Mathematics - 2016-10-06 (page 1 of 2)

Obliv

00:20

@sirC isn't that the same thing

Discovered as in you find an ancient stone tablet and uncover its secrets through an alien translation device? @sirC

Keith Yong

00:44

I have this homework problem:

Is there an integer x such that 2016x − 1 is divisible by 46466401? Find it, or prove that none exists.

I'm not sure what to do, I have thought about proving by x=even/odd but it doesnt matter because 2016x - 1 will always be odd

Obliv

oh wait I read that backwards lol

Keith Yong

I kinda cheated but I programmed a script and 2016 * 23048 - 1 = 4644767, and 2016 * 23048 - 1 = 46466783

That should count as a "proof" i think

Sir Cumference

@KeithYong Clever

What class is this

Keith Yong

Network Security, we're learning cryptography math atm

Obliv

01:18

I think I need to sleep lol.

Ted Shifrin

01:51

@KeithYong: Have you learned about the Euclidean algorithm? You're asking if you can write $1=2016x+46466401y$.

Adeek

03:37

Hey @TedShifrin here ?

would anyone like to discuss some presentation theory ?

Adeek

04:00

@SamuelYusim here ?

user228700

Hi :-) Can anybody help me with a bit of a homework-tsy question? I've been asked to find the range of a rational expression $y=(x+2)/(2x^2+3x+6)$. I multiplied both sides if this expression by $2x^2+3x+6$ to obtain a quadratic equation in x: $2yx^2+(3y-1)x+6y-2=0$

user228700

Since it's given that $x$ is real, this equation must have real roots. For the case in which $y≠0$, the discriminant, $D≥0$ Upon solving this, I find that $x$ lies in the range $[-13/3, 1/3]$-{0}

user228700

For the case that $y=0$, the equation becomes $x=-2$, which is possible, since $x$ is real.

user228700

So, is the total range of $x$ given by $[-13/3, 1/3]$-{0}+{-2}?

user228700

Please help if possible...

Balarka Sen

07:41

Hi @iwriteonbananas, @Danu

Danu

Hi @BalarkaSen

iwriteonbananas

08:03

@BalarkaSen What's up dude

Balarka Sen

@iwriteonbananas Killing time

what about ya

iwriteonbananas

Trying to see why the geometric realization functor $sSet\to Top$ preserves cofibrations

should be obvious I think

Balarka Sen

ah ok

iwriteonbananas

Do u know about model categories?

Balarka Sen

nope, not really

iwriteonbananas

08:06

I think you do

Given a category you just say what cofibrations, fibrations and weak equivalences are. Then these have to satisfy some basic axioms

In Top it's exactly what u tihnk it is

Balarka Sen

weak equivalence as in the analogue of weak homotopy equivalence?

iwriteonbananas

Yeah, iso on $\pi_n$

Balarka Sen

gotcha

why's equipping my category with such structures the "right" axiomatization of the homotopy category?

iwriteonbananas

No idea

Balarka Sen

oh ok. was just wondering why do we care

iwriteonbananas

08:11

Yeah, I haven't really gotten that far yet

Balarka Sen

fair enough

iwriteonbananas

What time is it in West Bengal?

Balarka Sen

like 2 PM

iwriteonbananas

Oh

You're only 4 timezones away from me

Balarka Sen

what's the time there?

iwriteonbananas

08:14

like 10am

Balarka Sen

ah

iwriteonbananas

I found this funny:

Given a diagram in Top, there is a spectral sequence computing the cohomology of the homotopy colimit in terms of the cohomology of the ingredients

If your diagram is a pushout diagram, this spectral sequence is basically the mayer-vietoris sequence lol

Balarka Sen

oh no, so all the years I fought with you about not saying the pushout when doing Mayer-Vietoris (partially because the standard proof doesn't really use the fact that it's a pushout) are all futile

great

cute though.

iwriteonbananas

haha

Balarka Sen

Ted challenged me to reformulate Guillemin-Pollack's proof of Borsuk-Ulam into the standard cup product (or transfer sequence) proof once upon a long time ago. I thought about it a little and I think I'd be hard-pushed to do that

curious

(yes, that's how I am killing time)

iwriteonbananas

08:26

What's Guillemin-Pollack's proof?

