@EricStucky What other information do we need?

You basically need the linear operator, but you can apparently get away with less information in some circumstances. In the link, you can see it's achieved with the characteristic polynomial, the minimal polynomial, and the nullity.

hi chat

Anyone familiar with statistics here?

Hi @Semiclassical

Hi @BalarkaSen

Heya @Danu, @SemiC. How's things?

Easy question (maybe?): Why do split short exact sequences remain exact* under tensoring (I need to understand my prof's proof of the universal coefficients theorem)?

I hate how little algebra I understand :(

sweaty. it's hot outside and stuffy in here

Well tensor product is functorial. So if your tensoring preserves shortness, it must also preserve splitness.

i.e., tensoring the section of the original sequence gives a section

@balarka what kind've stats? (I probably can't help, if it's really a stats thing versus say a probability thing)

Sorry.. remain exact

oh. but tensor product does not always preserve exactness

For split sequences

the kind of things which does are called flat modules

@BalarkaSen That's why I'm asking

@Semiclassical I want to understand clearly how the 4th moment has anything to do with peakedness of the frequency distribution of a given data set (context: kurtosis)

I have kind of a partial explanation, but I am not satisfied by it.

Hey guys! Good morning :)

hrm. yeah, that'll be beyond me

Actually, I guess the direct sum decomp. makes it kind of clear

besides the fact that a Gaussian has zero fourth moment

@MikeMiller How are ya?

@Danu Tensor product commutes with direct sum, no?

I think you should be able to use that to show that the tensored sequence is exact.

Sorry, that was a huge typo.

But I haven't checked this; better ask someone else. I am a bit pressed with nonmath things.

Like what?

My school work. Statistics and physics. I should also read chemistry but I won't.

I know physics ;D

Of course :) The stuff's I am reading is too basic though.

NEWTON

NEWTON EVERY TIME

Heh

I recently learned that Newton found some identities for symmetric polynomials

Oh yeah. $\sum x_n^k$.

Expressed in terms of elementary symmetric polys.

It's in general true that given any polynomial in $n$ variables, if it's symmetric (i.e., invariant under any permutation of the indices) then it can be written as a polynomial in terms of the elem. sym. polys.

This is the start of a cute topic of math called "invariant theory".

Oh, really?

Funny.

Mhm.

I mean the last thing---the other seems intuitive

'cute topic of math' fight me irl

you don't care about/like invariant theory? :D

no, I'm just being an ass about the diminuitive

ughhhh mathematica

^classic physicist ;D

you have failed me for the not-last time

(tbf, this isn't mathematica being dumb as much as the problem being a pain to compute)

(but mathematica is a convenient punching bag)

@Clarinetist I summon thee.

XD

Clarinetist Clarinetist Clarinetist?

Clarinetist is a satistics?

He's onto it I think, yeah.

more like sadistics, amirite

(no i am not)

I'll punch you in the teeth for saying that, and enjoy your pain amid laughter and merriment.

yeah. good thing ted's not here, or i'd have gotten a thwack

Hey @EricStucky :) how ya doin

Urgh... every book just says "the results is exact because the original splits" without further comment

What is the context?

I have: $0\to A\to A\oplus B \to B\to 0$ (exact)

(Abelian groups/R-modules)

I want $0\to A\otimes G \to (A\oplus B)\otimes G \to B\otimes G\to 0$ to be exact, too

Or rather, I want to understand why this is true

Is that really what the book says?

Yeah, several books.

Bredon, Rotman (has it as an exercise), etc.

My attempt... Assume $(a;g)$ is in the kernel

$(a;g)\mapsto (a',b';g')=(0,0;0)$

Maybe your A, B, G are something specific? I haven't checked this, but note that $(A \oplus B) \otimes G \cong (A \otimes G) \oplus (B \otimes G)$

No, any R-module works

Once you know that, note that $0 \to A \otimes G \to (A \otimes G) \oplus (B \otimes G) \to B \otimes G \to 0$ is exact.

Then the isomorphism gives a morphism of these two exact sequnces. Once you can show it really is a morphism of the two exact sequences, you have shown that it's an isom.

