Mathematics - 2019-05-14

Ted Shifrin

12:00 AM

That's a pretty standard problem. I recommend a more modern book these days — Friedberg, Insel, Spence.

Daminark

Hmm, not familiar with that one

Balarka Sen

Here's a problem involving similar ideas: if $T$ is a normal operator on a Hermitian space $V$, then $T^*$ is a polynomial in $T$

Ted Shifrin

It was disappointing when we put the simultaneous diagonalization of commuting diagonalizable operators on the algebra qual and NO ONE had a clue. I was livid. (It appears numerous times on previous quals, but students have a way of "praying" things they don't know won't show up.) Only one or two people even mumbled eigenvectors.

That was the last qual I co-wrote before becoming a bum.

Balarka Sen

Huh, strange

It's such a natural result

Ted Shifrin

Basically, the grad students hate/avoid linear algebra. Guess none of them want to do representation theory or geometry or applied math.

Balarka Sen

12:05 AM

I don't understand that sentiment. Linear algebra is so deep and vast. I remember there being a popular saying that one can never know enough of it

Ted Shifrin

I'm not condoning or explaining it.

Balarka Sen

It's everywhere

Daminark

I remember a professor told me once that in math the only things you know how to do are linear algebra and kinda combinatorics, and that you try to reduce other problems to those problems

Balarka Sen

was that Babai

Ted Shifrin

Sounds like it.

Daminark

12:07 AM

Nope, I guess it wasn't quite a professor, more a postdoc working in algebraic topology

Ted Shifrin

Those aren't the only things I know how to do, for sure. And I certainly don't do combinatorics very well.

Balarka Sen

I should learn combinatorics one of these days

Daminark

It was basically setting up to, in order to do topology you usually either put a manifold structure and do linear algebra or some complex to do combinatorics

Ted Shifrin

I'm not sure I consider the "linear algebra" of homology combinatorics at all.

Balarka Sen

simplicial sets :3

Ted Shifrin

12:09 AM

I mean, there is a "combinatorial" structure to the simplicial structure, but ...

Balarka Sen

@Daminark what about model categories tho

:3

Daminark

If you try hard enough that's also one of the two

Balarka Sen

does J P May agree

Daminark

Nah jk idk, but yeah I will say I did not pick out any of the combinatorics Babai taught me in AT, maybe I'm just not there yet or maybe it's only kinda combinatorics, sorta like what Ted said

Never asked him about that specifically

Earth Cracks

The only spaces that exist are CW complexes

Ted Shifrin

12:12 AM

Demonark: They refer to combinatorial structure if you can do things in terms of the simplicial structure. One of the big things when I was in grad school was a Russian paper on the combinatorial way to compute Pontryagin classes.

Balarka Sen

There are big function spaces which aren't CW complexes yet have good homotopical properties

Ted Shifrin

Are vector spaces CW complexes?

Balarka Sen

eg Whitehead's theorem pushes through for Frechet manifolds

@Ted Dami-level joke

Ted Shifrin

I wasn't meaning to be joking.

Daminark

They aren't finite CW complexes but I guess if you fill in a lattice maybe?

Ted Shifrin

12:16 AM

I'm not sure the topology works right, but I dunno.

Balarka Sen

What are we talking about? Vector spaces are cells

Ted Shifrin

I've gotten this far in my life not thinking about infinite CW complexes, so I'll stay content.

Balarka Sen

Oh, infinite dimensional

Ted Shifrin

OK, @Balarka, and in infinite dimensions?

Malice Vidrine

Is anyone here familiar with the Riemann-Roch theorem for curves?

Ted Shifrin

12:17 AM

Yes.

Balarka Sen

@TedShifrin Sure, you can realize it as a direct limit of finite-dimensional subspaces.

Ted Shifrin

Hmmm, how does that work as a CW complex?

For example, the unit ball will often be non-compact ...

Balarka Sen

Give $\Bbb R^{n+1}$ a CW structure in which $\Bbb R^n$ is a subcomplex (this is easy, Daminark's lattice construction works).

Then it's $\bigcup_{n \geq 0} \Bbb R^n$

Still a CW-complex

Ted Shifrin

Hmmm ... This seems to require CH or something.