Balarka Sen

It's an induction proof, using a version of the argument principle ($f : X \to \Bbb R^n$ be a map from an $n$-manifold $X$ with boundary s.t. $f \neq 0$ on $\partial X$. Then the number of roots of $f$ is the degree of $f/|f|$ on $\partial X$ -- either in the oriented or in the mod 2 category)

iwriteonbananas

Cool, I am totally unfamiliar with that proof of B.-U.

Little Alien

You want to approximate an integral with vertical bars. In simplest case you have only two bars. You divide the x-interval in the middle. Say, [0,1] is divided at 0.5. The question, what should I choose for the height of the bar? Should I sample the f(x) in the middle of the bars, i.e. at points .25 and .75?

I ask because it seems natural to me to sample the interval at .33 and .66. But, it seems incompatible with two bar separation.

Balarka Sen

@iwriteonbananas It's a neat proof.

I can sketch it in the alg-top room if you want.

iwriteonbananas

Yeah, please do

and link me to that room again

Balarka Sen

08:40

here

Mary Star

09:30

Hello!!

Let $F\subseteq E\subseteq K$ be consecutive field extensions and $f\in F[x]$ be non-constant.
I want to show that if $K$ is the splitting field of $f$ over $F$, then $K$ is the splitting field of $f$ also over $E$.

Since $K$ is the splitting field of $f$ over $F$, we have that $f(x) = c\prod (x-a_i)^{m_i}$, where $m_i$ are non-negative integers and $(x-a_i)\in K[x]$.
Since $F\subseteq E$ we have that $f\in E[x]$. Therefore, $K$ must be also the splitting field of $f$ also over $E$.

Is this correct?

Danu

09:59

22

Q: Has anyone ever actually seen this Daniel Biss paper?

A student asked me about a paper by Daniel Biss (MIT Ph.D. and Illinois state senator) proving that "circles are really just bloated triangles." The only published source I could find was the young adult novel An Abundance of Katherines by John Green, which includes the following sentence: Danie...

So unfortunate :P

Balarka Sen

Yikes

Danu

TFW you make a huge breakthrough, but all of it collapses based on a basic mistake.

Starfall

10:42

@MaryStar That is not quite correct.

Lozansky

11:21

If I find the power series solution to an ODE at the point $x_0=0$ in the form $a_0 y_1(x) + a_1y_2(x)$, will the solution to the same ODE at the point $x_0 =1$ be on the form $a_0y_1(x-1)+a_1y_2(x-1)$?

user228700

13:33

@mercio: Hi again :-) I've got a really small doubt along the lines of what we discussed yesterday...

Danu

When Huybrechts speaks of the degree of a line bundle on a complex curve, what does he mean?

I can imagine it makes sense if the line bundle arises from a divisor (which, on a curve, can be assigned a degree).

But without a divisor, what is the degree?

user228700

I'm essentially trying to find the conditions for the location of roots of a quadratic equation under different conditions. Eg: I need to find conditions on the coefficients, and $D$ if, say, both roots are greater than some number $p$. If my quadratic equation is $ax^2+bx+c=0$, then is it OK to rewrite this equation as $x^2+(b/a)x+c/a=0$ so that we're always dealing with an upward facing parabola (which may or may not cut the x-axis)? (These conditions vary; am I allowed to do this always?)

mercio

14:00

if $a$ is nonzero then $z= 0$ is equivalent to $z/a = 0$ yes

user228700

14:11

@mercio Yes, this I know but by rewriting it this way, I'm making the coefficient of $x^2$ positive and hence, the parabola will be concave. Is this correct?

Balarka Sen

14:38

@Danu Perhaps he cares about the topology only? Namely, the number a generic topological section represents in the top homology of the curve?

Aka the first Chern class

Danu

14:52

@BalarkaSen Wait, whaa

Balarka Sen

@Danu Hmm?

Danu

(i) what is a generic "topological" section?

(ii) how do I get it to represent a number in $H^2$?

Balarka Sen

@Danu A section which is not necessarily holomorphic, just a continuous map.

@Danu Intersection number with the zero section.

Danu

Hmm... Huybrechts has no information about transversality/intersection numbers so I'd be weirded out if he used it in a side remark

Balarka Sen

15:12

@Danu It agrees with your definition for bundles which do admit a holomorphic section.

But yeah, no idea.

Danu

OK, if you say so

Actually, what if the divisor is not effective @BalarkaSen

I guess then I don't get a holomorphic section anyways

Balarka Sen

Right.

Danu

OK, I can live with this :P

Balarka Sen

Excellent, me too.

