If you're doing trig, better do complex exponentials

We'll get to complex numbers later, so i'll certainly mention it.

And we do vectors and some linear algebra later, too, so that's some of my favorite geometric material.

Ooh linear algebra

Over finite fields? :^)

NO!

Lol, I know one professor who made a thing about finite fields but usually a linear algebra course is just done in char 0

Not my linear algebra course

If we could rustle up enough students, they do have an abstract algebra class I could teach. But the school just opened, so not too many advanced students. Plus high school kids tend to be way too busy with other things.

I am fine with real linear algebra that emphasizes dot products and projections, thank you.

That course sounds really cool (even if you do lin alg not over finite fields :P), I wish I would have taken something like that when I was in high school

I am in the weird part of linear algebra with all the ODE stuff lately

I guess the structure is just different. Here we have the LA class which I think uses Curtis, which I think focused on char 0 so it doesn't have to worry about it, and then algebra 2 does modules, and I think includes stuff over PID

Using module theory to do canonical forms is just ducky, Demonark.

My linear algebra course part 2 did actually proved all the module theory we needed

I loved it

I take it your linear algebra course was only for pure math majors?

yes

No engineers, physicists, applied math people?

I dunno how much stuff they do exactly. Especially because this year things have changed a good bit. Before we didn't even have a linear algebra class

Well, applied math people, yes we don't have distinct majors for pure/applied

Back in Herstein's day you did, Demonark. The last part of his book was exclusively linear algebra.

I know some physicists who took it, but they were doing so voluntarily

For most physicists, my approach would be more meaningful and useful, Mathei.

but the first part of lin alg was mandatory for physicists

We also did dot products and stuff like unitarity, orthogonality, self-adjointness etc.

Still worked over general fields most of the time

So the way it worked here before I arrived was that either in the intro to proofs class or the third quarter of honors calculus, you did a few weeks of basic linear algebra. Vector spaces, transformations, determinants, all that stuff. A bit came up in analysis when an eye toward multi, and then you'd really do it mostly in the second quarter of algebra. Not just as a special case of modules, like it used Curtis/Hoffman-Kunze for a good block, and did some multilinear algebra

we did motivate determinants geometrically, but defined them using exterior powers

multilinear algebra becomes just a special case of linear algebra if you have tensor products

Now they're probably cutting the linear algebra out of algebra because of the new class, which unfortunately doesn't usually get to normal form. So I'm not sure what's gonna happen

@MatheiBoulomenos Very true.

Sounds like it's a bit unstable, Demonark.

I watched a video solving $\sin(z)=2$ and it made sense and gave two solutions. In the comment section however, you can find the catchphrase "2 is pretty small so $\sin(2)=2$ amirite".

sounds like a physicists/engineer

2 radians isn't small :P

We should never teach degrees.

Yeah, I'm not sure how that's gonna go. Guess we'll have to wait and see. In the meantime, though, anyone want to talk about simplicial sets? Because I only sorta get them

By which I mean barely

Wow ... names from the past. Heya @Karl

@Daminark Bring it

Hi @Karl Cronenberg

Hey @Ted

Hey @Balarka

@Ted I read on your m.se profile that you're teaching class in spite of retirement. Are you in San Diago now?

Ah, your location on the profile says as much.

Okay so I know that you have the simplex category, objects are sets of the form {1,...,n}, morphisms are monotonic maps between them

Yep.

Morphisms are literally inclusions of the faces into ordered simplices.

If I think of [n] as an ordered n-simplex I mean

By Dold-Kan simplicial objects in Ab are just chain complexes, pretty cool stuff

lmao stop it

@Karl: Yeah, I've been in San Diego for 2 1/2 years. AoPS opened a brick-and-mortar school about 15 minutes from me.

Have you graduated yet? :)

@TedShifrin Oh!

@TedShifrin Nope :)

Lol my prof is less than fond of chain complexes, I think? But yeah so a simplicial set is supposed to be a contravariant functor from the simplex category to sets

Still an algebraist? You and Mathei should talk algebra.

I love to talk algebra :)

Demonark: You cannot NOT be fond of chain complexes.

@Daminark Yeah it's a presheaf on the simplex category

@TedShifrin Well, this is Peter May we are talking about, so...!

Maybe he likes model categories more

I would never have learned any topology this way.

Seriously.

lol me neither

Just abstract crap.

I'm so glad I had John Stallings as a prof.

Ah Stallings is great

Wait chain complexes are topology? I think of them more as a homological algebra thing

He was a bit ditzy sometimes, but I loved it.

They embody geometry/topology.

@MatheiBoulomenos I mean, they are the usual model for topological spaces when doing any specific homology theory.

