The point of a disjoint union is that we don't want to worry about these sorts of considerations, we take a separate copy of the skeleton and a copy of the disk and manipulate them as we please

Actually some further thoughts about it, if $D$ is infinite dedekind finite, and $C$ is countable, then $D \cap C$ must be finite. It cannot be uncountable since intersections never yield a set of larger cardinality than what you start with, and it cannot be countable as otherwise it will contain a countable subset and thus there exists a self injection that is not a surjection, and hence contradict dedekind finality

> Interestingly enough, Dedekind-finiteness can be graded into various level of finiteness, so some sets are more finite than others. For example, it is possible for a Dedekind-finite set to be mapped onto ℕ, which in some way makes it "less finite" than sets which cannot be mapped onto ℕ.

math.stackexchange.com/questions/1396676/….

Quick question: number of 4th powers in $\mathbb Z/p\mathbb Z$ is $(p-1)/2$ if 4 doesn't divide p-1, and $(p-1)/4$ if it does, correct?

yes

Great, thanks.

howdy, @anon ... Are you ready to teach me sphere geometry yet? :)

did a slight excursion with characterizing subspaces of pseudo-euclidean spaces

Hey @Ted!

hmm, what are those? I'm assuming not indefinite signature ...

hi Demonark!

will go back to lie sphere geometry right after

apparently you can use it to solve the appollonian problem though (descartes theorem)

@TedShifrin yes, indef signature

Ah, by subspaces you mean?

subspaces

I'm not sure what you're referring to with Apollonius and Descartes.

vector subspaces?

@anakhronizein, Hatcher.

@TedShifrin yeah

@TedShifrin problem of apollonius, descartes theorem on wikipeida

So you're asking — given a particular $(k,\ell)$ signature to start with and a subspace of dimension $m<n=k+\ell$, what signatures can we possibly get on that subspace?

OK, I'll look for that.

figured that out, but yeah

Ohhh ... yeah, I do know about that. Did I tell you to look at Pedoe's book Geometry: A Comprehensive Course? He has all that Apollonius stuff in there.

hmm, haven't seen that

Plus his is the only "elementary" book I've ever seen that does my favorite "How many lines in $\Bbb P^3$ meet four lines in general position?" without any Schubert calculus or nothing.

Dover book. Get it!

@Daminark, what do you mean "when the disk as defined in that space"?

Like imagine you construct a skeleton which ends up being set theoretically a subset of the unit disk

Say your skeleton is a tiny pentagon (just edges) and you take the union with a disk

You'd just get the disk back.

Or they'd overlap in some way.

That union will equal the disk, and you've just lost info on the skeleton now. The way around that is to take the disjoint union, so you can choose exactly how you want those two objects to interact with each other

I think another way of saying what Daminark is saying is that given an abstract topological space X, it doesn't make sense to say X cup D^n - what's the topology? what is the overlap? You haven't told me that

G'night, @MikeM.

You take a disjoint union and then make it no-longer-disjoint by gluing along some rule that says where the boundary of the disc goes

Because that's the only operation we could sensibly do

Has anyone studied the orbit method?

Context?

geometric quantization.

kirilov theory

oh, like symplectic reduction stuff ...

Yes.

if anybody i think tobias or anon but i don't think theyve thought about it

I know people who do that stuff, but I don't know it, sorry.

I am slowly getting into symplectic and contact geometry, despite most of my undergraduate career having done algebra.

It's beautiful stuff.

So I would not want to see the algebra courses go to waste.

Oh, you get to do coadjoint orbits and all sorts of representation theory. Not to worry.

Yeah, that's in particular why I was looking at the orbit method: a professor of mine suggested it as a way to not lose out with algebra.

I dunno about method. It shows up throughout the basic theory.

Yeah well the orbit method looks into studying the coadjoint orbits of a Lie group.

LOL, ok, sure.

And I guess each coadjoint orbit has a symplectic structure in its own right.

Which I find quite surprising, since I thought the constraints on dimension would not work out.

in $T^*G$, you mean.

So for instance let's say I have $X^{n-1} = S^1$, and I want to construct $X^n = S^2$. Thus, I need to glue two disks $D^2$ by the attachment $\phi: S^1 \rightarrow S^1$. However, this only makes sense if I originally took a disjoint union of S^1 with the disks. If I took a regular union, I'd just have a single disk.

