many thanks. this transaction will forever exist in the lesledger now

it is eternal

Another one falls for the ponzi scheme.

no no no

this isn't a ponzi scheme

a ponzi scheme is when you don't have a product

this clearly has a product

Oh yeah?

ted, if you're grumpy about your new yacht not arriving yet, it's because they thought it was an oligarch yacht and accidentally impounded it. the yacht was that opulent. we're working that out now.

i'll probably just have to give 500 lesliecoin to the harbormaster to clear this up. they aren't greedy, they just want to be respected.

yes, and, you spend compute on verifying and maintaining transactions, as well as generating new coin.

Well, my spot at the yacht club is only big enough for a dinghy, anyhow.

which controls inflation, and will inevitably justify the eventual $20,000 a lescoin

yes. we will also be selling a collection of poorly drawn pixel art as NFTs. it'll be portraits of mathematicians taken from the mactutorarchive, wearing dumb zoomer clothes and hats and stuff. gauss with glowing eyes, that kind of thing.

these items will make everyone even more fabulously wealthy.

and, speaking of ponzi schemes, you know, the US dollar (EVER SO RELEVANT TODAY. NOT FUNNY) is a ponzi scheme. Except the Fed has no boss, and can print $$$ to its hearts content

whereas lesliecoin is a finite resource, thus maintaining its value

yes. this problem has been solved. there's no fed. it's just big bags of wealth lying around (metaphorically) on the decentralized ledger.

you get a $, and you get $, and you get $. EVERYBODY GETS A $$$$$

fun fact, about a year ago someone on quora asked "why would just giving out money to everyone not solve poverty?"

do you mean here? i imagine people ask that every day out there in the world. i think i've been in a taxi where the driver asked me that.

ok, quora.

well done, quora. tackling the big issues.

so innovative with their solutions, too

Germany, Venezuela, and now, us, soon, too :-)

one time in one of my cases a guy on the other side had a side hobby as a crank financial advisor. it drove us nuts, because we wanted to use it in our case, but the case had nothing to do with taxes or fraud and it wasn't a situation where his credibility about anything even mattered. it was just a matter of contract interpretation.

man, we wanted to use that.

so thats where you get all your lines

i got some of them from him. that guy is actually in prison now for stuff unrelated to the contract.

r u a lawyer in your free time?

i also used to do pro bono work with people with interesting pasts. learned a lot of great stuff about how finance works.

not in my free time. lovely unscheduled afternoons like these are my free time. if the phone doesn't light up, i'm not a lawyer.

most tax crankery is not mathematically very interesting. it's like an algebra problem where the student is actually trying to double count or under count or make a sign error or whatever they need to do to get the wrong answer.

so you were a mathematician, and THEN decided to become a lawyer?

"I'm a mathematician and a physicist"

yes. who knows where the career train will head next. i joke with my wife, maybe med school?

you might as well

if there's something longer or more expensive, i'd consider that too. i'm not wedded to med school.

I get crap from my wife for the same thing

Im not a lawyer, or a doctor

Maybe mafioso next. Munchkin will train you,

but I was in grad school for some years

@TedShifrin Can I regard John Lee's intro. Smooth manifold textbook as a Riemann geometry textbook?

Definitely not.

I want to study about Riemannian metric. My topology professor said it's ok to study them through Lee's textbook.

Lee has 50 textbooks. He has one on Riemannian geometry specifically.

in the intro to my copy, lee says he prefers not to use the term "differential geometry" for the subject matter of his book because his focus is on smooth manifolds and while he gives examples of extra structures they are not a focus.

he does define 'riemannian metric' in one section of the book but definitely not a focus.

@TedShifrin You mean intro. to Riemannian manifolds. Are smooth manifold and Riemannian manifold different?

i was speaking specifically of "intro to smooth manifolds" here.

“Study about Riemannian metric” is vague.

