Hi, can someone tell me if P(AUBlC)=P(AlC)+P(BlC)-P(AnBlC)

@CarryonSmiling no worries

hello

How did you free your mind, free_mind?

well i grow up in a cult... lol

cool

i got out of the cult when i was about 25 (now 32) and this happened once i freed my mind

math and physics is what helped me to think for myself :)

the religion looked down on higher education and so my parents didnt allow me to attend college... which is why i'm no pursuing a math degree at 32

now*

great

so math question...

i'm trying to find the limit of a sequence

sequence of functions

mmhmm

$f_n(x)=\left(\frac{1-x}{2}\right)^n$ so I broke it up like this: $\lim \limits_{x \to \infty}\frac{1}{2^n}-\lim \limits_{x \to \infty}\frac{x^n}{2^n}$

the first limit is 0

but confused by the second one... is that also 0?

can't do $(\frac{1}{2}-\frac{x}{2})^n=(\frac{1}{2})^n-(\frac{x}{2})^n$

consider $a^n$ as $n\to\infty$ for fixed $a$. (you should split up into cases based on $a$'s size)

then apply with $a=(1-x)/2$

so $\lim \limits_{n \to \infty}a^n=\infty$

not always

depends on $a$'s size

my interval is also bounded between $[-1,1]$

you mean you're only considering $x$ on that interval?

right

if $x\in[-1,1]$ then $(1-x)/2\in [0,1]$ right?

if $a\in [0,1]$ then what is $\lim a^n$? (depends on whether $a=1$ or $a<1$)

1?

if $a=1$ then $\lim a^n=1$ sure. then what happens if $a\in [0,1)$?

0

and at $-1$ the limit is $-1$

piecewise function of 3 parts

when $x=-1$, the value of $(1-x)/2$ is $1$, so the limit is $1$

if the domain of $x$ is $[-1,1]$ there are two pieces to the graph

I am not sure whether there are many users interested in it, but there is area51 proposal Math Review and it needs more example questions (27 more, to be specific). This proposal has also been discussed on meta.

Although it seems that it was received a bit better on meta.MO than on our meta.

What is the parity problem in number theory?

Good sunday all users!

@BalarkaSen Is that first picture really a surface?

Yes.

How do I get rid of the single remaining edge?

What do you mean by the remaining edge?

If I identify the red edges

There is still a single edge left

It "sticks out" from the bouquet of two circles (which one subsequently identifies, yielding a single circle)

Eh?

you don't see what I mean?

If you cut along the diagonal, take out the triangle consisting only of the red edges. Then identifying the red edges on that single copy gives a Moebius strip.

I don't see what you mean by an edge sticking out.

Just fold a handkerchief maybe

if you just stick the two red edges together

where you glued them there will be a line sticking out

Oh wait, you're talking about the first picture. I am thinking of the second.

Alternatively, this graph (with no disk filled out) is just a bouquet of three circles where two of them have to still be identified, but the two halves of the third circle also have to be glued together to make a single line sticking out

Yeah, Lee said I was wrong about it not being a surface---I don't know if I believe it quite yet

He just "cancels" the red edges (?!?!)

Hey everyone. If i have column of 0's in a matrix, is the matrix invalid?

@betarunex Define invalid

@Danu I still don't see what's the edge sticking out. If you glue those two red edges you get a disk (forgetting about the other triangle consisting of blue edges).

That's what Lee Mosher's talking about when he says "cancelling"

@BalarkaSen I don't see how I get a disk

It looks like a cone to me

@Danu Is such a matrix possible/useful in calculations?

@betarunex Who knows?

A cone is topologically the same as a disk...

@BalarkaSen Oh lol I'm thinking smoothly :P

:)

Fail

I shouldn't post answers on MSE

You should. You gathered some intuition today, or rather, figured a flaw in your intuition.

This is learning.

@Danu I ask because if AC=BC, then A and B are identical right? But that's not true if I have a matrix with a column of 0's which would disprove that theorem. So is such a matrix possible or are matrices with a row of 0's an exception to that rule?

(agreed, one has to embarrass oneself a bit - as I did numerous times - but in the end it's productive)

@betarunex If $AC=BC$ then it does not follow that $A=B$. This is obvious: Take $C$ the zero matrix.