@Danu Hmm?

@BalarkaSen For $A,B$, I mean. $G$ is an Abelian group

Oh, I see what you were referring to. OK. I just mean you should start by using that isomorphism of distributivity of tensor product and direct sum and write down a commutative diagram using that.

Then show accordingly that the sequence is exact. It should not be too hard if it's true.

@Danu I'm ok.

@MikeMiller Not sick anymore?

Nah.

Back.

Speaking of, I am a bit sick today. Not particularly relevant though.

@Semiclassical On which of my problems did you work lately?

none, if i'm honest. been busy with mathematica

OK

going to have to go back to that soon and start another batch of calculations

@Semiclassical I found the solution to a class of very hard integrals (I didn't manage to work on it in the last days, but just a bit).

@Danu How are you?

nice

@Semiclassical The main work was finalized some days ago.

@MikeMiller Pretty excited. I had a long talk with the top2 TA (a postdoc) today.

What did he do wrong?

@MikeMiller Nothing (?)

Just a joke about the long talk phrase.

I see

It was real nice of him to make time for me (we talked for about 3.5 hours)

What about?

First we spent a bunch of time on some questions about the lecture, and after that we talked about different branches of mathematics, my choices regarding my future in academics, etc

heavy stuff

Kind of

I'm having a lot of doubts :P

I know the feeling. Feel free to email me if you want to talk about it.

I sound like an angsty teenager :P

@Semiclassical Also, almost forgot, I have received an amazing work offer (these days - on a project with some mathematicians).

Hey @OneRaynyDay: I'm doing well, but today is kind of a boring day; just getting done some little tasks that have piled up.

@MikeMiller the multiplication-by-g map is a function $R_g:P\to P$. If we can somehow obtain a function $f:P\to\mathfrak g$, then $f$ can be considered a Lie-algebra-valued 0-form, so $d_P f$ is indeed a Lie-algebra-valued one-form. So far I'm getting it. But how do you get an $f$ from $R_g$?

oh, nice!

Things move here with a very high speed lately.

(positively)

I'm out for now to take some sleep.

@Danu We all know the feeling. Don't feel that way.

@AndyMiles I don't remember the details anymore, I do suggest looking at Kobayashi-Nomizu. But a map to $G$ also induces a Lie algebra valued one-form, because the tangent bundle of $G$ is canonically isomorphic to the trivial $\mathfrak g$-bundle over it.

The horizontal tangent bundle of $P \to B$ is $\mathfrak g_P$.

when we say $\sigma$ is a permutation of length $n$ do we mean it is a cycle of length $n$ or its total length (between all of its disjoint cycles) is length $n$?

in the context of the symmetry group $S_n$

Getting closer. Thanks.

@EricStucky I see - gotcha. At least not too busy!

@MikeMiller chat.stackexchange.com/transcript/message/29135473#29135473

Would you be interested in answering this?

Does anyone know a reference that deals with Riemann surfaces in a concrete fashion? In particular, I'm looking for an explanation of the Riemann surfaces associated to algebraic functions, which would allow me to do computations (find poles and zeros etc.)

Hi @EricStucky.

@Danu maybe "Algebraic Curves and Riemann Surfaces" by Miranda?

He's not anymore. :p

@danu what kind've stuff are you trying to compute, out of curiousity?

heyy

Hello

wanna take this to the other chat?

Sure

@Semiclassical Zeros and poles :P

@MikeMiller I meant concrete in the sense of "enabling one to do computations"

@SteamyRoot Thanks; I'll have a look.

So do I. Multiple perspectives are often helpful.

Sure

What's the most elegant way to derive a formula for 1^k + ... + n^k

I found youtube.com/watch?v=8nUZaVCLgqA but I find it very hard to follow

In my opinion, reduction to degree one lower.

But it's a cute trick that's not so easy to come up with, perhaps.

@Danu, could you elaborate?

it'll be hard to find a nice explanation, given that the generic result involves Bernoulli polynomials: en.wikipedia.org/wiki/Faulhaber%27s_formula

But the video link I posted seems to bypass the need for Bernoulli polynomials...