Why is The Mean Value theorem named "Théorème des accroissement finis" in french ? I lost the meaning of both of them now..

Sure, but you can define and use them over any abelian category

So chain complexes did come up in our thing when he was talking about the simplicial sets stuff, because he was showing us how to link to homology, though I didn't completel y pick up on that because it was quite fast at the time, but he showed us how to define a simplicial set out of a small category by taking its nerves

That French is incorrect, @Fuzzy. Disagreement between noun and adjective :)

@MatheiBoulomenos Sure.

But I have no idea why they're calling it the theorem of finite increments.

Dieudonné calls it Le Théorème de la Moyenne.

@Daminark I think simplicial sets are the analogous (to chain complexes) models for topological spaces when you do homotopy theory.

But I know nothing of that

Meh, that gets even cringier when translated from french into arabic literally $:($

And he pulled Eilenberg-Maclane out of this, by letting a group be a groupoid with one object and having BG be the geometric realization of the simplicial set of nerves of that category

That sounds pretty cool

@TedShifrin abstract crap is the best kind of crap

You can take B$\mathcal{C}$ for any category $\mathcal{C}$, yes.

Crap it is.

Also are there any other similar theorems to The Mean Value and The Intermediate Value theorems ?

But yeah so I'm just trying to understand the link between simplicial sets, ordered simplicial complexes, and topological spaces

That would be useful to keep in mind?

There's a Mean Value Theorem for integrals, @Fuzzy.

There are actually a few of those. You'll find them in Spivak.

@Daminark The first two are pretty easy to understand, I think. What are you having trouble with?

Since we're moving on next week. We might do k theory next, or (dual) CW complexes

Just being sure what everything is exactly. So a simplicial complex is a set of subsets of a set which is closed under subset, yeah?

@Balarka: These people will never have any idea what simplicial/singular/cellular homology/cohomology mean, if they even see it.

Where each set in the collection is a face. Then you need to have an ordering on it that restricts to a total ordering on the faces

weewoo weewoo

Are you benapped, Meow?

quite fernappled

Now, I'm trying to just be sure, this has to do with triangles by having the set be a set of vertices, right? So then given a subset of the vertices, you should be able to construct one of the faces of the triangles?

@Daminark I never think of simplicial complexes this way, so let's see. My definition is it's a bunch of simplices glued togather in a way so that each simplex is determined uniquely by it's boundary. I think that's an equivalent imposition as each simplex being uniquely determined by their vertices.

i guess that means ts rational limit time

I'm not forcing you to do any of this, Meow. :)

So yes, if I look at the set of all 0-simplices (vertices)...

huh? no, i want to

i have nothing else to do

im waiting for my microsd card to come in the mail any day now

and i need practice anyways

@TedShifrin Do you mean me by "these people"? I took an algebraic topology course with A Geometric Approach$^{TM}$, light on "abstract crap", so I think I have some intuitive understanding of those things. I still prefer to use groupoids to prove Seifert-Van-Kampen like May does in his book

No, Mathei, I meant the students in Demonark's class.

I see

I think it was more directed at me, since I've only sorta seen the chain complexes, mostly axioms and E-M spaces

Mathei knows his shit, I think.

thanks, I guess

lmao

Of course, we're all trying to push Daminark to also know his shit and not end up like nlab people

Though for what it's worth this isn't meant to substitute for an actual AT class, where we're probably gonna do it the standard way.

You can be an algebraist and still have intuitive grasp of things, you know :)

Aka, be a "good algebraist"

I'll agree with that

not all, Balarka. I've thrown up my hands and given up on Demonark.

@Daminark Right, so a simplicial complex is a subset $X$ of $\mathcal{P}(S)$ such that for any $A \in X$, $B \subset A$ implies $B \in X$.

That is the abstract essence

I still sometimes make categorical arguments, but sometimes just for fun

But of course I found this from reverse manipulating the geometric meaning of simplicial complexes

And how does the ordering translate to triangles?

I mean a few times in my life I actually have used the language of categories and functors, but only a few times ... and I truly do not think that way.

My second abstract algebra course used categories and functors seriously

@Daminark Ah, so, Hatcher has a picture of this.

Adjoint functors are pretty ubiquitous in algebra

So think of [3] say

That is a triangle with vertices 1, 2 and 3

Order the triangle so that you have an arrow i --> j on the edge joining i and j if j > i

so you have the ordered edges 1 --> 2, 1 --> 3 and 2 --> 3

That is how you order a triangle from an ordering on the vertices

@TedShifrin the groupoid proof of Seifert-Van-Kampen is more intuitive to me than the standard one. You don't have to mess with the construction of pushouts in the category of groups aka amalgamated products. (It's still good to know how to compute it by generators and relations, I'll admit that)

I can only understand it in terms of the pictures.

