I feel like its not true that for a finite cw complex X, pi_n(X) is a fg pi_1(X)-module

definitely not for n>2

ah

but n=2 is, like, nice enough

youre exploiting pi_2(X)=H_2(Xcover)

thats pretty smart

yeah, and i explicitly reduce to the 2-skeleton

whose universal cover i can give explicitly

Nice I like that

That works

What about $\pi_3 = \Bbb Q$ lmfao

@Thorgott What's an example btw?

Maybe $\pi_3(S^2 \vee S^1)$?

@BalarkaSen absolutely clueless lol

this idea with the universal cover is something you can use to describe the general effect of cell attachment on homotopy groups

@BalarkaSen i think that's it

ok cool

i see why not

oh yeah

you get a lot of non-trivial Whitehead products

so all the examples are fg $\Bbb Z[t, t^{-1}]$-modules

for the $\pi_2$ torsion free but not free case

we have $\pi_3(S^2\vee S^1)=\pi_3(\vee_{n\in\mathbb{Z}}S^2)$ by the universal cover and the Whitehead products of various summands should embed the group of infinite matrices of finite rank into this or sth?

yeah thats what i had in mind

computing $\pi_3$ of a finite wedge of $2$-spheres is one of my favorite computations

homotopy classification of $1$-connected $4$-manifolds

@BalarkaSen idk you could get stranger $\pi_1$s, could you not

Ah OK fair enough

my gut says $\pi_3=\mathbb{Q}$ is possible, but im not gonna try and actually do it

@AlessandroCodenotti do you know someone that knows descriptive set theory?

It's honestly sort of shocking that there doesnt seem to be a Kahler manifold with $\pi_2 = \Bbb Z[1/p]$, given how easy it is to construct a manifold with that property.

i mean, being Kähler is super restrictive

my only intuition for Kähler manifolds is how rigid they are

You'd think you can produce some algebraic surfaces with oddly positioned divisors that lift to a necklace of $S^2$'s giving torsion-free $\pi_2$ with complicated relations nevertheless

the thing is that the relations consist of finitely many $\pi_1$-orbits lol

True, by what you said

a presentation of $\mathbb{Q}$ is, like, generators $x_n$ with relations $x_n=(n+1)x_{n+1}$

and that doesn't behave well with any action by homomorphisms

this train of thought is what made me convinced it wasn't possible lol

Q is an awful group yeah

I love the 8-element group $Q$.

What other non-abelian group has the property that every subgroup is normal? :D

i think i read the answer to that once

philosophical question: suppose you have an alternative, more elegant proof, or a significant simplification to/of an existing proof, that the original author had missed, but you are reasonably confident that the original author would eventually get to your argument by himself if you allow him some time. Is it "ethical" to publish your new proof?

strictly speaking, you're not doing anything wrong, of course.

thats my least favourite subject

Howdy, Joe Shmo. What do you mean by "publish"?

How substantial and how old is the original result?

original result brand new. considerable result. not earth shattering

how much time do you allow? how fast do patent rights decay?

asking for a friend

publish, as in publish a paper

friend named Joe?

Oh, brand new result. It's a question of how interesting the result and the new proof are, and up to the judgment of the referee(s). Have you asked your adviser, for example, for advice?

@Jakobian oh some joe schmo

Certainly independent and very different proofs can be published, if there's enough interest. One must reference the original and be honest about it.

Since your "friend" wouldn't have thought of this without the original work and result, one polite route might be to contact the original author and see if there's any interest in a joint paper. Again, this depends on how interesting to the "public" the results are in the first place and what level journals we're talking about.

that's what I'm thinking too. new result uses a completely different argument. more elegant. I think I'm gonna tell him to contact original author and offer a joint paper. its the right thing to do, and an easy way to make friends

How can I show a polynomial $x^4 + 2x^3 + x + 1$ is irreducible in $\mathbb{Q}[x]$ if the rational root test & eisenstein criterion don't apply?