@betarunex Ah, now that I read your whole message I see that you realized this.

@Danu Wait. It might just be if matrix C is invert able

Yup

Thanks

Can someone help me understand the what the parity problem in sieve theory is? Is it about understanding the signs of the Liouville function?

Just to check i'm not going mad or falling asleep: let F be the free group functor, Set->Grp. The theorem that "FX iso to FY iff X iso to Y" is just saying that F creates isomorphisms, right?

Back.

if 4 % is the interest per annum then would 4/12 monthly interest?

can someone help me with this ?

Hi @Krijn, @AndrewT.

How were your days?

@PatrickStevens Um, no, it doesn't mean F is an isomorphism between Set and Grp. For one, not all groups are free groups.

F isn't even a bijection between the set of objects of the relevant categories.

@Krijn More or less fine. How about you?

@BalarkaSen Great, weird and not too sober. All in all, okay.

Too little math though, I'm starting to miss it

@GeorgeJPadayatti 0.04/12 or (4/12)% would be monthly interest. so 0.33....% monthly.

@be

@Krijn The way we all (at least so I think) feel about math is strangely similar to how Stalker feels about the Zone, isn't it? :P

@betarunex so 1-monthly_interest will give what?

Not quite relevant, but a passing thought.

I think that applies more to you than to me

@GeorgeJPadayatti Compounded monthly or yearly?

Hah.

@betarunex yearly

@BalarkaSen I'm glad you didnt say similar to Requiem of a Dream, hah

Did you see that?

Nope, but I have heard of it. Should I see it?

You should!

Although maybe not until you're just a tad older

@GeorgeJPadayatti Then interest each month would just be Total*(0.04/12). Assuming all you wanted was simple interest (not getting interest on your previous interest).

Otherwise I feel irresponsible

:P

@betarunex actually i am trying to solve a question vgy.me/yDDcL0

@Krijn What kind of math do you plan to start on, then?

Bit of arithmetic geometry I think. Bit of Hartshorne. And I should do something about my knowledge of category theory.

Nice. But that's all algebra I have no idea about :)

Oh well.

Category theory?

I mean, all of that, not just category theory.

I did like commutative algebra, did you do that?

The amount I needed to learn the basics of algebraic geometry, yes.

MacDonald & Atiyah?

Mhm.

The whole thing or just some chapters?

Oh, just some chapters. The first 5, I think, and a bit of the last few.

Did you enjoy that?

You don't seem to approach things very algebraically, most often

Sure, but I prefer motivating commutative algebra with geometry.

That's one of the reasons I like Reid's book better, which covers more or less the same content as A-M.

Yeah, that's true

I missed that in the course as well, but doing alg. geom. at the same time made up for that

@BalarkaSen Oops, I think I might mean "reflects isos".

@PatrickStevens Not sure if I ever heard that terminology.

ncatlab.org/nlab/show/reflected+limit

I do not see how that "reflect" terminology is appropriate in this context.

But I am no category theorist.

What can one do with cyclotomic polynomials?

Are there any examples where the eigenvalues of a matrix are also zeros of the Fourier series that the matrix generates?

Or where the eigenvalues of a matrix are frequencies in a Dirac comb generated by the matrix?

Det frågtes blott men gavs ej svar.

Or if not zeros, at least related to the frequencies?

@TedShifrin Suppose $f : (X, \partial X) \to Y$ be a map transverse to $Z \subset Y$, and $\partial f \pitchfork Z$ too. Am I right in saying $f^{-1}(Z) \subset X$ is transverse to $\partial X$?

Hey @Akiva.

Hi!

I have a silly question related to Hardy-Littlewood inequality (en.wikipedia.org/wiki/Hardy%E2%80%93Littlewood_inequality)

Namely, does it still hold for sequences? If $a,b\in\ell_2(\mathbb{R})$ are non-negative, and $a*,b*$ denote a non-increasing rearrangement of the sequences, is it true that $$\sum_{k=1}^\infty a_k b_k \leq \sum_{k=1}^\infty a*_k b*_k$$?

@BalarkaSen I'm not staying now, but see section 1 of Chapter 2 of G&P.

