Mathematics - 2018-02-20 (page 2 of 5)

Secret

03:02

$\int_1^y \frac{1}{x} = \ln y$, want $\ln y \leq 1 \implies y \leq e$

nitsua60

Okay: there are 720 permutations of the numbers 1--6. I have a transformation on permutations that partitions those 720 into twelve 60-member equivalence classes.
I have another transformation that can take a permutation from one class into a permutation from another.
I'm trying to put together (a) sequence(s) of those transformations that will traverse all 720 permutations without repeat.
But I'm at a loss for notation. Anyone have any brilliant ideas?

mike239x

720 / 6 = 120, not 60. am I wrong?

nitsua60

You're not wrong. [hurriedly edits...]

mike239x

anyways, how does your second transformation work? does orbit of every permutation consist of exactly 12 permutations from different classes?

nitsua60

It's not a huge problem for the 1--6 case, but it is a huge problem, literally, when I need to work on the 1--8, 1--9, 1--10 cases =)

Second transformation, unfortunately, is only applicable to one-fifth of the permutations. I.e. if one started to use T_1 to walk around class_1 there are only twelve members on which T_2 is a valid operation.

mike239x

03:15

lol

nitsua60

(yeah, I'm not so good at division tonight =\ )

Semiclassical

This feels like s geometric group theory problem

Cayley graphs and such

nitsua60

@Semiclassical Yes. I'm told Cayley diagrams... [sniped]

Semiclassical

Hah. Not that I have anything intelligent to say about them

mike239x

what will happen if I try to use T_2 twice (or even more) on a permutation? assuming I can apply T_2 to it on the first time.

anon

03:19

what does it mean for a transformation on a set (in this case the set of permutations) to partition it?

mike239x

@anon well you take orbits of elements and get a partition

anon

elements of what?

nitsua60

@mike239x T_2(T_2(x))=x, if that's what you're asking?

mike239x

@anon of the set

@nitsua60 yeah, something like that

Semiclassical

Hmm: groupprops.subwiki.org/wiki/…

Daminark

03:21

@Ted so it turns out what I said was true and it shouldn't have taken me this long to see it

anon

of the set of permutations? but the set of permutations acts transitively on {1,..,6}, so it induces a trivial partition on {1,..,6}, not a partition of {720 permutations} into 12 parts

Semiclassical

With generators being (1,2) and (1,2,3,4,5,6)

Daminark

See, $(T-aI)^2$ is a positive operator (obviously symmetric since $T$ is, and $\langle (T-aI)^2x,x\rangle = \|(T-aI)x\|^2$)

nitsua60

So T_2 doesn't nicely create an easy path. T_2 may take element 12 of class_1 to an element of class_2, but might take element 24 of class_1 to an element of class_8, for instance.

Daminark

So its spectrum lies in $[0,\infty)$. Meaning, it excludes $-b^2$

nitsua60

03:22

@Semiclassical I do not have that second generator available to me.

(no element within the permutation is displaced more than one place from its previous position by either T_1 or T_2)

Daminark

Thus, $(T-aI)^2 - (-b^2)I = (T-aI)^2 + b^2I$ is invertible. So we're good

nitsua60

This is the context, if anyone cares: en.wikipedia.org/wiki/Campanology#Method_ringing

mike239x

@anon OP has a single action T_1 which acts on S6 (permutation set), partitioning it.

anon

@mike239x so the partitions are the cycles of T_1, i.e. the orbits of the cyclic subgroup <T_1> within Perm(S_6)?

nitsua60

@anon Yes.

Semiclassical

03:26

If you can’t generate the permutation (123456) by some method then I’m perplexed what you’re hoping for

anon

If you want <T1,T2> to act transitively on S_6 then you need <T_2> to act transitively on S_6/~

nitsua60

@Semiclassical Sorry--lemme check I'm reading that right. Is that an identity you're writing there, or a cyclic permutation of all six elements?

anon

I assume the cycles of T1 are a block system for <T2>

Semiclassical

Cyclic permutation

nitsua60

@anon "block system"?

anon

03:28

@nitsua60 if X is a cycle of T1, then so is T2(X)

nitsua60

@Semiclassical Then, yes, I can't have that. It requires one of the members to "jump" five places within the order, which is physically unmanageable.

anon

so if x,y are in the same cycle of T1, then T2(x),T2(y) cannot be in different cycles

nitsua60

I see what you're saying. It makes me fear I've described something imprecisely/incorrectly.
Let me present a few things I've worked out by hand:

I can take any (starting) permutation and generate an orbit of length 60, using T_1 repeatedly.