Danu

Now, when H. writes $H^1(X,\Bbb Z)\otimes_{\Bbb Z}\Bbb R)\otimes_{\Bbb R}\Bbb C$, does he mean tensor products as modules over given ring?

What kind of a $\Bbb Z$-module is $\Bbb R$?

Balarka Sen

15:18

Abelian group.

Just define action of Z on R by n * a = a + a + .. + a

Danu

Oh yeah, that was it, $\Bbb Z$-modules are abelian groups

but mostly what I was thinking about

what is its dimension?

infinite?

Balarka Sen

Yes.

Danu

Yeah, okay, and that's okay, I guess?

Balarka Sen

Why not?

Danu

I dunno, infinite-dimensional things are scary

Balarka Sen

15:22

$H^1(X, \Bbb Z) \otimes_{\Bbb Z} \Bbb R$ should be finite dimensional as an $\Bbb R$-module however, so that shouldn't be much of a problem (don't forget the second tensor product!)

Danu

Yeah

Mike Miller

15:39

@Danu all we're doing is changing what ring it's a module over

Balarka Sen

^

Danu

@MikeMiller Yeah

I don't have a big problem with it :)

A harder problem for me is the following:

Given that $H^1(X,\Bbb C)$ is the above tensor product, "this shows that $H^1(X,\Bbb Z)\to H^1(X,\Bbb C)\to H^{0,1}(X)$ is injective with discrete image that generates $H^{0,1}(X)$".

Mike Miller

Seems a little sketchy. Dimensions seem wrong.

$b^{0,1}$ should be $g$.

Danu

I don't know what $g$ is

Mike Miller

genus

Danu

15:46

Of an arbitrary complex manifold

?

Mike Miller

I'm talking about a curve here for convenience, sorry

For a Kahler manifold $H^1 = H^{0,1} \oplus H^{1,0}$

Danu

Yes, I know

Mike Miller

so you see the dimension problem?

Danu

No @MikeMiller

The image is supposed to be discrete

Dimensions of what?

bolbteppa

16:01

Lets say you were talking to a person who knew what a Fourier transform was, and you wanted to motivate the fact you can derive the quadratic formula by taking a discrete fourier transform of the roots of your quadratic, how would you motivate even taking that DFT? It works, but it's not like you're expanding a wave function in exponential waves, what are you doing

Mike Miller

$H^1$ and $H^{0,1}$

so I guess it could still be true that the full composite is injective, but that second map does kill a half-dimensional complex subspace

Danu

Yes,

I think the full composite being injective is what Huybrechts says

this is what I don't understand

Lozansky

Ugh

My book says "Note that if $r$ is not a rational number, then $x^r$ is defined by $x^r = e^{rln x}$"

Mike Miller

How do you want to define it?

Lozansky

Oh right, we might encounter infinite many values for $x^r$ if $r$ is rational

Mike Miller

16:12

Huh?

Lozansky

@MikeMiller Uh, I meant finitely many values*

Mike Miller

No. $x$ here is positive, and $x^r$ has a single well-defined positive value for any rational $r$.

Lozansky

So why does $r$ have to not be a rational number?

Mike Miller

You could define $x^r$ that way when $r$ is rational as well. It's just that when $r$ is rational, there's already a perfectly good way to define it: if $r=p/q$, take the positive $q$th root and then take it to the $p$th power.

Lozansky

In my complex analysis book they say "If $\alpha$ is a complex constant and $ z \neq 0$, then we define $z^{\alpha}$ by $z^{\alpha} := e^{\alpha log z}$

Danu

16:22

Anyways @MikeMiller I guess that $H^1(X,\Bbb Z)$ injects into $H^1(X,\Bbb R)$, which then injects into $H^1(X,\Bbb C)$, filling out $H^{0,1}(X)$

Mike Miller

@Danu Duh. Thanks.

Danu

Not super sure about how that works, but okay...

It should be something like that

Mike Miller

I'm forgetting all the math I ever pretended to know.