Malice Vidrine

@TedShifrin - Has anyone written a good "narrative" account of what it's saying? I understand sheaves from a logical point of view, but my broadly geometric experience stops at introductory differential equations and complex analysis; but I'm trying to understand Riemann-Roch against my better judgment....

Ted Shifrin

12:22 AM

What if my vector space isn't $\aleph_0$-dimensional?

@Malice: Phillip Griffiths has beautiful explanations in terms of the geometry of the osculating spaces to the canonical embedding of the curve.

You can find this on pp. 12 ff. of Geometry of Algebraic Curves, Volume I, by Arbarello, Cornalba, Griffiths, and Harris. It's also in Griffiths & Harris.

Malice Vidrine

Excellent, thank you. I'll give those a look.

Ted Shifrin

You're welcome.

Also, look at Griffiths's Introduction to Algebraic Curves. Very classical. Nothing sheaf-y.

Malice Vidrine

The classical stuff is actually my stumbling block :P

Balarka Sen

@TedShifrin Ok yeah the topology is a problem; I had standard $\Bbb R^{\Bbb N}$ in mind, no weird general sort of topological vector spaces. By what I said, a separable Hilbert space has a dense subspace which is a CW-complex. I don't know beyond that.

Ted Shifrin

LOL, @Balarka. I will continue not to think about it.

Balarka Sen

12:27 AM

Me neither. Subtle point. Interesting.

Malice Vidrine

I've done diffeq and complex, but my geometric intuition stopped at first year calculus. But I've done research in categorical logic, because that actually makes sense...

Daminark

So Peter subbed for the first day of Shmuel's AT class and in office hours I remember seeing them have a somewhat similar exchange

Ted Shifrin

Too bad you never learned multivariable calculus or differential geometry. Pretty hard to do higher dimensional geometric stuff without those.

Malice Vidrine

Multivariable I've done.

But not differential geometry.

Ted Shifrin

Geometric intuition has to come from multivariable.

Demonark: Similar to what Balarka and I were just mumbling about?

@Malice: You certainly don't need fancy differential geometry for this, but thinking about curves and surfaces in 3-space builds lots of intuition.

Daminark

12:30 AM

Peter said on day 1 that every group arises as the fundamental group of some space, by just attaching disks along relations to a bouquet of circles. Shmuel was like well that gives you finitely generated (presented?). For a second Peter was like, wait why? Shmuel asked "What's the topology on an infinite bouquet of circles?" and Peter was like oh

Ted Shifrin

Right.

Balarka Sen

Lol

user193319

Claim: If $Y$ is a proper linear subspace of a TVS $X$, then $Y$ has empty interior. Proof: By way of contradiction, suppose $y$ is some interior point of $Y$. Then there is some $U$ open such that $y \in U \in Y$ which implies $0 \in U-y = Y-y = Y$. Let $x \in X$ be arbitrary. Since $\{x\}$ is bounded, there is some $s > 0$ such that $\{x\} \subseteq s(U-y) \subseteq sY = Y$, so $x \in Y$, which is a contradiction.

How does that sound?

Ted Shifrin

I made a comment here about the Hawaiian earring not being an infinite bouquet of circles.

Huh? @user193319?

user193319

What's the problem?

Ted Shifrin

12:33 AM

What in the world is going on with $0\in U-y$, then $U-y = Y-y$, then $Y-y=Y$?

Daminark

Looks fine to me modulo some notation

user193319

I'm working in a topological vector space.

Ted Shifrin

And $U\in Y$ is of course wrong.

user193319

$Y-y = Y-\{y\}$

Ted Shifrin

I know that.

Interior is in terms of $X$, first of all. So $U$ is open in $X$. What is going on?

user193319

12:34 AM

Oh, sure...$U \in Y$ is a mistake.

Daminark

Wait aside from $U-y \subset Y-y$ and $U\subset Y$ what's wrong here?

He's using $A-B$ not as set difference but translating if that's the confusion?

user193319

Isn't it true that $y$ is an interior point of $Y$ if and only if there is some open set $U \subseteq X$ such that $y \in U \subseteq Y$?

