i have obtained a truly disgusting formula

Let $a(n,j,k,l)$ denote the number of solutions to $x_{1}+x_{2} + ... + x_{j} = n$, where $k \leq x_{i} \leq l$ $\forall i \in \{ 1,2,...,j \}$.

Oh, $k \geq 0$.

generating functions are powerful but so, so annoying

@MikeM: Some multiple of $\text{tr}(AB)$, of course.

@SAW: I don't find them annoying, but we've already established that you and I don't see eye-to-eye :P

well, to restate: the 3 hours i took to find that formula was annoying

@MikeMiller also ${\frak so}(\Bbb R^3)\cong(\Bbb R^3,\times)$, can use standard inner product on latter

especially because i doubt my prof will accept it as an answer hahaha

tern, is it obvious that one is biinvariant?

(of course, it's the same, I know)

@SAW: Mathematics that takes you months and then you realize it was obvious ... that's always annoying.

Three hours ... psssh.

hahahaha i am deeply familiar with that

I wasn't nearly so familiar with it until years into my Ph.D. and many years after, @SAW.

put my head down on my pillow at 4 a.m. and then solved a question in my research my prof and i had been working on for 3 months in 15 minutes

so immensely ... frustrating

LOL, welcome to mathematics, dear person!

yeah ..........

what a wonderful, relaxing field of work

@TedShifrin I mean, it's Ad-invariant because Ad in so(3) corresponds to usual applying-a-rotation in (R^3,x). (Not sure what bi-invariant on Lie group corresponds to for lie algebras.)

Well, I think it's ad-invariant. But, anyhow, we get to the same place pretty quickly. I think of it in terms of Maurer-Cartan forms on $SO(3)$, anyhow. :)

Hello! I recently started learning about tangent bundles and I was wondering what some of their uses are. That is, what new information about manifolds can we gain by looking at tangent bundles? This structure is given quite a lot of attention, so I just wanted some motivation for using it.

Vector bundles in general turn out to have lots of information, and there are important things called characteristic classes that can be computed. (For example, the Euler characteristic of a compact manifold is recovered from one of them.) But, more basically, all tensor quantities, differential forms, vector fields, etc., on manifolds are built out of the tangent bundle.

The very notion of derivative on manifolds is defined in terms of the tangent bundle.

G'night @MikeM

Hi.

Thank you! I will keep those things in mind. Are there any more basic ideas which take advantage of the tangent bundle? From what I know characteristic classes take quite a lot of background to develop. I think the book I'm looking at discusses vector fields as sections of the tangent bundle, so that seems like something feasible for me, although at a quick glance it seems like rewriting vector fields in a different language–is there a benefit to it?

As always, a more general language gives you more in the end, not at the beginning.

Yes, that is quite reasonable. Thanks for your help!

Come back anytime!

@O'''''''''''''''''''- (1) As already mentioned, derivative of a map of manifolds is literally a map from the corresponding tangent bundles. This puts the notion of maps between manifolds and derivatives of them on an equal footing: you can now work with derivatives coordinate-freely, like you sometimes for maps between manifolds. This makes a lot of things simple. (2) It's not rewriting vector fields in a different language, it's literally how one defines a vector field, as a map $M \to TM$.

(3) A simple, but perhaps indirect, application of tangent bundles is the Whitney embedding theorem. If $M$ is a (compact, say) $n$-dimensional manifold, it always embeds in $\Bbb R^{2n+1}$. The idea is to embed $M$ is some $\Bbb R^m$ for $m$ large, and "cutting down dimensions" by projecting it to hyperplanes.

To do this you need (1) injectivity (2) immersion of the projection of the embedded image. In the immersion bit a crucial step is to apply Sard's theorem to a map $TM \to \Bbb R^m$ to choose appropriately the hyperplane on which to project.

Injective immersion being embedding for compact domains, you can conclude the theorem for at least compact manifolds. Generalizing to noncompact is now just a matter of partition of unity.

The point is constructing $TM$ allows you to state genericity results for not only just maps, but derivatives of maps too. That's useful.

Are you awake too early again, @Balarka?

Yeah. My sleep cycle has split into two disjoint cycles, one from 1AM to 7AM and one from 2PM to 8PM. Not sure why this is happening; probably because of the antibiotics.

How do you work school into that schedule?

It doesn't help when your sleep cycle is homotopically distinct from the one you had earlier.

LOL @Balarka. Only you would indulge in such nerdy humor.