You want to compute $$ \sum_{m=1}^n m^k$$ Instead of approaching it directly, consider $$ \sum_{m=1}^n m^{k+1}=\sum_{m=0}^n (m+1)^{k+1}-(n+1)^{k+1}$$

@Semiclassical, yer ... I was looking at that page as well

Expand, and you'll find your sum in terms of all the lower ones.

Bernoulli numbers, sorry. but i'd be skeptical regardless

I guess this is how you get your Bernoulli numbers. The principle is very simple

Danu's trick is explained on there as well, btw

Okay.

main thing is that it gives you a convenient way to bootstrap, i.e. if you know the sum for $k$ you can use that approach to get it for $k+1$

You need all the lower ones, in fact.

hmm $foo$ is failing to render in Chrome

right

@Pi I see it fine, in Chrome

@Axoren, Did you have to do any fiddling? I have looked at the tinyurl link in the topic (math.ucla.edu/~robjohn/math/mathjax.html) & clicked ChatJax & refreshed, but no joy.

It's a bookmarklet. Drag it on to your bookmarks bar

Go to the tab you want to activate jax in

And click on that bookmark

It's a local script. It only activates for the tab you're currently open in.

oh!

Anyone know some unicode? I'm looking for a whitespace character that's really small, ideally nonexistent.

nonexistent?

lol

I mean, width 0

not sure if it's what you mean, but I think alt 0173 used to be invisible in many places

hmm I will try

either that or it shows up as a hyphen, I think

XD encyclopedia dramatica has an article about it wao

yep that works :)

apparently it is very widely non-rendered so here's hoping *crosses fingers*

@Danu, I can't see how that helps

I can see the idea of trying to decompose a k+1 problem into a k problem

some kind of telescoping..

But I can't see what the next move is

Expand the $\sum (m+1)^{k+1}$

You'll get it.

But the original problem was to compute $\sum m^k$, and this is more complex..

Just do it; there are cancellations.

if nothing else, take Wikipedia's equations and write it out for $k=1,2,3,\cdots$

but there's really no way around the fact that getting it for generic exponent either requires 1) knowing it for every smaller exponent first and using the telescoping approach, 2) using an approach that utilizes the Bernoulli numbers. It's not something that has a simple solution.

If you're looking for something with an easier approach, I think that series like $$m(m-1)+(m-1)(m-2)+\cdots+(2)(1)$$ or $$m(m-1)(m-2)+(m-1)(m-2)(m-3)+\cdots +(3)(2)(1)$$ are more tractable

I don't understand the craze about $\sum i^k$ in this discussion.

eh. someone asked, we answered

the discussion seemingly went on for an hour, is simply what i was referring to. i don't see why the sum is interesting enough to ponder on for hours.

but then it's just me

eh. comments may have been spread out over an hour, but the main back-and-forth was in the first five minutes.

if you say so

It was interesting enough for you to join in ;)

dunno why i bothered writing out an answer to this question

i doubt the OP will bother to understand it, and anyone else reading it will probably not either

lol @ the comment below the answer

yeah :/

the line "There is no royal road to Geometry" comes to mind.

in this case being, "There is no shortcut to understanding motion problems."

motion problems are confuzzling sometimes.

occasionally. but projectile motion problems are usually just plug-and-chug

right.

e.g. "it goes up and then comes down, so how long does it take to get back to a height of zero? And then how far does it go in that time?"

On the other hand, a problem that's tough if you try to handle it directly is what the maximum range of a projectile is if you allow a starting height.

Physics so confuz amirite?

the temptation is to get the horizontal range as a function of initial angle and maximize it. that way sucks.

@Danu "yeah. they call them point particles, but also assume they have nonzero mass. what a confuzzling assumption"

better way is to get the maximum height achievable at a given horizontal distance, and figure out when that max height goes to zero. :)

@Danu 100% agree. I still don't know how magnets work.

fun physics question: Can a magnet do work?

What's a magnet?

@Semiclassical I haven't had particular experience with projectiles yet. Soon, though.