Ted Shifrin

So we're supposing $U\subset X$ is open and also $U\subset Y$. So those equals are $\subset$s?

Balarka Sen

@Daminark Well, I don't see why Shmuel's comment is a problem, though. I can still make a cell complex with number of 1-cells and 2-cells being likely some infinite cardinal.

Malice Vidrine

@TedShifrin - I found multivariable strightforward so far as it went, but there must be some dot I never connected. Because complex was gibberish to me, even though I did reasonably well in it, and nothing's coming together for me with the algebraic geometry (I've been reading Vakil's notes and Hartshorne).

user193319

12:36 AM

Oh, crap...I made a lot of notational errors...

Ted Shifrin

No, @Malice. Those are orders of magnitude far more advanced and sophisticated. You need a year of undergraduate algebra and then some graduate algebra to do algebraic algebraic geometry.

Malice Vidrine

I've done a year of abstract algebra.

Ted Shifrin

Doing it with Griffiths's approach you need some complex analysis, topology, and a bit of differential geometry, but it's still very sophisticated.

Oh. Well, you didn't say that.

Balarka Sen

The fundamental group on the bouquet is still free, and the relators get killed anyhow. You just need some transfinite version of van Kampen which is annoying.

Malice Vidrine

I don't think of it as being geometric :P

user193319

12:37 AM

Here's a correction: By way of contradiction, suppose $y$ is some interior point of $Y$. Then there is some $U$ open such that $y \in U \subseteq Y$ which implies $0 \in U-y \subseteq Y-y = Y$. Let $x \in X$ be arbitrary. Since $\{x\}$ is bounded, there is some $s > 0$ such that $\{x\} \subseteq s(U-y) \subseteq sY = Y$, so $x \in Y$, which is a contradiction.

Is that better?

Ted Shifrin

Algebraic algebraic geometry is more algebra than geometry. That's not how I think of it, because I'm a follower of the Griffiths style.

$U$ open in $X$, not in $Y$, @user193319. That's the whole complaint.

user193319

I never claimed that $U$ was open in $Y$.

Ted Shifrin

But you're supposing that an open subset of $X$ is contained in $Y$. Make things clear.

Daminark

Oh I guess for what it's worth I don't know the technical definition of a topological vector space, in normed spaces I obviously see why $\{x\} \subset s(U-y)$ and I'm just taking it on faith here that this translates over to TVS

user193319

I just said it was open in the ambient space.

Ted Shifrin

12:39 AM

Actually, you never said anything of the sort.

This is why I lose patience.

user193319

Or, rather it was implicit that $U$ is open in $X$.

Ted Shifrin

"implicit" ... bah.

OK, I will quit for now.

Daminark

Whoa let's keep it friendly

Ted Shifrin

After I've spent hours and hours helping you, feel free to be a rude ass.

Malice Vidrine

Likewise, I know a reasonable amount of sheaf theory, but I know it as a logician, so thinking of them as geometric is weird to me. I think my instructor for this course was hoping the sheaf theoretic approach would play to my strengths, but it hasn't turned out that way.

user193319

12:40 AM

You never had to help. I was asking help from anyone who was willing to help.

Malice Vidrine

Anyway, I found a copy of Griffths's Introduction to Algebraic Curves, so I'll see if that's better...

N. Maneesh

math.stackexchange.com/questions/242450/… I know this function locally invertible at non zero elements of $\mathbb R^n$ using Jacobian. But proving function injective. I used the standard technique. $f(x)=f(y) \implies x||x||^2=y||y||^2 \implies ||x||^3=||y||^3\implies (||x||-||y||)(||x||^2+||x|| ||y||+||y||^2)=0$

Ted Shifrin

I don't know the right answer for you, @Malice.

Daminark

Algebraic curves seem fun, should probably learn about them soon

N. Maneesh

Form this How do I prove that $x=y$?

Ted Shifrin

12:44 AM

Oh, that works, @N.Maneesh. Unless $x=y=0$ the second factor can't be $0$.

loch

@MaliceVidrine My opinion is that it's probably a good idea to learn some manifolds first to have some sense of 'local properties' vs 'global properties'

Malice Vidrine

@TedShifrin - It's okay, I don't expect that there is one, much less that anyone would know it.

