Mathematics - 2017-03-14 (page 1 of 4)

Mike Miller

00:00

@arctictern 40% my new best

probably could have gotten the four corners if it wasn't for some pesky guy on the top wall

arctic tern

@MikeMiller cool

Akiva Weinberger

Reading the inverse function theorem proof right now. I guess the idea is that $f(x)$ is close to its tangent plane, which is invertible, so $f$ is invertible as well. As the inverse of the tangent plane is Lipschitz continuous, so the inverse of $f$ is as well.

Haven't gotten up to the differentiable part yet, but I think that's the intuition

Locally invertible and locally Lipschitz, at least.

So invertible functions whose inverses are not locally Lipschitz don't have invertible tangent planes. Which makes sense; an example would be $y=x^3$ at the origin.

Adeek

@MikeMiller yeah my answer was correct math.stackexchange.com/questions/2185588/…

Qiaochu Yuan has better argument though

very short

Akiva Weinberger

00:46

And it's differentiable because the inverse of the tangent is the tangent of the inverse, which seems plausible geometrically and can be proven rigorously after what looks like not too much effort

And the derivative is continuous, because the derivative of the original function is continuous and that doesn't change after inverting the planes

Secret

00:58

Hey guys, I have a question concerning multivariable functions and possibly group theory:

Meow Mix

@Akiva so do homeomorphisms preserve open sets?

Secret

Consider an antisymmetric function of the form $a(x,y)=-a(y,x)$. Suppose x,y are coordinates of the xy plane, how to work out the spatial group symmetry being induced by the antisymmetric relation above?

Akiva Weinberger

@MeowMix Yes

In fact, one could define a homeomorphism to be a bijection that doesn't destroy or create open sets

Secret

My initial suspicion is it is some kind of centrosymmetry, that is, the function flip sign whenever it is reflected about the origin

Akiva Weinberger

(i.e. $f(A)$ is open iff $A$ is open)

Meow Mix

01:00

And this is between two topological spaces, right?

Akiva Weinberger

Yeah

@Secret What do you mean by the spatial group symmetry

Like, the symmetry group of the function's graph?

Secret

yeah, something like translation, sheer, rotation, inversion etc.

a symmetry group that acts on the points of the function

Akiva Weinberger

Of the function's graph or the function's domain

Looking at the graph, we have that $(x,y,a)$ is in the graph iff $(y,x,-a)$ is

Meow Mix

@AkivaWeinberger I understand the topological limit definition

Akiva Weinberger

So the matrix $\begin{bmatrix}0&1&0\\1&0&0\\0&0&-1\end{bmatrix}$, I guess.

Which is a rotation.

Secret

01:03

I think the image of the function, that is when one plot it, the surface formed by the function (is that what we called the function's graph)

Akiva Weinberger

Yeah, that's the graph I'm talking about

It's a 180 degree rotation about the vector $(1,1,0)$, I think.

Meow Mix

Is a continuous function a homeomorphism from $\Bbb R$ to itself?

Yash Farooqui

there's only one non-identity symmetry in general. So your symmetry group would be isomorphic to Z_2.

Akiva Weinberger

All continuous bijection between $\Bbb R$ and itself is a homeomorphism, if that's what you mean

Meow Mix

Oh yeah :P

because it must have a continuous inverse

Akiva Weinberger

01:06

Normally, we require the inverse to be continuous also for it to count as a homeomorphism, but I think that in this case it follows

Meow Mix

By the way, all bijections of $\Bbb R$ are either increasing or decreasing?

That is, everywhere negative derivative or everywhere positive derivative

Akiva Weinberger

Yeah. (Continuous ones)

Yash Farooqui

Not in general.

Akiva Weinberger

@MeowMix Well, now you're assuming they're differentiable

Yash Farooqui

Continuity is sufficient yes.

to force bijections to be monotonic

Secret

01:08

If we instead seek for the symmetry group of that function's domain (which should be $\mathbb{R}^2$) is that symmetry "reflection about y=x"?