Use the right index for $n$, @gian.

Where did I go wrong?

I mean the Lie group acts on the dual of its Lie algebra, and you look at these coadjoint orbits.

I just wanted you to have $n=2$ and $n-1=1$.

We're learning now about 1-forms and I want to check that I understand a bit about the basics. So let $\alpha$ be a 1-form on $M$ and $\gamma$ a path in $M$. To do $\int_{\gamma} \alpha$, can I think about it as at each point $x \in \gamma$ theres a vector tangent to $\gamma$. So then $\alpha$ is a cotangent vector at each point. So then we put the 2 together at each $x$ and add them all up?

Oh, right, I've totally forgotten what little I knew. Best to start with basic examples, @anakhronizein.

Vaguely, yes. It's just the usual work line integral, @Kevin, if you think of the $1$-form as corresponding to a vector field.

Come to think of it, I have not seen a simple example of this.

But I have $X^n = S^2$

Right, so $n=2$.

Maybe I can work one out, but I am not sure which Lie group would yield the simplest.

Start with something like $SO(3)$, @anakhronizein.

Right.. I'm slightly confused what you meant by your original comment.

Perhaps SO(2) might be better, even? Or do you think SO(3) is simple enough?

@anakhronizein I assure you there are ways to find places to use algebra in symplectic geometry.

SO(3) is simple enough

Really? @gian ... I'm trying to make your $n$ align with your example. That's all.

And you're right that you need two disks if you're going to attach to a circle (as opposed to a point).

I see lots of very algebraic people doing work in the area with good success

@anakhronizein: There's nothing going on with $SO(2)$ ... abelian group, one-dimensional ... blah.

Certain tools are indispensable and algebraifally randy

fancy

Okay, so ultimately we just take a disjoint union to ensure our "ingredients" are separate and we are free to do as we like with them.

@TedShifrin Ya i was trying ot make that connection. But I didn't want to confuse myself between vector fields and covector fields.

I don't know what that means. You're talking about the different $2$-cells? @gian

It's literally dual stuff, @Kevin.

Using the Riemannian metric.

So on $\Bbb R^3$, you should either watch my lectures on this or go through how $d$ and forms align with vector fields and grad, curl, div.

When you take the disjoint union of $X^{n-1}$ and the disks $D^n$.

Well, disjoint until you make identifications, yes.

But it doesn't end up disjoint.

Right, so you just quotient by the relation defined by the attachment map.

Right.

Got it.

Oh okay actually this brings up a point I forgot to ask about a while back. When you were talking about the Hodge star and Laplacian, the reason we needed a metric was so we could talk about an "orthonormal basis" for the tangent space, right?

Do you see how to use only one $2$-cell if you use $X^0$ instead of $X^1$? @gian

Demonark, ayup.

Yeah, you identify the boundary of the 2-cell with just the single point in $X^0$.

Good.

Is it correct to say 2-cell though?

Yes.

if $a \cdot (r-b) = 0$ where $b$ is some vector dependent on $a$, what is the equation of $r$?

Aren't cells open and disks closed?

@TedShifrin Do you do 1-forms and higher forms in an undergrad class?

I would freely use either of those to mean either open or closed depending on how I feel (:

Usually with an "open" or "closed" in front

In my multivariable math class I sure did, @Kevin. We even did integrals over manifolds in $\Bbb R^4$ with Stokes's Theorem in one lecture :P

@gian: To me, ball is open unless you specifically say closed.

Okay, why do we need that?

@Ted Do you talk explicitly about tensors or save that for graduate stuff?

No, I did everything in terms of determinants. No tensors.

Demonark: To whom was that addressed, and what is "that"?

Is the distinction really a concern?

Most notation/vocabulary is local to a text or a course, but some is universal.

@Ted Ah ok. When you do get to tensors and the tensor product in the grad class, how do you introduce it?

In terms of the properties and the universal mapping property.

Orthonormality, I feel like I should see why it comes up in the later stuff but I can't as of yet

@Daminark if omega is a "unit length" 1-form, *omega is the unique n-k form with omega ^ * omega the volume form

you need a notion of unit length and volume form

oh i need some perpendicularity assumptions

so that's just all wrong

whatever bye

Nah, that's OK, @MikeM, but you haven't defined length. I gave him the careful definition yesterday.