Rational root test is not very helpful when degree $>3$, @Obliv.

My first try would be reduction mod some convenient primes. If $\bar f\in\Bbb Z_p[x]$ is irreducible, then $f\in\Bbb Z[x]$ is irreducible over $\Bbb Q$.

(removed)

nvm I can use that

If it has no roots, then, yes, you must show it cannot be factored as a product of quadratics. Try undetermined coefficients, then.

I'll do undetermined coefficients for now

@TedShifrin it seems to work out pretty effortlessly with $p=2$

I always told my students that's a last resort. (You can sometimes be tricky and shift and then use Eisenstein, but this one doesn't have that look.)

yeah I forgot how to shift

@Thor Yes. Particularly if one has a table of irreducibles mod 2 and 3. I have that in my book :)

But I agree that $x^4+x+1$ and $x^4+x^3+1$ are well-known irreducibles mod 2 ... or so I vaguely remember from 10 years ago.

FWIW wolfram alpha can also do this at least for specified values of p if you ask (the given one is irreducible mod 2 or 3 but not mod 5)

there's only so many quadratics you have to try

Right. Undetermined coefficients is easier mod 2 and 3. But I used to make my students make a table using the sieve of Erothostenes idea (misspelled).

@TedShifrin princiPAL. :P

Or maybe princiPLE.

@Xander Who's the principal, in principle?

Probably Gauss.

Well, we value him.

Euler

@Jakobian depends on what "know" and "knows" mean, but I'd say that I do, why?

is there a name for a natural analogue of an isomorphism at the level of categories of categories?

like i guess i am thinking of a functor that "bijective"?

Equivalence of categories?

oh i see

yes i think this is what i was looking for t hanks

I need some help proving this.

Hello everyone! Having a bit of trouble with the terminology here. What is a "Zero-mean random walk"? What does it mean for a stochastic process to have zero-mean? Math expectation to be zero? But it's true for any random walk, why clarify that? Or I'm missing something? Thanks

@AlessandroCodenotti I'm not very good with the theory so I am looking for someone to look at a question of mine concerning Borel maps

@SillyGoose isomorphism of categories

@MagnusAlexander not every random walk is symmetric

If you start at 0 and move forward with 1/3 probability and backward with 2/3 probability then expected value after n steps will be non-zero

The walk likes to move backward more

@ThomasFinley still need help?

@Jakobian the one from the other day?

I was asking a couple of them

I'm thinking about this one specifically

Its at a weird intersection of topology and measure theory that I'm not sure how to handle

Seems suspicious. What happens if X is not Borel in any of its Polish completions for example?

I have no idea but many maps can be extended, say the identity map from X to X

I barely have intuition about this

Actually the spaces I mentioned above are the only obstruction. If X is Borel in its completion just define the map to be constant in the remainder $\tilde{X}\setminus X$, without even extending the codomain

Yeah but thats the point of extendending the codomain, $X$ can be a set of size $\aleph_1 <\mathfrak{c}$ and then you need to work around this

If X is non-measurable

@TedShifrin If i understand correctly now, the dot product is key because, for a normal vector $\mathbf n$ to a hyper plane, any vector $\mathbf x$ with dot product $\mathbf n \cdot \mathbf x = 0$ must lie within the hyperplane perpendicular to $\mathbf n$ and therefore, since this is 3-space and $\mathbf n$ is perpendicular to my plane, $\mathbf x$ would lie in the plane.

@Jakobian unfortunately, yes.

The $\pi$ 's are not distinct, right?

@ThomasFinley they aren't

take $x = \pi_1...\pi_n$ and $y = \pi_1'...\pi_m'$

what is your idea on how to construct their gcd from this?

like, what would be your idea to construct for natural numbers lets say

you take prime numbers which divide both right

waiting for response

Is it correct to say that any injection of a set into itself must be a bijection?

