12:13 AM
Today I reminded that the free product $G * H$ has $G$ and $H$ as normal subgroups.
Though, that raises a question... The free product $G * H$ has the direct product $G \times H$ as a factor group. What is the corresponding normal subgroup?

@DannyuNDos If $x, y$ are generators of free group on two elements then does $yx\in \langle x\rangle \cdot y$?
It doesn't because then $yx = x^ny$ for some $n\in\mathbb{Z}$ which leads to a contradiction

Huh, nevermind.

so $G = \langle x\rangle$ and $H = \langle y\rangle$ are not normal subgroups of $G*H$ their free product

The source of confusion was this: Given a group $G$ and its subgroup $H$, I tried to find the pushout of the identity map $\mathrm{id}: H \to H$ and the inclusion map $i: H \to G$.
I thought the result would be $(G * H) / H$.
Which is... isomorphic to $G / H$, in my second thought?
And still, $G * H$ should have $G$ and $H$ as factor groups.
Wait, $G / H$ isn't a thing; $H$ might be not normal. Dang.

12:36 AM
pushing out along an identity map does nothing

12:59 AM
D:

1 hour later…
2:00 AM
trying to decide if the following sentence is grammatical: What radius circle will a particle (mass m, charge q) follow if moving at speed v in a magnetic field B?
specifically the "what radius circle" bit

it seems grammatical to me, but i have no idea. and, there is sometimes a gap between grammatical and good exposition.
this is something ted could have answered, because when he went to school they still taught grammar

(i'd hope that being grammatical is a necessary if not sufficient condition but i'm not confident of that)

part of the question is an implicit assertion, which is that a particle under those circumstances moves in a circle. if the point is that you don't want the student to prove/investigate/focus on that assertion, i might just state that fact - a particle doing bleh will move in a circle. now you, student, tell me the radius of that circle.

yeah, they're given the formula r=mv/qB on the equation sheet, and this is a multiple choice question

@Semiclassical it's not $\frac{mv^2}{qB}$?

2:06 AM
in the last physics class i took, i think we did some kind of experiment relating to this equation to compute the mass of an electron.

no. mv^2/r = qvB, hence r=mv/qB
i'd say something about units but that would require me thinking about Tesla as a unit

ah, forgot that qvB was force

and life's too short for that

2:37 AM
@robjohn ur mom is force!

2 hours later…
4:42 AM
Which branch does K-theory belong to, algebra or topology?

it began in algebraic geometry, in grothendieck's formulation of the riemann roch theorem, but topological K came pretty soon after
why not both

So it belongs to geometry then.

Too sadge geometry isn't broad enough to be considered the 4th branch.
Analysis, Algebra, and Topology are the three branches, but geometry?

4:59 AM
@leslietownes you're pretty good at finding stuff on the internet. Are you a detective?

nope

@leslietownes do you have PhD??

where is all of this going. what do you really want to know?

@leslietownes your profession? But you're not telling me :(

i am a lawyer
i guess i forgot to answer that

5:06 AM
@leslietownes so you have done math degree then you studied law?

yeah, why not

I think, you spend more time here than in court haha no offence :)

i don't know where you live or how the court system works there, but in the united states, it isn't exactly a badge of honor to have one's clients appearing in court all of the time
there's a good joke in the US about a country lawyer who says "i'm the greatest lawyer, every will i have written has been upheld by the supreme court"
the underlying premise of this joke is that you are f-ing up badly as an attorney if anything you do leads to a supreme court case

5:55 AM
I can't find any wiki or nlab articles on tame sets and wild sets. Can someone suggest some references where I can find the definitions and basic stuffs about those?

i never saw such things, what field do they arise in?

3 manifolds. I don't know if they appear in other fields or not.