Lozansky

I mean, $r \in \mathbf{R}$ so it's certainly complex and $x \neq 0$ so surely it holds even for rational $r$

Danu

I have this crippling insecurity when it comes down to writing out anything when it comes to "modules" (magic word)

The reason why I care is because Huybrechts proves that the kernel of $H^1(X,\mathcal O^*)\to H^2(X,\Bbb Z)$ (from the exponential sequence)---which is equal to $H^1(X,\mathcal O)/H^1(X,\Bbb Z)$ on a compact manifold (because the map between those two in the LES is injective)---is a complex torus. Which kind of sounds cool

So the point is to prove that the map from $H^1(X,\Bbb Z)$ into $H^1(X,\mathcal O)=H^{0,1}(X)$ has a lattice as image

Mike Miller

16:26

mhm

Danu

This is done by proving it for the map induced by inclusion $\Bbb Z\subset \Bbb C$ (post-composed with projection to $(0,1)$) and then noting the map agrees with the one from the exp. seq.

Mike Miller

that's called the Picard variety of your complex manifold

or rather, the identity component of the Picard variety

Danu

Eh... could be

it's the "continuous part of the Picard group" for Huybrechts

$\operatorname{Pic}^0(X)$

Mike Miller

sure

Danu

Identity component... so it's something like $\operatorname{Pic}(X)$ is $\operatorname{Pic}^0(X)$, plus somehow the discrete part generating further copies?

Mike Miller

16:31

yes, and the discrete part here is $H^2(X;\Bbb Z)$

Danu

Well, the image of Pic in it but yes

somethingsomething Chern class

I'm excited for ch. 4 which should clear some stuff up about chern classes

So could you perhaps tell me why $H^1(X,\Bbb R)$ injects into $H^{0,1}(X)$ (as opposed to...??) @MikeMiller?

Mambo

16:48

@MikeMiller Hi

Are you busy?

Danu

Oh, actually it's clear that the discrete part generates copies. Thanks for setting me onto that track Mike :)

Semiclassical

17:07

Hi chat

Alessandro

hi everyone

Today in class we saw that if $\mathfrak{M}_\mathcal{L^1}$ denotes the subsets of $\mathbb{R}$ which are Lebesgue measurable then we have $|\mathbb{R}|< |\mathfrak{M}_\mathcal{L^1}|\le |2^\mathbb{R}|$. Can we prove that there are as many measurable sets as there are subsets of the reals without assuming some form of GCH or could the second inequality be strict too in ZFC alone?

Semiclassical

@Danu I initially read the above as "something something Chem class" and wondered why you'd care about chemistry :p

theindigamer

Hi, I had this question math.stackexchange.com/questions/1924388/… closed as off-topic but I think it should be closed as duplicate. I don't quite understand the reason given for closing the question as off-topic.

Alessandro

actually I already know the answer to the question above now that I think of it, nevermind

Balarka Sen

17:38

@Krijn: Hey.

Krijn

Hey Bala!

Have you done some more algebra/number theory already? :D

Balarka Sen

None yet, and I don't plan to do so any soon.

I gave your Weil conjecture papers a quick look yesterday, by the way.

Krijn

Ah, yes

Although I don't know how much Elliptic Curve theory you know

Balarka Sen

You know more than me anyway

Krijn

Well, yeah, I took two courses in it ._.

Balarka Sen

17:42

Right

Krijn

One on elliptic curves and one on alg geom

But I want to understand why Algebraic Number Theory is sort of the same thing

Balarka Sen

Um, not sure what you mean by "the same thing"

The same thing as what?

Krijn

Why we can see it as a theory of algebraic curves

I sort of understand the laymen's explanation but I would like to understand it in full detail

Balarka Sen

something something Spec Z

Krijn

Jup

Balarka Sen

17:47

I don't know anything about it either :)

Alessandro

hi @Krijn and @balarka

Balarka Sen

Hi @Alessandro

Krijn

Hi @Alessandro

Alessandro

I'm looking for a movie to watch tonight, any suggestion @Krijn?

Krijn

@Balarka knows much more movies than me, I think

Balarka Sen

17:49

What kind of movie do you want to watch?

Nah

Krijn

Coincidentally, I'm also watching a movie tonight

Balarka Sen

You never finished the one I recommended :P But I won't push it.

Alessandro

Any genre is fine, as long as it is a good movie

around 2h30m at most as running time though

Mike Miller

define 'good movie'

Krijn

Imdb rating > 8

Mike Miller

17:51

lol

Krijn

@BalarkaSen I'll finish it!