Akiva Weinberger

Monotonic functions (aka increasing or decreasing) need not be differentiable

Yash Farooqui

yup yup :)

Akiva Weinberger

@Secret Well, I'm not actually sure what that would mean, to be honest

Secret

ok I see

Meow Mix

@AkivaWeinberger good point

they could be scattered everywhere

Yash Farooqui

01:10

wait. Are you sure that there exist nondifferentiable strictly monotonic continuous functions?

It sounded like a thing at first but upon further thought I'm not sure.

strict monotonicity is a pretty strong condition after all

(That are also bijective)

like I'm pretty sure that

strictly monotonic forces smoothness except on a set of measure zero

at max

Akiva Weinberger

@YashFarooqui Well, we could have the graph be the union of two rays

but I think there should be a nowhere differentiable example as well…

Yash Farooqui

that's still smooth. And you won't find one

I can prove that to you

Akiva Weinberger

The Cantor staircase function plus $x$ would also work

Yash Farooqui

Strictly monotonic

Akiva Weinberger

Plus $x$, I said.

@YashFarooqui How would it be smooth?

Yash Farooqui

01:16

except on a set of measure zero

Akiva Weinberger

Ah.

You could modify the Cantor staircase function for that.

Yash Farooqui

The except on a set of measure zero is like a pretty important thing.

Think about it

Akiva Weinberger

Modify the Cantor staircase to work on a fat Cantor set

Yash Farooqui

you won't be able to without violating strict monotonicity

Akiva Weinberger

(and then add $x$)

Meow Mix

01:18

@AkivaWeinberger do you have any cool analysis question?

Yash Farooqui

Here

suppose that f is finite on an open interval (a,b)

Akiva Weinberger

@YashFarooqui Hm, seems like you're right

Not sure what's wrong with the modified Cantor function thing, though

Yash Farooqui

oh okay so you probably saw the proof

I don't know either

we can think about it

Akiva Weinberger

@YashFarooqui I didn't see a proof,

Yash Farooqui

here's one linked to me by stackexchange

Akiva Weinberger

01:20

I saw someone on Stack Exchange say that a proof existed in a book somewhere

Yash Farooqui

books.google.com/…

take a minute or two and convince yourself?

differentiability is ugly...

but it's similar enough to the definition of integration that it's kinda ok I gues

so I think that

the construction of the staircase falls apart on nonmeasure zero sets

Akiva Weinberger

I think I'd want to read that book from the beginning to understand that

Yash Farooqui

math.ucla.edu/~ralston/245a.1.08f/Vitali.pdf

this is a cool lemma

I should study more real analysis XD

oh here's a quick soft question I had

Secret

Speaking about reading a book from the beginning. I think I need to reread my DFT book because I don't know what I have been reading in the past 3 days due to too many interruptions on the flow emotional state

Akiva Weinberger

DFT?

Yash Farooqui

01:28

I seem to remember in a - I want to say analysis? - class I learned about interesting results involving Taylor series with zero radius of convergence

and now I'm just wondering

Akiva Weinberger

Factorial coefficients

Yash Farooqui

supposing I have such a function, is there a way to get a handle on the taylor coefficients of that function?

the usual complex analytic way won't work

Akiva Weinberger

@MeowMix Can you find a continuous bijection between two subsets of $\Bbb R^2$ that is not a homeomorphism? (If an example has not already been given to you)

Yash Farooqui

and differentiation is ... well possible but eww

differentiation is not bounded and approximations become ugly.

Meow Mix

so a homeomorphism would also have a continuous inverse, right?