@onepotatotwopotato Yes, it’s like asking if every topological space is a metric space.

if you have the book because it's free (i think it used to be on his website?) it does contain relevant definitions and a few technical results but not much else. he proves that a riemannian manifold is among other things a metric space, and that any smooth manifold admits a riemannian metric. but that's just establishing some categorical connections between these notions. he doesn't do anything with them.

that i can see, anyway.

"connection" does not appear in the index, speaking of connections.

ha, ha.

You can start with undergraduate curves and surfaces. Download my free text.

@robjohn 2 chores done now :-).

on a totally unrelated sidenote: is it OK to say "granny" when referring to one's grandmother? Or does this have a pornographical connotation (like e.g. "redhead" AFAIK)? Not trying to be rude here.

NGL it was on a porn website where I first encountered that term..

it's getting hot in here

AFAIC neither of these terms has a connotation

but thanks for sharing your tastes :-)

landon: no such connotation outside of the world of porn.

same with redhead, incidentally. although i probably wouldn't personally use it except as a term of endearment for a child. maybe some would use it for a significant other, who knows with nicknames. it's just a normal word that also happens to be a porn category.

according to some law on the internet, everything is a porn category.

Silly me … I thought redhead was a fine word.

one of my friends in high school called me that, but she was the only person who called me that. "red" was more common with guys.

when my sister and i were little we used granny vs. grammy to distinguish our dad's mom from our mom's mom without naming either one of them.

my daughter uses "gramma [first name]" because of remarriages and stuff two words are not enough.

need to spend less time on redhead granny sites.

of course, one does not need to stray far to see discussions of red & black balls.

he clearly had many Fu:x to give

Part b) seems complicated.

Any shorter way?

Problem: "Show that $f(x)=x^3+x+1$ is bijective from $\mathbb{R}$ to $\mathbb{R}$ and evaluate $\lim_{y \to \infty} f^{-1} \left(\frac{3y}{y+4}\right)$". I proved that it is a bijection, but I don't understand if it is possible to evaluate explicitly the limit. So I said that since $\mathbb{R}$ is both open and closed (so in particular is open) and $f$ is continuous, then $f^{-1}$ is continuous and so the limit is $f^{-1}(3)$.

Is this correct? If there is a way to evaluate the limit explicitly I would prefer an hint than a full solution, thanks.

the said bijection is continuous as well, as you said.

$f^{-1}(3)=1$

because f(1)=1+1+1=3

please ignore my first comment. @Gwyn.

inverse of continuous bijection being continuous holds for compact sets.

@Koro: Yes, I see my mistake. Thank you!!

@Gwyn

@Koro, yes, there is a shorter sequence for (2) in your link. Try to figure it out if you want.. it's cute. Remind me to post an answer tomorrow. Ciao.

@JoeShmo thanks. I'll think about that and remind you.

'remind' as per IST.

I run on EST

It's 9:30 am Sunday here.

it's 12:07AM here, and I am half asleep

:)

@copper.hat oh my! Is there no stopping you?

@robjohn i'm not a shopping fan, but the staff at Trader Joes make it enjoyable

but two chores, amazing ;-)

:-). but the cycle to inspiration point was, well, inspiring

@copper.hat exhaustion is inspiring?

endorphins keep me alive

Or can I just proof it saying the following: Let us assume $f$ is not constantly zero. Then we know by a theorem that all the zeros of $f$ are isolated. But this contradicts the assumption since there is a sequence of zeros tending to a zero. So f is constantly zero on Omega?

Have you looked at the proof of the theorem?

So we haven't seen a proof of the statement above.

A TA has made this statement as a note on the solutions

@Wave This is circular. The fact that all zeroes of analytic function are isolated follows from the identity theorem.

@onepotatotwopotato hmm okey and what would you then do?

It's a famous theorem. You can find the proof on any CA textbook

Hmm strange I haven't found it in my text book and our TA wrote that's an easy consequence of being analytic. Sounds a bit strange that a famous theorem is an easy consequence of analyticity

@onepotatotwopotato Has this theorem a name?

Theorem 3.7 in Conway. It uses the power series in the proof.