Secret

hmm.... how to find a real interval $S$ such that any finite sum of the elements in the interval are bounded above by $1$ but the countable sum of the elements in the interval are exactly 1?

nitsua60

I can, at selected points in an orbit, execute T_2 and generate an orbit of length 120.

I can also, at selected points, execute T_2 and generate an orbit of length 360.

Semiclassical

03:35

What exactly is T1, T2 here?

nitsua60

In real-world terms?

mike239x

maps from S6 to S6, it seems

Semiclassical

In terms of permutations

nitsua60

(gimme a sec)

Secret

that is, how to find $(a,b)$ (where the ends can be open or closed) such that for any $c_i \in (a,b)$, $\sum_{i=0}^n c_i \leq 1$ for all $n \in \Bbb{N}$ and for any $c_i \in (a,b)$, $\sum_{i=0}^{\infty}c_i =1$?

mike239x

03:37

@secret as stated such an interval doesn't exist

nitsua60

@Semiclassical To make sure I've got the notation right (because permutation groups were, like, 20 years ago): (12)(34)(56) would indicate that three neighbor-pairs swap positions, right? And (123) would indicate the first three elements permute cyclically while elements 4-6 do nothing?

@Secret How do you have a countable number of elements in a real interval?

anon

@nitsua60 that's right

nitsua60

okay--one sec then [scribbles furiously]

Secret

@nitsua60 every real interval which a=/=b has uncountably many elements

@mike239x uh, I think I missed that bit of the conversation?

mike239x

@Secret not really, I just skipped the thinking part and told you only the conclusion, my bad

nitsua60

03:42

@Semiclassical T_1 = 5*{5*[(12)(34)(56),(23)(45)],(34)(56),(23)(45)}

T_2 is the choice to replace one of the (34)(56) permutations with (23)(56).

Here's a visualization of T_1:

(Ignore the last non-faded row--it's there to show one that the orbit's complete.)

mike239x

@anon Think of it, if you have a point x which is not 0 inside the interval, you can take sequence made of x only, and its sum will diverge. So the only possible value in the interval is 0. But this way you will not get sum over infinite sequence equal 1. Hence no such interval exists.

Secret

right, make sense

Ted Shifrin

Demonark @Daminark: Oh, the point is that, a posteriori, if $\lambda$ is an eigenvalue, then it's real, so $b=0$. You assumed $b\ne 0$, so, yes, of course it's right.

Secret

I am terrible at working with infinite sums intuitively, they are much harder to visualise than integrals

PVAL-inactive

They're blocky integrals.

Secret

03:51

That question by kanderson8 drove me thinking, on the more general scenario of the criteria for a finite and infinite sum to be bounded above by some fixed value

Daminark

Yeah

PVAL-inactive

if the set is positive the inductively defined sup series I described earlier being less than that value is a complete characterization.

Daminark

But yeah once I figured it out I was kinda annoyed because it probably took me just half an hour to an hour to realize that we'd be done if we knew $(T-aI)^2 + b^2I$ was invertible, and then it took me over an hour to realize that changing "+" to "--" did it

Secret

yeah, never thought of using sup to constrain the sum of areas of each of those rectangles

PVAL-inactive

I think when I first thought about this

Joe Shmo

04:00

@TedShifrin you talk about piecewise smoothness in lecture 33, green's theorem

PVAL-inactive

I was wondering why we do not allow some sort of uncountable subadditivity in the measure axioms.

