Mathematics - 2017-03-06 (page 1 of 5)

Mike Miller

00:00

So you just need to pick a basis for $H_1(D)$, which I descried above.

Daminark

I shall defer to @Mike, this is above my experience level

Mike Miller

@EricAuld I believe it. The point is that we're getting rid of the unnecessary degeneracies.

(I hope you don't find my bi-discussions disrespectful; if you do, I'll gladly stop.)

Topologicalife

But @MikeMiller, if I have infinite curves can be collected in a finite set? :P

Eric Auld

So for our equivalence $\textsf{sAb} \to \textsf{Ch}$

No, it's impressive!

Topologicalife

What do you mean by 'there is a finite set of curves'. Does this mean there is a finite number of curves?

Mike Miller

00:01

@Topologicalife You can do it with a finite set as long as $H_1$ is finitely generated.

Yes

Eric Auld

For our equivalence, we need a functor each way. The functor from $\textsf{sAb}$ to $\textsf{Ch}$ is the normalized functor.

Mike Miller

Sure. The Moore functor, I assume, is not actually an equivalence?

(Maybe it's an equivalence on homotopy categories, but not on Ch?)

Topologicalife

Mm, I was considering a infinite number of curves.

Eric Auld

That sounds right, because Moore and normalized are chain homotopy equivalent

Mike Miller

@Topologicalife That sort of thing only arises if $D$ is a complicated domain; for instance, the complement of $\Bbb Z$ inside of $\Bbb R^2$. Then you need countably many curves, one for each element of $\Bbb Z$.

Eric Auld

00:04

We want the composition $\textsf{sAb} \xrightarrow{N} \textsf{Ch} \to \textsf{sAb}$ to be naturally isomorphic to the identity on $\textsf{sAb}$, and $N$ took away our degenerate simplices.

So basically the other functor (called $\Gamma$, usually), throws them back in.

Mike Miller

Ah yeah it can't possibly be essentially surjective - if there's a term in degree $n$, then there's a term coming from degenerate simplices in all higher degrees, too. So there's no bounded chain complexes in the image of the Moore map!

Topologicalife

I see, that's interesting.

Mike Miller

@Topologicalife But if you're doing, say, $\Bbb R^2 \setminus \{0\}$, then a conservative vector field is just an irrotational vector field whose integral around the unit circle is zero.

I'm just generalizing that statement to arbitrary domains.

Eric Auld

Yes, that's right! Good thought

Simple

math.stackexchange.com/questions/2173711/… Is $T$ suppose to be real?

Mike Miller

00:06

@EricAuld OK, sure.

By the way, right now I'm thinking about this in terms of mapping spaces. Kom should be a simplicial category by following the Dold-Kan construction on the chain complex of maps, so I'm interested in following this along to understand what the n-simplices in Kom are.

(This might also help elucidate the Dold-Kan correspondence itself, perhaps.)

Dragneel

Hello. I have a small question here. Looking at gyazo.com/42718b3db7fa1755f98a13bb77d7bb71, why does $aC2$ equal $a(a−1)/2$ and not $a!/2!(a−2)!$?

Eric Auld

If you ever read about $\Gamma$, expect it to be totally opaque and defined in this weird way. But given a chain complex $C_\bullet$, I define my zeroth level of my $\textsf{sAb}$ to be just $C_0$. Then my first level is $C_0 \oplus$ this copy of $C_0$ that I'll call $s_0C_0$, to keep track of it.

Mike Miller

Those are the same thing. Can you prove it?

@EricAuld Maybe I did a million years ago.

Eric Auld

The third level is $C_2 \oplus s_0C_1 \oplus s_1C_1 \oplus s_0s_0C_0$.

Mike Miller

Wait, wait.

Eric Auld

00:09

Notice I didn't put in $s_1s_0C_0$, because one of the simplicial identities is $s_0s_1 = s_0s_0$.