Is there like a free version online because I don't have this book.

dpmms.cam.ac.uk/~agk22/id-thm.pdf

i don't know about conways Functions of one complex variable

There maths.ed.ac.uk/~v1ranick/papers/conwaycx2.pdf

Google is your friend

That an the open mapping theorem are corner stones.

Okey thank you I see. I will take a look at it!

I have been thinking of a way to solve this

$A \neq B$ as they will differ in their first column

any suggestions are welcome!

Does things change if we impose a condition all eigenvalues of $AB^2 < 1$ and all eigenvalues of $BA^2 < 1$

when was first proved that a cubic has 27 lines?

also 3^3=27 which is so satisfying is there a hidden intuitive thing to understand behind that coincidence?

wut

27 lines?

Why does a topological vector space have a completion?

And why is it unique up to isomorphism

lam.edu.ly/ar/images/acadj/issue13/EN02.pdf

that answers the existence question

@Jam where are you seeing that a cubic has 27 lines?

Famous fact about smooth complex projective cubics surfaces.

Ah, how could I have missed that?

I'm still tempted to reply to this comment with $\int_0^{\pi/2}\cos^n(x)\,\mathrm{d}x=\frac{\sqrt\pi\,\Gamma\left(\frac{n+1}2\right)}{2\,\Gamma\left(\frac{n+2}2\right)}$

😈

@robjohn Certainly reveals the structure more.

Above the level of the question, but it does show why the integral is in $\mathbb{Q}\pi$ for $n$ even and $\mathbb{Q}$ for $n$ odd.

of course the $n$ to $n+2$ recursion does that too

But it allows the integral to be evaluated for non integer $n$.

and, along with Gautschi's inequality, shows that the integral is asymptotic to $\sqrt{\frac\pi{4n}}$

@robjohn Thank you very much!

why you gotta go? cause I need you

here inside my soul is where I breathe you

no I don't really know why this road is so cold im alone

but I'll fight for you day and night you know ill bring you home

How can you show that the Hyperbolic distribution arises from a random mixing of normal distributions?

@BalarkaSen I don't have an answer to your visualization puzzle yet

but I do want to mention this

A variation of the square puzzle thingy gives us complementary dense connected subsets of the plane

That is, $A,B\subseteq\Bbb R^2$ with $A\cup B=\Bbb R^2$, $A\cap B=\varnothing$, and $A,B$ each dense and connected

Which is pretty neat, that we can throw dense/codense into the mix

@AkivaWeinberger That's pretty neat

I can give the construction (it's not too far from what we've already discussed) or you can figure it out for yourself

I'm thinking of the piecewiselinear construction you suggested using the topologists' broom, but use spokes at rational vs irrational places

I guess one has to be tricky about how to place the spokes

i.e. what height should it be

Maybe not

Let $Q := (\Bbb Q \setminus \{0\}) \cap [0, 1]$. Let $A = ([0, 1] \times \{0\}) \cup (Q \times [0, 1)) \cup (\{0\} \times (1/2, 1))$ and $B = ([0, 1] \times \{1\}) \cup (Q^c \times (0, 1]) \cup (\{0\} \times (0, 1/2))$.

You mean $Q=\Bbb Q\cap(0,1)$?

What's that last bit for ($\{0\}\times(1/2,1)$)?

Also note that I asked for complementary in the plane, not the square

I was partitioning the square into two complementary subsets which are both dense connected and contains antipodal pairs of corner points.

Didn't realize you asked for the plane.

Ah

Also I need to throw in $(0, 1), (0, 1/2)$ to $A$ and $(0, 0)$ to $B$

Yeah, that works for that

For some reason I thought the last bit made it disconnected, but of course "it's in the closure of the rest" was the main lesson from last time

Yeah

Yeah once you have a path connected dense thing, nothing you add on to it can make it disconnected

So really we just need two disjoint path connected dense things, and do whatever we want with the remainder

hi everyone

what is x bar in this proof

Is it the conjugate of x?

Complex conjugate, yeah

why we need to multiply by it?

yutsumura.com/…

this is the link for the problem.

this is the problem I need to solve.

the system is stable if we have complex pair with zero real value.

so it means I need to prove that A=-AT has complex pair with zero real part.

