12:56 AM
Currently trying to prove that all irreducibles of $\mathbb{Z}[\sqrt{-3}]$ are primes in $\mathbb{Z}[\frac{-1+\sqrt{-3}}{2}]$. It is showing itself to be difficult.
I know that I should use the units of the two rings, somehow

1:33 AM
@Rithaniel three steps:
- show that for all $a \in \Bbb Z[\frac{1+\sqrt{-3}}{2}]$, there is some unit $u \in \Bbb Z[\frac{1+\sqrt{-3}}{2}]^\times$ such that $ua \in \Bbb Z[\sqrt{-3}]$
- conclude from step 1 that all irreducible elements in $\Bbb Z[\sqrt{-3}]$ are irreducible in $\Bbb Z[\frac{1+\sqrt{-3}}{2}]$
- show that $\Bbb Z[\frac{1+\sqrt{-3}}{2}]$ is a PID and hence every irreducible is prime
which step do you have trouble with?

Ah, that'd be the first step, actually
With that outline, it becomes fairly straight forward, actually

oh for the second step, you also want to use that every unit $u \in \Bbb Z[\frac{1+\sqrt{-3}}{2}]$ that lies in $\Bbb Z[\sqrt{-3}]$ is actually a unit in $\Bbb Z[\sqrt{-3}]$

Yeah, I got that part, the first one is what I might have some trouble with. But I think I can whip up some algebra to figure it out

it's not really very conceptual, take an arbitrary element, make some case distinction and multiply with the right unit in each case

It seems like I should do cases based on the oddness/evenness of the components?

1:47 AM
right

Well, that sounds like I need to figure out the full list of cases. Could I ask for a how many I should expect to see?

2:04 AM
@Rithaniel two cases

2:26 AM
So, if b is even or if b is odd. If b is even, it's already in the overring. If b is odd, multiply by $1+(\frac{-1+\sqrt{-3}}{2})$. Correct?
That is, where $a+b\frac{-1+\sqrt{-3}}{2}$ is an element of $\mathbb{Z}[\frac{-1+\sqrt{-3}}{2}]$
Wait, no, what if both $a$ and $b$ are odd

yeah I didn't count the case b even, cause that's obvious
you just need to multiply with one unit if a is odd and with another one if a is even

Alright, yeah. I always get lost when distributing these things. Like $(\frac{-1+\sqrt{-3}}{2})^2=\frac{-1-\sqrt{-3}}{2}$ took me longer than I should probably admit

4 hours later…
6:39 AM
@LeakyNun Hello

2 hours later…
8:42 AM
hello?

2 hours later…
10:19 AM
What is the term used to describe when someone computes the single value of an arithmetic expression? e.g. someone finds out that 2*2+3-1 is 6? Is it called evaluate, simplify, solve or something else?

10:35 AM
evaluate

thanks

1 hour later…
11:39 AM

12:07 PM
Can anyone help me prove $\int_0^{\pi/2} (\tan x)^{2/3}dx=\pi$
I am able to find the indefinite integral but still am not able to prove this
Even WA only gives an approximation wolframalpha.com/input/…

12:36 PM
I have a couple of problems I could use some pointers on, if anyone could help:
First one is "Let $F\subseteq K$ be fields. Show that the domain $F+xK[x]$ is integrally closed if and only if $F$ is algebraically closed in $K$."
Second one is "Let $F\subseteq K$ be fields. Show that the domain $F+xK[x]$ is completely integrally closed if and only if $F=K$"

"$F+xK[x]$" being the polynomials over $K$ with constant coefficient in $F$?

Yeah, exactly

1:37 PM
Well, one thing of note is that any UFD is completely integrally closed, so I know that $F+xK[x]$ isn't a PID as I would have suspected. That's just a corollary to these results, though, and doesn't actually help me prove them

Hi
is it a convention
for $0$ (in the integers) and any element of a ring $r$, to satisfy $0r=0$?

2:14 PM
@topologicalmagician it is a theorem
@Rithaniel which implication are you stuck in?

Leaky would like to help me in propositional Logic?