Secret

Well, Alessandro, Leaky and I have explored that topic briefly, and our conclusion is that uncountable subadditivity screw up translational invariance of lebegue measure

PVAL-inactive

or at least what happens when you try and include that axiom.

well I think for the Lebesgue measure it implies everything is zero

Daminark

Yeah it would

PVAL-inactive

but i think if you consider point valued measures on R

i.e where singletons can have nonzero values

the solution to this exercise should imply that all uncountably infinite sets have infinite measure.

say each singleton has a nonzero value

for instance

is what I mean.

Secret

04:07

Yeah, that will effects my be summing uncountably many nonzero things, which is going to blow up

(And stupid autocorrect)

Daminark

I think an uncountable sum of positive terms is necessarily infinite, right?

PVAL-inactive

yeah thats essentially what this exercise shows

Daminark

Oh lol I didn't notice the exercise, sorry :P

PVAL-inactive

that when you take the sup over finite sums of distinct positive reals of an uncountable sum you get infinity.

Secret

I need to check again the transcript to be sure, but I recall leaky came up with a weird measure that is not translational invariant which can sum uncountable things

PVAL-inactive

04:10

the measure question is a little different than that (different points can have the same measure), but its easy to prove from that.

Daminark

Secret, you may like this exercise actually

Show that there is no finitely additive translation-invariant measure on $\ell^2$ for which the measure of every ball is positive and finite

Sister exercise: show that there's no finitely additive measure on $\ell^{\infty}$ for which the measure of every ball is positive and finite (here I do not assume even translation invariance)

PVAL-inactive

Sounds like something using the non-compactness of the closed unit ball.

Secret

Dec 5 '17 at 9:28, by Leaky Nun

@Secret for example here's an uncountably subadditive measure: enumerate the rationals $f:\Bbb N \to \Bbb Q$. Define $\mu(A) = \displaystyle \sum_{n=0}^\infty \begin{cases} 2^{-n} & f(n) \in A \\ 0 & f(n) \notin A \end{cases}$

Daminark: thonking (currently on mobile)

PVAL-inactive

maybe what I said about uncountable subadditivity is nonsense.

I think the thing I was thinking about was measures with nonzero values on singletons.

Secret

I think leaky might have cheated a bit, the measure is defined on a countable support

Since only the rational entries have positive value

PVAL-inactive

04:20

yeah for uncountable set of real numbers, the sup of finite sums is infinite.

so necessairly you can't have uncountable disjoint sets with nonzero measure.

note in such a collection, there is a finite number of sets with \mu=c for c any positive real (or else adding all those together produces something infinite), so it reduces to the exercise that was given,

Secret

04:39

@PVAL-inactive hmm... this is not very obvious to me. For example I cannot see how finite sums where elements are taken from $[1,2]$ will diverge?

PVAL-inactive

you let the number of things being summed be really large.

n things summed together in [1,2] is at least n

Secret

Ah I see

PVAL-inactive

of course this isn't completely obvious when it isn't bounded below.

Secret

e.g. 1+1.1+1.11+1.111+...+1.11...1 is definitely going to be more or less n

PVAL-inactive

like for [0,1] obviously i can take .9+.99+.999 ... or whatever

but for general uncountable sets the point is the solution to the original question.

Secret

04:44

I wonder if it is not just uncountable sets, but dense sets in the reals that tries to sum all its elements. For example, I will expect a finite sum over $[0,1] \cap \Bbb{Q}$ will become unbounded as $n\to \infty$

as one obvious candidate is the harmonic series and then of course the .9,.99,.999... etc.

So I am guessing if a measure is such that there are countably many singletons being assigned positive value and these form a dense set, then a blow up might still occurs

and such measure will think that all sets that contains said dense set will be of full measure

PVAL-inactive

density shouldn't matter

There's nothing about the geometry of the euclidean line being used

the values being summed are actually the point measures not the points themselves

you can take any positive convergent sequence and some enumeration of the rationals and assign the rationals the terms in that sequence and that will be all well and good.

Silent

05:34

Where in the proof of Legendre's formula do we use the fact that $p$ is prime?

I mean, if $c$ is a composite positive number, can we use the same formula to determine highest power dividing $n!$?

Daminark

05:59

Yo Mathein!

