Mathematics - 2019-01-18 (page 1 of 2)

12:00 AM

Like, the image above is drawn from an underneath perspective for a reason

well the resolution I suggested is that "the question is undefined", but I have another resolution now (typing)

a red paint and a blue paint mixed together then you see purple, and this occurs when the particles responsible for red and the particles responsible blue end up so close together that they produce a mixed effect

the same would happen here, as the separation between green and yellow become too little for our eyes to determine, and then too little for photons to determine

Akiva Weinberger

Ah neat that makes sense

Leaky Nun

ok I really need to sleep now, see you

Akiva Weinberger

Same

Secret

Physically, if it gets too thin, the layer of atoms that made up that layer will be too thin to even reflect visible light, and hence will become transparent

25 mins ago, by Rithaniel

Though, you could argue that the end result would be an artifact of the nature of the infinite process, in which case you'd need something like the Ramanujan sum to assign values to divergent series.

Kinda relevant to this in that there is no reason the infiniteth step has to behave like some smeared out version of the other steps. It also explains why in transfinite induction, we need to prove the limit case separately

Because technically we cannot reach it from beneath

Akiva Weinberger

12:10 AM

I have no idea what an orbifold is but I like the quote featured on the Wikipedia article

> This terminology should not be blamed on me. It was obtained by a democratic process in my course of 1976–77. An orbifold is something with many folds; unfortunately, the word "manifold" already has a different definition. I tried "foldamani", which was quickly displaced by the suggestion of "manifolded". After two months of patiently saying "no, not a manifold, a manifol*dead*," we held a vote, and "orbifold" won.
Thurston (1980, section 13.2) explaining the origin of the word "orbifold"

Secret

12:22 AM

manifol dead lol

Akiva Weinberger

Oh lawd more trefoils

(From the same guy who made the cube thing)

Secret

Looks like someone drilled a lot of trfoil shaped holes on some star polyhedron

Akiva Weinberger

From this guy

rinusroelofs.nl

Rinus Roelofs

Rithaniel

Oh, that's a cool one.

Secret

rinusroelofs.nl/projects/elevation-inf/pr-elev-inf-055.html

Akiva Weinberger

12:33 AM

@Secret If you click on the up arrow on the left, you can see the whole series

Secret

The non interwoven versions are probably common sights in the 4D world

A "trefoil prism"

Rithaniel

(Also, talking about how vision physically works is kind of side stepping the question. Physically speaking, such a cake isn't even possible, so we're naturally talking about things in an idealized world, and trying to evaluate what would the "uppermost layer" of the cake actually be.)
Can you buy these things as sculptures?

(Well, it wouldn't be anything, obviously, but we can still talk about how it might behave)

Akiva Weinberger

The problem is that our idealized world isn't well-defined

Rithaniel

Really, I like the question more so for the thoughts that it brings up more any potential answer.

Secret

We literally need to axiomise their existence for example, I don't think we humans really understood infinity

Akiva Weinberger

12:37 AM

I don't think so but Henry Segerman does a similar thing and sells 3D printed things on Shapeways

(re: buying the sculptures)

Also I'm pretty sure some are renderings

Rithaniel

Yeah, that trefoil prism is almost certainly a render.

Akiva Weinberger

Rithaniel

Whoa, trefoil torus?

Fargle

Suddenly I want these as car tires

Akiva Weinberger

And this one is an actual sculpture l

Reminds me of some Séquin/Collins stuff

Collins and Séquin also made this snow sculpture called Whirled White Web

Mike Miller

12:49 AM

@AkivaWeinberger You know how if you take the $\Bbb Z/2$ action on $S^2$ by reflection across some great circle, and take the quotient, you get a disc? Because everywhere but the equator, you just identify a neighborhood with its (disjoint) antipode, and near the equator, you fold your neighborhood in on itself

You could, just for fun, write down that the quotient is "$D^2$, where the boundary comes from quotienting a flip action"

Akiva Weinberger

(They won second place; compare with first and third place, which are much less geometric/Modern)

Mike Miller

That's an orbifold

Akiva Weinberger

@MikeMiller If I take the quotient I get a projective plane

Mike Miller

reflection along a great circle, not the antipodal map

Akiva Weinberger

Oh

So it's a manifold with boundary with some extra information around the boundary?