@TedShifrin Thanks Ted, it's from Hoffman - Kunze's Linear Algebra (page 195).

@StupidQuestionsInc Hi
Did you see my ping?

@adeshmishra Yes i'm writing a question here and then i'll check it

Okay no problem

2:29 PM
Another question from that book,the $T$-conductor of $\alpha$ into $W$ is defined as $S_T(\alpha,W)=\{\text{polynomial} g\in\mathbb{F}[x]: g(T)\alpha\in W\}$ where $W$ is an invariant subspace for $T$ and $\alpha\in V$. While I do understand why the minimum polynomial of $T$ will always lie in that set (since it maps to $0$), I don't see why a polynomial $f\in S_T(\alpha,W)\iff$ the minimal polynomial divides $f$?
@adeshmishra that's more abstract logic even for me ^^ But you can see what it's talking about, for instance for 3, if the set of propositions contains a proposition $\varphi$ then it must also contain its negation
@adeshmishra btw are u done with the logic parts of the book that I sent you?

@StupidQuestionsInc yes

@adeshmishra ok, will you then move to basic naive set theory?

@StupidQuestionsInc Yeah, we can prove the distribution of “or” by the truth tables, ha?

2:47 PM
@StupidQuestionsInc I saw your profile many times but then also I couldn’t understand the reason behind your user name
😁

@StupidQuestionsInc How can I fight with your description? :-)

> Don’t just read it; fight it! Ask your own questions, look for your own examples, discover your own proofs. Is the hypothesis necessary? Is the converse true? What happens in the classical special case? What about the degenerate cases? Where does the proof use the hypothesis? (Paul Halmos)

Yes, but how is it related to username in any way?
@StupidQuestionsInc

2:57 PM
@StupidQuestionsInc hahahaha
But why Inc, you opened a whole factory of questions?

@adeshmishra you can say it that way :P

@StupidQuestionsInc Why your profile picture shows only front part of your face and rest of it is a cartoon image?

3:44 PM
Can you map $\Bbb R^{1,1}$ to $T^2$

3:55 PM
What is $\mathbb R^{1,1}$ again? I always think it looks like the supermanifold notation. @geocalc33

@anakhro oh, it's a two dimensional pseudo riemannian manifold
In differential geometry, a pseudo-Riemannian manifold,[1][2] also called a semi-Riemannian manifold, is a differentiable manifold with a metric tensor that is everywhere nondegenerate. This is a generalization of a Riemannian manifold in which the requirement of positive-definiteness is relaxed.

Every tangent space of a pseudo-Riemannian manifold is a pseudo-Euclidean vector space.
I don't know if this question really makes a ton of sense
because a pseudo-riemannian manifold is inherently different from a riemannian manifold, specifically the requirement of positive definitenss is relaxed for the pseud-riemannian case.
Maybe, $T^2$ has to be a pseudo-riemannian manifold but I don't know if a torus can be a pseudo-riemannian manifold

4:12 PM
Any (smooth) manifold can be given a Riemmanian metric
Just pull back the Euclidean one through charts and use a partition of unity argument to glue them together into a global metric

oh nice :)
okay I'm going to assume that we're dealing with two Riemannian manifolds now. These manifolds are $\Bbb R^2$ and $T^2.$ Quotienting $\Bbb R^2$ into a torus will require some stretching, so the metrics will be different
say you decide to mark $\Bbb R^2$ with rectangular hyperbolas before stretching it into a torus via the equivalence relation.

4:43 PM
@geocalc33 so when you ask "can you map $\mathbb R^{1,1}\to T^2$" what do you mean?

@anakhro I mean is there any way to associate $\Bbb R^2$ to $T^2$. I don't know if I'm ready to think about semi-riemannian manifolds btw that's why I'm switching to regular riemannian manifolds
Is there any way to associate $\Bbb R^2$ to $T^2$ AND somehow counteract the "stretchhing" so that the metrics are equivalent?

