@BalarkaSen is the tangent microbundle related to the tangent bundle?

Seems fairly different

@MatheinBoulomenos If you multiply them they should be $(2)$, not if you square them

right

oh wait, there's a theorem linking the two later on

@MatheinBoulomenos Then you just need that $(2, \frac{1+ \sqrt{65}}{2})$ is itself not principal

or either of them

I mean

@Alex Last summer, I tried to read up group theory and also studied with Leaky because it is very interesting, but then later, chemistry PhD kicks in, and I am currently torn between my passion of algebraic structures and my passion on the chemistry. I have yet to figure out a good way to do abstract algebra and my chemistry PhD at the same time with finite time

which is why so many of my math posts are half baked because I am too eager to share about my incomplete findings and see what's the next step is

@ÍgjøgnumMeg the norm equation a^2+ab-16b^2=2 has no solutions mod $5$

oh and a^2+ab-16b^2=-2, too, there might be units with norm $-1$

@Secret Aren't you hurting your phd? You seem to be doing this stuff for many hours a day, and you are possibly doing neither properly?

......

@MatheinBoulomenos Ah, that's something I should look for lol

@Secret I just mean you should be putting 7 hours a day into your phd right?

yeah, I should... currently, I was a bit sidetracked as on average I only have put 7 hours for onyl half a week instead of a full week

The equation $(2,\frac{1+\sqrt{65}}{2}) \cdot (2,\frac{3+\sqrt{65}}{2})=(2)$ shows that $\frac{1+\sqrt{65}}{2}$ and $\frac{3+\sqrt{65}}{2}$ are inverses in the class group

@Secret So you're doing like 21 hours weeks?

That sounds risky man

I just hope you're not 'productive-procrastinating' where you think this is work, so you don't feel lazy with what matters

@MatheinBoulomenos Hang on, what is the significance of units of norm $-1$?

well, if there's a unit with norm $-1$, it's not enough to check that there is no element of norm $2$ to conclude that an ideal of norm $2$ is not prinicipal

there might be an element of norm $-2$

@Alex Well, one my my problem is that, maths is basically like procrastination to me: Compared to the data analysis in my PhD, doing maths is like browsing facebook. It is so much easier to write down algebraic structures and investigate its properties since all you need is a pen and paper, then the data analysis which is quite tedious because of all that coding, even if the outcomes that came out are quite interesting

Ah I see

I guess it's not so much unit of norm $-1$ but the fact that there may be elements with negative norm in general

@Secret Precisely the problem, you're wasting tonnes of time, you're basically browsing facebook, but telling yourself that it was time well spent

Yeah sure

... I will try to combat my abstract algebra procrastination now. I keep telling myself to go back to the chemistry but I seemed to keep stuck inside maths, I have to find more efficient ways to motivate myself pass the coding part of the analysis...

(Not trying to attack you @Secret, it's just a trap that I've fallen into many times before, and it seems you're in it from my point of view)

@Alex I understand, and it is not your garden variety of procrastination since it is not something meaningless like facebook, but still have the same detrimental effect

I think I am now made aware how destructive is productive procrastination is now...

espeiclaly when the subject of procrastination is actually an academic topic

@Secret To say one last thing before going back to work properly myself: If you study math from a textbook, it'll feel much more like work, and much less like browsing facebook - and you'll learn much faster, but be able to concentrate for much shorter periods (precisely the sort of thing that'll occur with any actual productive time).

@MatheinBoulomenos $7 + 2(\frac{1 + \sqrt{65}}{2})$ has norm $-1$ lol

$(2,\frac{1+\sqrt{65}}{2})^2=(4,1+\sqrt{65},\frac{33+\sqrt{65}}{2})=(4,\frac{1+\‌​sqrt{65}}{2})$, this contains the element $4+\frac{1+\sqrt{65}}{2}$ whch has norm $4$ (the same as the whole ideal), so it is in fact generated by $4+\frac{1+\sqrt{65}}{2}$

no idea what is wrong with the LaTeX there lol

@Slereah the tangent micrbundle is (germinally) equivalent to the tangent bundle if the manifold is smooth