Demonark, ultimately, an inner product on $V$ induces an inner product on $\Lambda^k V$ and $\Lambda^k V^*$.

@KevinDriscoll, I found it helpful to think about the tensor product's similarity to the free product. In a way, you're just modifying the free product so that the maps become bilinear.

@gian @Ted That's actually how we defined the tensor product, as a certain quotient space of the free product. And I found the whole thing just really bizarre. So I wasn't sure fi it was a standard wya to do it or not.

Demonark: In particular, for a general $k$-form $\phi$ once the length is defined using the inner product, you have $\phi\wedge\star\phi = \|\phi\|^2 \,dV$.

@KevinDriscoll, what do you find bizarre?

Yeah, that's standard, @Kevin, although what's important is the universal mapping property.

How accurate is it?

Demonark, and if $\phi$ and $\psi$ are two $k$-forms, $\phi\wedge\star\psi = \langle \phi,\psi\rangle\,dV$.

@studrayght5: How accurate is what?

@gian Well I've never heard of a free product before. And I'm still new to quotienting vector spaces. So its all just very unfamiliar to me

I don't understand the beginning graphics. How is it turning a sphere into a parabolic trough?

Ah gotcha. Have you seen things about universal mapping properties before?

Ya I saw htat graphic the other day and did not understand the steps either

Does't seem exactly trivial to find SO(3)'s coadjoint orbits. :P

Oh, I see, it's using spherical coordinates. A bit glib, but OK.

I will get back to you on it, @TedShifrin ;)

@anakhronizein: I didn't say it was trivial. I once saw this, but I don't remember. But I'll be glad if you want to explain it to me.

@gian No, this is the first real math class I've taken so there's lot sof things I ahvent seen before. Jumped in the deep end a bit.

@Kevin: I think it's using the spherical coordinates area form $dA = \sin\phi\,d\phi\,d\theta$.

Anyhow, @Kevin, there are lots of concrete things (too many for you) in my lectures, but fast-forwarding you might find a few things you want to look at.

@TedShifrin, you have algebra lectures?

No.

I now understand though why you don't understand what physicists are doing with tensor products, @Ted. I mean I thought I sort of knew what a tensor product was but apparently I had no idea!

so(3) is basically R^3. So I am guessing that the action of SO(3) via the adjoint is just the action on R^3.

So either you get fixed points as orbits.

it's not basically ... it is, and [,] corresponds to cross product.

Or you get like S^2?

Huh?

Oh sorry, that's not coadjoint

That's just adjoint orbits.

@TedShifrin The gif that's supposed to explain the area of a sphere. What's the slicing bit? Riemann sums? It looks cool, but I don't understand it.

SO(3) acting on so(3) via the adjoint is gives orbits that are either a point or a sphere.

So for coadjoint, hmm.

@KevinDriscoll, I think the hardest thing for me in algebra is jumping from strictly internal definitions that use the elements of structures to definitions that are completely external of structure (universal mapping properties).

Right, you're foliating $\Bbb R^3$ by spheres, agreed, @anakhronizein.

Except for the origin.

So I guess it is obvious how you get symplectic structures on them.

eh, but this is still not coadjoint, I am getting offtrack

Well, I don't think you're going to see much difference once you co- things.

But you want to understand where the symplectic structure comes from.

You are right. I think the coadjoint orbits are identical.

@Daminark: I hate calling you that. :) Did you see my various comments above .. since I can't ping with your correct name?

@anakhronizein: So of course we know that spheres have a symplectic structure. But how does it come (in an appropriately invariant way) from the group action?

I don't remember the answer.

Thinking about $SO(n)$ for higher $n$ will, of course, be harder.

It's the symplectic quotient of the trivialized cotangent bundle

Uh I think it was that if you are looking at the orbit of f in g*, then it's w(x,y) = f([x,y])

Right ... so this is back to that symplectic reduction I've forgotten.

And presumably checking non-degeneracy uses that $x,y$ are in the orbit of $f$, @anakhronizein?

Yeah, I'd have to look it up later for you.

Well, worry about you more than me. :)

Heh. Will do. :P

But, yeah, definitely do examples

My professor mentioned something about representatives of equivalence classes so I am unsure.

hi chat

But yeah, more examples, will do.

Hi Eric.