I came across this proof but I've got stumbled on x bar.

@AkivaWeinberger Seems trivial to upgrade the $A \cup B = [0, 1]^2$ example I gave to a dense connected partition of the whole plane.

So $A^T=-A$. Whenever dealing with transposes, one really useful fact to use is that $x^TAy=y^TA^Tx$

Or, said another way, $x\cdot(Ay)=(A^Tx)\cdot y$

The reason is that $x^TAy$ is a 1x1 matrix (aka a scalar), and taking the transpose of a 1x1 matrix (a scalar) doesn't change it

In this case, since $A^T=-A$, we know $x^TAy=-y^TAx$

What I have is a little annoying to write down, though, so I am interested in hearing what you have.

Another thing we know is that $A$ is real, so $\bar A=A$

@CroCo Here's a question. Is it always true that if $x^Tx=0$, then $x=0$ (here $x$ is a vector, possibly with complex entries)?

@BalarkaSen So here's my construction

Let $S=\{(x,\sin1/x):x\ne0\}\cup(0,0)$

Let $S+q\hat\jmath$ refer to shifting $S$ up by $q$ units

Let $A=\bigcup_{q\in\Bbb Q}S+q\hat\jmath$ and $B=\bigcup_{q\notin\Bbb Q}S+q\hat\jmath$

@BalarkaSen I only vaguely remember the details. Something something Seifert something matrix?

but I'm asking why we need to multiply by the complex conjugate in the proof?

@AkivaWeinberger How is $A$ connected?

@Silent Did you ever get your integral? Contour integration is probably not the approach I would use.

"Need" is a weird word to use. There are lots of things you could try. Turns out that this one works @CroCo

@BalarkaSen Take, say, $S_{>0}$, the right half of the sine curve. Its closure contains $\{0\}\times[-1,1]$. So that's all one component

Now take $S_{>0}+\frac12\hat\jmath$. The closure of that contains $\{0\}\times[-\frac12,\frac32]$.

Those overlap, so those are one component

Basically any two branches that are close enough end up being connected

Um. You have demonstrate union of the closures is connected. i.e., closure of $A$ is connected.

So?

@AkivaWeinberger but in control systems, in state space representation, x and A are always real.

@CroCo Sure. In the picture you shared, though, the goal was to show that the eigenvalues are all pure imaginary

so to do that you need to consider imaginary vectors

OK, your claim is that $S \cup (S + 1/2 \hat\jmath)$ is connected. Which is true.

@BalarkaSen What's the intersection of the closure of $S_{>0}$ with $A$

I understand.

thanks.

@BalarkaSen Right, yeah

I didn't even need to translate by rationals and irrationals, I could've done by integers and nonintegers, honestly

Right.

@BalarkaSen Here's another construction

Let $(a_n)$ and $(b_n)$ be enumerations of disjoint countable dense subsets of the plane

@AkivaWeinberger I am trying to understand that definition in contrast to the definition using the "Alexander module". Infinite cyclic cover, homology stuff.

Wait what I was about to say doesn't necessarily work

@BalarkaSen I guess a loop on the Seifert surface should correspond to a transformation of the cover?

Actually it's a meridian of the knot, which generates $H_1(S^3 \setminus K) = \Bbb Z$. Unfolding this meridian to a $\Bbb R$, you get an infinite cyclic cover of the knot complement.

Equivalently, it's like cutting the knot complement along the Seifert surface, and then pasting $\Bbb Z$-many copies of that end-to-end along the two Seifert surface boundaries.

(Cutting along the Seifert surface gives two boundary components because the Seifert surface is orientable -- funny to speak about boundary of noncompact manifolds, btw)

So how do you turn that into a module

Ah yeah then it's what you said. Let $t$ be the meridian; there's an action of $t$ on $S^3 \setminus K$ by "translating the $\Bbb Z$-many copies of $S^3 \setminus \Sigma$" ($\Sigma$ = Seifert surface) by adding $+1$ to the integer label each has. AKA, deck transformation of the cover.