It sounds like it should be immediate given the HL-inequality, by going through the piecewise constant functions $(f_a,f_b)$ associated to $a,b$. But then I'm unclear on whether it's actually true that $(f^*_a,f^*_b)=(f_{a^\ast},f_{b^\ast})$.

Hey @Ted! Bye @Ted? ;)

@TedShifrin Well, I have, I don't think they prove anything like that. But if I am right, it's just a simple consequence of preimage of tangent space being tangent space of preimage for transverse maps.

I just asked as a sanity-check.

hi

can you see my profile?

Hey @Teddy

(also, if $a,b$ have finite support, this is immediate by the rearrangement inequality (en.wikipedia.org/wiki/Rearrangement_inequality))

Nope @Teddy

I see "page not found" o.O

That's strange... Maybe post on Mathematics Meta?

maybe

It's not so bad

I have to go offline now, bye bye

Darn, I have to get a lot of silly work done, whereas I wanted to get some math done.

Sounds like me

225 pages of notes for my last exam to go through

sigh

Maybe it'd help to know that I have to draw a bunch of graphs and crap for my statistics project. Reading notes is at least better than that.

Hi @arctictern

hi

What are you working on?

netflix

@robjohn these days I obtained the most important results from all my mathematical activity that allow me to calculate integrals that I don't think they were ever done before, or even known.

Or even imagined by anyone.

:P I meant as in math.

I have a few things on pause

writing and reading about octonions and rotations

currently have SBS tabs open on "small finite sets" and "the outer automorphism of s6"

over christmas I found a geometrically-inspired proof that ${\frak so}(n)$ is simple for $n\ne4$ that I want to write about for example

also included things like coordinate-free proof that ${\frak so}(3)\cong(\Bbb R^3,\times)$ and $\widetilde{\Bbb G}(2,4)\cong\Bbb S^2\times\Bbb S^2$

if we denote by $R(\Pi,\theta)$ the rotation by angle $\theta$ in an oriented 2D subspace $\Pi$ of an inner product space $V$, then every rotation looks like $\prod R(\Pi_i,\theta_i)$ for some angles and orthogonal oriented 2D planes $\Pi$

the infinitessimal version in ${\frak so}(V)$ is $E=\sum c_iE(\Pi_i)$

the geometric idea is $RE(\Pi)R^{-1}=E(R\Pi)$

Im making a letter template in latex... I am a student in the Mathematics department. Do you guys think it is ok to have my name

oops

Didn't mean to send yet

combined with $I\triangleleft{\frak so}(V)$ an ideal iff $I$ an $SO(V)$ subrep

we can conjugate an arbitrary element of ${\frak so}(V)$ by various plane rotations and then add the results up to cancel the summands and get a single $E(\Pi)\in I$, from which it follows $I$ is all of ${\frak so}(V)$ (assuming $\dim V\ne4$)

(1) Name, (2), Student, (3) Department of Mathematics -- in my header? Or should I not associate myself with the math department in that way until Im actually a PhD candidate / assistant teacher etc. Something more official?

what kind of letter?

Right now I am just using it to send a slightly informal letter to a proff to get advice on a paper I am writing before sending it off...... but I guess I would be saving the template to use for future semi formal letters I need to write

@arctictern what do you mean by rotation of a 2d subspace in a 4d space by some angle?

Basically I'm done for the whole year (except that I cannot interrupt my work for some obvious reasons despite my saying).

@BalarkaSen $R(\Pi,\theta)$ restricted to the 2d subspace is an obvious 2d rotation, and restricted to its orthogonal complement is the identity map. in 3d for example, we usually think of rotations around axes, but that doesn't generalize to higher dimensions, but plane rotations do generalize.

ah, you mean, look at it's orthogonal complement, rotate the plane around it's orthocomplement by some angle?

I mean rotate the plane around by some angle, and do nothing on its orthogonal complement

(in 3D its orthogonal complement would be the axis of rotation)

right, right, that's how i interpreted you. gotcha.