@MatheinBoulomenos That's good then, we can use $2 = (a + b \frac{1+ \sqrt{65}}{2})^2u$ with $u$ a unit

@Alex I'm productive-procrastinating chemistry with math now

:S

@MatheinBoulomenos And the units of $\Bbb Z[\frac{1+\sqrt{65}}{2}]$ look like $\pm(8 +\sqrt{65})^n$ with $n \in \Bbb Z$, and since $N(8 + \sqrt{65}) = -1$, the units with positive norm are $\pm (8 + \sqrt{65})^{2k}$, so you can absorb this into the $(a + b\frac{1+ \sqrt{65}}{2})^2$ term, but that implies $\sqrt{2} \in \Bbb Z[\frac{1 +\sqrt{65}}{2}]$

so the primes above $2$ are non-principal and they generate the class group

yeah

Nice

Thanks for the help :)

np

My bad habit of late is putting too much care into the reference solutions I write up for student HW

If I’m getting more out of it than my students are, that’s probably not great

my tab is full of stuff i want to read but wonder if ever will

it's giving me a panic attack

mine is exactly the same lol

Mine too :'(

And eventually they all close by accident

And are lost to the void

@BalarkaSen rest in peae

Oh the horror

@Alex well, at least with chrome it includes a "recently closed" item in history. so that makes it easier to not lose stuff

@BalarkaSen

double monitor chrome power

yyyyy

holy jesus

hmmmmm

I'm trying to remember. Suppose I've got Laplace's equation on the semi-infinite domain $\mathbb{R}^+\times [0,1]$

Is there a smart way to map that to a finite domain? I want to do some numerical demonstrations via Mathematica but NDSolve requires a finite domain

(In the present case I want the solution to vanish at positive infinity, so as an approximation I could just truncate it and say it equals zero there.)

Hi,

One of the boys :')\

Ryan @0celo7!

I now revoke any claim I thought I had to having too many tabs open

Those are just two of my windows

You haven't even seen my final form

flexes

On that note, I better think about sleeping

okay, I did $u=e^{-x}$. that's not conformal, but it seems to work fine for my numerical purposes

@Secret I see you had some fun with my determinant problem :)

On math.stackexchange.com/questions/261896/…, for the first solution, I understand how Andre got y=7(mod 23), but how does he conclude that y is also = -7 (mod 23)? I tried adding and subtracting multiples of 23 to 7 but never got to -7 OR 16. If anyone could help that would be great

Im referring to this paragraph:

If anyone could help I'd really appreciate it

If $a^2 \equiv b \bmod p$ then $(-a)^2 = (-1)^2a^2 \equiv b \bmod p$

^ Add to that the fact that he's completed the square to get y^2=3 mod 7, thereby removing the linear term in the modular quadratic equation

The (-a)^2 becomes a^2, anyway, no?

That would be the point, yes.

So if $y=a$ is a solution to $y^2\equiv b$ mod p, then what about $y=-a$?

y=-a would also be a solution, since we're squaring the negative

right

So y=7 mod 23 means that y=-7 mod 23 is also a solution

Nice, so this holds for any mod p?

Yup

Great thx

Again, though, he had to complete the square in order to do that

Cuz $(-1)^2 = 1$ with integers means it works mod any integer $m$.

It's easier to see if you write out what it means; $a^2 \equiv b \bmod p$ means $a^2 = b + kp$ for some $k\in \Bbb Z$

^^

so, as @Ted remarks, since $(-1)^2 = 1$ in $\Bbb Z$, the same happens mod $p$

or mod $m$

Doing stuff mod m induces more relations between integers, it doesn't remove them

@TedShifrin not sure you saw, but I did figure out a nice answer to my determinant thing

or nice enough for what I wanted anyways

oh, no, I didn't see ... I'm pleased for you :)

It's surprisingly easy, I was just thinking about it from the wrong direction

So let's see if I recognize your clever solution. What was it?

