Err, that's not a proof.

is it that horned thing with the forking loops?

Hi again chat

Then the complement of both $K$ and $L$ is homotopy equivalent to $C$, and the complement of $K$ is homotopy equivalent to the mapping cone of a certain map $L \to C$. Similarly with $K$.

@ForeverMozart Oh, wait, I misunderstood you

Now by Alexander duality we know that $C$ has $H_1 = 0$. Lemme think more, I just wanted to write the setup down.

You mean the thing I posted earlier?

@ForeverMozart Yeah, I mean that

Salut @Astyx: Il est bien tard !

Rehi @Ted!

Rehi Demonark

I just googled the book "Counterexamples in topology", and google showed : "Genre : Non-fiction"

Fair

À peine 23h @Ted

@MikeMiller I think you can find $H_1=0$ just by Meyer-Vietoris even

Yeah, I wasn't using the setup for that.

Oh, right, we switched to daylight savings.

Oh, sure. That's fine.

@Astyx One would assume

Well is maths fiction ? You have 4 hours

@ForeverMozart It's not really meant to be a conjecture, it's meant to be a puzzle: "Find a counterexample"

Forget the setup earlier, there's better ways to phrase all of it. I should probably start grading instead of thinking.

When are you coming around here @Ted ? (I know I already have asked you twice at least, but I can't find where I wrote it, last time, I promise :p)

Grr ... le 5 juin jusqu'au 11 juin.

Does daylights saving mean that you can sleep in for one day or does it mean you'Ve gotta sleep less for that day?

Désolé :s

@AkivaWeinberger It should obviously never happen for smoothly embedded knots. If you take their disjoint union that's also a smooth 1-submanfold. Take an arbitrary loop; take the disk it bounds in R^3 and make it transverse to the union. That's a bunch of finitely many isolated points on the disk. Now draw a loop which encloses all the points coming from a specific knot; delete that bit of the disk and replace it by some other disk so it has no points coming from that knot

@Balarka It should knot happen? Lol sorry

@BalarkaSen Knots don't have simply connected complement...

You mean arcs?

Arcs, yes.

Less @s.harp unfortunately

Hello

@AlessandroCodenotti where do you study?

Could someone help me with the proof of the basis theorem for finite abelian groups?

At least when going from winter to summer

@s.harp in Trento, in the nort-east of Italy

@AkivaWeinberger If "nice" means they're compact, smoothly embedded, 3-dimensional submanifolds, then your situation is actually quite simple: your shapes are just disjoint unions of 3-dimensional balls. This is equivalent to being made out of minecraft blocks (unless you attach them by their corners, in which the situation is almost the same).

So taking the union of more balls still won't destroy simply connectedness, no.

He wants 1-dimensional (non-closed) submanifolds, probably. Arcs.

(Or edges)

Ahh, yeah, that makes sense

@BalarkaSen Doesn't matter. Thicken them.

If $AB = I$, then $BA = I$, right?

Ah, ok. Yes.

@Meow When talking about matrices, yes

@MeowMix yes, if $A$ and $B$ are matrices

In any ring, no

by the way, what is the transpose symbol in LaTeX

\perp

$^t A$

Thanks

No, only for SQUARE matrices, and that's a theorem!

Actually you could even just isotope a smoothly embedded arc to the standard interval. Because of existence of tubular neighborhood bump function tricks should be able to make all of them the trivial arc after an isotopy

oh transpose, i thought orthogonal sorry

^\top

@Ted My matrices are square

use $^T$

If they weren't they wouldn't multiply together and $BA$ wouldn't exist!

Fite me @Astyx (I write $A^T$)

With non square matrices it wouldn't make much sense anyway, would it ?

It's still a theorem you don't have til chapter 4.

jk

@BalarkaSen Sure, but imagine you had a lot of surfaces with boundary and arcs etc that all linked amongst one another or something without changing the fundamental group. It could, in principle, happen.

Worst possible notation for transpose: $TR(A)$

@Daminark I'm not talking to you any more

@s.harp how do you denote the trace then?

@Ted I was just using it for the exercise where you give me $A$ and $A^{-1}$, and I have to calculate $(A^T)^{-1}$

@AlessandroCodenotti $Tr(A)$

@MikeMiller Right, yeah, I agree.