I have a suspicion that yes

@EE18 no

it fails for infinite sets

i will have to think about that, would not have expected it

What does it mean for two sets to have the same cardinality?

there exists a bijection between them

and if you remove a point from an infinite set, then what cardinality does it have?

the very same

so if you take an infinite set $X$ and $x\in X$, then there exists a bijection from $X$ to $X\setminus\{x\}$

I see. And that's a bijection (and in particular an injection) which is not onto when you consider the induced map extending the codomain back to $X$

thanks very much Jakobian

you're welcome. There was probably some axiom of choice involved here

definitely

might be equivalent to choice even

This is all coming from me trying to stitch this (math.stackexchange.com/questions/3641467/…) up in my mind

The answerer seems to say things like "obviously $Y \subset X$" and obviously that's not true but I figure there is some manner of speaking where the image of $Y$ in $X$ is a subset of $X$ (of course)

Schroder-Berstain

lol

yes

that too

@EE18 I don't see the word obviously

case closed

@Jakobian in that case, we take the common factors to both and say their product as the gcd of the two nos.

But in case of general R[x], the prime factorizations are unique upto assiciates and this creates the problem.

Its only actually a problem

you take $\pi_m'$ first, check if it divides $x$. If yes, good, let $c_m = \pi_m'$. If not, let $c_m = 1$

@Jakobian Furthermore, when we are saying a=\pi_1\pi_2...\pi_m$ we may have $\pi_i=\pi_j$ for some $i\neq j$ and $i,j\in\{1,2,\cdots,m\}.$

Then take $\pi_{m-1}'$, check if it divides $x/c_m$. If yes, $c_{m-1} = c_m\cdot \pi_{m-1}'$, if not $c_{m-1} = c_m$

go on this way to exhaust all possible $\pi_1', ..., \pi_m'$

$c_1$ will be the gcd of $x$ and $y$

@EE18 Someone may have said this already, but actually you would have expected it. Think about the even integers inside the set of all integers.

@Jakobian that's actually looks pretty neat. Let me ponder upon it a bit

@Jakobian its not actually a problem*

is someone forcing on you the other notation?

Most of all I need to understand it, as he always uses leibnitz notation

what does differentiating with respect to ds(t) mean

@Jakobian but how to establish this claim?

@TedShifrin You're talking about the injection of the set of all integers into the set of all even integers, a subset of all integers?

@ThomasFinley first show it divides both $x$ and $y$. Can you do that?

@Jakobian What do you mean?

@Curio you wrote /ds(t), that's very abusive of notation

I mean, it's ds but s = s(t)

@Jakobian I guess the "obviously" was tacit in there being no further elaboration. I guess the thing is that I need to somehow "interpret" that to really mean something like $f(g(X)) \subset f(Y) \subset X$ where $f:Y \to X$ and $g:X \to Y$ are the two bijections into a subset of the codomain which obtain from the hypotheses

@Jakobian I can say, $c_m|x,y$ but then how to show that $c_{m-1}|x$ and $c_{m-1}|y$ ? If I am able to do this for all $c's$ until $c_2$ then I can show what you asked.

@Curio see, this is why people shouldn't use Leibniz notation

$c_{m-1}|x,y$ if $c_m=c_{m-1}.$

But what if $c_{m-1}=\pi'_{m-1}c_m$? We have, $\pi'_{m-1},c_m|a,b$ but then?

@ThomasFinley $c_1$ is a product of $\pi_i'$ for some $i\in \{1, ..., m\}$

so naturally it must divide $y$

@Jakobian yes.

look at the construction, if $c_k$ divides $x$, then $c_{k-1}$ was chosen to also divide $x$

by induction, $c_1$ divides $x$

@Jakobian this seems to be confusing me.

$c_{k-1} = \pi_{k-1}'\cdot c_k$ if $\pi_{k-1}'$ divides $x/c_k$

obviously in this case $c_{k-1}$ divides $x$

and it was chosen to be $c_k$ if not. So again, $c_{k-1} = c_k$ divides $x$

@ThomasFinley grab some paper

to keep track of the argument

I don't know. Maybe also think of why are we writing $c_m$ like this and why are we choosing them like this

think of natural numbers

the point is, you can do it precisely the same as how you operate on natural numbers and their divisors

@EE18 Actually, I was injecting the even integers into all integers.