Corellian

06:13

test $\lhd, \unlhd$

Corellian

06:37

Hmm... it is a theorem that if $\phi$ is a group homomorphism with domain $G$ then $\phi(G)$ is isomorphic to $G/\text{Ker}(\phi)$. But Pinter instead specifies $\phi$ surjective (a group epimorphism), not sure why

Pinter defines a group $H$ as a "homomorphic image" of $G$ if there exists a surjective homomorphism $\phi:G\to H$, but this onto criterion is not mentioned in other sources. So he's consistently given to this surjectivity thing and I'm confused why even

Alex K Chen

For a finite group $G$, for some nonempty subset $H$ of $G$, if we have $a,b \in H \Rightarrow ab \in H$, doesn't that imply $H$ is a subgroup of $G$ ?

Corellian

@AlexKChen Yes

Alex K Chen

So for infinite group $G$, we change the condition $ab \in H$ to $ab^{-1} \in H$ ?

Are there an example of a infinte group $G$ and $H$ an subset of $G$, for which $ab \in H$ for all $a, b \in H$ but $H$ is not a subgroup of $G$ ?

Corellian

My laptop literally died without warning me of the low battery heh

Yes, when the whole group is infinite we also require closure under inversion

Alex K Chen

OK thanks @Corellian

Corellian

06:50

Think of a counterexample for your question @Alex

More generally, $G$ can be an infinite group. Then any finite subset $H$ closed under $G$ is in fact a subgroup. Of course, if $G$ is finite, then so must be $H$

It gets more complicated considering infinite subsets of infinite groups @Alex

Alex K Chen

@Corellian I guess for a counterexample, $G = \mathbb{Q}^+$ (under multiplication) and $H = \{1, 2, 4, 8,16, \cdots \}$ works

Corellian

Right @Alex. Indeed any subgroup of $G$ should be a group in its right (with respect to the same operation) and that $H$ is not

Also, take the additive group $\Bbb Z$. The subset of positive integers is closed under sums but not negatives, so it is not a subgroup of $\Bbb Z$

(and a more alarming red flag, does not contain the identity! immediate failure)

^any subgroup of $G$ should be a group in its *own right

Tobias Kildetoft

07:16

@Corellian The way Pinter does it is essentially the same anyway. One can always make a map surjective by restricting the codomain, and the result only cares about the image anyway.

Corellian

@TobiasKildetoft Yeah, you're right. I suppose his way is more (aesthetically?) pleasing. The whole target group preserves the desired information under the homomorphism

Secret

I think I am missing something. An arbitrary finitely additive measure $\mu$ only need to fulfill 3 axioms (empty set is measure zero, finite additivity and nonnegative). Now a unit ball in $\ell^{\infty}$ is given by $B_1 = \{x_n \in \ell^{\infty} : \sup_n |x_n| \leq 1\}$. Plugging everything in we have:

$$\mu \left(B_1\right) = \mu \left( \bigcup_{i=0}^{n} A_i\right) = \sum_{i=0}^n \mu(A_i)$$.

$A_i$ for example, can be made by partitioning the set of all sequences bounded above by 1 into 5 portions. Now unless $\mu$ is defined as such that any sequence with the same sup has the same mea…

and thus, I cannot managed to make a proof by contradiction

Silent

07:42

Where in the proof of Legendre's formula do we use the fact that p is prime?
I mean, if c is a composite positive number, can we use the same formula to determine highest power dividing n!?

Tobias Kildetoft

@Silent Which one is Legendre's formula?

Silent

@TobiasKildetoft This one.

Tobias Kildetoft

Ohh, I didn't know that had a name

Silent

I also found out yesterday :)

Tobias Kildetoft

But no, we can't use the same formula for non-primes

In general, the formula will undercount (take for example $c=6$ in $3!$)

Silent

07:48

thanks

Secret

08:11

@LeakyNun @Alessandro @AkivaWeinberger Here's something for you guys to ponder about:

> Without Axiom of choice. Is there exists a partition of the reals into countably infinitely many uncountable sets such that none of them are cocountable?

The closest thing I can find is a partition of the reals into uncountably many cantor set like copies:

8

Q: Partition $\Bbb{R}$ into a family of sets each one homeomorphic to the Cantor set

It is known that there is no (nontrivial) partition of $\Bbb{R}$ into a countable number of closed set. But is there a partition of $\Bbb{R}$ into sets, each one homeomorphic to the cantor ternary set?