I don't get the groupoid proof. I mean, it seems you literally need the same fucking tools you use to prove it by hands like Hatcher does

Hmm, okay so why do we ask that the ordering on the whole simplex restricts to a total ordering on faces?

also Ted

@Balarka the point of the groupoid prove is, since youre dealing with paths and not necessarily closed paths, you can just subdivide the path into the open subsets of your covering

But that's the usual proof.

@MatheiBoulomenos You do the exact same thing in the usual proof.

Using Lebesgue # ...

Just Lebesgue covering lemma

LOL

Damn Ted you're sniping me!

in your examples it just happens to have a factor of $(x-a)$ inside the abs value

<-- goes back to thinking about a letter of recommendation

@Meow: Explain to me why (for rational functions) that must happen.

@Daminark Here's a picture from H-dawg

umm one second im gonna work it out

The last picture is a 3-simplex with an ordering on it. Notice that each of it's faces are ordered too.

does it suffice to show it for just a polynomial rather thn a rational function?

(The ordering convention is again that you have an arrow $v_i$ ---> $v_j$ iff $j > i$)

You tell me, @Meow.

Oh wait now I know why it has to be a total ordering

You have a bunch of simplices and then each face has to be a single simplex, and everything has to be ordered. Thanks!

let me try for just a binomial

Alright so, if we're trying to build a simplicial set out of that

Hint: Do not write out explicit formulas.

Yep. But notice that being a simplicial complex means a little more than that; every face has to be uniquely determined by the vertices.

oh i did

Ugh.

umm how else could i show this

So like, two 2-simplices $abc$ and $a'b'c'$ joined along $a \sim a'$, $b\sim b'$ and $c \sim c'$ is not a simplicial complex.

That's what is called as a $\Delta$-complex, or a semisimplicial complex.

How do we usually prove things in math? We use letters :P

wait what do you mean

i was writing like $|(ax + b) - (ax_0 + b)|$

which would give $|a(x-x_0)|$ so it works for binomials

Well, you could write $p(x)$ and $p(x_0)$.

OH

Or $\left(\frac pq\right)(x)-\left(\frac pq\right)(x_0)$.

i think i understand this

$|(ax^2 + bx + c) - (ax_0^2 + bx_0 + c)|$

is $|a(x^2 - x_0^2) + b(x-x_0)|$

Now explain why this has to work.

i will do so by proving each term has a factor of $(x-x_0)$

Can you prove that, in general, for a polynomial $p$ and $p(x)-p(x_0)$?

Without writing out formulas ...

alright let me think of this w/o formulas

having a factor of $(x - x_0)$ means theres a root at $x_0$

$p(x_0)$ is the value at $x_0$

Oh okay, that makes sense

so at $x_0$, $p(x) - p(x_0) = 0$

meaning its a root

boom

Yup. Now can you handle the $p/q$ ?

BTW, I'm glad you made this observation. I wondered if you would.

holey shite

If I'm asked to prove something and I think I have a proof of something stronger without seeing a way to only prove the weaker thing I was asked to prove, is this a sign that I did something wrong?

Wrong?

is this analogous to proving that $p(x) - q(x)p(x_0)$ has a root of $x-x_0$?

for some non-zero polynomial $q(x)$

I mean, should this ring some alarm bells concering the correctness of the proof?

not analogous, equal

wait a second ignore that

You mean root of $x_0$.

@Mathei: Not necessarily. It suggests you're missing an idea to do the simpler case.

One of the trig homework problems I did the other day (for my class) I worked out knowing that I was missing a totally easier way to do it, but it was a correct solution. :)

yes, that $p(x)q(x) - p(x_0)q(x_0)$ has a root of $x_0$

That's not right, @Meow.

my original was right???

No.

darn

let me write this out

@TedShifrin yeah, sometimes I just like to troll the tutor who needs to correct my coursework. I had a 5+ pages proof of something that could be proven in 2 lines

$|\frac{p(x)}{q(x)} - \frac{p(x_0)}{q(x_0)}|$

But in that case, I was aware of the 2 line proof

$= |\frac{p(x)}{q(x)} - \frac{q(x)\frac{p(x_0)}{q(x_0)}}{q(x)}|$

I would kill you if I had to read all that, Mathei. I'd rebel.

does that look more right-er

It looks horrid. What polynomial do you want to have a root at $x_0$?