@Jakobian yes, I get it.

I am convinced that $c_1|x,y$

alright, now take some $z$ such that $z| y$ and $z | x$

@Jakobian ok

what does $z| y$ tell you if you know that $y = \pi_1'...\pi_m'$?

$z|$ atleast one of $\pi'$

no

it tells you that $y = z\cdot y'$ for some $y'$

if you factorize $z$ and $y'$ into prime elements, what does it tell you?

that there is some unit $u$ and some subset $S\subseteq \{1, ..., m\}$ such that $z$ is a product of $u$ and $\pi_i'$ for $i\in S$

@Jakobian Wait, aren't we assuming that $z$ and $y'$ are non units then?

why does $a\mid b$ and $b \mid a$ imply $a = \pm b$ and not just $a =b$ for integers. I guess associates in general rings

we never assumed that, you made that assumption yourself

we didn't assume anything that I didn't say

@Jakobian But that's from the definition of UFD?

$z$ isn't prime, the only hidden assumption I made is for $z$ to be non-zero

@ThomasFinley what is

I don't know what you're saying right now

@Jakobian This is the defn of UFD I am using as in the picture.

@ThomasFinley okay?

I don't see the conclusion you're making out of this all

@Jakobian so, if we are factorizing something, then, it means that it is not a unit?

I knew what UFD is before this

@ThomasFinley you can make an empty factorization its not a point of interest

@Jakobian I thought there might be issues with the defn I was using. But apparently, it seems that is not the case

@Jakobian what is an "empty" factorization?

Empty product $\prod_{i\in \emptyset} p_i$ is just $1$

so empty factorization would mean to write something as product of a unit and $1$

@Jakobian okay, that sounds good

okay to be more careful about the empty case I should have wrote $x = u\pi_1...\pi_n$ and $y = w\pi_1'...\pi_m'$ where $u, w$ are units, my bad

but the argument stays the same

@Jakobian yes, now that sounds more clearer

@Jakobian Yes, up until $c_1|x,y$

$z$ is now a product of a unit and some of the $\pi_i'$

From unique factorization property of UFD's and equality $y = zy'$ for some $y'$

@Jakobian I have studied unique factorization property in case of non units only, so this again not seems good with me.

whats the difference

@Jakobian A quick question for you given what we were briefly discussing about the axiom of extension yesterday not being a trivial definition of equality like i had wrongly suspected

Halmos gives it as follows: "Axiom. of extension. Two sets are equal if and only if they have the same elements."

A good summary (I think) of our discussion would be that the "if" direction in the above is nontrivial/a fact about sets that we are taking/defining

My question is would you agree that the "only if" direction is trivial/not even needed?

it is trivial, yes

I guess this hinges on some deeper mathematical logic that I am not familiar with around what equality really means

@Jakobian if y' is a unit then how is this valid?

it follows from equality as a logical symbol and what not

gotcha thank you

What is the proof for alpha + dAlpha in the picture above?

Draw it as an actual triangle, @Curio.

Do you mean PP'C?

Yes, that's what I meant.

@EE18 first there's logic, natural numbers and things we don't question, then there's set theory

It would be better to do this with a blackboard.

Think about adding up the angles of a triangle and working with the total angle at $P'$.

agreed, definitely not gonna wade that deep at this point. but wouldn't you swap set theory and natural numbers in that order you gave?

the point of von Neumann defnition of natural numbers $0 := \emptyset, 1 = \{0\}, 2 = \{0,1\}$ isn't to define natural numbers, but to define them in set theory

Ah I see. But do we need to know what they are to do set theory in the first place?

yes we do

how do they crop up as primitives in set theory i am curious?

what do you mean

crop up as ... is very idiomatic English :)

show up as ... might be clearer

oh yes, thank you for clarifying Ted

@Jakobian I don't like the word trivial. It betrays a certain truth about understanding. After you understand something then it's common to think "OH OF COURSE"

even if I interpret that as show up I still don't understand the question

@Obliv Good. I made it a practice to use the word trivial only in its technical sense.