I wish I could see if all people that ever come here and said stuff about integration, like it is such a small area, if they could obtain some of the results within 50 years, say. That's annoying that many talk but have no idea about integration, they only think they know, but actually know nothing.

if the angles mod 2pi are distinct up to sign, the decomposition of an arbitrary R into orthogonal plane rotations is unique. the first case where nonuniqueness occurs is in dim=4, where we have left and right isoclinic rotations

Anyway.

I'm out.

for example multiplying $\Bbb C^2$ by $i$, viewed as a real 4-dim space, is the same as rotations in any two orthogonal complex 1-dim subspaces

right

this is what gave rise to my $G(2,4)\simeq S^2\times S^2$ interest

left isoclinic rotations of $\Bbb H$ (quaternions) are left multiplication by unit quaternions, and right isoclonic rotations are right multiplications

indeed $SO(4)=(Sp(1)\times Sp(1))/C_2$

(since multiplying by -1 on both left and right does nothing)

hmm

in higher dimensions though, there is no distinction between left vs right isoclinic rotations in odd dimension, and infinitessimal left isoclinic rotations in ${\frak so}(V)$ are not closed under addition (although they are closed under scalar multiplication)

I think some special stuff happens in $so(8)$ because of octonions but not fully grasping that yet

anyway, switching gears, a post I came across yesterday about "efficient version of cayley's theorem" had me thinking about functorial ways of constructing symmetric groups from a group. one way is $G\mapsto S_G$, i.e. just forget the group structure and build a symmetric group on it. but perhaps we want to go further and create a symmetric-group-valued functor $S_G$ along with a natural embedding $G\to S_G$. I think this is equivalent to embedding holomorphs $Hol(G)$s into symmetric groups.

then got to thinking about minimal degree permutation representations but didn't get anywhere with that

oh, that is an interesting question

so in the end you want to associate a group with a natural self-action, no?

not relevant, but because of logistical mishaps I now have 2.5 large boxes of caramel delight girl scout cookies. not the boxes of cookies, the boxes that contain boxes of cookies.

then the permutation representation of that action would give you an embedding $G \hookrightarrow S_G$.

@BalarkaSen I don't mean $S_G$ necessarily as the symmetric group on $G$ in my last comment, just a symmetric-group-valued functor applied to $G$

@arctictern yikes, burn them all

maybe use $Sym(G)$ and $S_G$ to distinguish the two

they're my favorite cookies

that's why you burn them!

I just don't think I can eat thousands of them

even with my god metabolism

@arctictern ah, got it

but anyway, there's an obvious self-action of a group on itself. multiplication. that immediately gives you a naturally associated $S_G$ together with an embedding of $G$ in it.

what's the problem with that?

tangential, and I think I remarked on it before, but the only "natural" permutation groups on a finite set are trivial, symmetric, alternating, and the klein four group on a four-element set

@BalarkaSen just curious if there is anything smaller for particular $G$

oh.

@arctictern i remember vaguely. what did "natural" mean, again?

functorial

aha, ok.

has to be normal subgroups of symmetric groups or something

err, so that bijections $X\to Y$ induce $Sym(X)\to Sym(Y)$

yeah

understood

here's an exercise. try to construct a klein-four group from a 3-element set {a,b,c} in a natural way

as in, as a permutation group of that?

as in, given a finite set $X$, construct a klein four group $V_X$ from it, so that any bijection $X\to Y$ induces an iso $V_X\to V_Y$

or more intuitively, don't do anything that depends on choices

aha, ok, ok.

of course. my previous question is nonsense, the 4-group can never appear as perm group of a 3-elt set. sorry.

here's a hint: let a,b,c be the nontrivial elts of the klein four group. figure out how to describe the addition table.

or multiplication table, whatever you want to call it

I basically gave you the answer

@arctictern Nice....