Sup chat

Heya Eric

So, to restate the problem: I had four 3-vectors $a,b,c,d$, and I form the matrix $M=[a,b]^T [c,d]$

so matrix elements are dot products between the sets {a,b} and {c,d}

So cool that the number of conjugacy classes of $\pi_1(M)$ is a lower bound on the number of geodesics of $M$ (where $M$ is closed)

Oh, sure ... this is a Gram determinant sorta thing.

Were those 2-vectors, then I could say $\det M = \det [a,b]\det [c,d]$

@Balarka: in a fixed homotopy class?

but they're not, so those matrices aren't rectangular and that won't work

But the standard thing to do is to put in vectors orthogonal to the columns to make the matrices square, @Semiclassic.

That's the trick for getting the Gram determinant.

Is it? I didn't know that. That'd probably make things easier.

@Ted I stated a completely numerological fact. But yes, every free homotopy class of loops in $M$ contains a geodesic representative, apparently

With curvature constraints, Balarka, yeah.

Yeah, with $k$ vectors in $\Bbb R^n$, add on an orthonormal basis for the perp, @Semiclassic.

There's no curvature constraints. I think it's just easy to prove if $M$ is negatively curved (by passing to $\Bbb H^n$ universal cover)

Well, it's false on a sphere, @Balarka, isn't it?

What I did (eventually) was recognize that $$\det{M}=\begin{vmatrix} a\cdot c & a\cdot d \\ b\cdot c & b \cdot d\end{vmatrix}$$ is just the scalar quadruple product

The sphere has no nontrivial free homotopy class of loops

I guess I didn't state "nontrivial"

sorry

But there are still geodesics!

and the identity for that is just $\det M = (a\times b)\cdot (c\times d)$

Cross products won't generalize unless you have $n-1$ vectors, sadly, Semiclassic.

Sure, that doesn't contradict what I stated

Yeah

Every nontrivial free homotopy class of loops in $M$ contains a geodesic representative

Though for now I'm perfectly happy with 3-vectors

But this is the usual way of showing in diff geo that $\sqrt{EG-F^2}$ gives the fudge factor for surface area.

If $\pi_1(M) = 0$ the theorem just doesn't apply

@TedShifrin nice

That should follow immediately from what I said, @Semiclassic. Let me check.

@TedShifrin I was about to say, yeah

i mean, a-cross-b and c-cross-d are exactly what you'd need to get those extra orthogonal columns

But the clever thing is to swap them in the two matrices.

@Ted withdrew considerations from the REUs I applied for that didn't send stuff out yet and declined one today. So looks like my summer is locked into Chicago again

hmm

Oh well, Eric. :(

Oh. In that 'future opportunities?' vein

The only offer I got just wasnt enough money to make it viable over staying unfortunately

There was a paper I wrote with my advisor and a visiting prof a while back that came out nicely

Sorry to disturb again, but how can I solve a congruence like x^2=k(mod p)? I have a very large non-prime p, and a large k, so I can't use inspection. It's part of a larger problem I'm trying to prove by contradiction

Since you have two different sets of vectors, you want to put $c\times d$ in the $a,b$ matrix and vice versa

Even if it maybe would've been fun and different

We were doing things using an approximate/asymptotic method in order to avoid the deep math involved

About a year later, another paper came out which went at the deep math full force (Fredholm determinants oh my) and corroborated our results in a nice way

Hmm, hold on, Semiclassic. I'm no longer convinced.

Well, apparently those guys have put out notice for a postdoc position on that very topic

So that's a very unexpected possibility that I hadn't at all considered

Oh, I see. My proof works. You need to check that the volume of the parallelepiped spanned by $a,b,c\times d$ is $(a\times b)\cdot (c\times d)$.

Heya DogAteMy

@TedShifrin hmm, nice

One rather significant aspect of that postdoc position: It's in France.

So are you learning French?

I guess I'd have to?

Practically it means that if I even want to interview I'd need to get my passport

Well, you should have a passport irregardless.

yeah

Where in France?