Akiva Weinberger

01:30

Yeah

Meow Mix

so i wanna find a continuous bijection whose inverse is not continuous

Akiva Weinberger

Yup

Yash Farooqui

yeah Akiva, the ogf for n! is f(x) = integral from t = 0 to inf of e^-t /(1-xt)

Meow Mix

Oh, $\Bbb R^2$

that's complicated

Yash Farooqui

Since you just take the Laplace Transform of the exponential generating function

Akiva Weinberger

01:31

They might not be complicated subsets @MeowMix

@YashFarooqui I thought $\sum n!x^n$ had zero radius of convergence, though

Yash Farooqui

it does

Forever Mozart

I constructed an example from hell

Yash Farooqui

check that the function I described

is not defined for x != 0

Forever Mozart

my finest construction yet

Yash Farooqui

err

x>0

Forever Mozart

01:32

they shall call me the "king of counterexample"

Meow Mix

What does continuous mean in $\Bbb R^2$? The image of the neighborhood around a point is a neighborhood around the image of that point?

Akiva Weinberger

@MeowMix That's not even true for continuous functions in $\Bbb R$. Take $y=x^2$; neighborhoods of zero don't map to neighborhoods of zero

Yash Farooqui

Forever Mozart, what is your counterxample? ^.^ since you seem like you want to share

Forever Mozart

takes many pages to describe

Yash Farooqui

yes meow mix, basically

Forever Mozart

01:34

i'm just here to brag :)

Akiva Weinberger

The real definition is that the inverse images of open sets are open (or the inverse images of neighborhoods are neighborhoods)

Yash Farooqui

okay! I just wanted to give you the outlet if you wanted to share. Congrats!

Forever Mozart

are you familiar with basic topology?

Yash Farooqui

Yeah, but in R^n it's not that big of a deal to mix those up

yes

Akiva Weinberger

but the intuitive definition ("points near each other map to points near each other") should be fine

Forever Mozart

01:35

I can describe the property of the example then

Yash Farooqui

Ok.

Secret

@ForeverMozart Is it possible to drew on paper, or it has too much fractal like structure?

Akiva Weinberger

@YashFarooqui See my $y=x^2$ example above.

Forever Mozart

If $X$ is a connected space and $p,q\in X$ then $X$ is irreducible between $p$ and $q$ if no proper closed connected subset of $X$ contains both $p$ and $q$.

Yash Farooqui

f(x) = x^2 is a 1 dimensional function and it is continuous everywhere in a topological sense (I am fairly sure - unless topologists are incredibly bad people). I don't see the problem.

Forever Mozart

01:36

I proved that there is a connected $X$ that is irreducible between every two of its points, but such that every compactification of $X$ is reducible between every two points of $X$

Yash Farooqui

ugh

Forever Mozart

this is a major breakthrough in my research

Yash Farooqui

yeah I mean congrats

but also ugh that counterexample must suck

Forever Mozart

it sits in $\mathbb R ^3$ but cannot be drawn

Akiva Weinberger

@YashFarooqui I mean, it's continuous but it doesn't satisfy Zach's (Meow Mix's) proposed definition.

Yash Farooqui

01:37

like not suck as in be a bad counterexample but suck as in be a horrible function

Forever Mozart

@YashFarooqui it has very strange structures

@MikeMiller does that sound interesting?

Yash Farooqui

ok. I agree. I will admit I didn't read the definition too closely - I misread as "the neighborhood around a point lies in the neighborhood of its image"

*"lies in a neighborhood"

it sounds cool Forever Mozart. I can't really imagine a space like that but I guess that's the point

Forever Mozart

its totally crazy. over a year of work to reach this point

and still the big conjecture remains...

unproved

Yash Farooqui

well. Research takes time and you made a step!

Forever Mozart

yes

Secret

01:42

[Random conjecture] Anything that cannot be drawn on a piece of paper must be either a dense set, a structure that has fractal like properties, a product of an infinite process (e.g. consider alexander horned sphere)

From what is described here, it might seemed this conjecture is false

Well, when you got the paper out, I would really love to have a look at that space you proved

I like pathological counterexamples

Forever Mozart

I have a feeling that the big conjecture is true now, cause I'm pretty good at finding counterexamples.

@Secret it is fractal, yes

@Secret if I sent you my papers then I would reveal my true identity ;)

Secret

ok then

Forever Mozart

but maybe

Akiva Weinberger

@ForeverMozart Jeremy from the park?