Mike Miller

12:52 AM

Ah so that specific example doesn't give all the intricacies, that's just a very simple case. Whenever you have a finite group $G$ acting on a manifold, $M/G$ is an orbifold

An orbifold has charts that look like $\Bbb R^n/G$, where $G \subset O(n)$ is a finite set of linear isometries

Akiva Weinberger

Interesting

Mike Miller

Part of the data includes that stabilizer $G$ at each point (zero on the interior in the disc example above but $\Bbb Z/2$ on the boundary), as well as its action on the tangent space "upstairs" (though there might not always be an "upstairs")

Akiva Weinberger

Interesting

So I was told that this thing has to do with orbifolds

^Same object topologically

and I'm not quite sure I see how that works

Mike Miller

Yeah I dunno what they mean by that

A construction I like which I'm sure is totally unrelated is the following

You can encode a pair of a 3-manifold and a knot inside it as an orbifold: given $(Y, K)$, use the usual manifold charts on $Y \setminus K$; on a neighborhood of $K$ diffeomorphic to $S^1 \times \Bbb R^2$, consider that second factor as being $\Bbb R^2/(C_2)$, where I'm acting by the antipodal map here, so this is again homeomorphic to $\Bbb R^2$

This produces an orbifold with the same underlying space as $Y$, so that the "points with stabilizer $C_2$" form the subspace $K \subset Y$"

There's something nontrivial here in that oftentimes if you can do some construction for a 3-manifold, and make it respect symmetry of some sort, you can do it for orbifolds

So that tells you that you can do it for knots as well

There's something similar for embedded trivalent graphs

Akiva Weinberger

It feels like you've just put 720 degrees of space around the knot

Rithaniel

1:09 AM

I can kind of envision the links phasing through each other as they move horizontally. Though, makes me wonder, if you make that 3D shape hollow, is it possible to topologically turn it inside out?

Mike Miller

@AkivaWeinberger That's about right, or rather you've remembered there used to be 720 degrees of space around the knot

Akiva Weinberger

@Rithaniel Which one? The double cube?

Rithaniel

Yeah, that one.

Mike Miller

You can also take $Y \setminus N(K)$ (delete a nbhd of the knot) to get a compact manifold with torus boundary. Take the double cover in which twice the meridian of the knot lifts, but the meridian fails to; this still has torus boundary, so glue in the solid torus again. This is called the branched double cover of $K$ and is I guess sometimes denoted $\Sigma(K)$

it has an involution given by swapping the two sheets of the double cover, and on the solid torus $S^1 \times D^2$, it's the identity on the first piece and $z \mapsto z^2$ on the second piece; so the fixed point set is just the knot itself, and the quotient is $Y$ again

if you think of $(Y,K)$ as an orbifold as above, it's the same orbifold structure as $\Sigma(K)/(\Bbb Z/2)$

Akiva Weinberger

@Rithaniel As in, without passing through itself?

I think… yeah

Just like you can turn a sphere with a hole punched in it inside-out

The trefoils will stay the same handedness though

Like, a left-handed trefoil will stay a left-handed trefoil and a right-handed trefoil will stay a right-handed trefoil

Rithaniel

1:18 AM

Ah, yeah, allowing for passing through itself. I'm thinking of the faces being spheres and the bridges between surfaces being hollow tubes. So the structure would be without holes.

Akiva Weinberger

(It's been proven that the trefoil can't be deformed into its mirror image. Hard puzzle: try to prove this. So there's technically two trefoils.)

@Rithaniel Ohh

Then I'm not sure I see what transformation you're imagining

@MikeMiller Oh, I never did know what "branched" meant for that

Rithaniel

This video is terrible, but it gets the idea across: youtube.com/watch?v=-tj190Lcw48

Akiva Weinberger

That's not terrible, it's one of my favorite videos, and also that link only has the first part

(The same people also made "Not Knot")

But yeah you want to evert it then

I think, topologically, it's a 13-holed torus when you view it like that

Rithaniel

Well, it's terrible as far as video quality. Audio is out of sync and the resolution is capped at 240p

Akiva Weinberger

Oh well there are other links

I believe you can evert any surface no matter its genus

Rithaniel

1:25 AM

(Yeah, I couldn't find a better video off hand.)