The "equivalence" in Riemannian manifolds is called "isometry", and this requires a diffeomorphism first and foremost.
You won't get that in this case since there is no diffeomorphism from $\mathbb R^2\to T^2$.
What you can do is take a smooth map $\mathbb R^2\to T^2$ and find the metric on $\mathbb R^2$ induced by a metric on $T^2$.
This is done in the usual way for "pullback" structures.
But the problem is this isn't always a metric.

okay cool. I don't really care for an isometry, but that's good to know. I want some super weak form of equivalence

Since you can run into problems depending on the properties of your smooth map.

4:51 PM
In general, you will want something like an immersion at least.
I don't know if that is necessary and sufficient, but an immersion is definitely sufficient.

oh. so if you have some marked curves on $\Bbb R^2$ like the picture above (the hyperbolas), that are integral curves, specifically, the vector field is given by: $X=(x,-y),$ you can associate $X$ with $T^2$?

Hello guys! I am trying to answer How many isomorphins are between $\Bbb{Z}_n\times\Bbb{Z}_m$ and $\Bbb{Z}_{nm}$?

*you can associate $X$ on $\Bbb R^2$ to another vector field $X^*$ on $T^2$?

Part of the answer is that when $\gcd(n,m)=1$
@AlessandroCodenotti what do you mean? This is the definition of isomporphins between groups: sharmaeklavya2.github.io/theoremdep/nodes/abstract-algebra/…

@manooooh what's the isomorphism?

4:58 PM
I'm reading Ch. 14 in "Intro to Manifolds" (it's about vector fields)

@geocalc33 in general, I know it is very difficult to pushforward a vector field without a diffeomorphism.
There is probably in that chapter a note on pushing forward vector fields.

Sorry my last edited message was: But I don't know if there are other isomporphins. For example, for $\Bbb{Z}_3\times\Bbb{Z}_6$ and $\Bbb{Z}_{18}$ there is an isomorphism, although $\gcd(3,6)\neq1$

You said there is an isomorphism between $\Bbb Z/(3)\times\Bbb Z/(6)$ and $\Bbb Z/(18)$. I'm asking you ti describe this isomorphism

@anakhro yeah it says that it is very difficult to pushforward vector fields between manifolds, but there's a notion of "related vector fields" which im reading about in the section

5:02 PM
Good, that will help you with regards to this question of inducing a vector field on T^2 (or not).

@AlessandroCodenotti sorry, I don't know if there is an isomorphism between them. At least the cardinals match, since $|\Bbb{Z}_3\times\Bbb{Z}_6|=|\Bbb{Z}_{18}|$. I have not analyzed the other structural properties

@anakhro yes I'm curious about this mainly because of intrinsic interest. I can't really give you an amazing reason why you should care or some incredible application lol, but I'm just interested!

@geocalc33 Hi, how are you?

Well it's just that you don't have a clear enough goal if you do have one. If you are just learning what interests you, then that's fine. I just wondered if you had a goal to this that maybe would be helpful to know about in order to further help you.

@manooooh think about it then, try sending the generator of $\Bbb Z/(18)$ somewhere and see what happens

5:16 PM
@AlessandroCodenotti hm, a generator of $\Bbb{Z}/(18)$ can be $1$, because $\langle1\rangle=\Bbb{Z}/(18)$, right?

@anakhro You're absolutely right I don't really have a clear goal and that would be useful. I think I would really enjoy understanding the basics of special and general relativity, and see how they work on $T^2.$ Like spacetime geometry and stuff like that I guess

You might like Baez & Munian's book then: worldscientific.com/worldscibooks/10.1142/2324
Either that, or Wald's general relativity book.

simple example of two non-zero ideals such that their intersection is 0?

Is it true that $f:D\to\Bbb R$ uniformly continuous (where $D\in \Bbb R$) iff for any two sequences $(a_n),(b_n)$ with $\lim_{n\to0}|a_n-b_n|=0$, we have $\lim_{n\to0}|f(a_n)-f(b_n)|=0$?

@manooooh yes
@topologicalmagician $3\Bbb Z$ and $5\Bbb Z$

5:28 PM
@AlessandroCodenotti so what is the next step? How do I find $f([1]_{18})$?