The trivial solution of $Ax=0$ is the $0$ solution, etc.

@Obliv its just a word

in that case trivial meant, always true

even outside of given axioms of set theory

I guess what I mean is that I can see how mathematical logic is absolutely necessary to do set theory (the logical meaning of = as we discussed and much else I am sure). But why are the natural numbers necessary in this same way? Real numbers certainly aren't, for example, so why N? Do they crop up in some recursive definition that we can't make within set theory and without N?

You don't have to talk about the object $\mathbb{N}$ to talk about natural numbers. You can even be a finitist

they aren't?

@Thorgott wrong

@Jakobian Can I be a finitist who doesn't believe in a successor function or well-ordering

look at Godel incompleteness theorems for example. How did he prove them? Using natural numbers. Not as some god given concept, but as how humans understand them

@Obliv. Only if you allow sets no larger than 100 elements.

the natural numbers I'm talking about here are not a defined concept, but something every human being knows

@TedShifrin but how can I count to 100 if I don't believe in counting D:

You can use your fingers.

No, each individual finger represents a totally different thing. I would never treat them as objects of the same set.

Its kind of outside my scope to talk about those things, but I know natural numbers come before set theory, because thats what Kunnen tells me

not after set theory

also there was a post here talking about how we use natural numbers to define set theory somewhere on the site

@Obliv shows Obliv a certain middle finger :D

you can't just be this full-fledged formalist you need to start with something

@TedShifrin lol but I wonder if such a system can exist. You can still have bijections even in a non-ordered set, but not being able to count means you don't know the size/cardinality

That's why there are cardinal numbers ... for when you can't "count."

How come you can map $\mathbb{R} \to \mathbb{R}^n$ again? It depends on the function right

continuously and surjectively?

bijectively

not sure what you mean by continuously, that's an analysis term I haven't taken that yet

It certainly won't be continuous and bijective.

I know @XanderHenderson mentioned it wouldn't be an interesting mapping because it doesn't preserve certain structures

Do you know how to do it with $\Bbb N\to\Bbb N^2$? That is very concrete.

Xander probably meant things like continuity and so on

What does "interesting" mean? I don't think I can write down an explicit map.

Idk how to do piece-wise function notation but basically you take $2n \in \mathbb{N}$ and map that to $a \in (a,b)$ and the odds to $b$ or something right

I was thinking of the usual weaving through the checkerboard argument.

I don't know if you need choice for $\mathfrak{c}^n = \mathfrak{c}$ or not

is that $n+1 \to a$ and $n+2 \to b$

You can give an injection $\Bbb N^2\to\Bbb N$ by using primes, but I don't know how to get a bijection in a similar manner.

Isn't it just a variation of Hilbert's hotel

I don't see that.

oh wait $\mathbb{N}^2$ to $\mathbb{N}$

By the way, @Obliv, there's a very powerful result called the Schroeder-Bernstein Theorem. If you can give injections A\to B and B\to A, then there exists a bijection. The proof is quite beautiful, but it shows you cannot just write down naive arguments.

@TedShifrin map $x\in\Bbb N$ to $(n,m)$, where $n$ is the highest power of $2$ dividing $x$ and $m$ is such that $x/2^n=2m+1$

Ah, of course, so that's an arithmetic bijection.

@Alessandro Does one know how to write this down at all for $\Bbb R\to\Bbb R^2$ or vice-versa?

can you map $\mathbb{R}^2$ injectively into $\mathbb{R}$ though

@Curio Did you figure it out?

The usual "zig-zag on the infinite checkerboard" can also be written as an explicit formula but it is a bit annoying

Right. I have on occasion written the formula for that, but why bother.

why would it generalize

That certainly does not, @Jakobian.