Let's call this infinite cyclic cover $Y$.

Then $t$ acts on $H_1(Y)$. So does $t^{-1}$. This makes $H_1(Y)$ a $\Bbb Z[t, t^{-1}]$-module.

If this module was of the form $\Bbb Z[t]/(P(t))$, say, then I am told $P(t)$ is the Alexander polynomial. The general case I still have to work out.

@BalarkaSen Action of $t$ on $S^3\setminus K$ or on $Y$?

$Y$, thanks.

$H_1(Y)$ is like a quotient of $\bigoplus_{\Bbb Z} H_1(S^3 \setminus \Sigma)$. And $H_1(S^3 \setminus \Sigma) \cong H_1(\Sigma)$, by Alexander duality. Intuitively: cycles in $\Sigma$ can be pushed off $\Sigma$ by pushing them normally up using the oriented unit normal vector to the surface $\Sigma$. This gives rise to cycles in $H_1(S^3 \setminus \Sigma)$.

So $H_1(Y)$ is a quotient of $\bigoplus_{\Bbb Z} H_1(\Sigma)$. So it remains to see how to translate the $t$-action to that of the $\Bbb Z$-many copies of $H_1(\Sigma)$. On the indices, it acts by adding $+1$.

@BalarkaSen What do I get if I take an element of $H_1(\Sigma)$ and push it off $\Sigma$ the other way (the other unit normal)

Yeah, that ought to be the action by $t$

I think it is this: $\pi_1(Y)$ is certainly a normal subgroup of $\pi_1(S^3 \setminus \Sigma)$ because $Y \to S^3 \setminus \Sigma$ is an abelian cover, and $t$ is an element of $\pi_1(S^3 \setminus \Sigma)$. The deck transformation action corresponds to acting by conjuguation by $t$.

So $t$ acts on $\pi_1(Y)$. Abelianize this. $t$ acts on $H_1(Y)$.

Normal subgroup of $\pi_1(S^3\setminus K)$, not $\pi_1(S^3\setminus\Sigma)$, no?

Sorry, $K$ is indeed what I meant.

In fact, $\pi_1(Y)$ should be the *commutator subgroup of $\pi_1(S^3\setminus K)$

Right.

In other words, the kernel of $\pi_1(K')\to\Bbb Z$

So we're looking at $[G, G]/[[G, G], [G, G]]$ lol

where $G = \pi_1(S^3 \setminus K)$.

The abelianization of the commutator subgroup?

Yeah.

Makes me wonder if you can extract more invariants out from the derived series or whatever it is called.

Doesn't this all go out the window for links?

'Cause $H_1$ of a link is $\Bbb Z^{the}$ number of components

$G > [G, G] > [[G, G], [G, G]] > \cdots$, the thing which terminates iff $G$ is solvable

@AkivaWeinberger I guess $[G, G]/[[G, G], [G, G]]$ still makes sense.

Lol

Something something group cohomology

The geometry of the maximal abelian cover gets more confusing

@AkivaWeinberger Anyway yeah this is correct, by just thinking about the infinite cyclic cover in this case. How do I get the usual Seifert matrix definition out of this rigmarole though?

I suppose you oughta read Alexander's original paper

Basically because in $Y$, for every adjacent pair of $S^3 \setminus \Sigma$ pieces, there's an inward normal for the shared Seifert surface $\Sigma$ coming from one piece, and outward for the other.

so like if $h$ is a homology class of $\Sigma$, and $h_-$ and $h_+$ are the two pushoffs, then like $(h_-,t)\simeq(h_+,t+1)$ in $H_1(Y)$

or something

Yeah, that's correct.

so the real question is how do you relate $h_-$ and $h_+$ in $\Sigma'$

I mean, $S^3\setminus\Sigma$

Sigman't

Right. Have to chase some identifications, because $H_1(S^3 \setminus \Sigma) \cong H_1(\Sigma)$.

There's two isomorphisms basically.

