Mathematics - 2024-04-23

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?

Jakobian

@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

Dannyu NDos

Huh, nevermind.

Jakobian

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

Dannyu NDos

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.

Thorgott

12:36 AM

pushing out along an identity map does nothing

Dannyu NDos

12:59 AM

D:

Semiclassical

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

leslie townes

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

Semiclassical

yeah, grammatical != "reads well"

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

leslie townes

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.

Semiclassical

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

robjohn

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

leslie townes

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.

Semiclassical

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

robjohn

ah, forgot that qvB was force

Semiclassical

and life's too short for that

Xander Henderson

2:37 AM

@robjohn ur mom is force!

Dannyu NDos

4:42 AM

Which branch does K-theory belong to, algebra or topology?

leslie townes

it began in algebraic geometry, in grothendieck's formulation of the riemann roch theorem, but topological K came pretty soon after

why not both

Dannyu NDos

So it belongs to geometry then.

leslie townes

arxiv.org/abs/math/0012213

Dannyu NDos

Too sadge geometry isn't broad enough to be considered the 4th branch.

Analysis, Algebra, and Topology are the three branches, but geometry?

Lucky Chouhan

4:59 AM

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

leslie townes

nope

Lucky Chouhan

@leslietownes do you have PhD??

leslie townes

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

Lucky Chouhan

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

leslie townes

i am a lawyer

i guess i forgot to answer that

Lucky Chouhan

5:06 AM

@leslietownes so you have done math degree then you studied law?

leslie townes

yeah, why not

Lucky Chouhan

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

leslie townes

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

Soumik Mukherjee

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?

leslie townes

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

Soumik Mukherjee

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

Jakobian

8:49 AM

@SoumikMukherjee @BalarkaSen

Sounds like parts of what you study

one potato two potato

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

Jakobian

10:28 AM

@onepotatotwopotato does 3-manifold mean manifold with boundary here?

one potato two potato

yes possibly nonempty boundary

Soumik Mukherjee

12:19 PM

@onepotatotwopotato Thanks

YourLordJoyBoy

1:35 PM

@onepotatotwopotato MASHED POTATOES! SWEET POTATOES! BAKED POTATO! POTATO CHIPS!

YourLordJoyBoy

2:31 PM

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

EE18

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).

Soumik Mukherjee

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)

EE18

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?

Soumik Mukherjee

@YourLordJoyBoy what do you mean by opposite?

@EE18 That each non empty subset has glb and lub

Which is not true for R

EE18

Ohhhhh

oooofffff

Soumik Mukherjee

So they added that condition

EE18

Boundedness, of course

Thanks for setting me straight, much appreciated

Soumik Mukherjee

3:03 PM

Welcome

EE18

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

ZaWarudo

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?

Thorgott

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 $-1\le x\le0$

Jakobian

$(1+x)(1+y)\geq 1+(x+y)$ for $x, y\geq -1$ is a very simple observation to make

induction seems to be the most natural

Obliv

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

Alek Murt

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…

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$Mathematics$ Mathematics