Balarka Sen

C be a Cantor set, consider {R - C, C}

Oh you said countably infinitely many not countably many

Secret

yeah, I am one of those few people who like to use countable when I mean "countably infinite" and "finite" to mean finite

Balarka Sen

It should still be something dumb, consider cantor sets C_i of width 1/3 at each integer i. Then think about {C_1, C_2, C_3, ..., R - (C_1 \cup C_2 \cup ...)}

Rick

How is $f(x) = \displaystyle\frac{\cos^2 x}{1+a^x}$ an even function? $a>0$

Secret

hmm... right....

Rick

08:19

I got $f(-x) = \displaystyle\frac{a^x \cos^2 x}{1+a^x}$

Secret

so that means a countable partition of the reals into non cocountable sets is not sufficient to conclude non lebesgue measurability...

Rick

Oh wait it's not even

Alessandro Codenotti

$\Bbb R\times \Bbb N$ should biject with $\Bbb R$ even without AC, pick $f:\Bbb R\times \Bbb N\to \Bbb R$ such a bijection and partition the latter as $f(\Bbb R\times\{n\})$

Tobias Kildetoft

@AlessandroCodenotti I think you need countable choice for that bijection

Rick

But how is $\displaystyle\int_{-\pi}^{\pi}\frac{\cos^2 x}{1+a^x}$ always $\pi/2$?

a>0

Tobias Kildetoft

08:21

ahh, no, you don't need anything like that

Balarka Sen

I am not sure why I took Cantor sets, really. Just take intervals I_i = (i-1/3, i+1/3) around each integer i. Then think about {I_1, I_2, ..., R - (I_1 \cup I_2 \cup ...)}

Alessandro Codenotti

Can't you explicitely construct a bijection $\Bbb R^2\to\Bbb R$?

Tobias Kildetoft

@AlessandroCodenotti yes

Secret

Your cantor set example is better fit, cause I initially thought cocountable is enough to exclude any set that contains intervals

Btw, the underlying context for that pondering is as follows:

So I have been rereading the proof of the nonmeasurability of the vitali set. Eventually I noticed how the construction of the vitali set lead to the reals to be partitioned into countably many portions. Then since lesbegue measure is translationally invariant and all measures have to obey countable additivity, there is no positive number such that a countable sum of it can give a finite number, thus result in the vitali set to be nonmeasurable.

This caused me to wonder what is the minimal criteria for a set to become nonmeasurable. My initial guess is outlined in the question above, but balarka have got a counterexample for me, thus I am currently back to the drawing board

Balarka Sen

I have totally forgotten the Vitali construction

(Hey that rhymes)

Rick

08:29

Anyone?

Alessandro Codenotti

@BalarkaSen it doesn't to me!

Tobias Kildetoft

@BalarkaSen You mean totally and Vitali?

Balarka Sen

Spell "totally" and "Vitali" homonymously

Tobias Kildetoft

those don't rhyme to me either. The "a" in Vitali is longer

Secret

The investigation thus boils down to what kind of special property does axiom of choice gave to us that allow sets to become nonmeasurable, which at this moment, I so far only pin it down to a necessary condition of making countable partitions of an interval into uncountable sets (which causes the lesbegue measure to fail to give a finite number). Still trying to dig deeper atm

Balarka Sen

08:31

I spell it as Viteli because I'm a snub

Alessandro Codenotti

@BalarkaSen aesop rock has a weird influence on you. Waiting for your first math rap album

Shobhit

Hello. In some video on youtube some proffessor said that number theory cannot be done without algebra, so i started searching for use of algebra in number theory. One method of solving i got was using gaussian integers, it was cool. The second i got was gauss reduction of quadratic forms, but could not understand it. What are the other possible things?

Alessandro Codenotti

Algebraic number theory is a whole branch of maths

But so is analytic number theory to be fair

Shobhit

yes, but i cannot study them right now, so i was just looking at methods of solving using algebra

any you know

?