I take that as a compliment.

@TedShifrin I push the second one to start from 3 or 4PM.

Or I don't go to school that day.

You are allowed to skip school all that much?

It's been less than a week since that's happening though.

Sure, we're fine as long as it's 4 days a week.

And the holidays are coming fast in any case.

As long as what is 4 days a week?

Attendance, I meant, sorry.

Oh ...

Anyhow, the interrogation is over for now.

heh

Generally, when people starting to study something want motivation, they don't necessarily want super-technical motivation. But what you said was cool.

Right, I was wondering after writing all that if it got a bit technical.

nods

That's not how I think about Whitney in any case. The secant variety comes to mind before TM.

sure, non-injective is more evident than non-immersive.

Right.

but no need to mention partitions of unity (which s/he probably doesn't know yet)

true

The tangent bundle is more important than just what you said. You can define the notion of immersion simply as "derivative is injective in each chart." But the tangent bundle is where the actual derivative lives.

It's like the difference between "A function is nowhere zero" and giving the function. You can say the former without the tangent bundle.

Well, it's like intrinsic vs extrinsic, MikeM. You don't need the tangent bundle for it to make sense, but of course I get your point.

That's a pretty nice point.

@MikeMiller What's the "easiest" example of two nonisomorphic vector bundles with homeomorphic total spaces? If $E_1$ and $E_2$ are vector bundles with different Euler classes, their total spaces can never be homeomorphic due to cohomology (cup-square of the zero section). That's an obstruction. The simplest thing which breaks that is the tangent bundle of $S^5$ and the trivial 5-plane bundle on $S^5$. Are their total spaces homeomorphic?

I am guessing no, for no reason at all :P

You should be more careful about what you say. Do you really want to say "cohomology"?

because the cohomology is the same as the base for both... :)

Cohomology of their fiberwise compactification, I guess.

Sorry.

Compactly supported cohomology works too.

Right.

I shouldn't say simplest. There are simpler examples of that in dimension 3, where we already proved every oriented 3-plane bundle admits a global nowhere zero section. I should have tried to cook up examples with that.

E.g. T(RP^3) and the trivial 3-plane bundle on RP^3. Should be much easier.

How about T(RP^3) $\oplus$ R and the trivial 4-plane bundle on RP^3 instead? The total space of the former is homeomorphic to the total space of direct sum of O(1)'s over RP^3. The total space of O(1) is simple enough: RP^4 - {p}.

What's the total space of direct sum of bundles? Hmm. (I don't expect this will give me an example, but at least I'd be able to prove it for a case where the Euler class argument doesn't work).

Eh, I guess I don't see if the total space of that would be anything nice.

@BalarkaSen Thank you!!!

No problem.

@MikeMiller DO you teach an AOPS class?

(lol, I tried to find a few different examples - failed - but then googled to see that all orientable 3-manifolds are parallelizable so there isn't many examples in dimension 3 of the sort (aka comparing with the trivial bundle) I am looking for anyway)

Just a tick. T(RP^3) is actually diffeomorphic to the trivial 3-plane bundle though, I realized (RP^3 is a Lie group, duh). So if I could make that work that would actually give me an example, not a non-example.

Funny I didn't see that earlier.

@Human I do! Are you in it?

Alright, gotta go. I'll think about it later.

@MikeMiller Yeah! I'm in your Friday PreCal class. I'm Sem2000

how do I learn to prove the borsuk-ulam theorem painlessly

or at least, as painlessly as possible

given I at least have an intuitive understanding of some algebraic topology, I guess

man all my stuff is coming down to topology lately

@SamuelYusim Borsuk-Ulam in full generality requires some understanding of (co)homology. Do you want to see it done in 2 dimension?

@BalarkaSen Can you prove it smoothly, then use an approximation scheme to get the general result?

The approximation bit is a little bit unclear, but yes, I think so.

I know Brouwer can be proved smoothly, then Stone-Weierstrass leads to the general result.

(That is, can you smoothly approximate a continuous antipodal-preserving map by antipodal-preserving maps?)

@BalarkaSen Right. Something like that.

It's not clear that can be done.

Hmm, I guess it can be. Pass to projective spaces.

That's what you want to do. You need Whitney approximation for maps between projective spaces, not spheres.

@BalarkaSen I see. But one might as well learn (co)homology.

The smooth proof is itself nontrivial.

It's not distinct from the cohomological proof.