@manooooh the image of a generator determines the whole group homomorphism

@anakhro nice, thanks for the reference. I'm going to attempt to push-forward $X$ to $X^*$ today. So I have the vector field $X=(x,-y)$ and I have the vector field $X^*=(x\log(x),-y\log(y)).$ My goal is to take $X^*$ and push-forward the vector field to $T^2$ or if that can't be done, look into the concept of "related vector fields" as talked about in the book

@AlessandroCodenotti sorry I don't understand. Could you fill in details, please?

@geocalc33 what are the prerequisites for loring tu's book?

If you know the image of $1$ you know the image of every element of $\Bbb Z/(18)$. So send $1$ wherever you want and see what happens
Hi @Ted

5:35 PM
@AlessandroCodenotti I mean, I take $[1]_{18}$ which is a generator, then I apply $f\colon\Bbb{Z}_{18}\to\Bbb{Z}_{3}\times\Bbb{Z}_{6}$ i.e. $f([1]_{18})=\text{??}$

Hi, demonic @Alessandro

Hello Ted!

Hi, manoooo.

@manooooh You choose what $f(1)$ is and try to find a choice that gives an isomorphism

I am dealing with some group theory question, haha

5:36 PM
@topologicalmagician the prerequisites are one semester of abstract algebra and a year of real analysis

@geocalc33 You need $X$ to be invariant under the $\Bbb Z^2$ action on $\Bbb R^2$ to get something well-defined.

@AlessandroCodenotti I can chose $f(1)=(1_3,1_6)$, since $\langle1_3\rangle=Z_3$ and $\langle1_6\rangle=Z_6$, right?

Hi @TedShifrin !

Hi, @topologicalmagician.

So there is an isomoprhism between those elements, but actually there is no an isomoprhism

5:38 PM
@geocalc33 multivariable analysis as well? or is that contained?

Sure but you could also choose something different, it doesn't really matter
So let's try it out. $f(1)=(1,1)$. So can you tell me now what $f(2)$ is?

@AlessandroCodenotti well, why should I pick something different? I pick the elements that can make an isomophism. Why should I pick other element?

multivariable analysis for sure, plus some topology, I imagine, @topologicalmagician.

@AlessandroCodenotti it can be $f(2)=f(2,0)$

No it can't. If $f(1)=(1,1)$ there's only one option for $f(2)$.

5:39 PM
@TedShifrin do you know multivariable analysis up to what? I have the top pre reqs, I'm quite sure.

@manooooh Because you don't know a priori which choice will give an isomorphism. $f(1)=(2,3)$ determines an isomorphism of $\Bbb Z/(15)$ and $\Bbb Z/(3)\times\Bbb Z/(5)$ (and so does $f(1)=(1,1)$)

Inverse and implicit function theorems.
You have to be very familiar with the derivative as a linear map, chain rule.

@AlessandroCodenotti sorry I meant $f(2)=(2,0)$

@TedShifrin oh alright.

@AlessandroCodenotti yes, because $\gcd(3,5)=1$

5:41 PM
@manooooh There's no difference with what you wrote earlier

guys, what does the notation $\mathbb{Z}[X,Y]$ mean?

@AlessandroCodenotti I added an $f$ to $(2,0)$, I thought it was wrong. Why it can be $f(2)=(2,0)$? It should be $f(2)=(2,2)$?

What do you think?

Z appended with two indeterminants, X and Y, @topologicalmagician

Polynomials in $X,Y$ with integer coefficients.

5:43 PM
@manooooh Yes, $f(2)=f(1+1)=f(1)+f(1)$ if $f$ has to a group homeomorphism

@AlessandroCodenotti ok, so at the moment we have $f(1)=(1,1)$ and $f(2)=(2,2)$

What's $f(3)$?

What would be another example to show that $\{$ ij $:$ i,j are in ideals, I and J $\}$ is not an ideal that doesn't require polynomials?

@AlessandroCodenotti easy pizi, it should be $f(3)=(3,3)=(0,3)$

Right. So what is $f(6)$?

5:45 PM
To show what, @topologicalmagician?
Oh, I see.
What's wrong with polynomials?