@Semiclassical That is a terrible question. You glorious monster.

mmkay.

snerk.

the answer I'd give: it depends on how the magnetic moment of the object being acted on arises

if it comes from currents, then the answer is no: an induced electric field ultimately is what does the work. but if it exists as a property of the object---say, the magnetic moment of an electron---then sure, magnets can do work.

a quote from a book(let) i am reading right now "Once one of my students in his test on probability theory managed to obtain $P(A) = 4$ and wrote the following explanation: "this means that the event is more than probable""

ow

stuff like that is why I'm always happy when a student includes some check on their work

such as "do the units work out" or "does this number seem at all plausible"

units are ugh

units are essential

the only place where they become ugh is when magnetism is involved, coincidentally

the two statements above are not exclusive

Does anyone here have a neat proof that the symmetric polynomials over an arbitrary commutative ring are algebraically independent?

let me see: F=qE=qvB, so [E]=[vB]=Tesla*meters/second. But E=V/d, so [E]=Volts/meter. So Tesla = volt-seconds/meters^2.

and quickly one realizes how annouying that gets :p

plus one has amps = coulombs per seconds

Units pfft

so yeah, yaaaay EM units

HEP for life

lol

what I find more amusing than the HEP tendency to use hbar= c =1 is the condensed-matter habit of referring to temperatures in terms of energy scales and vice versa

Also very normal in cosmology

yeah, that's true

condensed matter people care about very low temperature physics etc.

i do find it funny that we can achieve laboratory temperatures so close to absolute zero that the very notion of temperature becomes kind've wonky

hard to talk about temperature when it's hard to talk about thermal equillibrium

@PedroTamaroff: Not all the symmetric polynomials surely! The elementary symmetric polynomials, perhaps?

small and silly thought experiment for you physicists: i am in outer space, and there's only my spaceship and nothing else is out there. not even a particle. my spaceship is moving with constant velocity (which is equivalent to not moving at all because there's nothing outside to compare motion with).

i am inside the spaceship, and set that as a reference frame. if, now, i accelerate the spaceship a bit then inside the spaceship i lean backwards a little bit. thus, some force has been applied. but where did this come from? how can i comprehend this force sitting inside the spaceship with that as a reference frame? newton's 1st law is apparently contradicted.

@Danu, gottit thanks

the spaceship hasn't moved either relative to anything in the outerspace, because there's nothing out there

@BalarkaSen You actually contradict yourself before then.

To propel your ship, you place particles in the space around you.

So there's no longer no particles in space.

ok, that's fair but it's a technicality. inside the spaceship i can still set it up as a reference frame.

newton's 1st law is still apparently contradicted relative to that frame.

what's wrong?

A spaceship is the same, just lots of really tiny particles going fast rather than one football going slowly.

@BalarkaSen Err.. how is this different from feeling the "force" inside a normal car that accelerates?

You're being pressed upon by the atmosphere inside the space ship because you're leaning.

Not because the acceleration is being applied.

I still have trouble believing you can accelerate the ship without the particles from your ship having an affect.

@EricStucky ¬¬

Yes.

@Danu dunno, maybe it's not. i tried to avoid frictions, etc., so set it up in outerspace.

I found one, though.

Which one do you know?

@Axoren I mean you can probably burn fuel etc but that still produces particles outside the space. But this is a technicality and not sure if relevant to what I want to ask.

Yeah I suppose to accelerate you need some other particle. Because you apply force to those, and they apply the force back to you by 3rd law which enables you to accelerate.

If there's nothing you cannot apply force to anybody and nobody applies any force to you.

OK, so abstractly I suppose I am asking if I have a frame inside a frame which turns and twists and loop-da-loops what happens to the physics inside that moving frame. Evidently the newton's laws do not work.

I am not sure if this is an interesting question anymore. But bwatever.

@BalarkaSen General relativity doesn't work in this specific instance. Guess we have to abolish it and start from scratch.

@BalarkaSen Well, then, here we go: en.wikipedia.org/wiki/Fictitious_force

@BalarkaSen Yeah, this is exactly the concept of "fictitious force" or "inertial force"

@Axoren This is not related to GR

That's a shame, a mistake on my part.

No worries