I Will

Mike Miller

mad max: fury road

Balarka Sen

hachiko

Alessandro

A movie which is good in the opinion of the person suggesting it to me I suppose @Mike whether I find it good too decides whether I'll ask that person for suggestions again :P

I've seen both

Mike Miller

the godfather

synecdoche, new york

Balarka Sen

17:52

have you seen any of Tarkovsky? those are my personal favorites

Mike Miller

the man who fell to earth

@BalarkaSen too long

Balarka Sen

fair enough

shcolf

Hi everybody,
Can anyone explain me the assumption below in answer
http://math.stackexchange.com/questions/25241/proof-there-is-a-1-1-correspondence-between-an-uncountable-set-and-itself-minus?rq=1

>>assuming countable choice it has a countably infinite subset B`

Krijn

The Grand Budapest Hotel

Alessandro

Not yet, but Solaris has been on my "list of movies to watch" for quite some time @balarka

Krijn

17:54

Amélie?

Balarka Sen

i have stalker and nostalghia on top of solarys, just so you know (in that order, yes)

Krijn

J'ai tué ma mere

Mike Miller

Solaris is a good movie but there's too much going on, somehow

theindigamer

@shc

@shcolf, it means en.wikipedia.org/wiki/Axiom_of_countable_choice

Balarka Sen

have any of you seen breathless, by Goddard? I have been planning to watch it

Mike Miller

17:55

Of course, Balarka and I are referring to the American version starring everyone's favorite, George Clooney, instead of the unwatchable Russian.

Balarka Sen

lol

Alessandro

I'll add those to the list too @balarka

I've heard breathless is good, but I didn't see it

Mike Miller

@Alessandro I gave you some suggestions; I can give more if you give me a feeling you're looking for tonight.

Alessandro

I think those were enough, I really should watch the godfather sooner or later and tonight might be a good moment to do so

shcolf

@theindigamer thanks

Mike Miller

17:57

That might pass your time cutoff.

Alessandro

@shcolf are you asking what the axiom of countable choice is or why do you need to assume it there?

Hm, it's a bit too long at 2h55m

I might have enough time but I can't be sure and I hate having to leave a movie halfway through to finish it later

shcolf

@Alessandro actually I'm wondering why do I need to assume it there, because I haven't studied it

Balarka Sen

@MikeM: I didn't watch naked lunch after seeing a film by Cronenberg. I can't take that much gore :/

Mike Miller

What did you see?

Balarka Sen

The Fly

Mike Miller

17:59

Also, naked lunch is rather bizarre.

The Fly is great. Not my favorite Cronenberg.

Balarka Sen

it's a nice film, yeah. just too visually disturbing

Mike Miller

It's just a movie about an annoying fly. Dunno what you could possibly find upsetting.

Alessandro

I think it's needed to be sure that $A\setminus B$ isn't an amorphous set and so it has a countable subset. Those are kinda technical details you don't really need to worry about, especially because you are already assuming the axiom of choice to prove that countable union of countable sets is countable @shcolf

Balarka Sen

well, my notions of disturbing are rather odd. I found Lynch's stuff (mentally) disturbing :P

Alessandro

I think most people would agree that Lynch is at least extremely weird, if not disturbing?

Mike Miller

18:05

@BalarkaSen Here's a problem I heard recently that you might like. Let $C_p$ be the $\Bbb R$ vector space with basis given by the set of collections of disjoint simple closed curves in $S^2$ that don't intersect a given set of $p$ points, and such that no curve bounds a disc disjoint from the other curves and the points.

This forms a chain complex, with $\partial: C_p \to C_{p-1}$ given by taking the alternating sum over what you get when you delete one of the $p$ points.

shcolf

@Alessandro can you please tell me, at which moment we assume axiom of choice in proving countable union of countable sets is countable, I think I've missed that also.

Mike Miller

The only thing is that sometimes, when you delete a point, you might get a circle left that does bound a disc (if that circle winds in a small loop around the point, eg). In that case, you delete that circle and multiply by $\delta$, where $\delta > 0$ is a real number.

So for instance if for $x \in C_2$ the two points are the north and south pole and the circle is the equator, $\partial(x) = \delta y - \delta y$ where $y$ is the configuration of no circles and one point.

What's the homology of this chain complex?

Balarka Sen

Hmm, alright (I don't see why I'd multiply by $\delta$)

Mike Miller

It's just the definition. It gives a family of chain complexes. It's related to something in operator algebras somehow.