...

$Tr(TR(A)) = Tr(A)$

$\mathbb{transpose}(A)$

@AkivaWeinberger The proof is simple. Let $M$ be the (closure of the) complement, a manifold with boundary. There is a general principle: the map $H_1(\partial M) \to H_1(M)$ is injective on half of the homology and kills the other half. This just comes from messing with duality. So the boundary has to be a bunch of spheres by assumption. Alexander proved that each of those spheres bounds a ball, and necessarily that ball is either $M$ or in the complement of $M$. So you can take your subset

You shouldn't need to use anything unproven, Zach.

and cap off boundary components.

$\mathfrak{transpose}(A)$

OK I shouldn't release these monstrosities on the world

Yeah, I suppose it's pretty elementary

@Astyx D: my favorite notational abomination: $$\frac{\widehat\Xi}{\overline \Xi}$$

I used $\Xi$ in a paper recently

@s.harp What does this even...?

What does it stand for ?

You're supposed to think about how matrix multiplication works, Zach.

Of course this means that if it's the complement of finitely many disjoint submanifolds, those submanifolds could only possibly be balls of various dimensions.

If I transpose $A^{-1}$ then it will just be the identity

Its Big Xi hat divided by Big Xi bar

Yeah, I realized

Because the spots of the zeroes will not change, and neither will the ones

Think about rows and columns respectively.

Yesterday I learned that $d\det_A(H) = Tr(^t Com(A)H)$

You can verify it by noticing $A^T(A^{-1})^T = (A^{-1}A)^T = I^T = I$

And that amazed me

what is det_A and Com(A)

derivative of determinant at A in the direction H

ah

not sure what com(A) is

What Mike said

Comatrice

i dunno what that is

"Free abelian group" needs a shorter name

An acronym isn't so appropriate @Akiva...

They're arguably simpler concepts than either abelian groups or groups

everyone's observed the thought you're having, @Akiva

Classical adjoint?

Adjugate matrix

gotcha

I didn't realise $Com$ was very (too?) french

Maybe call it "the set of chains", borrowing from algebraic topology

@AkivaWeinberger free Z-module.

less characters

I don't think anything involving the adjoint matrix is surprising, I have absolutely no understanding of it so that anything mysterious can be packed into it rendering the expression "probably plausible if only I understood what the adjoint is"

@BalarkaSen Now I wonder - if I just call it the "free module", will anyone yell at me

Yes

yes

OK then.

Lol

the obvious groups

(nvm balarka said it above)

The adjoint matrix is sort of like Hodge star — signed volumes of various $(n-1)$ dim parallelepipeds.

Free groups and free abelian groups are very different right ?

Yup.

Dat terminology ..

latter is former abelianized

except in the case that they have one generator (/smartass)

or zero

nice

Go grade!

give someone a zero

i'm just reading about clifford algebras instead

the grading will get done today

a zero mark on homework is also known as the "tautological hand-in"

Someone explain this to me

lol

a sequence of definitions of continuous, each considered more abstract than the last

The fourth one looks like I'd understand it if I knew the stuff

@Akiva brain explodes with category theory

Abstract math abstract art.

@MikeMiller Yeah, I know, I'm just curious what the fourth (and fifth) ones mean

We should rename "commutative algebra" to "abelian algebra"

I also had a question, is the $k$-th coefficient of the characteristic polynomial the sum of the determinants the submatrices of size $k$ taking out lines and collumns of same number (with $\pm 1$ as factor each time) ?

the fifth one is the second except replacing functions with functors

fourth is a commutative diagram. fifth is inverse limit and limit-preserving functors

$(A_f(x_n) )_n = f(x_n)$?

Principal minors

and limits being taken over diagrams

No signs.

the fourth one has the banach spaces $c$ and $\ell^\infty$, and s.harp just said what $A_f$ is

the fourth one is also the same as the second one

everything's the same

I think only number 3 works in a topological setting?