Mostly I'm just trying to track down any resource that has a story about this stuff. Like calculus has the story of looking at infinitesimal slivers of stuff, group theory has the story of symmetries and permutations.

I don't know what the story is here, yet.

Ted Shifrin

I stand by my original suggestion, @Malice, but still you need to know about the canonical embedding (and it fails for hyperelliptic curves).

N. Maneesh

@TedShifrin From that I got $||x||=||y||$, But If $x, y\in $ a same sphere. This happens to be true. Right?

Ted Shifrin

Now put that into your first equality.

Malice Vidrine

12:46 AM

@loch - You're probably correct. What I'm in now is a directed study course described as "applied abstract algebra" for maths seniors, and I have not been pleased that it has turned out to be algebraic geometry.

@TedShifrin - I will definitely check it out. I do appreciate the suggestion.

Ted Shifrin

It gives a concrete intuition, at the very least, @Malice.

N. Maneesh

Okay $x||x||^2=y||x||^2\implies x=y$

Ted Shifrin

Right, @N.Maneesh.

N. Maneesh

Thank you@ Tedshifrin :)

Ted Shifrin

(Assuming $\|x\|\ne 0$, but we already have no problem with $0$.)

N. Maneesh

12:48 AM

mmm

Ultradark

What does everyone think of the new proof by Maryna Viazovska and co.?

loch

Perhaps here is some part of the story:
- Algebraic curves = 1-dimensional algebraic varieties (shapes cut out by polynomial equations) - in some sense they're the simplest (but non-trivial!) shapes you can study. (1-dimensional objects is easier than higher dimensional objects! )

- In studying geometry, one common thing to look at is the collection of functions (satisfying some properties) that you can define on your object. For algebraic varieties, the functions you care about are functions given by polynomials. If you're working over \C, you can think of holomorphic functions instead.

Daminark

@Ultradark haven't heard of it, what was proven?

Ted Shifrin

+1 @loch

Daminark

Ah apparently a few years ago she seemed to solve sphere packing in dimensions 8 and 24 and has now generalized that result a fair bit

Ultradark

12:53 AM

@Daminark They proved that the configurations that solve the sphere packing problem in dimensions $8$ and $24$ also solve an infinite number of other problems about the best arrangement for points that are trying to avoid each other. The points could be an infinite collection of electrons repelling each other and

Ted Shifrin

I don't know how many electrons are in dimension 8 or 24 ...

Daminark

I'm not terribly familiar with sphere packing and related problems to be giving much of my sought after commentary but this sounds cool for sure

Ted Shifrin

Sphere-packing certainly is a classic hard question. I haven't kept up with the progress. Some false results were obtained in the 70s-80s, if I remember correctly.

Daminark

It seems like this is the paper in question: arxiv.org/pdf/1902.05438.pdf

Malice Vidrine

@loch - Are you able to elaborate more on "look[ing at] the collection of functions ... that you can define on your object"? I think part of this is I don't really know what kinds of questions one would even ask about curves or functions on them.

(And I definitely appreciate that outline above. Even that last point about what Riemann-Roch is relating is more than was clear to me before)

GFauxPas

1:06 AM

Well I think I passed

L

I got the questions that were obviously based on material we covered

loch

Hmm maybe an application would help: if you're ever heard of elliptic curves - they tend to be introduced as curves defined by the Weierstrass equation y^2=x^3+ax+b, (in fact its homogenized version Y^2Z = X^3 + aXZ^2 + bZ^3) satisfying some conditions (to ensure that the curve is smooth).

But that's now how geometers think of elliptic curves - because having this dependence on the presenting equation can be awkward. For instance, if you take the chart Y=1, your equation now looks like z = x^3 + axz^2 + bz^3 - it's not clear now why this should be an elliptic curve using that definition!

Joe Shmo

@GFauxPas congrats!