These are all just markers: $s_0s_0C_0$ is really just a copy of $C_0$.

Mike Miller

You meant $C_1 \oplus s_0 C_0$, right?

It looks like you're trying to write down a homotopy colimit of a constant diagram, almost.

Eric Auld

Yes!

Sorry

Also I meant the "second" level above, not the third

Mike Miller

Yup

Eric Auld

I wouldn't be surprised. I need to understand homotopy colimits better.

Mike Miller

arxiv.org/pdf/1609.09132.pdf see definition 2.7

They're not quite the same but they look quite similar...

Eric Auld

00:13

The degeneracy maps for this thing are almost built in. For instance the degeneracy map $s_1$ acts level 1 by taking $C_1$ to $s_1C_1$ and $s_0C_0$ to $s_0s_0C_0$. Because $s_1s_0 = s_0s_0$.

Mike Miller

Yup

Eric Auld

Let me come back to that if I have time

Mike Miller

To what?

Topologicalife

Thanks @MikeMiller

Eric Auld

To that paper you just sent

Mike Miller

00:13

K

Eric Auld

Let's illustrate the boundary maps on level two, which is C2⊕s0C1⊕s1C1⊕s0s0C0C2⊕s0C1⊕s1C1⊕s0s0C0.

Let me just retype that

$C_2 \oplus s_0 C_1 \oplus s_1 C_1 \oplus s_0 s_0 C_0$

I need to define $d_0, d_1$, and $d_2$.

Mike Miller

You can edit or delete messages using a little dropdown menu to the left of your messages.

Eric Auld

Cool

I declare that everything except $d_0$ shall act as zero on nondegenerate terms.

Just like before I only took the stuff on which everything but $d_0$ acted as zero.

Mike Miller

Let's think about this in terms of homotopies. The one-homotopies between maps of chain complexes are given by degree -1 maps and a degree 0 map. Presumably this tells me a "reference map" that I'm chain homotoping from - that's $s_0$ - and then the chain homotopy, from which I can also determine what the other boundary is

Eric Auld

Now I figure out what the boundary maps do to the other terms by applying the simplicial identities.

Mike Miller

00:18

Ah, so $d_1$ on $s_0 C_0$ is going to be the identity map

Oh

Other way around

Eric Auld

Yes

because $d_1s_0 = \text{Id}$

And it will go to the $C_0$ below.

Mike Miller

So I think my picture of chain homotopies up there is right

Eric Auld

Cool, I wish I had more mental energy to parse that right now.

Mike Miller

Sorry about that. Is this a bad time?

Eric Auld

No, no, this is fun

Mike Miller

00:22

Gotcha. Let's keep going then. Thanks for helping me out, by the way.

Eric Auld

So if I have a boundary map $d_i$, one aspect of the way we're defining them is that they'll land entirely in one summand of the target. Not spread between summands.

Let me rephrase that

When I apply boundary maps to $C_2 \oplus s_0 C_1 \oplus s_1 C_1 \oplus s_0 s_0 C_0$, if I focus on $s_0C_1$, say, and look at what the boundary map does to that term, it will land entirely inside of one term in the target. (Or it will be zero)

what a particular boundary map does to that term. There are more than one

Mike Miller

So if I write it as a block matrix, the columns will have only one term in them

Eric Auld

Yeah, that's right

Right

So, say I look at what $d_2$ does to the $s_0C_1$ term in level two.

Mike Miller

I thought the higher $d_i$ killed the degeneracies?

Or do they just send degeneracies to degeneracies?

(So that they're zero on the quotient complex?)

Eric Auld

Neither. What I said was that everything but $d_0$ acts as zero on the nondegenerate terms.

How the boundary maps act on the degenerate terms is a little more subtle.

As an example, let me look at what $d_2$ does to $s_0C_1$.

Mike Miller

00:29

That's the opposite thing as I said! Whoops!