Balarka Sen

@AlessandroCodenotti I have totally forgotten the Vitally construction though I have residually stored the memories neurally gathered aurally through conversations Uhhh -- starts beatboxing

Alessandro Codenotti

08:38

lol

Balarka Sen

The last bit doesn't fit in the beat quite as well as I want

I am not a rappa

Tobias Kildetoft

@Shobhit Some basic stuff would be the proof of Euler's theorem, generalizing Fermat's little theorem, as a consequence of Lagrange's index theorem

But using Gaussian integers is cool too, for example allowing one to count the number of ways to write a number as a sum of two squares

Secret

bascornelissen.nl/media/bcornelissen-nonmeasurable-sets.pdf

hmm...

Now to figure how to zoom further in...

Shobhit

ooohh that was cool,found many links, ty @TobiasKildetoft

Secret

In other words, a system where non measurable sets can be proved to exists involve countably additive measures where singletons are of zero measure, and a choice stronger than dependent choice

this is consistent with what we knew about the vitali construction

So I guess, the next step is to figure out how the DC+ choice generates non measurability...

Balarka Sen

08:47

My internet is broken today

Secret

So progress on the quest for infinity so far:

Naturals -> visualised
Integers -> Visualised
Rationals -> Visualised ... sort of
Irrationals -> Partially visualised
Reals -> Visualised
$\omega_1$ -> Visualised
Cantor family of sets -> Visulised
Infinite dedekind finite sets -> Visualised .. sort of
Amorphous sets -> Visualised ... sort of
Vitali set -> Viauslised .. sort of
Bernstein set -> NEXT
Non measurable sets -> In progress
Perfect set -> Not yet visualised
$G_{\delta}$ sets -> Not yet comprehended
Borel algebra -> Not yet fully comprehended

Now, better get back to my chemistry stuff...

mike239x

09:13

morning

@secret I see you'r still here ^^

Secret

I am still online, but I am doing things in the background

my chemistry stuff atm is data analysis, thus it is done on my comp anyway...

Tobias Kildetoft

09:28

Neat. MathSciNet explicitly allows you to include an evaluation of the quality of a paper in your reviews, as long as it is stated in a professional manner.

Balarka Sen

09:58

Is "yo dawg this is lit fam" professional?

Tobias Kildetoft

@BalarkaSen I am not sure I even understand what it means

Balarka Sen

lol

Daminark

10:13

@TobiasKildetoft I'm surprised you haven't seen that phrase in papers

Tobias Kildetoft

@Daminark Yeah, it does seem like the sort of thing that would be in most papers

Balarka Sen

I was eating berries. I think I swallowed one accidentally.

Tobias Kildetoft

@BalarkaSen Is that what the phrase means? Or is it just the sounds you make in that situation?

Daminark

Yeah exactly, if that's how you present a theorem in this paper then surely that's how some evaluations can be included

Secret

Do you guys happened to know any important theorems that requires additive (instead of multiplicative) absorbers?

MatheinBoulomenos

10:27

@TobiasKildetoft have you ever thought about how much of a group G is determined by the category of G-sets?

Tobias Kildetoft

@MatheinBoulomenos No, I haven't. You mean a sort of non-linear Morita equivalence?

MatheinBoulomenos

yeah, that was what I was thinking about

Tobias Kildetoft

So my first place to look would be at the various groups of automorphisms in the category

MatheinBoulomenos

Since G-modules are the abelian group objects in the category of G-modules, this in particular implies that the category of G-modules are equivalent and if you tensor with a field k, you get that the group algebras over any field are Morita equivalent

Gasparo

@Secret I really enjoy reading your long posts in this chat.

Tobias Kildetoft

10:35

@MatheinBoulomenos You only get that the group algebras are Morita equivalent, right?

MatheinBoulomenos

yeah I noticed it, too

Secret

@TobiasKildetoft any ideas on my question?

Tobias Kildetoft

@MatheinBoulomenos So that means groups which are equivalent like this have the same block structure.