Alessandro

@shcolf I don't know the details of the proof in your book or the one a professor showed you, but you have a countable family of countable sets, $\{S_n\}_{n\in\mathbb{N}}$ and at some point in the proof you're probably going to consider a sequence $(s_n)_{n\in\mathbb{N}}$, such that $s_n\in S_n\forall n$, to state that such a sequence exists you need the axiom of (countable) choice

Mike Miller

18:10

It's known that it vanishes for $\delta \leq 2$, I think (or maybe just $\delta < 2)$. People think it vanishes for all $\delta$, but aren't sure.

Balarka Sen

interesting

I'll read the problem once again carefully

Alessandro

worrying about those details is probably just confusing right now so I wouldn't mind too much @shcolf

shcolf

@Alessandro thanks alot man. After reading your answer and some answers

from math.stackexchange.com/questions/603456/… here, I think I'm getting an untuition

Balarka Sen

@MikeMiller Ah, so the objects of $C_p$ are formal linear combinations of configurations of such curves and the $p$ points.

Mike Miller

@BalarkaSen Equivalently, $C_p$ is generated by the set of "collections of simple closed curves on $S^2 \setminus x_1, \dots, x_p$ up to isotopy, modulo the relation that you can delete a circle that bounds, as long as you multiply the resulting thing by $\delta$"

Yes.

Balarka Sen

18:14

Got it, I was a bit confused about what to take alternating sum over after deleting a point, in the defn of $\partial$.

Alessandro

The axiom of choice is almost always (implicitely) assumed anyway, most of the people not assuming it are doing so on purpose to study what happens without @shcolf

Does someone have any idea for the general case of this question math.stackexchange.com/questions/1954108/… ?

Balarka Sen

so your "north pole, south pole, equator" configuration is a closed chain/cycle, yup?

Mike Miller

yeah

Balarka Sen

very interesting

Mike Miller

There is a subtlety here in that it seems like you need to pick identifications with what happens when you delete the north pole or the south pole, given that it's a point deleted in both,b ut in different places. I'm not sure what you're supposed to do for that.

Balarka Sen

18:21

Why can't you do an ambient-isotopy of the whole configuration?

Mike Miller

I'm not sure if that's allowed.

Balarka Sen

I don't see what can go wrong. I want to say "Two configurations in $S^2$ are equivalent if there is a diffeotopy of $S^2$ starting at the identity, taking points to points, curves to curves.", just to clarify

Mike Miller

No, it makes perfect sense. I just don't know if that's the definition of the chain complex, or if the points are demanded fixed.

Well, you would have to sign the points, I think, too. Which might be weird.

Or I guess I don't orient the curves in the first place. That could be OK.

Maybe I should find somewhere this is actually defined. :P

Balarka Sen

heh

Mike Miller

I find this to be really good working music.

Balarka Sen

18:30

speaking of, I seem to have lost my headphones

Adeek

hey @BalarkaSen just good question

If I have group of order pqr where p < q < r

which I proved isn't simple

then I showed that any two distinict prime pq can't be simple

Balarka Sen

nice

Adeek

but then

I was wondering why does this finish it ?

Balarka Sen

finish which?

Adeek

by lagrange we know that normal subgroup can have order p,q,r or pq,pr,qr

why can't our normal subgroup be product of primes ?

Balarka Sen

18:37

"our" normal subgroup? I am confused about what you are trying to show. Can you precisely state it?

Adeek

ok so what I am trying to show is the following I have a group of order |G| = pqr where p < q < r. I am trying to show it must have a normal subgroup of order p,q, or r.

I showed that such group can't be simple right ?

I showed it by direct counting argument

Now I would like to show that it must have order p,q or r

Balarka Sen

@Adeek Not "must" though. Just that there is a normal subgroup of prime order.

You've got quantifiers wrong

Adeek

I would like to show that it must be of prime order and not product of primes

Balarka Sen

You wrote here that you have to show there exists a normal subgroup of prime order. Now you're claiming you also have to show there is no normal subgroup of semiprime (aka product of primes) order?

Adeek

yeah

dustin

19:01

What is the interpretation of R^2 for random forest regression? With a linear regression, I understand R^2 is the explained variance in the dependent variable predicted by the independent variable.

Balarka Sen

19:41

@Adeek Sorry, I left. I am a bit rusty on group theory, but I don't think existence isn't hard. Take an $r$-Sylow subgroup of $G$. Number of such subgroups divides $pq$, and is $1\pmod{r}$. It can't be $p$ or $q$ because they are both less than $r$, hence dividing will give residue exactly $p$ and $q$ respectively. If it's $pq$, the total number of elements these subgroups occupy is $pq(r - 1) + 1 = pqr - (pq - 1)$ (prime order subgroups, so intersect trivially).