3 hours later…
8:49 AM
@SoumikMukherjee @BalarkaSen
Sounds like parts of what you study

topologically tame 3-manifold is a 3-manifold whose interior is homeomorphic to a compact 3-manifold. wild otherwise
I mean interior of a compact 3-manifold

1 hour later…
10:28 AM
@onepotatotwopotato does 3-manifold mean manifold with boundary here?

yes possibly nonempty boundary

2 hours later…
12:19 PM
@onepotatotwopotato Thanks

1 hour later…
1:35 PM
@onepotatotwopotato MASHED POTATOES! SWEET POTATOES! BAKED POTATO! POTATO CHIPS!

2:31 PM
Um, what would be the opposite of $3y^2-2xy+6x+2$?

Hoping to get some help with the following problem:
I've found this and intend to follow the proof: math.stackexchange.com/questions/791411/…
Thus my main question for the chat here is why would my text include the hypothesis about there existing $a,b$ such that $a<b$ and $a<f(a)$ and $b<f(b)$ (it does not appear in the linked question)?
My suspicion is that the existence of these $a,b$ is already implied by the other facts and so is redundant but perhaps there in order to have me do a little less work?
If so, what I am hoping for is help in showing how the existence of such $a,b$ is indeed implied by simply the fact about $f$ being an increasing function from $\Bbb R \to \Bbb R$ (and proeprties of $\Bbb R$, naturally).

3:01 PM
@EE18 Otherwise it is not true, f(x)=x+1 is an increasing function without any fixed point
Also you mean b>f(b)

I do indeed, yes
Thanks for the comment Soumik
What then am I missing from the linked answer? The hypothesis seems not to be there?

@YourLordJoyBoy what do you mean by opposite?
@EE18 That each non empty subset has glb and lub
Which is not true for R

Ohhhhh
oooofffff

Boundedness, of course
Thanks for setting me straight, much appreciated

3:03 PM
Welcome

I was wasting lots of time trying to prove that hypothesis unnecessary
Another question for folks
I've just proved Bernoulli's equality $(1+x)^n \geq 1+nx$ for $x \geq -1$ by induction but (as with most induction proofs, they feel mechanical) don't have good intuition for it
I vaguely recall Baby Rudin using it and "proving" it via the binomial theorem somehow
Something to effect of $(1+x)^n = \sum_{k = 0}^n{n \choose k}1^kx^{n-k} = \sum_{k = 0}^n{n \choose k}x^{n-k} ... \geq 1+nx$
But I can't see how one would go through the dots here
Any tips possible?
I guess I want to be able to show that $\sum_{k = 0}^{n-1}{n \choose k}x^{n-k} \geq nx$ but can't quite see how. I guess this leads to induction again so I'm in no better place

3:29 PM
Studying the local extrema in $\mathbb{R}$ of $f_{m,n}(x)=x^m(1-x)^n$ for positive integers $m$ and $n$, I noticed that $f_{m,n}(x)=f_{n,m}(1-x)$. Can I use this information to assume something and simplify the study?

4:16 PM
@EE18 if $x\ge0$, all terms are positive and you're just omitting all but $2$ of them and it's trivial
not sure this perspective is helpful for $-1\le x\le0$

$(1+x)(1+y)\geq 1+(x+y)$ for $x, y\geq -1$ is a very simple observation to make
induction seems to be the most natural

What are some very obvious situations where every subgroup of a group is normal
I thought that the alternating subgroup was one but apparently not
in this definition of a near-ring spawned from this answer, how can a group be non-abelian under addition
addition is by definition a commutative binary operation

4:38 PM
Hey guys I want to clarify something so If someone could help me would be great

For a function $F(U, x, t): \mathbb{R}^2 \times \mathbb{R} \times[0, \infty) \rightarrow \mathbb{R}^2$, the following inequality is proving that $F$ is Lipschitz with respect the variable x
$$\|V\|_{\infty} \leq M \text { and }\|W\|_{\infty} \leq M \Rightarrow\|F(V, \cdot, t)-F(W, \cdot, t)\|_{\infty} \leq k_3\|V-W\|_{\infty}.$$
have in mind that $U:\mathbb{R} \times[0, \infty) \rightarrow \mathbb{R}$. Can someone helpe with this please, since the $\cdot$ are confusing me or is the inequality only proving Lip