Eric Auld

By simplicial identities, $d_2s_0 = s_0d_2$. So I get $s_0d_2C_1$, but all higher things act as zero on nondegenerate stuff, so this map is zero

Now what about $d_0$ acting on $s_0s_0C_0$?

$d_0 s_0$ is identity, so we're left with $s_0C_0$, and this should be the map that takes $s_0s_0C_0$ to $s_0C_0$ as the identity.

(remember, these are all just copies of $C_0$, just labelled differently)

Mike Miller

Yup, I'm fine with that.

Eric Auld

The tricky thing can be to remember the simplicial identities. I guess everyone has a way of remembering them.

Mike Miller

Wikipedia ;)

Hi @Akiva

Akiva Weinberger

@MikeMiller I see you thought of a whale skeleton covered in a coat of polar bear fur.

Mike Miller

00:33

No, I just liked the suggestion

Eric Auld

I can tell you a way to remember them yourself really easily if you ever want. Probably easier as a picture. Basically just think of a row of dots. $d_2 s_1 = \text{Id}$ says that doubling the first dot, relabeling the dots, and then removing the second one just gets you where you started.

OK, so back to the equivalence of functors.

BAYMAX

Yes it's Pinter @Daminark , thanks!

Eric Auld

If I go $N \circ \Gamma$, I'm taking that weird simplicial thing I just built from a chain complex, and then applying $N$ to it.

I ask the question, on level two say, "what are the things in $C_2 \oplus s_0 C_1 \oplus s_1 C_1 \oplus s_0 s_0 C_0$ so that $d_1$ acts as zero and $d_2$ acts as zero? I claim the answer is exactly the nondegenerate term $C_2$.

Mike Miller

Is one of these compositions going to be the identity on the nose?

Eric Auld

This means that $N \circ \Gamma$ of my chain complex is just literally the chain complex I started with

Mike Miller

00:37

Yeah, that's what I was thinking.

Seems clearly the same even functorially

The real spooky part is probably going to be the other direction, right? I killed my degeneracies and added in new ones, and I need to check that the new degeneracies can be wiggled into the old ones

Eric Auld

No, it's not that bad! Coming in a sec

Now if I do $\Gamma \circ N$, suppose I started with a $\textsf{sAb}$ called $A$. My simplicial abelian group $\Gamma \circ N (A)$ has a second level, say, that looks like $N(A_2) \oplus \texttt{all the degeneracies}$. So since as I said before, $N(A_2) \oplus \texttt{degenerate stuff} \cong A_2$, it's not too bad to convince yourself that the levels are all isomorphic, and the boundary maps if you follow them, are just the ones you started with.

You've just labelled everything specifically by the thing it's a degeneracy of.

Mike Miller

Surely this isn't an isomorphism of categories?

Eric Auld

No, I think not. This one above I would say is not on the nose. I've sort of broken it up and put it back together.

Hard to call that thing above the identity.

Mike Miller

Gotcha

@EricAuld This makes a lot of sense, thank you. I think I understand what's going on much better now.

Eric Auld

That's awesome.

Fargle

00:52

Hiya chat.

Alias User

Hiya!

Mike Miller

Hi @Fargle

Fargle

How goes it?

Mike Miller

It's ok. Trying to understand things, which is hard forever.

4

Alias User

I'm trying to show that for a group $G$ with subgroup $H$, that $|gH| = |Hg|$. It looks like the map $k \in gH$ to $k^{-1} \in Hg$ will work, but I don't know how to prove that's a bijection. Any tips?

Fargle

00:56

@MikeMiller Jesus, I know the feeling.

Mike Miller

Note that there's a map $|Hg| \to |gH|$ given by the same formula, and the maps are (obviously!) inverse to one another - because $(k^{-1})^{-1}$ is $k$

arctic tern

$k\mapsto k^{-1}$ is a bijection $gH\to Hg^{-1}$, not $gH\to Hg$

Alias User

arctic tern: thanks!