Forever Mozart

huh?

Akiva Weinberger

01:45

Just guessing at your identity

By your reaction I see I must have been right

Forever Mozart

lol

sounds like an idiot savant

Jeremy from the park who swindles people for money over chess

some of the best chess players in the world are homeless in parks

fact

Secret

[Now formalising the pictorial representation conjecture] There exists no undrewable mathematical objects $X$ that is not a result of an infinite process, that is not infinite, that is not a fractal or fractal like or that is not a dense set

Definition: A mathematical object $M$ can be drawn on a piece of paper if its entire mathematical description (given as a class or propositions, relations, equations, axiomatic systems and so on) and properties can be mapped to corresponding pictograms that are subsets of $\mathbb{R}^2$ such that there is no missing information

I need to figure out how to define "fractal like", though: It covers many pathological examples such as the weistrass function, the devil staircase, the alexander horned sphere, nowhere continous functions, the set of rationals and irrationals etc.

Akiva Weinberger

You'd then need to define pictogram, but I kinda like where you're going with this

Secret

A pictogram is any subset of $\mathbb{R}^2$ that can be (...)

O that is not as trivial, I cannot just say "can be drawn on a piece of paper" as that will make the definition cyclic. I need to think about that...

Which does brought out an interesting question, that might be at the intersection of art and maths: What do all physically or digitally representable objects have in common

Akiva Weinberger

02:00

The acronym PICNIC is kind of funny

It's a type of computer problem

"Problem in chair, not in computer"

DHMO

02:38

@AkivaWeinberger The one I heard is PEBKAC

"problem exists between keyboard and chair"

Secret

[Random] To differentiate a function that is a sequence, use finite difference. Now what about differentiate a function f(n) that is a series where n is the number of terms. To be investigated

DHMO

@AkivaWeinberger @Secret basically I only thought about the elements of order 2, and completely forgot about the other elements.

@Secret you mean a function $\Bbb Z \to \BbbR$?

Akiva Weinberger

02:53

A sequence of sequences? @Secret

BAYMAX

Let's all go on a PICNIC :)

DHMO

@AkivaWeinberger I think he just means $f(n) = g(1) + g(2) + \cdots + g(n)$

Forever Mozart

i am preparing such a great paper

going to change life on Earth

BAYMAX

Wow!

Forever Mozart

haha yes

BAYMAX

02:55

What's that paper about?

Forever Mozart

scroll up to around 20:30

DHMO

@ForeverMozart you do know that local time doesn't work

in an international chatroom

BAYMAX

Ok..all the best for your paper... Beautiful mind

Forever Mozart

does anyone watch that show

Halt and Catch fire?

BAYMAX

Also you can be revealed once your paper comes to the wider bulkier internet ... he he :)

Forever Mozart

03:00

I'm binge watching and it's really amazing

BAYMAX

I thnk you must be Vendetta

@ForeverMozart

Just kidding!

Forever Mozart

i'm not looking to create anarchy

PhiNotPi

checks calendar tomorrow's gonna be a rough day

BAYMAX

Is it related to Mathematics Halt and catch fire ?

Forever Mozart

it is related to computers in the 1980s

BAYMAX

03:02

Any Mathematical show guys , any1?

Is it a movie?

Forever Mozart

no a tv show on AMC

3 seasons total

its sort of nerdy so you might like it

BAYMAX

Oh..nice will give a try

Forever Mozart

it's very good

has anyone seen this? imdb.com/title/tt1801123

looks interesting

BAYMAX

Just heard that it's related to LPP Travelling Salesman problem..

Forever Mozart

P=NP i cannot prove it though

too hard for me

BAYMAX

03:08

What's that?

Conjecture

Forever Mozart

yes

famous problem

BAYMAX

Happy $\pi$ day to all!