@Obliv if the group is abelian
another example is the quaternion group
and IIRC you can construct every group satisfying the property from these example in some sense
@Obliv theirs isn't

who is theirs? Why use the same word "addition" if it means something else

words typically don't have a single meaning

@Obliv commutative

@Jakobian what do you think? Should addition have one meaning in abstract algebra*?

4:46 PM
@Obliv near-ring is a more general of a concept than a ring
in this case addition just means another operation on a ring, as to distinguish it from multiplication
@Obliv yes we shouldn't be possessive of our nomenclature when it doesn't make sense to be

why call a potato a tomato? We're talking about binary operations on sets

Why call hydrogen oxide simply water?
you see why are you being ridiculous?

Not really, it's taught in primary school that addition is commutative. I think it makes a hell of a lot more sense to just call the operations something else that reflects their freedom to be more than addition & mult.

its taught in university that singletons are closed
but thats only because they are discussing metric spaces
why are you being so possessive of this?
we can still call it the same even if concept is more general
and doesn't have all the properties we were told before

in primary school you are taught addition of natural numbers

4:55 PM
well yeah, we're not going to be discussing matrices/permutations at that level
or whatever non-commutative stuff is going on in abstract algebra

You weren't protesting that binary operation on a group is called multiplication from what I know

out of curiosity, what are some non-commutative addition situations for near-rings

thats generally non-commutative

@Jakobian I was actually about to :D

so why are you objecting to near-rings but not groups?

4:57 PM
I mentioned to thorgott a couple weeks ago that I wished we had more symbols for each operation on sets that met their respective axioms

Are you objecting to cross product too?

en.wikipedia.org/wiki/… like the table here for groups, we could develop some sort of babylonian symbols lolol

God forbid we call it product

@Jakobian yes, throw it all out

@Jakobian Hydrogen Hydroxide or Hydroxic Acid

4:59 PM
do it properly, make everything abstract :D

shrug Do whatever you want, I'll still call it product or multiplication, or addition of near-ring an addition

well, there are plenty of good reasons why you don't see this opinion shared among professionals

Hello. Any idea how to solve the following limit? $$\displaystyle\lim_{(x,y) \to{(1,1)}}\left ( 1-\dfrac{\log^2(1+x-y)}{\sin(x-y)}\right)^{\dfrac{2}{(x-y)^3}}$$

Point of abstractifying things is to relate them to known concepts. So if you're trying to strip that familiarity away, you're doing it wrong

I am not sure if i can do the subst z=x-y and compute the one variable limit when z goes to 0

5:04 PM
@Odestheory12 yeah you can

@Odestheory12 since $x$ and $y$ are only seen in the expression $x-y$, you can definitely use $z=x-y$ and evaluate $\lim\limits_{z\to0}$
The only problem is that when $x=y\ne1$, that is you are approaching on a diagonal, you are using $z=0$, and the expression does not have a value on that diagonal, so the two variable limit won't exist on that diagonal.

@robjohn I'm not sure if its right to say that the limit won't exists on that diagonal, we simply don't consider those points since the function is not defined there

5:23 PM
@Jakobian Look at the definition of a limit, the function is supposed to exist in a neighborhood of the point in question, so you need to explicitly ignore that diagonal.

this is that old split between continental people and calculus books

clearly, it's impossible that different conventions exist for such an everyday object

i think u linked the same thing

5:33 PM
yes because the link didn't update when I clicked on it

@leslietownes I literally just vented about this. I don't think there should be so many "conventions" for math. looks around AM I WRONG?
Maybe I am..

obliv: hahaha the static around limits is one of the most annoying forms of this
there are only a few that genuinely annoy me as opposed to being "ok, whatever"