Akiva Weinberger

Torus eversion: turning the torus inside out

►

^Example with torus

Dunno how it works for a double torus. I imagine you first morph it into a torus with a tiny handle

and then do the motion in the video and just let the tiny handle go along for the ride

and then I suppose you end up with a torus with a handle facing the wrong way at the end, but a handle facing the wrong way is the same as a handle on the outside rotated 90 degrees to the side

So yeah you should be able to evert it

Rithaniel

Alright, then that's pretty awesome. (Never heard the term evert before)

Akiva Weinberger

Other related terms: immersion, regular homotopy

Basically when you have a surface intersecting itself in space, that arrangement is called an immersion

Rithaniel

Ah, like the Klein bottle in 3D space.

Akiva Weinberger

Technically it's a smooth map from the surface to the space such that the tangent planes don't collapse into lines anywhere

Yeah

And a regular homotopy is basically a transformation from one immersion to another immersion

so that it stays as an immersion all along the way (doesn't develop any creases for example)

See the last paragraph of the introduction here

Rithaniel

1:32 AM

What if it somehow stops self-intersecting along the way?

Akiva Weinberger

That's fine, if it doesn't intersect itself it's still an immersion

It's just that immersions allow self-intersection

If it doesn't allow self-intersection then it's called a (smooth) embedding

Rithaniel

Okay, I'll add these concepts to my volcabulary.

Akiva Weinberger

These concepts come from differential topology

which studies "differential manifolds"

Rithaniel

Maybe I can take a course on differential topology down the line, because this stuff is so freakin' cool.

Akiva Weinberger

Smale actually found an existence proof of the sphere eversion before an actual explicit eversion was found. I looked at Smale's original paper, and I didn't understand a word because it used a lot of concepts that I didn't understand

but he basically proved that there was an eversion without finding what the eversion was

Insane stuff

I have a theory on how these abstract concepts and high-powered tools come to be

Rithaniel

1:43 AM

Yeah, I recall something about that. It blows my mind that you can prove that something exists without actually finding it. Kind of like how showing the Riemann hypothesis to be undecidable would actually prove it true.

Akiva Weinberger

which is, sometimes you have a concept X that you're trying to study, and so you develop a tool to study it

and you also have a concept Y that you're trying to study, and you develop a tool to help you there, too

So someone comes along and says, hm, what happens if we try to apply the X-tool to concept Y? Even though that's not what the X-tool was designed for

Rithaniel

and you consider the intersection of X and Y?

Akiva Weinberger

and probably usually it doesn't work, but sometimes it does work and you have no idea why

and you discover new things about Y that you didn't know before

and new things about the X-tool that you didn't know before

and so you try to get to the bottom of this, try using the Y-tool on concept X, maybe try using the X-tool and the Y-tool on each other

and eventually you understand X and Y so much better now because the X-tool just happened to work on concept Y

so then eventually they write the textbooks on concept Y

and the textbook tells you what concept Y is, and it tells you what the X-tool is and how it can be used on concept Y

and the student is left scratching their head, thinking, "How did anyone ever think to come up with such a strange-looking tool?"

Because they don't know that it was originally developed for a completely different thing

Does that make any sense?

I might be spouting gibberish to be honest

Rithaniel

Yeah, absolutely.

Akiva Weinberger

Absolutely making sense or absolutely spouting gibberish

Rithaniel

1:50 AM

Though, I think it's more a matter of regression, people striving to simplify things to the least common elements among all these tools people have developed.

Absolutely it makes sense (sorry, should be more clear.)

Akiva Weinberger

Right yeah so all of the elements of the X-tool that aren't relevant to concept Y get stripped away

and now it's even less obvious to the student how anyone came up with that tool

Rithaniel

Yeah, like talking about real numbers and integers and matrices, people trying to unify their study, they come up with the more basic types of structure, and start studying the structures themselves.

Akiva Weinberger

Like rings and fields

Rithaniel

Exactly

Category theory is the most direct form of this that I know, but I know next to nothing about the field except that it's one of the most abstracted forms of math.

Akiva Weinberger

Same

Rithaniel

1:54 AM

You know, I had a thought about Godel's Incompleteness Theorem.

Akiva Weinberger

In knot theory there's something called the "Jones polynomial"

which was discovered in the 1980s, basically a century after knot theory was founded

Rithaniel

So every system is going to have hole in it: statements which the system can't decide. But could you say that, for every statement, there exists some system which is capable of deciding it? Is there a statement which is undecidable in any system at all?