What was the Seifert matrix again? Pick $\alpha, \beta \in H_1(\Sigma)$, push $\alpha$ up by the unit outward normal, compute linking number of $\alpha^+, \beta$?

I think so

and then do $\det(M-tM^T)$ or something?

Yeah

So weird

Alexander must have had a lot of free time.

Looks like it's $t^{-k/2}\det(M-tM^T)$ where $k$ is the order of the Seifert matrix $M$

That's just to normalize so that the Alexander polynomial is symmetric, I think

The Seifert matrix is probably unimodular. I think?

Wait, is that $\det(t^{-1/2}M-t^{1/2}M^T)$?

Remind me what unimodular means?

As in, pick a basis $\alpha_1, \cdots, \alpha_k$ of $H_1(\Sigma)$. Then for any other loop $\alpha$, $\alpha = \sum_i \ell(\alpha^+, \alpha_i) \alpha_i$.

$\ell$ = linking number

thinking in terms of the billinear form here

@AkivaWeinberger That is to say, it's a form $H_1(\Sigma) \times H_1(\Sigma) \to \Bbb Z$. Does that give you an iso $H_1(\Sigma) \to H_1(\Sigma)^*$?

Dual in the sense of abelian groups

Some kind of Poincare duality

Not sure

Maybe not

Here's a question I thought of a while ago

Let $K$ be an oriented knot with a specified crossing

We can denote by $K_+$, $K_0$, and $K_-$ the three possible resolutions of that crossing

Easier to see for links instead of knots. Consider the Hopf link but linking number of the pair of components is 2. Then you can choose an oriented Seifert surface which is homeomorphic to $S^1 \times I$ (Mobius strip with 2 full twists). This has a single homology class, the central circle $C$. A normal pushoff of $C$ is going to link with $C$ exactly twice.

So the Seifert matrix is... the $1\times 1$ matrix $[2]$. This is not a unimodular matrix. Lol.

@BalarkaSen Ah whoop

Anyway

Yeah listening

Continue

Suppose we associate every knot and link in the world with a point in $\Bbb R^3$

and say this association (this function from links to R3) satisfies the colinearity condition if $K_0$, $K_-$, and $K_+$ are always colinear

For example, this certainly is satisfied if every point associated to a link is colinear

Is this all that's possible?

Or is there a nontrivial function from links to points satisfying the colinearity condition

From what little I've thought about it, I feel like the answer is probably "no, there's no nontrivial solution"

What about $\Bbb R^n$?

Yeah, you could ask it there. I doubt it would change things

There should not be a "finite dimensional" knot invariant which satisfies the skein relations, maybe

Bonus question: Why do they call these relations skein?

Is there any arrangement of colinearity conditions that is satisfiable in R^n, n>3, but not R^3?

@BalarkaSen Isn't skein a Scots word for a ball of yarn or something?

I wasn't aware of that. A ball of yarn is too complicated a knot

Oh, not even Scots, just an English word

coincidence. they were discovered by Wayne Skein

Skein Wayne, a knot theorist in the day, a rogue vigilante at night

The knotman

Because he's, um, knot a man

but he kept going until the bitter end

rumor has it that the knittler, along with the HOMFLY and an infamous chiral knot, is threading a rather tangled conspiracy against him

On his way down the bunny slope? @leslie

yes

they have a plaque there, like broom bridge

ahh, the old country

@BalarkaSen They had him hanging by a thread, but at the end the three of them ended up in stitches

Neat problem from main

Suppose, in a group, $k$ is the unique nth root of $g$ (that is, $k^n=g$ and if $k'^n=g$ then $k'=k$)

Further, suppose that $gh=hg$

Prove $kh=hk$

In other notation: if $g$ has a unique nth root $g^{1/n}$, then if $gh=hg$ then $g^{1/n}h=hg^{1/n}$

What makes the first statement difficult?

unique seems to say it all, no?

Oh, you're just explaining the def nvm

Yeah that first line isn't the puzzle, that's just me trying to be clear

It's not a difficult puzzle anyway, I just thought it was a cute result