@AlessandroCodenotti $f(6)=(6,6)=(0,0)$. Oh but then $f(0)=(0,0)$ hence $f(6)=f(0)$ meaning that $6=0$ in $\Bbb{Z}_{18}$, which is a contradiction. Right?

You need something that's not a principal ideal domain. Do you know examples?

@manooooh No, it just means that your map isn't injective
Hence not an isomorphism

@AlessandroCodenotti right, thank you!

@TedShifrin No, I don't know many examples yet. I was able to find one that required polynomials, I was wondering if there's one which doesn't require polynomials

5:48 PM
Of course there could be other functions between those two groups which are isomorphisms

I do know what a principal ideal domain is though

You're going to have to work with adjoining things, which amounts to polynomials.

(Spoiler there is none. Hint: what is the highest order of an element in the two groups?)

@TedShifrin oh, I see.

26 mins ago, by Silent
Is it true that $f:D\to\Bbb R$ uniformly continuous (where $D\in \Bbb R$) iff for any two sequences $(a_n),(b_n)$ with $\lim_{n\to0}|a_n-b_n|=0$, we have $\lim_{n\to0}|f(a_n)-f(b_n)|=0$?

5:50 PM
@Silent: $D\in\Bbb R$ is wrong, of course.
And $n\to\infty$.

@AlessandroCodenotti that's precisely my question; I can pick other ordering pairs and see if meets the conditions of a group isomorphism. We picked one ordering, what happens with the others possibilities is me question

The answer is in my next message

Except for your mistakes, it looks correct, @Silent. Where do you get stuck?

@TedShifrin Most math looks correct except for mistakes :P

@AlessandroCodenotti the highest order of $\Bbb{Z}/(18)$ and $\Bbb{Z}/(3)\times\Bbb{Z}/(6)$ is $18$

5:52 PM
smacks particularly demonic @Alessandro

What's an element of order $18$ in the latter?

@AlessandroCodenotti I don't know, how do I find it?

I don't know, you claimed that there is one, what's your claim based upon?

$3\cdot 6 = 18$

@TedShifrin Oh! really sorry for those silly mistakes! I am sure I would not have committed them had i written on paper :) No, i just wanted to verify. I proved myself, but was not quite sure.

5:53 PM
@AlessandroCodenotti intuition based on $3\cdot6=18$

Well try to find such an element then

@Silent: The proof I think of naturally for $\impliedby$ is by contrapositive.
@manooo: What about $\Bbb Z/(2)\times \Bbb Z/(2)$?

@AlessandroCodenotti $3\mathbb{Z}$ $\cap$ $5\mathbb{Z}$ is $15 \mathbb{Z}$

@AlessandroCodenotti I have seen on Internet that the order of an orderer pair is the lcm of the elements, but I don't understand this

@TedShifrin yes, i got it, thank you very much.

5:55 PM
@topologicalmagician oh yeah sorry

@TedShifrin there are four elements: $(0,0),(0,1),(1,0),(1,1)$. How to find such orders?

Well, what are the orders of those elements?

@TedShifrin $o(0,0)=1$, $o(1,0)=2$, $o(1,1)=1$ and $o(0,1)=2$ right?

$(1,1)$?

$(1,1)+(1,1)=(0,0)$ and $(1,1)+(0,0)=(1,1)$ so it has order $1$?

6:00 PM
Huh?

How would you find the order of $(1,1)$?

What's the definition of order?

@TedShifrin sorry it would be $o(1,1)=2$ right?

Yup.

Nice

6:02 PM
So notice that the maximum is $2$, not $4$. Now go back to your question with Alessandro.

@TedShifrin I can't realize what would be the max order, should I find the order of $(2,5)$?
I can find the order of all elements, but that would be tedious
the order of the generator i.e. $o(1,1)$ is $6$

the generator?

So the highest order of $\Bbb{Z}/(18)$ and $\Bbb{Z}/(3)\times\Bbb{Z}/(6)$ is $6$, right?

If $(1,1)$ has order $6$ it generates a subgroup with $6$ elements, not the whole group

@TedShifrin Can you help me with Logic?

6:08 PM
I'm not fond of logic.
Depends what that question means.