Akiva Weinberger

and Jones discovered it completely accidentally - he was studying some stuff in statistical mechanics when he stumbled upon an equation that seemed similar to an equation in knot theory

and so after a bit of work he was able to get the ideas from statistical mechanics to work in knot theory, and the result was the Jones polynomial

and then this guy Kauffman came along and found a way to describe the Jones polynomial in a much simpler way, without having to use statistical mechanics to define it

and in retrospect the Jones polynomial kinda seems obvious, Kauffman's definition is actually really simple, but people didn't discover it for a hundred years until Jones

Rithaniel

What's the Kauffman definition?

Akiva Weinberger

@Rithaniel You can't prove the consistency of anything stronger than your metalogic

@Rithaniel I dunno how well I can describe it without drawing pictures

Well actually I have a picture somewhere but I dunno if it's enough without context

Secret

2:01 AM

16 mins ago, by Akiva Weinberger

and probably usually it doesn't work, but sometimes it does work and you have no idea why

Akiva Weinberger

A knot diagram is a picture of a knot. There are certain moves called "Reidemeister moves" that turn a diagram into another diagram without changing the knot

Secret

Pretty much how I generate all those weird ideas

Akiva Weinberger

These things

Secret

My current challenge is: Figuring out why it works and then explosively generalising from that understanding

Akiva Weinberger

So what Kauffman does is, define a function $\langle K\rangle$ on the knot diagram satisfying those things I wrote in the box in that picture

Secret

2:02 AM

And the holy grail is to transcend creativity itself by directly create something that is unknown unknown in order to understand when the next technological breakthourhg will occur and how

Akiva Weinberger

which you interpret the first line of the picture as, if three knots are the same everywhere except for a small piece, where one of them has a crossing like that, one has a crossing like that, and one has no crossing like in the picture, then the bracket of the first is $A$ times the bracket of the second plus $B$ times the bracket of he third

(The second line is just the first line but rotated ninety degrees, so they're really the same)

The question is, can you choose values of $A$, $B$, and $C$ such that it doesn't change under the Reidemeister moves?

And it turns out you can get it so that it doesn't change under Reidemeister moves II and III

and unfortunately you need to put in a corrective term to make it not change under Reidemeister move I

Rithaniel

Knot theory is like category theory to me. I have only the smallest degree of knowledge on it.

There's still so much stuff I have to catch up on.

Akiva Weinberger

If I was sitting next to you with a pencil and paper and able to draw then I would probably be able to explain it better

I dunno how much you can get from staring at that one picture I drew

Rithaniel

Is this sort of fundamental knot theory stuff?

Akiva Weinberger

It doesn't need a lot of prerequisites

but there's lots of stuff that doesn't build on this

Secret

2:07 AM

Is there actually a maths domain that study when putting X-tool Y-tool X and Y together, which of these works, which of these don't work and the underlying reasons and mechanisms that governs those relations?

Akiva Weinberger

(like, for example, the theory of Seifert surfaces)

(which we were discussing a while ago)

@Secret That would be a really powerful tool.

Rithaniel

I would wager a guess that category theory would be the best bet there, @Secret, but take that with a grain of salt, because my interpretation of category theory could be very incorrect.

I've gotta look up "metalogic" now.

Akiva Weinberger

That might've been the wrong word, it might've needed to be "metatheory"

Not sure

But basically when you're studying logic systems you have to separate the logic system you're studying and the logic system you're using to reason about it

Rithaniel

Well, it seems that metalogic is directly related to this sort of stuff. It seems a good place to start, at least to get an introduction to these concepts.

Akiva Weinberger

See also metalanguage

when you're using a language to talk about a language

Rithaniel

2:13 AM

I like to say to people "I enjoy things that give me a headache."

Akiva Weinberger

I never finished the book Gödel Escher Bach but it talked about this sort of stuff

Secret

@Rithaniel For me, I enjoy explosively generalising anything I came across

Akiva Weinberger

Have you heard of it?

Secret

nope

Let me check...

Fargle

Hofstadter writes some neat stuff. I first got introduced to his work through his most recent Surfaces and Essences

Rithaniel

2:15 AM

I have not. Currently I'm reading a couple of textbooks for classes, but if Godel Escher Bach is good for relaxing, I could pick it up for the summer.