@arctictern is there anything wrong with the standard multiplication table? don't think so.

yeah, adding anything to itself gives you 0, and adding two distinct things gives you the third

that's the answer

right

or equivalently, the free $\Bbb F_2$ vector space on $\{a,b,c\}$ mod the element $a+b+c$

oh, clever way to say it

it's obvious from that that it doesn't depend on anything

same as $({\cal P}(X),\Delta)/\langle X\rangle$

where $\Delta$ is symmetric difference

Terminology question: I see in what sense the condition $\phi_{kj}\circ \phi_{ji}=\phi_{ki}$ for transition functions on e.g. bundles is the same as the cocycle condition in Cech cohomology, but is there something more to that name?

transition maps is pretty universal terminology I imagine, since that's what you call maps satisfying that condition in e.g. direct limits and inverse limits in category theory

@Danu something something sheaf cohomology of the sheaf of sections of that bundle

maybe you'd want to try it out, see if that's relevant

@arctictern I don't know anything about category theory though

But thanks for mentioning it

@BalarkaSen But what about transition functions between charts in a manifold? Or do they not satisfy it?

@arctictern he's asking why it's called the "cocycle condition" though, not why they are called transition functions, i think.

oh

derp

okay, another exercise

let Lin(X) be the set of linear orderings on X, and Perm(X) the set of permutations on X

obviously |Lin(X)|=|Perm(X)|

show that there is no natural isomorphism Lin->Perm

(say our domain is the category of finite sets with bijections)

@Danu not entirely sure how to make them satisfy something similar to the cocycle condition.

@BalarkaSen Probably not :)

i mean, if three charts overlap, $\varphi_{ij} \circ \varphi_{jk} = \varphi_{ik}$. For sure.

i guess that'd be your cocycle condition. i dunno how all of these has anything to do with cocycles though, but you're the one who knows cech and sheaf cohomology :) think about it

So I got bored and answered this nonsense:

Any ideas on special properties/something else of the matrix $$ \begin{matrix} a & b & c \\ b & c & a \\ c & a & b\end{matrix}$$?

ask anon/arctic tern.

?

@arctictern i'll think about it.

@Danu circulant matrix

related to cyclotomic polynomials and group determinants of cyclic groups

@arctictern Nice, thanks.

But it's even stronger than that, isn't it?

It's circulant in the columns as well as the rows

not stronger. comes automatically.

D'oh hahahaha

actually I guess yours is like circulant but in the other direction

Yeah, but it doesn't really matter.

actually I guess conventions differ on which direction to cycle the things

yeah, things are equivalent

Nice that it actually turns out to be something that has been studied

I think they were studied hundreds of years ago but became obselete in light of other things we developed

That turned out to be a surprisingly fruitful waste of time :)

@arctictern There is a 1979 book on them!

Reviewed here.

page 7 of circulants.org/circ/circall.pdf states they originally came about in early 1800s

before abstract algebra or even matrices

The review mentions some stuff

Catalan (1846)

I gotta grab something to eat.

@arctic Lend me some of your cookies.

@user1618033 very nice. what kinds?

Uck

That's hell of a weird of a topological space in Brian Scott's answer

@BalarkaSen It's a Cech compactification.

So yes, weird.

Agreed.

@anon Re coordinate free proof of oriented Gr(2, 4) being S^2 x S^2. Identify R^4 naturally with C^2. The subspace of Gr(2, 4) classifying the real 2 dimensional complex subspaces of C^2 is precisely CP^1. Is there a way to see Gr(2, 4) is a P^1-bundle over that, so that it's a zero section of that bundle? I can't see immediately how to get, for every complex line, a P^1's worth of real 2 dimensional subspaces.

Passing thought.

@BalarkaSen yeah, I talked about CP^1->Gr(2,4)->S^2, where S^2 is the space of left-isoclinic rotations of R^4 with unit norm in so(4).

I'd have to google what isoclinic rotations mean

if I understand correctly, if we call $S_L^2$ the space of infinitessimal left-isoclinic rotations with unit norm in ${\frak so}(V)$, and similarly $S_R^2$ for right-isoclinic, then $G(2,4)\to S_L^2\times S_R^2$ is what we want

basically, an infinitessimal isoclinic rotation is two right angle rotations in two oriented orthogonal planes. if the planes' orientations add up to that of the ambient 4-dim space, then it's left isoclinic, otherwise it's right isoclinic.

wow, @robjohn, @arctictern, @Pedro all here at once !!

and hi, @Balarka

Hello @Ted.

I still claim that if you understand what's in that section of G&P, you'll see it addresses your question.

Remember that transversality is a rephrasing of a regular value statement.

Yes, it's funny.