I came across a cool analysis problem today: If $f :[0, 1] \to \mathbb{R}$ in in $C^{\alpha}$ (Holder) and the "difference quotients" $f_{h}(x) = \frac{f(x+ h) - f(h)}{|h|^{\beta}}$ are uniformly $C^{\beta}$ then what else can we say about $f$?

How are $\alpha,\beta$ related?

no relations

Oh, so I'll guess $C^{\alpha+\beta}$.

if $\alpha + \beta <1$ youre correct

if $\alpha + \beta > 1$ it's Lipschitz, and if $\alpha + \beta = 1$ then you get log-lipschitz

which is intuitive to me, but it's surprisingly tricky

Interesting.

it came up today because the $\alpha + \beta < 1$ case + the $\alpha + \beta > 1$ case implies that if $f$ solves a fully nonlinear uniformly elliptic equation then $f \in C^{1, \alpha}$ for some Holder exponent

but it ended up taking me a few hours to show this lemma

twas cute

Yeah, I knew something was wrong with my answer when $\alpha>1$ and $\alpha>\beta$.

You're forsaking geometry for analysis ... sigh. :P

@anakhronizein I find SO(3), SO(4), SU(3) to be small enough that I can think about this

Those are usually my limit

You haven't done $E_8$, @MikeM? :)

wee, exceptional Lie grourps

I'm sure this must all be written down in some text somewhere. If only I hadn't given away all my books.

I like the torus as a Lie group.

is the definition of torus $\Bbb R^n/\Bbb Z^n$?

only cause my hardest class this quarter is a heavy analysis class @Ted :P

@anakhronizein: I think abelian groups will turn out boring.

Think about what the adjoint representation is.

Likely.

Tori are nice but Ted's got it right

I usually only work with SO(3) anyway, though, so that's good enough for me

Hello handsome peeps

@Ted hey, I just got back, sorry

What sort of math do you do, @MikeMiller ?

OK, Demonark. Just read the various comments. :)

Hi, Kasmir.

@TedShifrin Hi Ted

OK, time for me to go cook dinner. Bye.

[Random]

Some ponderings about properties of infinite dedekind finite sets:

Recall a set $D$ is infinite dedekind finite if it doe not biject with any sections of $\Bbb{N}$ and any self injection $f : D \to D$ is a bijection (alternately, it does not contain a countable subset or it does not inject into any of its proper subsets)

1. Any subset of $D$ is infinite dedekind finite:

what do you call $\operatorname{Hom}(\Bbb R^n,\Bbb R^m)$?

@TedShifrin

@leaky i got it

@Faust nice

Leaky: The set of all homomorphisms from $\Bbb{R}^n$ to $\Bbb{R}^m$?

@Daminark wanna do my algebraic topology hw

@Secret the vector space thereof

(set woudn't be interesting!)

Lol, I've got to get sources for my bio stuff but I'd be interested to at least see it

@Eric

What are you guys doing in class anyway?

I am inclinded to answer $\text{GL}(n,m)$ but I forgot whether nonlinear maps can be homomorphic

@Secret homomorphism is essentially linear maps

but that notation isn't right

$GL(n,\Bbb F)$ is the group of invertible n-by-n matrix with field $\Bbb F$

@BalarkaSen you might have an idea

@Daminark hOmOOlOgYYY~~~~

@LeakyNun $M_{m \times n}(\Bbb R)$

@BalarkaSen as a vector space?

Yes.

does it have a name?

Space of $m \times n$ matrices

It is vector space isomorphic to $\Bbb R^{m n}$

just space? I mean, $\operatorname{Hom}(\Bbb R^n,\Bbb R)$ is called the dual vector space

@BalarkaSen don't get me started :P

Ah that's fun stuff

because that's the space of all linear functionals of $\Bbb{R}^n$

i just like

dont wanna use my hand

to write things

$\hom(A,B)$ is called a hom-space

use your feet?

I assume you mean general transcription by writing? E.g. LaTeX isn't an alternative

@Leaky genius

ya

I dont like to tex my alg top psets cause of all the pictures i draw

@anon no full name?

I mean aren't you gonna be doing less of that so your grader doesn't rekk you?

NB Before reading on group theory, symbols like $\text{Hom}(A,B), \text{Aut}(A,B),\text{End}(A,B)$ are like alien language to me

what do you think Aut(A,B) or End(A,B) mean?