$p(x) - q(x) \frac{p(x_0)}{q(x_0)}$ is what i get from the above

but you said it was horrid so i dont think thats right

@TedShifrin that tutor was always pleased by my solutions

OK, @Meow, or, slightly nicer, $q(x_0)p(x)-p(x_0)q(x)$.

right

@Mathei: In my experience, most graduate student graders (and lots of faculty, too) don't grade very carefully and certainly don't read solutions that go more than a page.

hmm

so why does that have a root at $x_0$

Duh.

Hello Ted :D

Heya @anon!

Hi, Kasmir.

heya

Anon :D

hey toon link self-caffeinating

Hi Kasmir & anon

Matheiboulom :D

when i here kasmir i always think cashmere

hear*

oh wow im dumb

@KasmirKhaan how's the group theory going?

i just realize, at $x_0$ we have $q(x_0)p(x_0) - q(x_0)p(x_0) = 0$

duh

so that makes sense

Whence my "duh." :)

guten Tag alle, bonjour a tous, hola todos, buongiorno tutti

@MatheiBoulomenos Well , it is alot better since like 4days ago, was thinking it was hopeless now I understood most of group theorey part =P

Only part left is sylow =P

You'll see tricks sorta like that later in Spivak when you prove things like the generalized Mean Value Theorem, @Meow.

guten Tag @Leaky

@KasmirKhaan give me more time

ooh generalized

@LeakyNun exam is on monday >< but its ok if you cant do it =p thanks anyway :)

What is this exam about?

@LeakyNun Dont want to make pressure or something , it is totally fine if you cant do it ill work with old exams that we got

well it is about ring theorey and group theory =p

I only got the group theory part haha ><

was too fast for me

anywya we also have a reexam

But at least now I feel I understand these stuff =p

anyways its

burrito baturday

Well, @Meow, you took all that time without doing another $\delta$ :)

i did half a delta

but it kinda looks the same as the other one

so i stopped and went to ask you that question

They all seem similar once you get the hang of it.

@Kasmir I would not object to helping you with ring theory, is there anything in particular that's causing trouble?

Maybe 14 would be different.

let me try 14

before burrito time

Late for burritos!

@MatheiBoulomenos That is very nice of you ! :D but how long will you be online ? =p

the salsa can wait

@MatheiBoulomenos so i can prepare what I study :>, am planning to study all night

It's weekend now, so I don't have to maintain a regular sleep schedule right now :D

No, I meant that the hour is late for eating burritos :)

english can be ambiguous at times

Haha well Am planning to stay up untill 7 pm , so I sleep and wake up 5 am on monday =p

because my sleep paddern now is crazy ><

anyways, its never too late to eat burritos

@MatheiBoulomenos Ill prepare some questions and then text you :) ill start working now !

in the words of the great patrick star

"OH BOY 3AM!"

I think it's @Balarka's fault that the sleep cycles of everyone under 23 here is a total mess.

sleeping is overrated

Make that under 25 :^)

okay so the thing i end up having to minimize is $|x+1|$ and $1/|2x^2 + 2|$

$|x+1|$ is easy, with the default $|x-1| < 1$ you have an upper bound of $3$

the second is interesting

i guess its a burrito intermission time

@Ted I'm a good influence, what can I say?

Ted needs to stop by new jersey some time'

He does?

indeed

I have friends (and a sister) in Pennsylvania and in NY and DC who expect me first.

The advantage of having a bad sleep cycle is that you eventually learn to function with very few hours of sleep each night

Or you get run down and get the 'flu or pneumonia.

maybe some time when im in college

just 4 more years

If we're still alive then ...

well thats dark

think on the bright side, that means i have no more nightly 2 page algebra 2 packets :)

hi chat

That's a good bright side.

hi @Semiclassical

Wait, you're 4 years away from college and you're already taking algebra 2? How does that work? @MeowMix

high school algebra, Mathei.

Ah, I'm dumb

Algebra 2 is stuff like exponents and all

Lol sniped by far

We don't have fancy names for high school math, we just call it "math", that's why I was confused

I can't handle all these headshots anymore

its like

conic sections, complex numbers

Gotta retire from this chat/war

See it on the bright side

Most people can't handle their first headshot

its basically the study of $^2$

everything is about $^2$

$x^2$, $y^2$, and more!

featuring: $x$ and $y$ of Algebra 1

the esteemed $+$ and $\cdot$

that does sound dull

right now were studying $^2$ in graph form

the parabola

and the $^2$-meister, the circle

anyways i wont postpone the burrito any longer

@TastyRomeo now I'm functionally identical to them though, I've used up my lifetime quota

@Daminark imma 360 noscope the shit out of you

get $2\pi$'d

psh, amatuer. gotta get that 4pi solid angle

Is "$2\pi$ noscope" a thing yet outside this chat?

god i hope not

resigns from the chatroom

Ted is too old for this

damn right