Speaking of Borsuk-Ulam, my linear algebra professor's advisor was Borsuk.

nice

I hope we get a formula sheet for my QM exam

Especially with trig formulas

I can't remember complicated trig formulas either. I've always had to derive it from scratch.

Interesting. Greg Kuperberg's parents were both students of Borsuk.

Trig formulas are just multiplication of complex numbers.

In essence, yes. However, I do not agree that DeMoivre always helps.

Once you learn the latter, you'll always be able to derive the former.

@MikeMiller Derive, yeah. But I don't always know what identity I'm looking for.

Fair enough.

Aha. I foiled wrong, but had the right trig identity.

So, apparently we're going to be in war with a neighboring country soon. Shit.

@BalarkaSen Hopefully not Pakistan.

At least you're sickly, you won't be drafted.

:P

@0celo7 It is that.

@BalarkaSen They have nukes.

Of course.

Why are you going to war?

Hell if I know. There has been a tension about who gets Kashmir for a long time.

Also, China seems to support Pakistan. I hope we'd drop it.

what's there?

Shrug. Just a bit of the Himalayas. :P

oil?

goats?

something?

Literally nothing, AFAIK.

There has to be some reason.

Good Tourism spot? lol.

foreignpolicy.com/2014/02/04/…

let's see

"Kashmir remained part of India, despite its Muslim majority, and the rest is history, or rather, rivalry."

And with India having a pro-Hindu Government, that's not going to get any better.

@MAFIA36790 5 minutes ago.

And I don't want to compute horrible QM commutators.

Look at a new set which has the same values as your set, multiplied by -1?

Yes.

$\inf (A)=-\sup(-A)$.

@0celo7 You mean $-\sup(-A)$.

@s.harp sorry I meant that fixed points exist along the y-axis

@BalarkaSen Yes, thanks.

@s.harp I didn't get how looking at those cases of $y<0,x>0$ or $y>0,x<0$ would help me with what's going on in the case of $y=0$

hey @BalarkaSen could you help me with a dynamical systems problem?

Nope.

damn you were my last hope

@SoumyoB Why? Balarka knows nothing about ODEs

^

@0celo7 I'm really hoping you do then?

Depends how easy the problem is.

well I'll tell you the exact problem instead of beating about the bush, it's as follows-

I found out the obvious that the y-axis is the set of fixed points, and that the eigenvalues are $0$ and $y$

so for $y>0$, we have unstable fixed points, for $y<0$ we have stable ones, but what happens at $y = 0$, when both the eigenvalues are zero boggles my mind

OHHH WAAIIT I've made a mistake in calculating the eigenvalues, dammit

uhmm well I had actually done it correctly

well so I'm back to square one

could anyone tell me what happens then when both the eigenvalues are zero?

@SoumyoB in the case $y=0$ take a look at the equations, they are

oops didnt mean to hit enter

Hi guys! Can someone help me in understanding the 8 queens puzzle?

@s.harp thanks I think this should clear it up

oops meant to post the link in a more formatted way

hello @MikeMiller!

Hi @Danu.

@MAFIA36790 you mean upload to the chat on here?

wow you learn something new everyday

so I just drag the image to the typing space?

@MAFIA36790 I have forgotten, what's the defn of order-completeness?

where do you upload the pic though

@MAFIA36790 consider $\mathbb R -\{0\}$, here the set $(0,1)$ has a lower bound but no infimum

@MAFIA36790 I don't see any upload button :/

mine doesn't have an upload button there

perhaps it's an option unlocked when more reputation points are earned

I only have 46

@BalarkaSen hi

I almost understood what Ted told me yesterday

I just don't see how homogeneous polynomials restrict to k-linear maps on a ray

@MAFIA36790 are you referring to the real numbers? this is not the only set on which one has a total (or partial) order, and you may want to speak about other sets that have an order that is order complete

Obviously :'(

I'm thinking of this example, for instance: $f:\Bbb C^2\to\Bbb C$, $f(z_0,z_1)=z_0^3$. How does this restrict to a $3$-linear map on e.g. $\ell=\lambda(1,0)\subset\Bbb C^2$?

@MAFIA36790 I don't quite understand your question

@MAFIA36790 so it would not be wrong to reformulate the question as: "Why is it not redundant to define "order-completeness", since it is automatically true for every totally ordered set?"