@s.harp What does that mean

@Balarka sequential continuity is morally not the same as continuity

2 works with nets

Can "principal minors" mean two different things ? I think I've seen it used with another meaning

@s.harp disagree

Also, what's $c$

And no signs ? That's weird

"Now look at this net, that I just found!" - Robby Rotten

The fifth is the usual continuity in the topology of open sets in R I believe

@MikeMiller it says $\Bbb N$ not $J$/$I$ etc

space of convergent sequences

@s.harp yeah but spaces are small

Like what, Astyx?

teeny weeny

is non-first countable necessarily "big"?

@Ted I know a solution using $(AB)^T = B^TA^T$ but I'm not satisfied because it doesn't have to do with the rows and columns

Like the $k$-th principal minor being the determinant of $(a_{i,j})_{1\le i,j\le k}$

I know what a commutative diagram is and what $\ell_\infty$ is, I just don't know what $c$, $A_f$, and $L_n$ are

now we need to add a picture defining continuity on $\infty, 1$-topoi and a blank blueish box next to it, with the man not present

@AkivaWeinberger $c$ is the space of convergent sequences, $A_f$ on a sequence is the sequence with members $f(x_n)$, $L_n$ is the evaluation of a sequence at $n$

@s.harp I disagree, that obviously commutes

Actually yes

now im convfused

$L_n$ up top is limit and $L_n$ on the bottom should be a choice of Banach limit

I automatically distrust the intelligence of people that make these kinds of images so I must default to "the image is wrong"!

Ah, OK. So it's just a rephrasing of sequential continuity

yeah

Wikipedia says these are the principal dominant minors, that might be it (although I'm quite sure "dominant" wasn't there where I read that) @ted

@Mike not a choice, but it should commute for any possible limit point (including $\pm\infty$)

It should, Zach. Dot product of row and column is either 0 or 1.

so they're wrong ;)

ok I have to go, dont want to miss my bus and be stranded in the airport until 6 in the morning!

I've never heard or said dominant for that, Astyx.

wonders why @s.harp is in the airport

well at least you'd have more time to sort out that picture

So the fourth one doesn't make sense ? What a disappointement

And what do the notations of the last one mean ?

hi chat

Hi @Semi

Actually rather bye chat

@MeowMix Mind helping me with Assembly really quick? :P

Sure, what processor?

assembly hard

What's the difference between getting a derivative of a function and differentiating the function?

Solution to all your problems: stop using assembly

Welcome back daminark

@CausingUnderflowsEverywhere No difference

Assembly is the solution to all your problems. The solution is running your own compiler in assembly

Hey @Causing!

Oh. Why the two different names then o.o is a derivative the result of a differentation?

Yes

And bleh, C was too low level for me

@Dami LOL When I hacked OoT I had to work with machine code.

Good thing MIPS is fixed width instructions...

Dang you didnt even assemble it?

What?

I was working with the machine code the developers already put in the game

So, for example, if I had an indirect instruction like beq in some replaced area of code, it wouldn't work. So I had to hardcode it to jump to the specific location I predicted it to be.

Half the time, if I saw a segfault I'd just be like "Eh screw this I'm done"

Also keeping track of mallocs was annoying

Write in an interrupt handler to give you valuable debug information when you have a segfault :]

Yeah in a lab we eventually were given this lldb thing or whatever

Too many teachers use "Whom" instead of "who" in the wrong spots to sound smarter.

"Whomst"

E.g. e-mail from teacher started with "To whomever tutors [person], Blah blah blah"

Whomst've'd

Whomst'd've'ff

(I'm so happy you know this meme @Meow)

For which whomst'd've it bequeaths, I convey that thus a contradictory plethora of cajolements is rubbish and henceforth, abolished and viewed as abhorrent from this vicinity.

^ Total trash

I use whom, but I think I've been using it correctly

Whom is only used when it's referring to an object pronoun.

Tip: replace it with "him" or "her" and see if it makes sense

(as does Ted)

To whom I murdered and sliced up, I'm sorry $\to$ I murdered and sliced up him

To whoever murdered and sliced me up, you're mean $\to$ He murdered and sliced me up

Who is to He as Whom is to Him

@MeowMix Quiz: What's right here? "I know who(m) killed your father." (Note: It's "I know whom" and yet "Who killed your father")

Because you have "I know" before it

@MeowMix What's your answer

I think "whom" but I'm probably wrong