Look at the $q$-Sylow subgroups. Number of such things divide $pr$, and it can't be $p$ as before ($p$ is not 1 mod $q$). So is at least $r$ (if not normal!). That gives a total of $r(q - 1)$ nonidentity elements. Also the $p$-Sylow subgroup gives $p - 1$ nonidentity elements. So you have a total of $r(q - 1) + p - 1$ nonidentity elements. This is larger than $p(q - 1) + p - 1 = pq - 1$

That implies either the number of $r$-Sylow groups is 1, or the number of $q$-Sylow groups is $1$, so you get a normal subgroup in any case.

I think the other stronger result that such things are the only possible orders of normal subgroups should be provable from similar counting arguments.

Actually the stronger result is trivially false. Look at $\Bbb Z_{15} \times \Bbb Z_2$. $\Bbb Z_{15}$ is a fine normal subgroup of semiprime order :P

Hi @arctictern

arctic tern

19:59

hi

Balarka Sen

what's up?

arctic tern

looking at past putnam problems

just did this one: count the triples (A,B,C) of sets with A∪B∪C={1,...,10} and A∩B∩C empty.

Danu

@MikeMiller this is high on my list...is it as good as it looks?

Balarka Sen

@arctictern nice

Mike Miller

@Danu It's in my top 5.

Danu

20:11

The actor is so great. The master was masterful

Seymour Hoffman

Andrew Thompson

but it only got a 7.5 on imdb and hence can't be that good because that's how movies work.

Balarka Sen

@AndrewThompson bleh

Danu

Also, this result that the Picard group is a torus fibration is awesome

@AndrewThompson hahaha

Imbd rating correlateds with quality in a... Complicated way

Balarka Sen

i prefer rotten tomatoes' rating over imdb's.

but i do not consider any of their ratings overly relevant to the film's quality

Andrew Thompson

yeah, people tend to say that. i don't really watch movies a lot so i wouldn't really know.

Danu

20:15

Yeah

Alessandro

I usually see what Roger Ebert thinks about the movie

Danu

Best movie I've seen in a long time was "Embrace of the Serpent"

Andrew Thompson

@MikeMiller Did you get through the (parts you were interested in in the) paper?

Danu

(Original title is Spanish)

The ending is a bit of a matter of taste but everything leading up to it is great

The settlement gone astray is one of the best scenes I've seen

(Intentionally cryptic... Don't wanna spoiler)

Balarka Sen

I plan to watch some of Kiarostami's stuff soon, especially since I have more time now

Danu

20:22

Once upon a time in Anatolia was also good

Mike Miller

@Danu I saw that out here before the oscars. It was amazing.

@AndrewT Not yet. I'm working through a web of related stuff too. I'm about halfway through it.

But right now I'm trying to find my screwdriver.

Danu

@MikeMiller embrace of the serpent? (I'm on phone, can't see what you replied to)

How did you like the ending?

Mike Miller

I didn't reply to any message. But yes.

Semiclassical

@arctictern shouldn't that just be $6^{10}$? i feel like i'm somehow making it too easy

arctic tern

@Semiclassical yes

Semiclassical

20:25

mmkay

Danu

My girlfriend thought it was weak. I liked it

arctic tern

nother one I liked: compute average number of "local maximums" of a permutation in S_n (define appropriately)

Danu

Wait, they got an Oscar?

Semiclassical

huh

would a local max be something with a->b->c and b>a,c?

arctic tern

right, or edge cases just bigger than its one immediate neighbor

Semiclassical

20:27

not sure what you mean by that

each element has both a predecessor and a successor

arctic tern

like f(1) bigger than f(2) (but there is no f(0) since 0's not in {1,...,n})

Semiclassical

how could there be no f(0)?

Mike Miller

I don't remember. They were nominated, and there was a viewing for members of the academy.

Semiclassical

well

hmm

arctic tern

can use symmetry to arrive at the solution relatively quickly

Semiclassical

20:29

sure

i'm tempted to think of it diagramatically

i.e. draw $n$ dots on a circle and then $n$ arrows between them

but i can't be arsed, if i'm honest

arctic tern

nah, you want to see how often k is a local max for each k=1,...,n

split into cases k=1, 1<k<n and k=n

Danu

@MikeMiller oh, nice

Mike Miller

20:43

Man I hate time limits

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$Mathematics$ Mathematics