Right, I looked. How they prove $\partial X \cap f^{-1}(Z)$ is a manifold is essentially transversality.

Back to what... 2012?

oh, and belated hi @Danu, whom I missed.

@TedShifrin Yup.

@Pedro, sorry about misconstruing your "just" for a "the" :D

No problemo.

Well, that statement isn't good enough. You need that you get a neat intersection of $f^{-1}(Z)$ and the boundary, which is in fact what transversality means. You rule out things like a parabola meeting the $x$-axis.

BTW, @Pedro, you caught on that I have been pretending to know a little math? :D

@TedShifrin Oh, you.

of course, but they prove it by locally cutting $Z$ and etc. looks like a reproduction of the transversality condition in local coordinates.

Yes, and the picture looks like projection onto the $z$-axis in the upper half-space, for example, @Balarka.

In terms of point-set, this was a sort-of-cool question, although the OP has yet to respond to my answer.

However, $df_p(T_p \partial X) + T_q Z = T_q Y$ (because $\partial f : \partial X \to Y \pitchfork Z$), so taking preimage by $df_p$, I get $T_p \partial X + df_p^{-1} T_q Z = T_p X$. Now $df_p^{-1} T_q Z \cong T_p f^{-1}(Z)$ (because $f \pitchfork Z$), so isn't that a coordinate-free proof of the fact?

$q = f(p)$.

That doesn't address whether $f^{-1}(Z)$ is meeting the boundary too tangentially.

At least I don't see it at the end ...

@TedShifrin hi!

It proves that $\partial X$ is transverse to $f^{-1}(Z)$. Not sure what you mean by "too tangentially".

I'm so hungry but I don't have any food at home :(

Oh, you mean, upto change of charts it should be $\Bbb H^n$ with a half-ball (or something) hitting the boundary sphere-at-infinity.

Right, I don't get any local information from this.

Oh, I see @Balarka. G&P's theorem could be stronger, is the point. It's not just that the boundary of the preimage is the intersection with the boundary; it's that it's a "neat" intersection. Yes.

I just took some almond cakes out of the oven, @Danu. I can feed you.

(Well, there's other food, too.)

Cool point. That's thrice this week.

Can you guys help me figure out what I'm missing here? math.stackexchange.com/questions/1869679/…

So I got that part of the problem, but I need to write the problem using summation in terms of $n$ and then calculate the limit

@free_mind: What you put in the question is correct. I didn't look at anything else.

However, when I do that and plug in $n=6$ I don't produce the correct values...

I often wondered about that. We have a local description of submersions, immersions. Is there a local description of transverse intersection of submanifolds?

Taking the limit gives you the true area; having $n=6$ is only an approximation. You can't take a limit as 6 goes to infinity.

ahhhh

Like, upto change of charts, it should be something canonical like intersection of two vector subspaces transversely.

I dunno how to prove this.

@Balarka: Sure, $k$- plane crossing $\ell$-plane transversely in $\Bbb R^n$.

@TedShifrin Urgh

It's actually hurting my stomach

Go out and buy something, @Danu.

It's 1:30 AM and that means everything is dead here

Find a good friend?

Nobody lives close to me

I give up.

I guess I have to go to sleep :(

Schlaf gut.

@Danu drink some water

Anyways, how are you @Ted?

I am fair, @Danu, thanks.

You really should go to sleep.

So should Balarka (hours ago). ... I'm like a den mother here.

haha

Okie

Fijne avond verder nog

^Dutch

@TedShifrin Oh, nevermind, that's pretty easy to prove now that I sit down and think for a minute. Use immersion theorem to change charts and make $X$ and $Z$ look like two subspaces of $\Bbb R^n$ locally in $Y$. Then the tangent spaces "coincide" with the subspaces, so the transversality condition just tells me what they do.

Boo. Or am I being too hurried?

Well, you can't apply the immersion theorem to both simultaneously?

You have to do your own re-proof of the immersion theorem, but it's easy.

@TedShifrin Oh, right.

There are probably other ways.

I'll do this tomorrow. I do not believe it's non-obvious anymore.

Good night.

I didn't say I was going to bed :)

Note that once this is done, my transversality thing would yield a proof of G&P's local thing.