Mike Miller

OOPS

arctic tern

need to use conjugation for a bijection $gH\to Hg$

Alias User

00:58

@MikeMiller -- would I just show that both of the maps are injective and then use Schroder-Bernstein?

arctic tern

heh

Mike Miller

No. We have a map $f$ and a map $g$ such that $gf = \text{id}$ and $fg = \text{id}$

The first equation implies that $f$ is injective (why?) and the second that $f$ is surjective

Of course the formulas I gave were wrong, but as a proof of concept pretend I was showing that $gH$ is in bijection with $Hg^{-1}$

Alias User

@MikeMiller ahh, if $f$ wasn't injective, we'd have two distinct elements that would map to the same $y$, which would map then to the same $z$, implying equality?

Mike Miller

Yup

arctic tern

@Alias Suppose you have $gh\in gH$. What can you multiply this by on the left and right to get an element of $Hg$?

(Given that $g$ is a known, fixed element but $h$ is an unknown element of $H$)

Alias User

01:02

@arctictern it'd be $h^{-1}g^{-1}$, right?

oops, hold on

arctic tern

Write $x=gh$. What do you do to $x$?

have to leave

later

Alias User

Take the inverse.

Thanks!

Mike Miller

No! You don't want $h^{-1}g^{-1}$

You want $hg$

Alias User

What can I multiply by then to get that?

Fargle

You don't have to have $hg$. $h^{-1}g$ is enough.

Mike Miller

01:05

You can't multiply by a single thing on the left - or a single thing on the right - to do this for all $h$. So how do you get from $gh$ to $hg$ by multiplying on the left and on the right?

@Fargle Yes, but that's not what we get.

Fargle

OH. Fair. I was reading it wrong.

Alias User

@MikeMiller that makes more sense. We left-multiply by $hg^{-1}$ and right multiply by $h^{-1}g$.

How does that help us?

Mike Miller

I suggest left-multiplying by $g^{-1}$ and right multiplying by $g$ :)

Alias User

Haha, that makes more sense.

DHMO

@MikeMiller well played

Mike Miller

01:07

I need to go - @Fargle, want to take over?

Alias User

So we use that as our map instead?

@MikeMiller -- awesome, thank you!

I apologize -- I have to run as well. Thank you all for your help!

stantheman

Wasn;t sure if this question would be appropriate for the stack so I want to briefly ask it here guys!

I am given that a function $f(z)$ has all poles $s$ on the left-hand side of the complex plane ($Re(s) < 0$) and the fact that $Re(f(iw)) > 0$, $w$ an element of the reals. What new additional information can be concluded about this function.

I have taken complex analysis a while back and don't remember all the theorems too well. First thing that comes to mind is the Cauchy Integral Formula

Dragneel

May someone please assist me with this chat.stackexchange.com/transcript/message/35846658#35846658

MATH ASKER

$√ 3 + 3/ 1-√ 3/3$

Compulsive Mathurbator

@Dragneel What's the problem?

MATH ASKER

01:23

how do u solve the following

Dragneel

It's Discrete math. I have to prove something. It's on the first line.

The answer was given, but I cannot understand it.

Compulsive Mathurbator

You said earlier "why does $aC2$ equal $a(a−1)/2$ and not $a!/2!(a−2)!$? "

But they are the same

Dragneel

Are they?

Compulsive Mathurbator

$a!=a(a-1)(a-2)!$

Dragneel

Is this an identity I'm unaware of?

Compulsive Mathurbator

01:24

2!=2

That's how we define factorials

n!=n*(n-1)!

0!=1

Dragneel

What if $a$ was some other integer?

Compulsive Mathurbator

for example?

Dragneel

$aC2$ where $a$ is $4$.

Compulsive Mathurbator

Its still going to be the same, whatever the integer is.

Dragneel

Really? $a(a-1)$ is a replacement for $a!$?