@AlessandroCodenotti right, so? Finding the order of all element would be tedious

:D why $\neg \neg \bot$ doesn’t belong to PROP ?

Yeah, not a question for me. I don't even know what it means. Alessandro is the logician.

If I would call him by the name that gave him he might get very very angry :)

@manooooh: You'd better first understand why $(1,1)$ does not generate the group :P

6:11 PM
Alessandro hi

What's PROP?

See this
0

I'm studying Logic and Structure by Van Dalen and I came across this The set PROP of propositions is the smallest set $X$ with the properties $p_i \in X (i \in N),~ \bot \in X,$ $\varphi, \psi \in X \implies \left( \varphi \land \psi \right), \left(\varphi \lor \psi \right), \le... Why do you think that$\neg\neg\bot$is not in PROP? Van Dalen says that And I want to know why? What does he say exactly 6:15 PM @TedShifrin yes I understand it, as Alessandro kindly said, it generates a subgroup of$6$elements, not of$18$elements But do you understand why it goes wrong? @Mathphile If you have the indefinite integral, then it works. It's not bad: a substitution and then partial fractions. @AlessandroCodenotti He goes on with something like this $$\neg \neg \bot \notin X$$ Suppose$\neg \neg \bot \in X$and X satisfies i, ii, iii. We claim that Y = X -$\{ \neg \neg \bot \}$What in the world is$\perp$? It's a symbol for false It refers to any contradiction statement @AlessandroCodenotti Should I send you the image of that page of Van Dalen? 6:21 PM Yes, I don't have it available right now Just a sec Sorry for being late$\neg\neg\bot$is not in PROP,$(\neg(\neg\bot))$is Yes, but why? What's not clear to you in the proof? @AlessandroCodenotti He never says in the proof that $$\neg \neg \bot$$ is syntactically incorrect. It is you who has helped me in seeing that. 6:38 PM I wondered if associativity was to be taken for granted. Suppose that$\mu$is the measure on$(\mathbb{Z}^+,2^{\mathbb{Z}^+})$defined by $$\mu(E)=\sum_{n\in\,E}\frac{1}{2^n}$$ Prove that for every$\epsilon>0$, there exists a set$E\subset\mathbb{Z}^+$with$\mu(\mathbb{Z}^+\setminus\,E)<\epsilon$such that$f_1,f_2,\dots$converges uniformly on$E$for every sequence of functions$f_1,f_2,\dots$from$\mathbb{Z}^+$to$\mathbb{R}$that converges pointwise on$\mathbb{Z}^+$. looking for hints to start the problem @TedShifrin That's why nobody likes working on the syntactical side of logic. Or at least why I don't 6:56 PM @AlessandroCodenotti Can you please tell me what exactly he (Van Dalen) proves in the page that I have given a shot of ? 7:12 PM @adeshmishra PROP is defined as the smallest set satisfying some properties. He shows that if$\neg\neg\bot$were in PROP then there would a smaller set satisfying the same properties, which is a contradiction @AlessandroCodenotti How does those properties defined the syntactical rule? so basically infinite descent for sets neat What? What do you mean with syntactical rule The problem I am facing is to show there exists$E$such that$\mu(\mathbb{Z}^+\setminus\,E)<\epsilon$. Since the measure is a geometric sum, there exists a sufficiently large integer such that$1/2^n<\epsilon/n$and$n\notin\,E$7:18 PM @AlessandroCodenotti Well you told me that$\neg \neg \bot$is wrong because the right way of writing it is$\neg (\neg \bot)$and Mauro told me that anything which is not syntactically correct doesn’t belong to PROP Actually the right way is$(\neg(\neg\bot))$Being syntactically correct is just another way to say that you have an element of PROP, I'm not sure what you're asking Okay, then? Can you please recommend me a simple book for learning these things? I’m finding Van Dalen quite hard for beginners I'm afraid Van Dalen is already a simple book for logic. It's an hard subject at the beginning When he writes the first property as $$p_i \in X , \bot \in X$$ What is he trying to say? Why a contradiction needs to be a an element of a PROP set and what does$p_i \in X$means?$p_i$Are just propositions.$p_i$are just variables 7:28 PM So, what does it signify by saying they belong to X. Well if X is a set of propositions then surely propositions will belong to it. "$\phi$is a proposition" is defined to mean$\phi\in X$You have an intuitive idea of what a proposition should be, but if you want to study propositions as mathematical objects you need a formal definition, which is what Van Dalen is doing by defining this set$X$Ciao @Lukas Oh wow! Ciao @Alessandro Thank you, really thank so much Alessandro Ciao @Lukas und Guten Abend. 