Malice Vidrine

@loch - Interesting. And, strangely, I understood most of that.

loch

It helps to have examples

GFauxPas

We'll see

I didn't know how to find the normal vector to a surface of a curvy line y=f(x) rotated around the x axis. And that ruined a hugely weighted question

Malice Vidrine

1:24 AM

I hope I can end this course without being entirely scared away from the subject; it seems like it might have high points. But I'm pretty sure I'm going to fail unless we're only graded for effort... :P

Adeek

2:13 AM

hi @loch

attending first conference ever today

very exciting

loch

hi @Adeek
exciting indeed - what's the topic?

Adeek

2:28 AM

@loch algebraic cycles

@loch pims.math.ca/scientific-event/190513-shtam

loch

2:38 AM

i see

i hope you get to learn a bunch of cool math !

Adeek

me 2. Just have to learn how to socialize properly haha

N. Maneesh

3:12 AM

Let $f:\mathbb R^n \to \mathbb R$ be a function defined by $L(x)=\langle x,y \rangle$ for $x\in \mathbb R^n$. I know that $DL(x)=L,\forall x\in \mathbb R^n$. right?

GFauxPas

3:39 AM

What's $DL$ and what's $y$?

N. Maneesh

@GFauxPas $y$ is some fixed vector.

Daminark

Presumably $y$ is a fixed vector, $f = L$, and $D$ is the derivative

GFauxPas

ah

N. Maneesh

So, $DL(u)=DL(v),\forall u,v \in \mathbb R^n.$ Right?

GFauxPas

I forgot the definition of a derivative of a functional. Is it supposed to produce a functional?

N. Maneesh

3:48 AM

Here, $L$ is a linear transformation. If $T$ is linear transformation. Then $DT=T$

GFauxPas

oh its the best linear approximation operator

$f(x + h) = f(x) + Df(x)h + o(\Vert h \Vert)$ ?

Daminark

Just to make sure it's conceptually clear

GFauxPas

@Semiclassical @TedShifrin I got this question wrong on the final and I'd like to know how to do it

Daminark

So if you have a function $f:\mathbb{R}^n\to \mathbb{R}$, then given $x_0 \in \mathbb{R}^n$, the derivative of $f$ at $x_0$, called $Df_{x_0}$, is going to be a linear map $Df_{x_0}:\mathbb{R}^n\to \mathbb{R}$ such that $f(x_0 + h) = f(x_0) + Df_{x_0}(h) + o(\|h\|)$

GFauxPas

let $f: \mathbb R \supseteq [a..b] \to \mathbb R$ be a positive smooth function

let $E$ be the set obtained by rotating the curve $f([a..b])$ around the $x$-axis in $\mathbb R^3$.

What's the normal vector to $E$ at a point $(x,y,z) \in E$?

Your help is appreciated tia

Daminark

4:00 AM

When you're thinking of the derivative as an operator, the idea is that you have a map $Df:\mathbb{R}^n \to L(\mathbb{R}^n,\mathbb{R}) = (\mathbb{R}^n)^*$ given by $Df(x) = Df_x$

GFauxPas

well sure, the best linear approximation to a linear functional is the linear functional

I'm given an embedding

for $t \in [a..b], \theta \in [0..2\pi]$,

$\begin{pmatrix} t \\ \theta \end{pmatrix} \mapsto \begin{pmatrix} x = t \\ y = f(t) \cos \theta \\ z = f(t) \sin \theta \end{pmatrix}$

but not sure if the embedding is meant to help for this part or a later part of the problem

Isa

4:19 AM

What is $f^-$ in R? I know in $\overline{R}$ is the maximum of {0,-f(x)}

GFauxPas

is the context integration?

Isa

yes, Lebesgue integral

GFauxPas

then it's the negative part of $f$

Positive and negative parts

In mathematics, the positive part of a real or extended real-valued function is defined by the formula f + ( x ) = max ( f ( x ) , 0 ) = { f ( x ) if...

oh you have the definition right there

it's the same thing

Isa

oh thank you. I was confused because for R, $f^+=\frac{f+|f|}{2}$

so I thought something similar would be for $f^-$

GFauxPas

The WP page I posted has some identities

$f^- = \dfrac {f-|f|}{2}$

Alessandro Codenotti

7:50 AM

@EarthCracks 1 is true, I don't know what 2 means and 3 is true

More generally in 3 if you have $X=\mathrm{Spec} R$ and an $R$-module $M$ you can define a (quasicoherent) sheaf $\widetilde{M}$ as $\widetilde{M}(D(f))=M_f$ for all $f\in R$ ($D(f)$ is the usual basic open set $\{f\neq 0\}$, you get restriction maps from the universal property of the localization and a sheaf defined on a basis can be extended to a full sheaf