Secret

the intro of that book sounds like something right up my alley

Fargle

I do, however, find it hard to assess to what degree his takes are meaningful versus to what degree he's, so to speak, smelling his own farts.

Akiva Weinberger

Yeah check it out when you can

Alright, I should go to bed now

Night

Rithaniel

G'night Akiva

Fargle

Goodnight

Rithaniel

2:17 AM

I should probably go do some stuff, too.

Secret

probably should get back to my PhD stuff

(Chats prepares to go to sleep)

also gdnight Akiva

Piyush Divyanakar

7:05 AM

I have a homogeneous pde with characteristic equation that gives complex roots. How do I handle that

Ex $$\frac{\partial^2 z}{\partial x^2} + \frac{\partial^2 z}{\partial y^2} =0 $$

Leaky Nun

7:43 AM

$A=Q^tDQ$, $\langle v,Aw\rangle = \langle Qv, DQw\rangle$

Vrouvrou

12:17 PM

Any help for this please: math.stackexchange.com/questions/3078125/…

nbro

12:34 PM

What does this notation mean $(X, Y) \sim \mathcal{N} \left( {0 \choose 1} , { 1 \, 0 \choose 0 \, 2 } \right)$, where X and Y are random variables?

I think this has something to do with the joint distribution of X and Y and it is related to the mean and variance and the covariance. But what is what?

student

it's a bivariate normal distribution with mean $(0, 1)$ and covariance matrix as given, I'd assume

normal, $\mu = (0, 1)$ and $\Sigma = \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}$

nbro

@student is the covariance matrix always given when the joint distribution is specified like that? I guess the answer is no, because the way the variables change together may not be always known.

student

if it's not known then you cannot give the joint distribution

here $X$ and $Y$ are independent in any case

take the famous example of $X$ normal, $W = \pm 1$ with equal probability, $Y = W X$. Then both $X$ and $Y$ are individually normal and uncorrelated, but they are not jointly normal and not independent.

nbro

@student so when you write (X, Y) ~ ... you assume that the joint distribution is known and thus the covariance must also be known?

@student Why do you say that? How do you know they are independent? From the covariance? How can you visually see that?

student

12:50 PM

when you write $(X, Y) \sim \ldots$ the $\ldots$ specify the joint distribution, so I am not sure what you are asking

Secret

0

Q: Is is possible to have a commuting observables only in a single direction?

In quantum mechanics, for two observables to be compatible, successive measurements of the observables, say A and B, should yield the same result as earlier, i.e if we do the measurements with the order $A \to B \to A$, the result from the first A and the last A should be the same, and similarly,...

hmm...

student

it literally says "the pair of X and Y, together, is distributed as ..."

Secret

So let some quantum state be v, then the question is basically saying:

Bv=BABv
Av=/=ABAv

nbro

@student I understood that. I just didn't realize that to specify the joint distribution of two random variables you needed to specify the covariance of these two same variables too

student

@nbro the covariance matrix is one of the parameters of the multivariate normal distribution, that's why you need to give it (the other is the mean)

nbro

12:54 PM

@student Ok. I am just not very familiar with this.

@student But I have another question. The first column and row usually corresponds to variable X and the second column and second row to variable Y, right?

student

If e.g. you have a multinomial distribution, your parameter would be a vector $(p_i)_{i = 1, \dots, n}$ so you would not specify the covariance matrix to write the joint distribution (but you can still compute the covariance matrix from the distribution function, of course)

@nbro do you know how the covariance matrix is calculated from the distribution?

nbro

By looking at that covariance matrix, we can say that X and Y are independent because of the zeros in the upper right and lower left corners, which are the covariances, right? The diagonal entries should be the variances of the individual variables

@student I don't think so. What do you mean from the distribution?

student

let's do the independence first, and there is a very important step in the middle: from the covariance matrix, you can see that $X$ and $Y$ are uncorrelated

that does not necessarily mean they are independent

but $X$ and $Y$ are jointly normal and uncorrelated, which indeed implies independence

(you can see this because the probability distribution function factors)

nbro

@student Hm... I am not sure I understand your explanation. If two r.v.s are jointly normal and uncorrelated, then this implies independence. Why is this the case?

student

@nbro well, what does independence mean?

nbro

1:07 PM

It means P(X, Y) = P(X)*P(Y) or P(X|Y) = P(X). In English, knowledge of Y does not say anything about X

student

so take a multivariate normal (just two variates, say) with no correlations (diagonal covariance matrix), what happens to the p.d.f. ?

you will find that p(x, y) = p(x) p(y) happens

nbro

But how do you get to p(x, y) = p(x)p(y)? I'm not asking you to do it, but I am just wondering how you show it

student

stick a diagonal covariance matrix into the p.d.f.

nbro

@student And you would obtain two univariate normals multiplied?

student

only one way to find out

nbro

1:20 PM

@student Ok, thanks a lot for your explanations!