LPTMS-Orsay and LPTENS-Paris

Oh, Orsay is just a 30-minute train ride from downtown Paris. I spent some time there years ago. Very cool.

ahh

it says one year at each, so two years total

not a small commitment

Formidable!

yeah

It'll probably be super-competitive.

I'm not sure how I feel about it. On the one hand, the topic is one I've had some direct contact with through that paper, and it's one I'd be interested in.

Yeah

I mean, there's two levels to this

Hmm, two years in France with money. Yumm.

(1) Should I interview for it? (2) If I were offered it, should I accept it?

Would you get good letters of support from your folks here?

From my advisor, I'd say yes.

Would need two more, though

hey guys, how would u find max path in 2d array? so you have starting and end points.You have k specified path length.You can move horizontally && vertically.Each cell contains positive number.

I think it's pretty unlikely that I'd get it if I'm honest.

I say yes to (1) and (2) ... but you have to be committed to pursuing research. You sound like this really interests you.

Yes, it is unlikely, but you should still go for it.

@sparrow2: There are algorithms for such things, but I sure don't know anything about it.

It does. The problem underlying (2), beyond just the stress that such a transition would entail, is that I remain skeptical about staying in academia.

Thing is, of course, that (2) is a bridge that I can cross when I comem to it.

Even if you don't, this will be a great experience and will look great on your résumé.

Applying for it?

oh, staying in academia

I mean ... France! Out of Trompolini hell for a few years.

hah, yeah

that does sound appealing

I'm coming to visit :)

And then when it ends the 2020 election will have happened

So I'd get to decide at that point whether I need to stay the hell out of the US for longer :P

I vote that you throw caution to the winds and go for it.

In any case, because I don't have a passport it'd be about a month before I could go for an interview.

You can expedite the passport if you pay extra bucks.

Right.

Get on it!

I do feel like I want to apply to it, at least. Even if I don't succeed (and I think it's unlikely I would) it'd be a good kick in the butt.

You know my vote.

Right.

I'm not going to be able to start the passport process until at least Monday (the passport office I stopped by at was way too busy and I wasn't in a position to wait forever)

But I can at least get started on thinking about this.

I'll probably want two of my letters of recc to be my advisor and the visiting professor who we collaborated with (who both received the ad for the postdoc position)

Wow, this is an overly specific lemma, I don't think I'll need this one again: "If $R \hookrightarrow A$ is an integral extension of one-dimensional Noetherian rings and $A$ is equidimensional and a finite product of domains, then $R$ is also a finite product of domains"

Not so sure about the third yet.

Anyways. Let me make sure I can reproduce your method for the determinant problem, since this is something I want to be able to a prof I know

Use my proof :)

Oh, that's what you said.

If you'd asked me a few years ago, I would have added this as an exercise to my linear algebra book :)

hahhh

fyi, another version of this problem would be to do $[a,b,c]^T [a,b,c]$ i.e. three sets of identical vectors

that case is easier, since they're square to begin with

That's the standard Gram determinant I mentioned.

Right.

That's classic and well known.

Hm, I am forgetting how what I said is proved for negatively curved spaces. $M$ be a negatively curved manifold; then Cartan-Hadamard says I get the universal cover $p : \Bbb H^n \to M$. If $\gamma$ is a closed loop on $M$ with $[\gamma] \neq 0$, it lifts to a path $\tilde{\gamma}$ on $\Bbb H^n$. Now cojoin the translates of $\tilde{\gamma}$ under the $\pi_1(M)$-action on $\Bbb H^n$ by deck transformations, endpoint-by-endpoint

This should (I think) give a long ass path $\iota$ on $\Bbb H^n$ limiting to two unique points $x, y \in \partial \Bbb H^n$. So $p(\iota) = \gamma$. Let $\gamma_g$ be the unique geodesic between $x, y \in \partial \Bbb H^n$. $h_t$ be the isotopy of $\Bbb H^n$ taking $\iota$ to $\gamma_g$. This gives an isotopy of $M$ taking $\gamma$ to $p(\gamma_g)$, I think, which is a geodesic representative.