Forever Mozart

everyone thinks its true

but nobody can prove it, which sucks

PhiNotPi

Everyone thinks P != NP

Yash Farooqui

A lot of people thought bpp != p too

Forever Mozart

03:10

haha pipe dreams

Akiva Weinberger

P=NP when P=0 or N=1

Where's my prize money

PhiNotPi

@YashFarooqui I thought that was also unsolved?

BAYMAX

Ha Ha ... :)

Yash Farooqui

But idk much about the problem aside from the fact that it's provable that basically every method anyone has thought of will fail

Forever Mozart

@AkivaWeinberger wow

Yash Farooqui

03:11

BPP is P, it was proveb

Forever Mozart

very deep

Yash Farooqui

But anyways

Akiva Weinberger

(Whoops)

PhiNotPi

My solution is to use time dilation to speed up algorithms.

Yash Farooqui

Quick question. A book I was reading said that \log(|\zeta(\sigma + iy)|) = sum over k>=1 sum over n>= 1 of (p_k^(-nsigma))/n cos(ny log p_k)

My question is where the isin(ny log p_k) goes

BAYMAX

03:13

Hey guys..everybody relax as I am BAY_the_MAX. :)

Yash Farooqui

Sigma >= 1 so everything's nice except at s=1

Like why does it contribute nothing to the absolute value?

Oh writing that out helped

No nvm it didn't. Where did the sin go?

Forever Mozart

greatest paper ever arxiv.org/abs/1004.4206

read the abstract lol

Yash Farooqui

So basically quantum immortality.paper?

Forever Mozart

`Many Cats' interpretation of Quantum Mechanics LOL

I think it claims to prove mortality

Yash Farooqui

Seriously though can someone help with the zeta sum or is it just an error in the book? It seems too minor for a full question, plus I'm on mobile

Secret

03:17

@DHMO yup, that's how a sequence is defined

Yash Farooqui

Is there a way to attach pictures?

BAYMAX

click on upload button

Aiden Grossman

@ForeverMozart that abstract is pretty funny! the paper though actually looks fairly interesting.

Forever Mozart

@AidenGrossman 12 pages... someone had too much spare time

Secret

@AkivaWeinberger Well the thing about series is that it basically act like an integral, except with a counting measure. I do vaguely recall there's a fundemental theroem of finite difference that allow you to take finite diference of series. What I had not checked yet is whether they work on series which converges for all n finite and diverge otherwise

The harmonic number, fibonacci number and other related series defined numbers are interesting in that they only diverge if n is infinite

Yash Farooqui

03:22

What is the conversation about? Can't you just consider the partial sums or Cauchy sums?

DHMO

@Secret can you diverge if n is finite, for any sequence at all?

Yash Farooqui

You can

*can't

Unless you sum infinite things?

Secret

@YashFarooqui Well I am thinking about the notion of differentiating a series. I knew derivatives are well defined for functions and sequences (where it is known as finite difference), what I am not sure is whether they are also well defined for series that diverge only at infinity

Yash Farooqui

Finite differences is the way to go. The finite differences of infinite series behave like you'd expect- consider the sequences of partial sums, you can't converge to anything else.

Secret

So one can talk about something like $\Delta f(n)=\sum_{i=1}^n f(i) - \sum_{i=1}^{n-1} f(i)$ for any $n$, without need to worry about ordering issues of the terms which commonly plague divergent series?

Yash Farooqui

03:28

You mean conditionally convergent? And I don't believe so but let me try proving it in case I'm wrong

Also those are finite sums

Aiden Grossman

@DHMO you can diverge for some specific sequences. For example alternating the signs of the recirpocals of posotive integers produces a convergent series. en.wikipedia.org/wiki/Convergent_series

Yash Farooqui

Every convergent series has to have the sequence you are summing go to zero

So the sequence of partial sums of delta f(n) goes to f(0)

Err -f(0)

Secret

@AidenGrossman I think he is asking whether you can have a series consists of finite terms and can still diverge

Yash Farooqui

well you can have a sum of two variables, but a finite sum of finite sequences won't diverge.

Aiden Grossman

@Secret Okay, I guess I did not read the part about n being finite.