For all primes

And obviously have the same order

MatheinBoulomenos

But if we look at, say group algebras of finite groups over $\Bbb{C}$ is it even possible that $\Bbb C[G]$ and $\Bbb C[H]$ are Morita equivalent, but not isomorphic? In the Wedderburn decomposition of $\Bbb C[G]$, we always have a factor of $\Bbb C$ due to the trivial representation, but if we take a matrix ring $M_n(\Bbb C[G])$, there is no factor of $\Bbb C$ for $n\geq 2$ (due to uniqueness in Artin-Wedderburn)

this argument should work over any field

at least with semisimple group rings

Tobias Kildetoft

@MatheinBoulomenos The algebras being Morita equivalent just means that the groups have the same number of irreducibles

Whereas the algebras being isomorphic means that these irreducibles have the same dimensions

MatheinBoulomenos

10:42

Oh of course, I was thinking about Brauer groups, where Morita equivalence can be characterized via matrix ring

that was nonsense yeah

okay so they have the same number of conjugacy classes

Tobias Kildetoft

and the same number of conjugacy classes of elements of any given prime power order

sorry, of any given order not divisible by a given prime

Secret

$$bu+a \implies b(u + a) + a \implies bu + ba +a \implies bu + (b+1) a$$

MatheinBoulomenos

Hmm, seems implausible that they could be equivalent without being isomorphic

Tobias Kildetoft

I agree

Secret

This is the structure in question:

1. $(a + b) + c = a + (b + c)$
2. $0 + a = a + 0 = a$
3. $a + b = b + a$
4. $(ab)c = a(bc)$
5. $1a = a1 =a$
6. $a (b + c) = ab + ac$
7. $(a + b)c = ac + bc$
8. $0a = a0 = 0$
9. $u + a = a + u = u$

Tobias Kildetoft

10:46

I forget if the integral group algebra already determines the group

I think it does, whereas we obviously need the structure as a Hopf algebra to get the group from the algebra over the complex numbers

MatheinBoulomenos

jstor.org/stable/3062112

Tobias Kildetoft

I see

MatheinBoulomenos

But we only have Morita equivalence of the integral group algebras

Tobias Kildetoft

Right, which is even weaker

So we will need some of the non-linear stuff as well

Secret

Other results:

$bu + uu = (b+u)u=uu$

$uuu+uuu=u(uu+uu)=u(u(u+u))=u(u(u))=uuu$

but otherwise, the significance is still not clear

MatheinBoulomenos

10:55

Some random thoughts: the functor that takes each G-set to its invariants is represented by Hom(pt,-) where a one-point set is characterized as the final object, so we get the invariants functor from the category. Is G isomorphic to the automorphism group of that functor?

Secret

Silent: There is silence because calculations have shown that at least one user here actually does not see my messages

MatheinBoulomenos

I think the automorphism group of the forgetful functor from G-set to set is isomorphic to G, but I'm not sure how to get that from the category

Tobias Kildetoft

@MatheinBoulomenos You can get the size of the set as the number of hom's from the terminal object

Not sure if that is quite enough to get the forgetful functor though

MatheinBoulomenos

@TobiasKildetoft no, you can only send the point in terminal object to a fix-point under the G-action

Tobias Kildetoft

Ohh, right

As you just said

MatheinBoulomenos

11:00

No wait, the fix-point functor doesn't lead anywhere. By Yoneda Hom(Hom(pt,-),Hom(pt,-)) is isomorphic to Hom(pt,pt) which is trivial

Akiva Weinberger

3 hours ago, by Secret

@LeakyNun @Alessandro @AkivaWeinberger Here's something for you guys to ponder about:

3 hours ago, by Secret

> Without Axiom of choice. Is there exists a partition of the reals into countably infinitely many uncountable sets such that none of them are cocountable?

@Secret Uh… $[n,n+1)$?

(@LeakyNun @AlessandroCodenotti)

Secret

@AkivaWeinberger yeah, I made a mistake as balarka implicitly pointed out, I forgot to excluded intervals

Akiva Weinberger

Oh

$([n,n+1)\cap\Bbb Q)\cup([n+1,n+2)\setminus\Bbb Q)$

Something stupid like that

Secret

yeah, it turns out (a few messages later down the transcript) that underlying question is not a trivial one and that non cocountable sets is not sufficient to ensure non measurability

MatheinBoulomenos

@TobiasKildetoft From the fixed-points functor to set, we can also get the functor from set to G-set that gives each set a trivial G-action, because that's the left adjoint to the fixed-points-functor, maybe we can get the forgetful functor from that?