7:37 PM Bonsoir @Ted Здравствуйте I think I misspelled that. Maybe that's better. Still studying for exams? @Lukas @Alessandro no, avrò un esame dei funzioni L, ma probabilmente non sarà difficile. Sto scrivendo la mia tesi. E tu? 7:53 PM mumbles to self that he must learn yet another language for this chatroom I'm also studying for an exam and working on my thesis Right now I'm stuck on a topology problem What sort of topology? Can we say that$\Bbb{Z}_{18}$and$\Bbb{Z}_{3}\times\Bbb{Z}_{6}$are not isomorphic because$\gcd(3,6)\neq1$? @TedShifrin italiano è piu facile che francese o russo :P und auch als deutsch? 7:57 PM ich denke schon @manooooh: As my previous question to you suggested, the simplest case to understand is why$\Bbb Z_2\times\Bbb Z_2$and$\Bbb Z_4$cannot be isomorphic. @TedShifrin yes because$\gcd(2,2)\neq1$Well, that's not a justification unless the question you just asked has a theorem to go with it. It being sufficient is different than you knowing why it's sufficient @TedShifrin yes, my book states: "In general,$\Bbb{Z}_n\times\Bbb{Z}_m$is isomorphic to$\Bbb{Z}_{nm}$if and only if$\gcd(n,m)=1$". I have used this theorem 8:01 PM "Das Bekannte ist darum, weil es bekannt ist, nicht erkannt." - Hegel If you have that theorem, make sure you understand the proof and then stop asking us over and over! 5 @Lukas: Is this now the philosophy chat? :D I just thought it was a fitting quote to the situation at hand @TedShifrin the problem is that if the proposition holds, it is so easy! So I wanted to know other ways to prove that those groups are not isomorphic Understand the proof. @manooooh well, if you understand the proof, you can apply it to concrete examples 8:03 PM I give that comment another star because I think it is funny; nothing more than that We already discussed the fact that the product group cannot be cyclic, by inspection. Well, I meant it as a serious comment. @LukasHeger my book does not prove it. I have to search on the Internet for a proof. I am not a math student; the subject is called "Discrete Mathematics" @TedShifrin I guess it's called either geometric topology or dimension theory. I'm trying to see if what I suspect to be a characterization of Lebesgue covering dimension is in fact a characterization Oh ... I won't know that. I don't understand that quote 8:05 PM @LukasHeger Once you get the conjugation of verbs down the rest should be easy in Italian I guess that's true of most Romance languages though @TedShifrin see? I could not see that point. I did not know that the product group were not cyclic. Sorry for not seeing that No, grammar is a lot more than conjugation, @Alessandro. @manooooh: That was the point of our discussion about$(1,1)$earlier today. You never could give us an element that generates. I know, but I feel like that's the hard part of Italian grammar and the rest is comparatively easy @manooooh just quoting a theorem that you don't understand the proof of, especially if it trivially implies the problem, is only a solution in a very weak sense. You're essentially blackboxing everything that that there is to do @AlessandroCodenotti yeah I agree with that as an Italian learner by comparision, in German, adjective declination is horribly complicated nouns and adjectives in Italian are pretty easy @Thorgott it means that just because you know something, you don't necessarily understand it Hmm, I'd agree with that, but I'd say there's a difference between "something isn't necessarily understood, because it is known" and "something isn't understood, because it is known" 8:23 PM tbf, black-boxing is a necessary life strategy I don't need to know how an airplane works in its entirety, to know that I'll need to charter one to get from A to B in a reasonable time @Thorgott I guess the point that Hegel is making (as I understand it from the context) is that in order to understand something, you have to approach it as if it was unknown, because if you just take it for granted, there is no point in