I am mathematically not very well educated

student

@nbro you might like Jaynes' book, it's quite argumentative and starts from nothing

nbro

I often forget all these concepts and, especially in mathematics, if you do not have a solid understanding of something, then it is very hard to understand a concept that builds on top of the previous one (which very often the case)

Silent

1:41 PM

Any thoughts on this:

1

Q: Determinant definition from recursion

I know this definition of Principle of Recursive Definition from Munkres Topology, Ch 1: Principle of Recursive Definition: Let $A$ be a set; let $a_0$ be an element of $A$. Suppose $\rho$ is a function that assigns, to each function $f$ mapping a nonempty section of the positive integers int...

definition determinant recursion

nbro

Can we perform causal inference on variables which are not random?

Emilio Pisanty

2:16 PM

yo folks, if you've got time

-1

Q: Wat would happen if I choose 1 = 0 when i make a field in algebra?

What would happen if I choose 1 = 0 when I make a field in algebra ? I mean $1$ is the neutral element for $\times$ and $0$ is the neutral element for $+$. So, what would happen if they are the same number?

mathematics

I imagine this has a duplicate on MSE but my searching-for-formulas skills aren't quite enough to find it

Akiva Weinberger

Neat pic

@EmilioPisanty Why did you post this in physics stack exchange and not math stack exchange

nbro

@AkivaWeinberger Maybe it wasn't him who asked the question...

Akiva Weinberger

Also, as a commenter points out, if you do that, then all of the elements of the field are equal to each other

and you end up with a "field" with one element

(I put "field" in quotation marks because technically the definition of a field includes $0\ne1$)

Emilio Pisanty

2:47 PM

@AkivaWeinberger why do you think it was me posting this? Particularly when I'm in the list of close voters

Silent

Please help me with this Rudin proof:

I can't see how $f_i$ continuous at $x_0$, because we are talking about only one sequence of set, converging to $x_0$

mrnovice

how do u find the rss of a linear model when you dont have any data?

Mats Granvik

4:49 PM

$e^x=\lim_{n\to \infty } \, \left(\frac{x}{n}+1\right)^n$
$\log(x)=\lim_{n\to \infty } \, \left(x^{1/n}-1\right)n$

Nicholas Roberts

5:48 PM

Suppose I have the sum $\sum_{n=1}^{\infty}(\frac{1}{3})^{n(1-\frac{p}{q})}$ where $ 0 < \frac{p}{q} < 1$. I have a feeling this sum converges, but can anyone give me solid reasoning as to why?

Zee

It’s a geometric series

Nicholas Roberts

yeah but with non-integer power, so idk how the usual rules apply

the power is not the index

Thorgott

6:16 PM

$a^{bc}=(a^b)^c$

Nicholas Roberts

oh ok, so its a geometric series with ratio $(\frac{1}{3})^{1-\frac{p}{q}}$

Rithaniel

Is it true that $\sum_{n=0}^\infty a^{ln(n)}$ converges for $0\leq a<\frac{1}{e}$?

Yeah, you can show that with the squeeze rule, right?

If $a<\frac{1}{e}$ then $a=\frac{1}{e^b}$ for $b>1$ and then we have that $\sum_{n=1}^\infty a^{ln(n)}=\sum_{n=1}^\infty \frac{1}{n^b}$ and we have the zeta function.

Easy enough.

nbro

8:15 PM

If A is dependent on B, then B is dependent on A, and if A is independent of B, then B is independent of A.

Tobias Kildetoft

8:59 PM

Hmm, one of the students found a really clever solution to one of the exam problems I had not thought of myself.

Alessandro Codenotti

What was the question?