@anon lol

Automorophism spaces and endomorphism spaces?

@Daminark some of the problems Calegari asked us to draw pics for

Aut and End are associated to single objects, not pairs of objects. so Aut(A) and End(A). and Aut(A) generally isn't called a space (nor is End(A), but it at least technically is a vector space if A is one, unlike Aut(A))

automorphism endomorphism

I am not going to spend time on making up terminology for a rather natural object

I never seen that notation however :D

@Balarka if i ever have to name a thing im gonna give it a dumb name

Aut (G) i seen , but not Aut(A,B)

like zippityzappity or something

Automorphism from the Latin 'auto-' meaning self-

@KevinDriscoll it's Greek, and we can stop teasing Secret now

@KasmirKhaan Because I forgot the -self and made a careless mistake

btw I suppose $\text{Iso}(A)$ is not very useful?

Can "picture" stand in for commutative diagram?

@Daminark if i meant commutative diagram i wouldve said commutative diagram

Commutative diagrams are not bad

i aint got no problem with em

they're efficient ways to encode info

tru

@Secret isomorphisms from A to itself are called automorphisms, and these are collected in Aut(A), so what do you intend Iso(A) to be?

lol

@LeakyNun Sup leaky

As far as Peter is concerned, the two are synonymous :P

two wrongs don't make a right

in scratchwork I've used Iso(A) to be the group of isotopies of an algebra

@KasmirKhaan inf

@LeakyNun >< not that kind of sup

:P

@leaky well thats what I get for trying to be cheeky

What's an easy way, given a presentation of a group, to show that two elements are distinct?

@anakhronizein there's no easy way

But yeah you did seem pretty good at pictures when I remember your Q&A session in the bootcamp so that's something

Hmm, I wonder if the ordering is like this: Relations/Predicate -> Maps -> Homomorphism -> Endormorphism -> Automorphism -> Isomorphism

my pics aint bad

i give em 8/10

Is there another way other than showing they do not give the same action on the group?

@Secret hom -> iso, and hom -> end -> aut

two branches

Homomorphism , isomorphism , automorphism. they are a bit similar

and then you also have monomorphisms and epimorphisms and surjective homomorphisms

@KasmirKhaan each one is the previous with added property

Mine tend to be... A bit sloppy

i actually dont draw pictures that often when i do stuff

nvm, epimorphism is surjective homomorphism

i was about to tell u that leaky

><

Not always @LeakyNun

Like I'll draw a horribly distorted oval and be like "K so we got S^1"

epimorphisms and surjections do not always line up.

DCal makes some fine ass pics @Daminark

Similar for monomorphisms and injections.

There's also $\text{Out}(A)$ for outer auto (or is it homo?) morphism. It is found in the context of various groups such as $S_5$ having a nontrivial of such is the reason why it is unsolvable

and so far my understanding on that is still very poor, because I am not very good at semidirect and direct products

I'll start my day by doing the Laplacian computation I should have done yesterday

Ugh

outer automorphism group, the group Aut(A) / Inn(A)

@Secret you dont need direct products for this =p

note that Out(A) is not the group of outer automorphisms.

@Balarka classic computation my d00d

yeah it's good for me

like vegetables

bleh, I fail group theory. Oh well...

@Secret a function f : G--->G' is a homomorpshim if f(ab) = f(a)f(b) , first product is in G , and second product is in G'

vegetables can be tasty and good for u at the same time @Balarka. It's wild

@Secret a function f is an isomorphism , if it is a homomorphism and bijective ( 1-1 and onto )

that is quite true my friend

it just looks repelling

but the principle is to not make that stop u

@Secret the last special case is the automorphism part, f is an automorphism if it is an isomorphism and f :G-->G ( from the the group onto itself )

i think i did the laplacian computation in my chem class in hs and my teacher was like "oh ya dude check this out" and she handed me a book on pde in quantum chem @Balarka was fuckin crazy

@Secret @Semiclassical @LeakyNun Divergent series: Integral function means

hahah

i wish i had a chemistry teacher like that

@Secret sometimes we can find more than 1 automorphism from the group G to itself, but the set of all automorphisms from the G to itself, forms a group , where the elements now are the automorphism functions

@Eric yeah I've heard

@Daminark i guess it makes sense considering his teacher