CausingUnderflowsEverywhere

01:28

INTERESTING @ZachHauk it's n^2 subtract the previous term because the previous term is whats missing well thats what subtraction is about, but it's because if you were to make each of the terms be that number you're squaring, well... and find the amount you're missing, it's quite the reverse n -1 n -2 n-3 ..

gosh im stupid

Compulsive Mathurbator

No, $a(a-1)=(a!)/(a-2)!$

Dragneel

Whoa

I see now. It was canceled with the denominator.

Compulsive Mathurbator

Yeah

Dragneel

I am so silly, thanks so much @CompulsiveMathurbator

Zach Hauk

do I see a wild @Ted?

CausingUnderflowsEverywhere

01:30

welcome back friend

Compulsive Mathurbator

We knew you would be coming

Zach Hauk

i had some more italian meatballs

Compulsive Mathurbator

insert creepy laugh here

CausingUnderflowsEverywhere

sounds spicy

Zach Hauk

@CompulsiveMathurbator well, he did nickflash me

CausingUnderflowsEverywhere

01:31

flash? I did not flash you

Zach Hauk

I said nickflash

as in

Nick flashed me

CausingUnderflowsEverywhere

as in flash my shaving nicks

Zach Hauk

I guess you could say I came here

in the NICK of time

CausingUnderflowsEverywhere

to explain how the summation formula for natural numbers is derived*

Zach Hauk

Bleh

use induction

CausingUnderflowsEverywhere

01:33

or you just came here in the NICK of time, no worries

Compulsive Mathurbator

I'm going to assume that urbandictionary's definition of nick flash is wrong

Now I have seen my ignorance

Zach Hauk

Base case: n=1, so we have 1 = 1(2)/2 which is correct

Compulsive Mathurbator

My pride is NICKed

Zach Hauk

Induction: assume 1 + 2 + ... + k = k(k+1)/2. We have that 1 + 2 + ... + k + k + 1 = k(k+1)/2 + k + 1 = (k(k+1) + 2k + 2)/2

CausingUnderflowsEverywhere

would you reckon a mathematician used trial and error or their intelligent pattern identifiying skills to just see that pattern and come up with the formula

Compulsive Mathurbator

01:35

@CausingUnderflowsEverywhere No, its actually an arithmetic progression

CausingUnderflowsEverywhere

okay let me re-check my code and try again

Compulsive Mathurbator

Or I assume that's how they discovered it first

Zach Hauk

= (k^2 + 3k + 2) / 2 = (k+1)(k+2)/2

thus the equation holds for k+1

Apply induction, Q.E.D.

@CausingUnderflowsEverywhere draw it on a grid, you'll see why we have it

CausingUnderflowsEverywhere

what was the nickname they gave me in the other chat..

Zach Hauk

(that is, what is the rectangle with area n(n+1)? and why is half of it equal to the area of that staircase looking thing?

CausingUnderflowsEverywhere

01:47

it sounds familiar, but I cant find a picture of it

and whats QED?

Zach Hauk

Quod Erat Demonstrandum

Q.E.D.

Q.E.D. (also written QED) is an initialism of the Latin phrase quod erat demonstrandum, meaning "what was to be demonstrated", or, less formally, "thus it has been demonstrated". The phrase is traditionally placed in its abbreviated form at the end of a mathematical proof or philosophical argument when the original proposition has been exactly restated as the conclusion of the demonstration. The abbreviation thus signals the completion of the proof. == Etymology and early use == The phrase quod erat demonstrandum is a translation into Latin from the Greek ὅπερ ἔδει δεῖξαι (hoper edei deix...

CausingUnderflowsEverywhere

Demonstratum sounds my kind of stuff

Compulsive Mathurbator

I've thought it was 'demonstatum' for the longest of times. RIP latin

Zach Hauk

Yeah me too

then I was looking it up to post a link

and it was Demonstrandum :(

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$Mathematics$ Mathematics