Tobias Kildetoft

The exercise was to show that $x^2+1$ is a prime element in $\mathbb{F}_p[x]$ iff $p\equiv 3\pmod 4$.

This student noted that this is the $4$'th cyclotomic polynomial, and we know from a remark in the book (that we also proved as an exercise in class) that the $n$'th cyclotomic polynomial is irreducible mod $p$ iff $p$ is a generator for $(\mathbb{Z}/n\mathbb{Z})^*$.

(and we know that irreducible and prime are the same for polynomials mod $p$ of course)

Mike Miller

Of course this is equivalent to the statement that -1 is a quadratic residue iff p-1 is singly even

and p not 2

Tobias Kildetoft

singly even?

Mike Miller

4k+2

Tobias Kildetoft

9:08 PM

then you need non-residue

Mike Miller

Sorry, that's what I meant but failed to say :)

Tobias Kildetoft

(they also have that exact statement in the book, and I this is what I expected them to use)

so far nobody has gotten full marks for the last part of this exercise, which is to show that if $p$ is an odd prime and $F$ is a field with $p^2$ element then $x^2+1$ is not a prime element in $F[x]$.

Mike Miller

Misread again

I'm being too sloppy today

Tobias Kildetoft

there are a couple of ways to do it. The one most directly related to the exercise and one more related to group theory

MatheinBoulomenos

@TobiasKildetoft that's pretty immediate, isn't it? $4$ divides $p^2-1$

Mike Miller

9:19 PM

@TobiasKildetoft I was being a little slow. The point is that this is equivalent to the existence of $a$ with $a^2 = -1$, and -

Yeah

One needs that $F^\times$ is cyclic

MatheinBoulomenos

right

Tobias Kildetoft

@MatheinBoulomenos right

so far nobody has even started that direction

Mike Miller

wow!

That's surprising to me

Tobias Kildetoft

Several have claimed that $F$ will be isomorphic to $\mathbb{F}_p[x]/\langle x^2+1\rangle$, without noting that they just showed this to only be the case half the time

MatheinBoulomenos

or you can say that if $x^2+1$ doesn't have a root already in $\Bbb F_p$, then $F=\Bbb{F}_p[x]/(x^2+1)$

Tobias Kildetoft

9:21 PM

yeah, that is the more immediate solution given the rest of the problem

but even those who get that far then confuse themselves over naming the variable $x$ twice

(which really disappoints me)

Mike Miller

Yeah

Tobias Kildetoft

But I take comfort in the fact that I am still going through those whose id put them as having started studying 2015 or earlier, which means that they are "irregular" in some way.

Alessandro Codenotti

Hi @Mathei

MatheinBoulomenos

there was one exercise where they had to compute the Galois group of the splitting field of $x^4+2x^2+2$ over $\Bbb F_3$. A lot of people got it right, but some claimed it was $S_4$ or $D_4$ or $\Bbb Z/2\times \Bbb Z/2$ or even $(\Bbb Z/2\Bbb Z)^3$

hi @Alessandro

Alessandro Codenotti

I was looking for you a couple of days ago

ÍgjøgnumMeg

9:24 PM

Hey y'all

MatheinBoulomenos

despite the fact that they showed that Galois groups of extensions of finite fields are always cyclic in the lecture

Hi @ÍgjøgnumMeg

ÍgjøgnumMeg

I didn't get my application into Heidelberg on time so now I'm not doing a Masters until like 2020

I think

wasting my L I F E

MatheinBoulomenos

damn

ÍgjøgnumMeg

my fault but damn, disappointed

MatheinBoulomenos

there's nothing you can do?

Mike Miller

9:26 PM

Yeah man life sucks

ÍgjøgnumMeg

I rang up to try and appeal and they just said "No sorry, there's no other reason, you just got your application in late"

MatheinBoulomenos

:(

ÍgjøgnumMeg

I have an interview to work at a company near my hometown developing mathematics education software

and it's a fixed term contract until March 2020 so I guess I can work there (if I get it) until Sommersemester 2020 and just.. self-study for a year

MatheinBoulomenos

why can't you apply for Wintersemester 2019/20?

ÍgjøgnumMeg

Well I can, but if I enjoy this job then I guess I can save more money if I work until the summer

Alessandro Codenotti

9:29 PM

What about the applications you sent to other places?