@leslietownes Obliv is objecting to calling operation of a group multiplication, to calling cross product a product and to calling addition of a near-ring an addition

the 2nd one of those follows from the first

Is what he was venting about

5:35 PM
i'm completely at peace with 0 maybe or maybe not being a natural number, for example, to the point that i forget which alternative i prefer
jakobian: those objections are interesting because of how novel they are. i kind of like them

@leslietownes I don't know, if in your definition of a limit function is supposed to exist around the point $(1, 1)$ then clearly its ill applicable to the current problem. So its not a conventional issue imo

@Jakobian I do like this, but I think it also obfuscates what you're actually saying by making things sound similar (or in this case identical)

yes, you would have a strongly held opinion that your opinion isn't even an opinion :)
i kind of like the idea of not calling the operation in a group a multiplication
i don't like the other ones, but i like that one
where does it stop, though? is there a limiting principle? can you use juxtaposition to denote the operation in an abstract group

38 mins ago, by Jakobian
Point of abstractifying things is to relate them to known concepts. So if you're trying to strip that familiarity away, you're doing it wrong
while there is nothing wrong in not calling a group operation "multiplication", objecting to others doing so is ridiculous

wdym by that? I'm just saying don't call $+,\cdot$ the operations for a structure if they don't mean what we mean for $\mathbb{C}$ (I use $\mathbb{C}$ since $\mathbb{R},\mathbb{N},..$ are all contained within it)
actually I shouldn't use uncountable sets
I heard just using $\mathbb{R}$ means ur talking about topology

5:43 PM
that's sort of what i'm concerned about. if this leads me to having to use funny notation for the group operation, i can't be on board with it
i like the background philosophical position, however

@Obliv not correct
and not what the post is saying

@Jakobian How is that ridiculous? Why would I call it something else just for myself? lol

the question of how much structure comes along with a single letter is an interesting one, but not (as far as i can tell) one that is raised by that post

well in this context I'm using $\mathbb{R}$ to mean a group under $+$ or $\cdot$ which the user said involves topology

5:46 PM
HE ISN'T SAYING THAT

@Obliv No, that isn't what the answer says.

we should make a list of all groups and invent a different notation for the binary-operation-that-I'm-not-allowed-to-name for every entry individually
2

The answer says that you can generate a subgroup in a particular way if you are only considering the algebraic structure (e.g. the additive group structure).

@Jakobian In Wikipedia it says $$\lim_{\substack{x\to p\\x\in T}}f(x)=L$$ For every $\epsilon\gt0$, there exists a $\delta\gt0$ such that $\color{red}{\text{for all }x\in T}$, $0\lt|x-p|\lt\delta$ implies that $|f(x)-L|\lt\epsilon$ (I added the red for emphasis).

But if you want $\mathbb{R}$ to have the structure of a topological group, then the same object generates the entire thing, as a topological group.

5:48 PM
So $T$ must avoid $x=y$

@robjohn Yes, so you want to let $T = \mathbb{R}^2\setminus \{(x, x) : x\in \mathbb{R}\}$

And even that explanation is deeply caveated.

I was being pedantic

@leslietownes IT ISN'T A NATUAL NUMBER!!!!!
PISTOLS AT DAWN!

i WILL fight you

5:51 PM
@Thorgott u only need one notation and u just have to specify what it does on the set

@Thorgott The gloves are off, man! Let's go!