@TedShifrin The most concrete one I know, from $(0,1)^2$ to $(0,1)$ is to take two reals $0.a_0a_1a_2\ldots$ and $0.b_0b_1b_2\ldots$ and map them to the real $0.a_0b_0a_1b_1a_2b_2\ldots$

Oh, I guess if we work with dyadic decimals in $[0,1]$, we can give a bijection $[0,1]^2\to [0,1]$.

Right. I just thought of that as you typed that, Alessandro.

but one needs to be a little careful with reals that have two decimal expansions

Right.

But that's the standard argument that $\{0,1\}^\omega \cong \{0,1\}^{2\omega}$, etc.

Whence the Cantor set is in bijection with any power of itself.

I am briefly reading Halmos right now Jakobian and it is rewarding to see that two of the things we were discussing yesterday come up within his first 6 pages! Reading this would not have made sense if not for our discussion so thank you very much again

More generally the Gödel pairing functions gives a bijection $\kappa\to\kappa^2$ for all infinite cardinals

Now you're over my head, @Alessandro. What is that?

just the function from the argument that $\kappa^2 = \kappa$

Well, that was enlightening.

Order $\kappa^2$ by $(\alpha,\beta)<(\gamma,\delta)$ if ($\max\{\alpha,\beta\}<\max\{\gamma,\delta\}$) or (the maximums are equal and $\alpha<\gamma$) or (the maximum and first coordinates are equal and $\beta<\delta$)

so first order by maximum, then by first coordinate, then by second coordinate

@AlessandroCodenotti why would a number have 2 decimal expansions?

Ah, sorta like the lexicographic ordering.

@Obliv TWO numbers.

Oh, you meant non-uniqueness.

Look at 0.09999999999 ...

looks at it

$0.999999....\neq 1$

@Jakobian LOL

clearly they differ by $0.00000000....1$

Right but doing it in this order ensures that the result order on $\kappa^2$ is a well order

@Jakobian are you going to try to convince me that .9999... = 1? and so they are two decimal expansions of the same number or something

And it turns out that it has order type $\kappa$, so mapping every pair in $\kappa^2$ to its position in this order gives a bijection $\kappa^2\to\kappa$

I haven't thought about such things in 50+ years, but thanks, Alessandro.

@Obliv All numbers whose decimal expansion is finite also have an infinite expansion ending with an infinite string of $9$s

@Obliv No. I'm just making fun of the $0.9999... = 1$ deniers

So we do not allow an ending infinite sequence of 9s.

I don't even know what territory we're in in mathematics right now but I don't like it.

I would say elementary set theory and I'm sad you don't like it

high school pretty much

@Jakobian I don't know the formal argument for why .9999... = 1 so I assumed they weren't equal lol

@Obliv whats the definition of $0.9999....$?

Well you need to define what .99999.... means, once you have a proper definition it should follow very quickly

a person who advocates for they are different will never tell you what 0.9999... is

@Obliv: Hint: This is a geometric series. To what does it converge?

I didn't take analysis yet so I don't know anything about $\mathbb{R}$ tbh

once you define it, you see its just a geometric series $9/10 + 9/10^2 + ...$ by definition

That's nonsense. This is just basic high school/calculus.

and so, it equals $9/(10(1-1/10)) = 1$

You can consider expressions of the form $0.a_1a_2...$ and give them a natural ordering

but the problem is that if you don't do identifications that we do for real numbers, then this order has successors

I understand that $9/10 + 9/10^2 + ...$ "converges" because of calc 2 but is that the definition of equality?

its not an ordering we would like to have

Yes, the meaning of the series $\sum a_n$ as a number is the value of the limit of the partial sums.

@Obliv I don't know what you mean by definition of equality

Obliv: in high school this all isn't formalized

so technically you do need to define stuff about $\mathbb{R}$ and what $\mathbb{R}$ is, yes