ÍgjøgnumMeg

@Alessandro I haven't sent them yet, and the applications aren't closed yet either

but I'm not sure if I want to go to Frankfurt?

It depends if I get this job on Tuesday; if I do then I'll do that, if I don't I think I'll get my application into Frankfurt and go there instead

MatheinBoulomenos

Frankfurt is the most expensive city in Germany (probably tied with Munich)

ÍgjøgnumMeg

sad

Alessandro Codenotti

Hmm I see

MatheinBoulomenos

Heidelberg isn't exactly cheap either, but not as bad as Frankfurt

ÍgjøgnumMeg

9:31 PM

Can't stand to live at my parents' place for another year or however long lol

ERGH

kms

Alessandro Codenotti

I wanted to study in Göttingen for my bachelor but that didn't work out because of deadlines and language tests, so I decided to do my bachelor somewhere else instead of waiting another year and it turned out well anyway

ÍgjøgnumMeg

How is Frankfurt for Number Theory?

Or Göttingen in fact

Leaky Nun

@MatheinBoulomenos I'm in love with modular representation theory

ÍgjøgnumMeg

I know they both have big names working there

but that doesn't necessarily determine the size of the departments lol

MatheinBoulomenos

@LeakyNun wow really cool

Alessandro Codenotti

9:34 PM

But I wanted to ask whether you took (or know someone who took) symmetric spaces with Beatrice Pozzetti this semester @Mathei

MatheinBoulomenos

no, I don't know anyone who took that

@ÍgjøgnumMeg I'm not sure, but from the website, they have 5 profs working on NT/AG which seems pretty good

Alessandro Codenotti

I see (she's doing the same course in Bonn in the summer semester so I was curious)

ÍgjøgnumMeg

@Mathein Frankfurt seems like the better place, they have a module ANT2 that has a big variety of interesting stuff (I really wanna take Iwasawa Theory)

MatheinBoulomenos

for Iwasawa Theory, Heidelberg seems hard to beat with Venjakob being a big name in that area

ÍgjøgnumMeg

Yeah I know that :(

or I could just go to Oxford/Cambridge or Imperial

MatheinBoulomenos

9:36 PM

@LeakyNun modular representation theory is really cool indeed

ÍgjøgnumMeg

lol

MatheinBoulomenos

these are really good universities ofc

higher in international rankings than German ones, probabbly (haven't checked)

ÍgjøgnumMeg

Cambridge Part III is weirdly specific

they don't have many generally named courses

nothing like "Algebraic Number Theory I/II"

Leaky Nun

@MatheinBoulomenos let's see, it doesn't have any roots, so it would have to have quadratic factors. its derivative is $x^3+x$ which splits into linear factors, so it can't have any common factor, so the quartic is separable. monic irreducible quadratic polynomials are all factors of $x^6+x^4+x^2+1$; we know $x^2+1$ is irreducible and does not divide the quartic; $(x^6+x^4+x^2+1)/(x^2+1)=x^4-x^2+1$ is not our quartic. So the quartic is irreducible and the group is Z/4Z

MatheinBoulomenos

@LeakyNun yeah

Leaky Nun

9:40 PM

do you have a faster way?

ÍgjøgnumMeg

well anyway, I have Serre's local fields arriving soon, I will try and work through as much of this as I can in the downtime lol

Leaky Nun

I think I came up with the separable trick earlier today, for no reason at all

I was thinking about how to write an algorithm to factor polynomials

MatheinBoulomenos

that's pretty fast already, there are other ways to show irreducibility ofc

Leaky Nun

how would you do it?

MatheinBoulomenos

when you try to factor into quadratics, since the only way to write $2$ as a product of two terms is $2=2 \cdot 1=1 \cdot 2$ you can suppose that $x^4+2x^2+2=(x^2+ax+1)(x^2+bx+2)$ then you compare coefficients and get a contradiction

user76284

9:45 PM

Anyone know where I can find the paper cited in this answer? math.stackexchange.com/a/437873/76284. Dana S. Scott. More on the axiom of extensionality, in Essays on the foundations of mathematics.

Leaky Nun

@MatheinBoulomenos did I outsmart you :P jks

MatheinBoulomenos

the contradiction is really quick. you get $a+b=0$, $2a+b=0$ and $ab=2$

Leaky Nun

interesting

so you outsmarted me! :P

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$Mathematics$ Mathematics