This is actually an equivalence of categories between quasicoherent $\mathcal O_{\mathrm{Spec R}}$-modules and $R$-modules (the inverse functor is the global sections functor $\Gamma(\mathrm{Spec} R,-)$), and you can check it sends free modules to free sheaves

Secret

8:19 AM

googology.wikia.org/wiki/First_uncountable_ordinal

$\omega_1$ first ordinal that has a fundamental sequence longer than countable

Damien

Are you guys able to post pictures on the website?

Secret

use the upload button

Damien

it keeps failing "Failed to upload image; couldn't reach imgur"

Secret

Then try go to imgur and post the link from the picture there instead, it might be just some system hiccup

Damien

it doesn't allow, it needs to be on i.stack.imgur.com

just wondering if it's just me having the issue or if there is a problem with the website

Secret

8:27 AM

ok it does not work for me also, probably a website problem

Damien

ok, thanks for checking

Damien

8:55 AM

any pros of audio amp in here?

Silent

9:11 AM

How to see that $4x^2+6x+3$ not zero divisor in $\Bbb Z_8[x]$?

Rithaniel

12:28 PM

So, given the set of equivalence classes of finite groups (related by being isomorphic to each other), define a binary operation on this set to be where, given two groups, combining them gives you some type of product of the two groups (direct product, semidirect product, wreathe product). In the direct product case, this resultant structure is a monoid, if I'm not mistaken. Though, is there a type of product which would make the resultant structure a group in it's own right?

Secret

12:39 PM

Jun 27 '18 at 2:58, by Alex

You should really go about fixing a set of groups, and giving the set magma,semigroup, monoid structure etc. What do you want as an inverse?

Your biggest challenge is to define some kind of product such that you can produce an identity when you multiply two groups together. It is not clear how this can be done

N. Maneesh

2:24 PM

I know that 1,2 are false.

I know that $(\nabla f)(x)=2(a_1x_1,a_2x_2,...,a_n x_n)=\vec 0\implies either a_ix_i=0\implies a_i=0 $or $x_i=0$. Hence $f=0$

(4) is false. right?

consider $f(x_1,x_2)=x_1^2-x_2^2$. $f(1,-1)=\vec 0$

but $(\nabla f)(1,-1)\neq 0$

Am I correct?

Thorgott

3:06 PM

Looks correct to me

N. Maneesh

okay. Thanks

Silent

3:25 PM

Does there exist a function $f:(0,1)\to\Bbb R$ that is twice differentiable such that $f$ bounded, $f'$ bounded but $f''$ unbounded?

N. Maneesh

$f'$ is bounded means uniformly continuos

Silent

omg! i did not know that!

So, can f'' be unbounded in my question, @N.Maneesh?

N. Maneesh

I am thinking

This was the firt thing I got .

Silent

alright

Thorgott

3:55 PM

I think $f(x)=\int_0^x\sin\left(\frac{1}{t}\right)\mathrm{d}t$ should work, though I also think there should be a less contrived example

user193319

4:08 PM

If $\{f_n\}$ is bounded in $L^1[0,1]$, is $\{f_n\}$ uniformly integrable over $[0,1]$? If it isn't, please don't give a counterexample.

Nick

4:46 PM

These are neat:

Narcissistic number

In recreational number theory, a narcissistic number (also known as a pluperfect digital invariant (PPDI), an Armstrong number (after Michael F. Armstrong) or a plus perfect number) is a number that is the sum of its own digits each raised to the power of the number of digits. This definition depends on the base b of the number system used, e.g., b = 10 for the decimal system or b = 2 for the binary system. == Definition == The definition of a narcissistic number relies on the decimal representation n = dkdk-1...d1 of a natural number n, i.e., n = dk·10k-1 + dk-1·10k-2 + ... + d2·10 + d1,with...

user280247

6:04 PM

Hi, im with a naive question:

user280247

suppose have no knowledge of maths and for practical purposes measure with a stick three values

user280247

a,b,c

user280247

a is the angle of a right angled triangle, b and c the sides

user280247

suppose we measure it for a particular case $a_1$, $b_1$, $c_1$, is there any way to predict mathematically that for the same relation $b_i$,$c_i$ $a_1$ is the same?