Secret

11:07

also... a continuity analysis suggest that tobias actually cannot see my messages

it does not seemed so just because alessandro happened to relay part of it outside

resulting in a false positive continuity reading

Tobias Kildetoft

@Secret Or maybe I just ignore random questions that happen to be addressed to me

2

Secret

well, that question on MSE had been tumbleweed for a few days

and there is not progress so far, and other algebra guys don't have much clue on it so far

Perhaps I should wait longer...

Balarka Sen

Anyone wants to recommend me some c l a s s i c a l music?

Tobias Kildetoft

@MatheinBoulomenos Hmm, so we can get the number of orbits by considering hom's to those with trivial action and more than one object

But I don't see how to get better than that

MatheinBoulomenos

@TobiasKildetoft I'm off now, I'll probably ask this on main later, thanks for your input

Gasparo

11:39

@BalarkaSen Yeah, I just did some today: youtube.com/watch?v=DRChXlr57GI

@Secret It's OK to be ignored, though this may not be the case here. Never let others have mental leverage over you and affect how you feel.

Secret

What if we told you, they will no longer have that option by 2020...

Di Castle 2 (enhanced)

►

Abcd: THAT, is The Plan, in its full glory

Secret

12:09

Gasparo: The thing I hate the most, is not even a human being, and that is is the frustrating part

Gasparo

@Secret I do not understand. What is it that you hate the most?

Secret

Ghosting and ignorance (and at the social level: Apathy)

Anything the perpetrator that does, and nobody can be able to brought them to justice, a karma houdini

Gasparo

@Secret In life, we just need to do our best, and kamma will take care of the rest. Don't struggle.

Secret

karma?... too slow

and so many have got away, mao zhe dong have mascarred many people, and he only died of old age

Gasparo

Kamma may take many lifetimes to come to fruition, but it will come indeed.

Secret

12:13

and what about the help vampires, thinking there are ways to stop them, they just keep coming back

Those who are being ignored, don't even knew they are being ignored, and have their lives slowly wasted to nothingness

Gasparo

This present life is just one in a whole series of lives we have. This universe is beginningless and endless.

Secret

And let's not get to apathy itself...

I don't care what the universe think, this is the one and only thing I hate, and in order to prove that a concept is not invulnerable, the universe will be erased, and stay erased

Balarka Sen

steps in Whoops, wrong chat room steps out

Secret

Feb 9 at 13:52, by 0celo7

I think Secret has finally snapped

What 0celo saw is nothing

COMAPRED TO THIS!!!

Balarka: Please made your way to shelter the h bar, thank you

Let us show you what our true form is

(This is going to get interesting..., lol)

Balarka Sen

Please dont

Im a humble man who comes here to shitpost and do math

3

this conversation is too much for my brain

have some mercy on the innocents fam

\o @AkivaWeinberger

Akiva Weinberger

12:33

I'm feeling a bit good because I found a Japanese sentence in the wild that I actually understood the grammar of. "未来の文字コード体系に私達は不安をもっています。" Literally "Future no character code system ni we wa anxiety o have", translating to "We are feeling anxious for the future character encoding system"

although technically there's only one conjugation in that sentence and it's one I haven't learned yet

(Also I like how the Japanese word for "code" is cōdo, clearly a loanword)

Secret

there are many loanwords in japanese, such as the japanese for chocolate

Akiva Weinberger

And "table", would you believe

Secret

(and so far the only one I can recognise without looking up a translation)

Leaky Nun

@AkivaWeinberger well there's tsukue for table

Akiva Weinberger

They have native words for table but apparently they also use tēburu (from English)

Leaky Nun

12:35

sniped

Akiva Weinberger

because English is fancy or something

Secret

japanese have a strong fascination of the western culture

they will go O_O if you tell them you are from australia

Transcript for

$Mathematics$ Mathematics