In this case you must note that it is not actually true for every totally ordered set, the completeness axiom is a statement about the reals. But it is easy to construct also subsets of the reals that are not order complete, for example $\mathbb R-\{0\}$ is one such set that is not order complete

as the set $(0,1)$ is bounded in $\mathbb R-\{0\}$ but it does not have a infimum in $\mathbb R-\{0\}$

@Danu By 3-linear he means homogeneous of degree 3, yes?

In which case it's obvious.

Oh, sorry, I see. You're identifying $O(3)$ with $O(1) \otimes O(1) \otimes O(1)$ and asking why a section aka homogeneous polynomial restrict on a fiber to a $3$-linear map.

Every set of real numbers (ie every subset of $\mathbb R$) that is bounded from below (ie there exists an $a\in\mathbb R$ s.t. $a≤x$ for all $x$ in the set we are considering) has an infimum in $\mathbb R$ (ie there exists a $b\in\mathbb R$ so that $b≤x$ for all $x$ in the set and if $a≤x$ for all $x$ in the set then also $a≤b$).

This refers always to a space we are considering: the situation is that a set is a subset of the space and the condition is that the set is bounded in the space and then the consequence is that the infimum exists in the space

so while $(0,1)$ has lower bound in $\mathbb R$ and a lower bound in $\mathbb R-\{0\}$, it only has an infimum in one of them

also note that $(0,1)$ is not bounded from below if you consider it as a subset of $\mathbb R_{>0}$

so these notions are always relative to some space, the axiom of completeness says that $\mathbb R$ is order complete, if you consider a modification of $\mathbb R$ by removing some points or adding some points then the statement does not apply

@BalarkaSen Yea!

And I'm getting real confused

Do you see what's going on?

Haven't thought about it much. I am not really familiar with this language, I think topologically. One direction is clear: if you have a linear functional $f$ on $O(1) \otimes O(1) \otimes O(1)$, on a fiber $f(\lambda x, \lambda y, \lambda z) = \lambda ^3 f(x, y, z)$ by linearity on each component, so that's homogeneous.

But that homogeneous ones are always multilinear...

Right, so section gives a hom poly

But why hom poly's give a section...

Not sure

@Danu OK, no, I see what's up. A homogeneous polynomial is a multilinear map $\Bbb C^{n+1} \otimes \cdots \otimes \Bbb C^{n+1} \to \Bbb C$, is the point.

The number of copies in that tensor product is the degree of the homogeneous polynomial.

So it takes k vectors as input??

And how is it linear in each argument (in my example, for instance)?

A homogeneous polynomial is of the form $\sum z_0^{i_0} \cdots z_n^{i_n}$ where $i_0 + \cdots + i_n = k$. Think of $z_p$'s as elements of $(\Bbb C^{n+1})^*$ (namely, the coordinate function).

@Danu Your example is not relevant because you seemed to have misunderstood the domain of the k-linear map. $z_0 \cdot z_0 \cdot z_0$ is a fine element of $(\Bbb C^2)^* \otimes (\Bbb C^2)^* \otimes (\Bbb C^2)^*$.

@BalarkaSen Hmmmm that seems sensible-ish

Think about it algebraically.

@BalarkaSen Yeah, I sorta see what you're saying

Thanks :)

I feel so bad about asking everybody for help all the time :(

It's fine. I can rarely help you anyway, so I'd probably do the same thing.

Had I been studying this stuff, I meant. Which I am not, so annoying troll grin towards you.

^^

(not really; I am kind of envious)

Hey yo yo yo

Is anyone here?

Just study with me

OMG PEOPLE AR EHERE!

I wrote a summary of what I did so far (about 1/3 the length)

Can I ask a combinatorics questions here? I don't think it merits asking on the SE

I'll read it if you'd send it.

Yeah, sure

@Danu Was that to me?

@VermillionAzure No. But don't ask to ask

@Danu Well okay

I have negative probability in my equation to solve a certain problem but that doesn't seem valid

I'm in an introductory level probability class and we're in distributions but I can't understand what distribution fits the problem

But I came up with a formula already

But the formula involves negative probabilities but it's canceled out by zeroes after being intersected with the probabilities of impossible/null conditional trials

Is that... okay?

...

yeah maybe I should just ask this

We can't help you if you don't give the formula

@MAFIA36790 let me know if there is something dodgy or I'm sort of missing the point with my explanation, I'm not the best at constructive pedagogy

Hey, is anyone here ?

yup

Okay nice, can you help me with something?

I want the number of combinations for a puzzle where each row has 6 choices. There are 4 rows, and each row must have a unique choice. How many different puzzles can there be?