Rithik Kapoor

6:40 PM

is $\lim_{n \to 0} \frac{1}{n}\sin \left( \frac{x}{n} \right)$ a function?

also is it a space filling curve?

Balarka Sen

10:36 PM

Hi @Ted, @PaulPlummer

Paul Plummer

Hi @BalarkaSen

Balarka Sen

How's it going

Paul Plummer

Things are going okay, not being very productive though. You? how is college life?

Balarka Sen

meh @ college

The semester's over, I'm back home for some time.

Paul Plummer

I guess I had a small paper(and even smaller result) show up on the arxiv a few weeks ago

Balarka Sen

10:45 PM

Oh nice congrats

Ultradark

11:04 PM

What's repulsive but also attractive

user76284

11:18 PM

An electric charge.

user193319

Let $M$ is a field, $\alpha$ an element in some larger field, and $E = M[\alpha]$. If $f(X) \in M[X]$ is a minimal polynomial for $\alpha$, does it follow that $|E: M| = \text{deg } f$?

ÍgjøgnumMeg

Yes; $E = M[X]/(f)$, which is an $\deg f$-dimensional $M$-vector space

user193319

Ah, that's good to know. I thought it was true. Thanks!

ÍgjøgnumMeg

@user193319 Note that the usual notation is $E = M(\alpha)$. The notations $M[\alpha]$ and $M(\alpha)$ are equivalent when $\alpha$ is algebraic over $M$, while if $\alpha$ is transcendental over $M$ you have $M[\alpha] \cong M[X]$ and $M(\alpha) \cong M(X)$, which is a cool fact I guess

user193319

Oh, yes. You are right. I should be more careful.

Balarka Sen

11:40 PM

@ÍgjøgnumMeg I advocate thinking about $|E : M|$ as the number of $\overline{M}$-valued points in the scheme $\text{Spec} E$ lying over a fixed $\overline{M}$-valued basepoints in $\text{Spec} M$ instead of the dimension of the $M$-vector space $E$ :3

Also what a horrible notation for field, $M$

Use $F$ or $K$ or $k$ for Christ's sake

ÍgjøgnumMeg

Obviously it should be $k$

$F/k$ is a finite extension, $K/k$ is infinite

Balarka Sen

Very natural

F for Finite

K for ... Knot finite

Ha

Gottem

woppah

11:43 PM

u r reading Schneps right

i want to read it too

ÍgjøgnumMeg

Nooo I am reading Tamás Szamuely's book

Daminark

Hey guys

Balarka Sen

Oh I forgot the author

"Galois Groups and Fundamental Groups"?

ÍgjøgnumMeg

Hey @Daminark

Yeah that one

Balarka Sen

K that's the one I had in mind

Daminark

11:45 PM

Ah yeah I remember I was gonna read that at some point, but as with many other things I intended to read, I sadly didn't

ÍgjøgnumMeg

One of my old lecturers recommended it to me for some direction over the summer

Balarka Sen

Alas those reading lists that keeps getting exponentially bigger

I might read it next year second semester during my field and Galois theory course

Daminark

Okay so funny thing actually, I was talking to a friend and have thought of an evil idea

If I'm teaching a class I'll try to assign a problem which for a nearly unreasonable task but the condition is subtly vacuous

For example, in differential topology I'd say something like "Let $M$ be a hypersurface in $\mathbb{R}^4$ with $\chi(M) = 2$. Show that $\pi_3(M) = \mathbb{Z}\times \mathbb{Z}/2$"

And just watch to see how many people freak before they get it

Balarka Sen

I hope you mean closed hypersurface!

Daminark

Oh yeah closed lmao

Adeek

11:51 PM

@Daminark hi

Ultradark

The Greeks used two tools to make different geometric constructions. Name the tools.

Adeek

@Daminark I was wondering if you could help me with something ?