(assuming u dont wanna go for my babylonian symbols idea)

I've been both considering and not considering $0$ to be a natural number in the past

like $(G,\star)$ and $(R,\star_1,\star_2)$

but now I just assume both based on what's more convenient for the moment

5:53 PM
The Natural Zeroists will be the first up against the wall when the Revolution comes.

thank you, xander, for reminding me what the correct position is

Heh.

and only say $(\mathbb{N},+)$ when u mean addition. Use something else when u mean a near-ring or a non-abelian group

@Obliv Meh. + is a single chord on the keyboard, while \star is five keystrokes. Ain't nobody got time for that.

actually + is two if u count shift

6:00 PM
@Obliv It is a single chord.

what kinda keyboard does not have a + key??

oh i see what u mean, i thought u mean stroke
i thought most had + be the shift+=
shift with = **

@Thorgott The + key is all the way over on the numeric keypad. If you don't want to move your finders from home, it is a chord, i.e. [shift]+[=].

oh damn, QWERTY is wild

(On the American QWERTY).

6:01 PM
what do u use..

+ right next to enter
this is a culture shock moment for me

WHAT... is that standard
it is for me too

@Thorgott I have none of those symbols

@Thorgott LEARN TO TYPE AMERICAN, JACKASS! :P

its bizarre seeing y and z swapped
We're so different..
@Jakobian so is it not a european standard then

6:04 PM
@XanderHenderson no I will commit an offence
@Obliv QWERTZ is the German standard

@Thorgott We're already scheduled for pistols at dawn. I am not sure how to respond to this new offense.

my mind is being blown at the thought of each country having their own standardized keyboard. I guess binary operations are the least of my concerns now
I'm out of my depth

same

Same

6:06 PM
why is everyone posting communist keyboards to the chat all of a sudden

@leslietownes 'Cause we're being overrun by commies, no doubt.

we just use the American keyboard with right alt for special letters

We all got scammed, where are those extra 50 keys in our boards?

@leslietownes oh... if you want to call America a communist country shrug

6:07 PM
imagine if germans used qwerty maybe we would have called $\mathbb{Z}$ yahlen
ok im being silly

@leslietownes WOLVERINES FOREVER!

@SoumikMukherjee its a dlc apparently.

@SoumikMukherjee this is the advanced one

There are 4 optional keys as well, who knows what those contain.
@Jakobian wow

jesus used a qwerty keyboard without any of those weird accent marks, that's good enough for me

6:12 PM
I've never seen the advanced one before
it looks like the standard German one but slightly more complicated

Imagine having a keyboard battle against that keyboard

Only true jedi masters know the ascii codes for all of these symbols and can bridge the worlds
Any keyboard with an alt+numpad is then equivalent just with potential for rsi/carpal tunnel

one of my friends did give himself a repetitive stress injury when his keyboard slowly stopped working and his workaround involved alt codes
to his defense, it is surprising how quickly you can pick it up if you absolutely need to

All the other keys are so congested, yet the space key won't help them, true capitalist

@leslietownes keyboards are cheap though

6:18 PM
they are cheaper now than when he did this to himself, but, your point is well taken

@robjohn hey thanks that is exactly what was making me doubt about the substitution
That*

6:36 PM
@Jakobian Oh, no. Alt-right keyboards, now?
They are missing out on a fourth set of characters: no alt, alt, alt gr, and alt+alt gr.

7:03 PM
@Thorgott on italian keyboards too (confirming this required immense mental effort since my laptop has a german keyboard)

> Claim Let there be two separable metric spaces $E$ and $F$. Then $\mathcal{B}(E\times F)=\mathcal{B}(E)\otimes \mathcal{B}(F)$.
Here $\mathcal{B}$ is the Borel sigma algebra and $\otimes$ denotes the operation of product $\sigma$-algebra. In showing $\mathcal{B}(E\times F)\supset\mathcal{B}(E)\otimes \mathcal{B}(F)$, it says that it is easily verified that for a fixed open $B\subset F$, then $$\mathcal C_B=\{A\in \mathcal{B}(E): A\times B\in\mathcal{B}(E\times F)\}$$ is a $\sigma$-algebra.
I struggle with this. The empty set is in $\mathcal C_B$ since $\mathcal{B}(E)\ni\emptyset=\emptyset\times B\in \mathcal{B}(E\times F)$, but why, if $A\in\mathcal C_B$, is $A^c\in\mathcal{C}_B$?
$E\times F$ is equipped with the product topology by the way.