understanding why it holds Anyway my problem is the following: Is is true that if every map$f:X\to\Delta^n$is inessential, then the Lebesgue covering dimension of$X$is less than$n$? "Sometimes it doesn't matter how, it matters when, why, where AND how you get there"-anonymous (Inessential usually means homotopic to a map$X\to\partial\Delta^n$relative to$f^{-1}(\partial\Delta^n)$, but the author of this paper seems to be using it to simply mean that there is a continuous function$X\to\partial\Delta^n$which agrees with$f$on$f^{-1}(\partial\Delta^n)$) reminds me a bit of a Pauli quote: “Like an ultimate fact without any cause, the individual outcome of a measurement is . . . in general not comprehended by laws" 8:31 PM I have a good quote coming up the point being that, in the orthodox interpretation of QM, you never seek to "understand" the result of a given measurement you can understand the statistics of many such measurements, and perhaps understand what outcomes are possible/not possible yeah, that makes sense these are all really good points but if QM says that there's a 50-50 chance of getting heads/tails, then there's simply nothing else to say in the orthodox account "Nothing is accomplished all at once. And it is one of my great maxims, and one of the most completely verified, that Nature makes no leaps: A maxim which I have called the law of continuity." Gottfried Leibniz But doesn't nature make leaps relative to different observers perspectives? 8:41 PM well, presumably Nature itself would have a perspective under Leibniz for example, if I invent and patent a new kind of tea called Tasty Tea, and it astonishes the world... from the world's perspective I have totally revolutionized the tea industry(a discrete jump), but from my perspective nothing has really changed because it's been a continuous transition of thought and perseverance to invent such a tasty tea and the lack of leaps is according to this universal perspective okay that makes insane sense that said, if you reject such a perspective, then Leibniz's claim is not going to be particularly plausible I don't reject it on a global level but the way he defines nature on a global level bugs me I think it should be like a field in physics or something nature exists at every point idk 8:47 PM I'm not a particular expert on Leibniz, so I won't say anything further haha okay Can anything curve spacetime? mass curves spacetime Can anything curve momentum space? that's what I meant to say not sure that's a meaningful phrase. one uses "momentum space" in the context of quantum mechanics, and in particular non-relatistivistic QM so it's not going to be curvature in the GR sense it's potentially meaningful in the context of "Berry curvature" in QM, but I couldn't really say I read an article months ago about this but it was written in 2011 so I'm wondering if there's been any research or progress made on the topic probably only a handful of people actually study this 9:42 PM @TedShifrin does the hausdorff measure contain a divergent series? what does it mean for a measure to contain a series? @Thorgott sorry, I meant can the set which the infimum is being taken of, contain a divergent series? sure @Thorgott then there's an issue I'm having with a proof, may I tell you what it is? The claim is the following: Let$F\subseteq \mathbb{R}^n$be bounded and$f: F\rightarrow \mathbb{R}^n$be Lipschitz (where c is the constant). Then for all$s\geq 0$.$H^s(f(F))\leq c^s H^s(F)$The proof goes as follows Let$(U_i)_i$be a$\delta$cover of$F$Then$(f(U_i\cap F))_i$is a$c\delta$cover of$f(F)$. (This part is obvious). Then$\sum_{i=1}diam(f \cap U_i)^s \leq c^s \sum_{i=1}diam(U_i)^s$You can only have an an inequality if the series converges. Only that case was considered. Why is that assumption being made? @TedShifrin @Thorgott 10:20 PM Suppose$b<c$and$A\subset(b,c)$. Prove that$A$is Lebesgue measurable if and only if$|A|+|(b,c)\setminus\,A|=c-b$. I am stuck on the reverse direction.$|\cdot|\$ is outer measure

1 hour later…
11:21 PM
yesterday, by Ted Shifrin
@Akiva DogAteMy: Have you heard the Brentano Quartet? They're in residence at Yale. I just heard them in concert tonight.
@TedShifrin No

11:57 PM
@topologicalmagician this inequality always holds