Yash Farooqui

03:31

Anyways like if you just stick to partial sums you can reafrange all you like

Secret

Ah I see

Another question about derivatives is the following: How will one compute:

$$\frac{d}{dn}f^n(x)$$ where $f^n$ is acting $f$ n times

BAYMAX

still @YashFarooqui You can check out Paul's Online notes on sequences and series they are nice.

Rate of change of derivatives with order of derivatives?

@Secret

Yash Farooqui

Speaking of sequences, for some reason the book I'm reading says log|zeta(sigma + i y) | = sum over k sum over n p_k^(-ns)/n cos(ny log p_k)

Secret

@BAYMAX rate of change of f wrt number of iterations

BAYMAX

ok

$\frac{df}{dx}$ is rate of change of $f$ wrt x right?

Secret

03:37

yes

Yash Farooqui

Now. The log expansion makes sense but I don't get why the imaginary part of the sum is basically irrelevant. Shouldn't |(p^k)^(-n(sigma +iy)| just be |(p_k^nsigma)|?

This is obviously true which makes me feel like I am missing something very obvious

BAYMAX

$\frac{d^{2}f}{dx^{2}}$ is rate of change of $f^{'}$ wrt $x$ , right?

Yash Farooqui

Yes

Secret

@YashFarooqui My first guess will be the imaginary part somehow cancel out, they often do in many cases

as they often end up as a phase factor

Yash Farooqui

Zeta is the riemann zeta function as usual. And yes, that was my first thought too, but it doesn't seem to be the case

I believe it's something to do with |log(1-z)| not being quite what I expect- otherwise it's an error in the book

Log is here

The logarithm excluding negative real axis where log(1)=0

Secret

03:40

@BAYMAX For a bit of a context, imagine computing the rate of change of the function along the k direction in this question

Yash Farooqui

I think that was implicit but never hurts to be explicit. Anyways the imaginary part doesn't magically cancel or the real part would too

BAYMAX

nice @secret you are reviewing your questions

interesting though

Secret

yeah, sometimes looking back at questions you ask will generate a new question or idea to be tested

that's really one of the ways I came up with so many weird ideas

Yash Farooqui

That's cool .

Secret

For example: This PSE question is actually inspired from watching Doctor Strange

1

Q: Experimental test for a universal scale closed timelike curve?

So recently, in chat, I was interested in about what if our universe contains only one CTC (thus making it unlike the Gödel metric, van Stockum dust where there can be more than one CTCs found) and that is the time dimension of the whole universe wrapped back onto itself (meanwhile the spatial di...

Yash Farooqui

03:46

Anyways cancellation can't occur since log p_k is irrational

Since p_k is the kth prime

I'm wondering if the author meant log|Re(zeta)| but that wouldn't make much sense either

Adeek

@arctictern are you here ?

arctic tern

ish

satisfied with my 67%

Adeek

I am proving the following Let $f : A \rightarrow B$ be a morphism of rings. Let M,N be A-modules prove that $(M \otimes_A N) \otimes_A B \cong (M \otimes_A B) \otimes_B (N \otimes_A B)$

Okay I proved that the map given by

arctic tern

left, right or bimodules?

Adeek

$\phi : (M \otimes_A N) \otimes B) \rightarrow (M \otimes_A B) \otimes_B (N \otimes_A B) $ given by $(m \otimes_A n) \otimes_A b \mapsto (m \otimes_A b) \otimes_b (n \otimes_A b)$ is well defined

left

arctic tern

03:51

@Adeek that doesn't strike me as well-defined

Adeek

I proved that this map is well defined I am having troubles constructing the inverse map

why ? Fix arbitrarily $b \in B$ consider $\phi_b : M \times_A N \rightarrow (M \otimes_A B) \otimes_B (N \otimes_A B)$ given by $(m,n) \mapsto (m \otimes_A b) \otimes_B (n \otimes_A b)$ this map is bilinear so it induces

arctic tern

it induces a map $M\otimes_A N\to (M\otimes_AB)\otimes_B(N\otimes_AB)$, sure, but then that is not linear in $b$ so it fails to futher induce a map out of $(M\otimes_AN)\otimes_AB$