@psie $E\times B$ is Borel as its open
so if $A\times B$ is Borel, then so is $(E\times B)\setminus (A\times B) = A^c\times B$

ok, thanks, I'll make a note. Is there any special property of cartesian products that comes in handy when showing that if $A_n\in\mathcal C_B$ for $n\in\mathbb N$, then $\bigcup_n A_n\in\mathcal C_B$?
I guess just distributive property

7:42 PM
@Jakobian Ya this is sort of what I did, I guess it just didn't feel like something I could connect to my grade school intuitions. Will think about what @Thorgott said here

7:53 PM
@psie yeah

8:39 PM
@SoumikMukherjee EXTRA 50 KEYS!? :D
@SoumikMukherjee There's no hidden meaning. I mean literally the opposite.

I don't know what opposite of a polynomial function means
Okay do you mean -f(x,y)?

Just multiply by -1 then

9:20 PM
opposite isn't a great word for that, at least not in english
the "negative" polynomial, maybe, but tbh i don't see a reason to use words rather than algebra for that

9:56 PM
I wish baseball season was back
oh it is back I didn't realize it had started.

10:33 PM
@JohnZimmerman college baseball is almost over.

would it be in bad faith if for the answer to this problem
I write $\mathbb{Z}_{12}\oplus \{e\}$
I can't tell if I'd get scolded/get marks off. Personally, I think you shouldn't take marks off because the question wasn't specific..

10:49 PM
it's a poor question, but that answer would still be in bad faith

Why does it say "Write $\mathbb{Z}_{12}$ as a direct sum of two of its subgroups"? Shouldn't it be "Write two subgroups of $\mathbb{Z}_{12}$ as a direct sum"
how can I write $\mathbb{Z}_n$ as a direct sum/product of anything if doing so gives me n-tuples

@Obliv if this is the type of text to make a big deal out of distinguishing between "internal" and "external" direct sums, that could be the reason. to a working mathematician, those two phrases mean the same thing.

Are you saying it's okay to write $\mathbb{Z}_{12} = G_1\times G_2$?
eh I guess I should refer to another text

I'm saying it depends on what the text means by "direct sum"

I'm a paraphrasing from Gamelin's and Greene's book:
> Let $(X_i,d_i), i=1,2,\dots,n$ be metric spaces. Let $X=\prod_{i=1}^{n}X_i$ and let $(X,d)$ be the metric space where $d$ satisfies certain nice properties, so that componentwise convergence of sequences holds. For $i=1,2,\dots,n$, let $U_i$ be an open subset of $X_i$. Prove that the subset $U_1 \times U_2 \times \dots \times U_n$ of $X$ is open and that each open subset of $X$ is a union of sets of this form.
I struggle with the second part. In the book they say it suffices to show that if $U$ is open in $X$ and if $x\in U$, then there exists open sets $U_k$ in $X_k$, $1\leq k\leq n$, such that $x\in U_1 \times U_2 \times \dots \times U_n\subseteq U$. How does this establish that $U$ is the union of open sets of the form $U_1 \times U_2 \times \dots \times U_n$?

11:13 PM
pick such neighborhoods for every point of $U$, take their union

ah ok, I think I understand. That's a nice property open sets enjoy I guess

is $[x,y]_n$ bad notation to denote a pair of congruence classes both from $\mathbb{Z}_n$
or is $([x]_n,[y]_n)$ better

tge former is wrong, the latter is correct

also how can I write $\langle a\rangle \pmod n$ for a cyclic subgroup of $\mathbb{Z}_n$
can I do $\langle [a]_n\rangle$
has anyone tried writefull on overleaf? I wonder if it's any good

11:55 PM
I mean, my general advice would be to just not write elements of $\mathbb{Z}_m$ as equivalence classes unless there is a need to