the set-up for connections in a $G$-bundle.

11:27 PM

Oh OK

At the end they apply to $3$-manifolds.

LOL, the last sentence I'd forgotten. "We do not see how this helps to settle the Poincaré conjecture."

@MikeMiller this is a nice analogy but is $45\sigma + p_1^2$ actually an understandable geometric quantity?

@BalarkaSen Is the Hirzebruch signature formula geometrically comprehensible?

or does it pop out of calculations

i don't know the signature formula but ill look that up

That's where I got it from

In dimension 8 it says $\sigma(M) = \frac{1}{45}(7p_2(M) - p_1^2(M))$

11:30 PM

ohh i see

it kind of makes sense, $\sigma$ is a cobordism invariant

Genus of a multiplicative sequence

In mathematics, a genus of a multiplicative sequence is a ring homomorphism from the ring of smooth compact manifolds up to the equivalence of bounding a smooth manifold with boundary (i.e. up to suitable cobordism) to another ring, usually the rational numbers, having the property that they are constructed from a sequence of polynomials in characteristic classes that arise as coefficients in formal power series with good multiplicative properties. == Definition == A genus φ {\displaystyle \varphi } assigns a number Φ (...

I've been compiling the write-up from the beginning. My supervisor said it's a great habit of mine. All I need to do is reorganise then add a few tables summarising my results. Perhaps I'm asking too much of myself, then, @TedShifrin; I'm no Nash. 2019 was an awful year for me. I was hospitalised twice for my mental health, for about two months at a time. I could have died a couple of times too. (I won't go into it.)

It's related to $\frac{\sqrt{z}}{\tanh(\sqrt{z})}$ sorry to say

lol

thats awful wtf

@BalarkaSen apparently he read hartshorne while sitting in high school classes. pure beast

11:32 PM

@Shaun: I do not know you, of course. But of course I sympathize and I would urge you to think about how your health must come first. If the stress of graduate school is making it worse, then you should think about healthier options.

All these multiplicative sequence things come from crazy Laurent series, @Balarka. :P

@user2103480 i dont think thats such an unusual thing anymore

ive met lots of people who are/were like that

uhhh what's your standard

that's like 0.1% of math students tops

he's always had high standards :P

Alessandro Codenotti

@LeakyNun The title is amazing

In my long career, I have taught a few brilliant students. But, even counting Berkeley and MIT, not just UGA, no one quite in that league. I taught a high school kid in AoPS who might be close to that league.

11:35 PM

plenty of high school kids read unusual things for their age; a friend of mine in high school used to bring Schopenhauer in class

In my days, the odd-ball kids read LOTR.

@AlessandroCodenotti so are condensed sets

avid literature fanatics are pretty up there as well

true

its something to do. young people are bored and wile time away in obsessive things

Alessandro Codenotti

11:38 PM

@TedShifrin I read LOTR in high school

I have never been a fantasy sort.

Alessandro Codenotti

I wanted to see the movies, but my mother was always a big fan of reading the book first

Turns out that she was right though

@BalarkaSen fair enough, but I was an outlier for doing some uni mathematics while still in high school, and it was far from anything on that level. I can't really weigh in on the difficulty of hartshorne, but people don't call it easy if I recall correctly

That's a helpful perspective, @TedShifrin; thank you. I'll have some fun with some group theory for 'rest o' night. Oh, and I've read LOTR too.

people like kripke, scholze, wolfram are immense outliers

11:40 PM

i wasnt really particularly thinking about math

LOTR is great

@Shaun: I truly wish you nothing but the best. :)

4

Alessandro Codenotti

All of Tolkien's work are tbf

Take care @Shaun

2

i truly miss being in 11th grade, in which there's no serious exams you have to give unlike 10th (middle-to-high graduation) or 12th (uni admission exams); i remember reading so much (and not just mathwise) and listening to so much music and watching so many different movies

it was an obsessive thing, because there was so much time

Alessandro Codenotti

I bought a copy of Dune recently but haven't started it yet

11:44 PM

there's no time anymore. these things wear off when you have no time

ultimately, we are all destined to run out of time

Kreck told us a story a while ago about how he gave a lecture at Bonn aimed at postdocs where he summarized every single piece of work in topology that earned a Fields medal. Apparently most people quit the lecture cause it was too tough, but there was one person who not only followed it, but even corrected Kreck on every of his mistakes on the board.
One day, Kreck approached him after the lecture, asking if he was a topologist. He replied that no, he was an algebraic geometer. Of course you know who he was; I believe it was before he got his PhD even.

I don't know who. Who?

the now-fields medalist

Alessandro Codenotti

That boy? He was Albert Einstein

11:52 PM

Oh, Scholze, whom people were referencing earlier. I know none of these people, even with Wikipedia.

Alessandro Codenotti

@Thorgott that sounds like an interesting lecture, but it would be rather short if done for set theory/logic instead

@AlessandroCodenotti lmao

I don't think I know any set theory fields medal result

what's an example?

@Thorgott there's only one

CH

11:53 PM

cohen

oh, duh

ok, I do know that result

Alessandro Codenotti

@user2103480 and the fact that Shelah didn't get a fields medal is a strong indicator that there will never be another one

suck it, set theorists

your spaces=no extra structure=boring, our spaces=lots of structure=exciting+cool

Alessandro Codenotti

Yet as soon as set theorists do some topology you complain it's awful

Highly technical point-set topology is awful.

If there's some interesting differential topology, let me know.

Alessandro Codenotti

11:57 PM

Grrrrr

Puzzle: Given a configuration of spheres in $\Bbb R^3$, call it's nerve graph to be the one obtained by taking a vertex for each sphere, and an edge for every touching pair of spheres.

ok done, I called it

Are all graphs nerve graphs of some config of spheres?

Discrete ricci flow probably proves it

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