Adeek

$\bar{\phi_b} : M \otimes_A N \rightarrow (M \otimes_A B) \otimes_B (N \otimes_A B)$ given by $m \otimes_A n \mapsto (m \otimes_A b) \otimes_B (n \otimes_A b)$ Then we can let b vary and we construct now $\psi : (M \otimes_A N) \times_A B \rightarrow (M \otimes_A B) \otimes_B (N \otimes_A B)$ given by $( (m \otimes_A n),b) \mapsto \bar{\phi_b}(m \otimes_A n)$

this is linear in b as well as linear in $M \otimes_A N$ no ?

arctic tern

$(m\otimes b)\otimes(n\otimes b)$ is not linear in $b$

Adeek

but why is my map $\psi$ above not bilinear ?

arctic tern

03:58

does $\psi(m\otimes n,2b)=2\psi(m\otimes n,b)$?

Adeek

ohh no

that doesn't work

we get 4

arctic tern

because you have two $b$s being tensored ("multiplied"), so it's not linear (it's "quadratic")

Adeek

oh I see

I thought that was very natural map I guess not !

that is why I can't construct a natural inverse

maybe something like this will work ?

$(m \otimes n) \otimes b \mapsto (m \otimes 1) \otimes (n \otimes b)$ ?

@arctictern ?

arctic tern

dunno, I don't know how to see surjectivity of that

Adeek

hmm I will try bunch of maps and see

arctic tern

04:08

what are the $B$-module structures on $M\otimes_A B$ and $N\otimes_A B$?

Adeek

The B-module structure on $M \otimes_A B$ is defined as $b (m \otimes b_1) \mapsto m \otimes (b * b_1)$

arctic tern

oh, then goodie

Adeek

why ?

ohh

ohh yess then the map that I defined works just fine

arctic tern

that's kind of weird though, because it seems to collaps $(M\otimes_AB)\otimes_B(N\otimes_AB)$ quite a bit

Adeek

Yeah that makes sense because otherwise I think that if we define other B-module structure then $(M \otimes_A B) \otimes_B (N \otimes_A B)$ then it will be "bigger" than $(M \otimes_A N) \otimes_A B$

@ramanujan_dirac here ?

ramanujan_dirac

04:15

Yup

Adeek

So the map $(m \otimes n) \otimes b \mapsto (m \otimes 1) \otimes (n \otimes b)$ works

based on how we just define the B-module structure

I just have to go through the details now

arctic tern

note $(m\otimes b)\otimes(n\otimes 1)=(m\otimes 1)\otimes(n\otimes b)$ so it's actually symmetrical

ramanujan_dirac

ohh i see okay

Ahh

Adeek

yeah I was just gonna say @arctictern

that is very cool

it is kinda weird that some objects in tensor product collapse in some non-trivial manner if we consider different build up.

btw @arctictern I am currently learning homological algebra it is so amazing with what I have seen so far.

I can't wait to explore more of it.

Adeek

04:50

@arctictern This multiplication aspect of tensor product is really evident in the example above. For example surjectivity is clear since $(n_1 \otimes_A b_1) \otimes_B (n_2 \otimes_A b_2) = b_1(n_1 \otimes_A 1) \otimes_B (n_2 \otimes_A b_2) = n_1 \otimes_A 1 \otimes_B (n_2 \otimes b_1b_2)$

arctic tern

mmhmm

Mike Miller

how would you want to do #7 here?

Adeek

@MikeMiller are you TAing for calculus ?

Mike Miller

nope, but a friend is

i have no idea how to do that reasonably

Adeek

I wouldn't do it using power series I would just compute the derivative notice a pattern and get explicit formula I think for this one it is very doable to do it like that

Mike Miller

04:59

i don't think that's true for this one at all.

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$Mathematics$ Mathematics