I've got: The sum from x = 0 to 4 of (6 - x)

@BalarkaSen I need it in full generality. I know what homology and cohomology are, I've just never used them for anything in a topological context

the proof in Hatcher seems to want me to read the entire homology chapter which is perhaps a little much

What do you mean by unique choice?

Oh, is it just that you can only choose one thing from each row?

If the rows are different and the choices are different, it's just $4^{6}$.

Oh.

Why would it not be 6^4 + 5^4 + 4^4 +3^4?

You have $6$ choices in each row, correct?

Also, my mistake, should be $6^{4}$. xD

Your first row has six, your second has five, your third has four, your fourth has three.

Ah, I see. I misunderstood what you meant by unique choice then.

Maybe I'm still misunderstanding though, hold on.

Then it should be $6 \times 5 \times 4 \times 3 = 360$.

Your bank of available choices for the rows is 6. There are four rows and each must be unique.

That is all.

Oh, alright.

That was my first suggestion!

Then I got confused.

Well not really wait, I said sum.

Should be product.

I hope I don't get accused of spamming MSE with duplicate questions again...

I haven't seen another one like it. Jessy Cat grin

Who in here is a number theorist?

Help!

math.stackexchange.com/questions/1940878/…

Interesting how that doesn't link ...

Hey guys, if I simplify $(\sqrt[3]{a^2})^2$ would it be $a^\frac{-4}{3}$

$a^{\frac{4}{3}}$

oh yea, the minus was a mistake on my working out, thanks

What's up with the notation of $\nabla \times \mathbf{v}$ for the curl of $\mathbf{v}$?

I understand that this matrix can be likened to the curl of a vector field $\mathbf{v} = (a,b,c)$ in $\mathbb{R}^3$ but it seems too coincidental: $$ \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \partial_{x} & \partial_{y} & \partial_{z} \\ a & b & c \end{vmatrix} $$

@Kari Precisely that is what's with the notation.

Ah, I see. So it's not exactly a legitimate vector product because you have differential operators as one of the tuples$^{T}$, @TedShifrin?

Also hi! Haven't been on here in a while. How are ya going?

If $H,K \leq G$, for $|G| = p^{k}$, and $|H| < |K|$, does that imply that $H \leq K$?

Also, if $|a^{k}| = |a|$, does $a^{k} = a$?

Ah wait no, that's not true.

@Kari: Right, it's not a literal vector cross product, but all the usual properties of dot and cross work out just ducky. (curl grad = 0 is $\nabla\times\nabla\mathbf F =\mathbf 0$, div grad = 0 is $\nabla\cdot(\nabla\times\mathbf F) = 0) ... I've been around — where've you been hiding?

@SAWblade True indeed that it's not true.

yes, that was a severe typo ... wish we could repair

you're probably good at lazing :)

I'm a veteran :-b

I imagine you've been doing a bit of that recently?

yeah, retirement leads to a bit more laziness

(Not sure if it was you, but I recall somebody retiring!) It was you :-)

although I still taught (long distance) two former students a topology course last spring

Oh wow! That's dedication!

@SAWblade Why would you think this is true? Something conjugate to $H$ will be in $K$, but ...

hi

hey hey

g'night, @MikeM

I'm asking because I haven't had algebra in a while and this homework is kinda destroying me. xD

That made me laugh, @SAWblade.

You pride yourself on being destroyed, @SAW, so stop whining :)

I can totally relate to the sentiment of being rusty and subsequently destroyed :-)

I'm not ... whining ... ? xD

Trying to re-familiarize myself with the feel of abstract algebra proofs. xD

well, do not forget group actions, @SAW. Super powerful and important stuff.

I don't understand those too well. xD

Good time to learn, @SAW.

@MaryStar: When is the reciprocal of a polynomial again a polynomial?

Or did they actually write $g(u)\in F(u)$ instead?

Mm, true.

Groups and their actions are the most beautiful things I know.

But don't tell my dates that.

Is that a vacuous reference again, @MikeM?

Boo.

hi

I am question regarding the direction of gradient vector.

Why does it mean to say "gradient to be always pointing to to the maximum increasing direction"?

From this math.stackexchange.com/a/222020/200649 I understand it will be max. magnitude. But the direction whether or decreasing should on partial derivatives?

what am I missing here?

@TedShifrin When there is a polynomial such that the product is equal to $1$, or not?

@TedShifrin No, it is $F[u]$.

hi chat