multiplication is the group operation

ok i'm done

@GFauxPas go on

was that not a good choise

you keep explaining things by bringing in new words

and explaining those words with more new words

so it never ends

how do you define "new words"?

geocalc when you have only a handful of elements in a group you can describe them by just listing a Cayley table, which is like a multiplication table but the "multiplication" is the group operations

so Leaky's $G$ would be

$\begin{bmatrix} \circ & a \\ a & e \end{bmatrix}$

not really the notation I wanted to do but i dont know how to do it in mathjax

@geocalc33 ok let's do this

so your new group is {1-e,1-a}

now what is (1-e) multiplied by (1-e)?

is it in the group?

is your group closed under multiplication?

so you see Leaky's rule tells you everything about the group: $a\circ a = e$

hmm

yes it's closed under multiplication

what is $(1-e) \circ (1-e)$?

it's not closed under multiplication

$\begin{bmatrix} \circ & 1-a & 1-e \\ 1-a & ? & ? \\ 1-e & ? & ? \end{bmatrix}$

if you claim your thing is a group, you need to tell me how to multiply things together and to fill out that table

if any of those question marks are "undefined" then it's not a group

tell me

@geocalc33 so is the issue resolved?

tell you what?

yeah it's not closed under multiplication

ok nice

so it's not a group

its no group

now maybe you mean that you invert all elements of the group, meaning you replace $a$ with $a^{-1}$ and $e$ with $e^{-1}$

is that what you mean?

There is a certain concept in point groups I'd like to better understand. Its called "descent in symmetry", anyone here who could help me with that?

@GFauxPas it's just a reflection of the group I was talking about earlier

It looks pretty much exactly the same that's why I thought it was a group

I don't know what the reflection of a group is

here we go again

hi @Daminark

I took my group and reflected each element over the line y=1-x

facepalm

geocalc33 it's great to learn new things but you need to learn the terminology associated with new things to talk about it with other people

Hi @LeakyNun @Daminark and everyone else

@Rudi_Birnbaum looks like taking a subgroup to me

hi @MatheinBoulomenos

is there a term for that

a subgroup?

No it isn't.

subgroup is called a subgroup

but i recommend you reading up on basic group theory terminology

is there a term for reflecting a group

@GFauxPas who are you talking to

geocalc

@Rudi_Birnbaum maybe you should use the physics chat room

Well, it is kind of.

you really should learn the basic vocabulary of groups

Well I'd like to understand who to formulate that concept mathematically correctly.

wait i was talking to geocalc

is Rudi geocalc

@GFauxPas I'm a statistics major that's why I don't know anything

@LeakyNun: In particular I'd like to understand if it can be formulated without recurring to a basis.

@Rudi_Birnbaum I know nothing about descent in symmetry, sorry

@geocalc33 you can learn it on your own

i'm sure people here can recommend sources to you on group theory

@LeakyNun: But that could be good. Since I then have to explain it to you.

ok go ahead then

@MatheinBoulomenos ????

you're confusing two lines of conversation

oh, that's possible

nevermind

lmao

The descent in symmetry (DS) indeed brings you from a group $G$ to subgroups of $G$.

@geocalc33 it's important to always be learning and it's great that you're enthusiastic. but I'm not sure why the functions you're interested in are interesting

Hey @Leaky and @Mathein, how's it going?

@MatheinBoulomenos so let's say I have an infinite algebraic extension K/F and I want to define a topology on Gal(K/F) using as little field theory / galois theory as possible, and maybe more technology on the topology side. how do you suggest I do it?

The only thing you have to specify is an IRREP of $G$.

Also hey everyone!

learning about branches of the logarithm IS interesting and important, for example

and the corresponding equivalent ways to define complex powers

so focusing on $f^{1/\log x}$ I don't think will give you insight

@LeakyNun well, I'd say you better have K/F Galois

right, Galois

@Daminark pretty well thanks

but studying $f^g$ for arbitary complex functions $f,g$ is interesting and insightful

I'm helping a friend with his bachelor thesis on rep theory, so that's fun

@GFauxPas do you doubt the power of 1/lnx ??

@LeakyNun: And now to be very specific I must recurre to a representation of the group elements in $\Bbb R^3$. But maybe you can comment already now.

I doubt the usefulness of considering the power $1/\ln x$ over any other power

@Rudi_Birnbaum I still don't know what descent in symmetry is. So I have a point group and I specify an irrep, now what?

why not just consider the powers of $f^z$ for complex $z$ constants

of which $\ln x$ will be an example

@LeakyNun: My first question would be could you develop an idea on your own how to uniquely define a concept which brings you from $G$ to a set of subgroups of $G$ by specifying a certain irep of $G$?

that's something that's interesting, important, and includes functions of the type you're interested in

specifically, it will lead you to the study of logarithms

which are super important and more nuanced on the complex plane than on the real plane

AND it includes your case

@MatheinBoulomenos what does $\operatorname{Gal}(\overline F/F)$ mean if $F$ is not perfect?

@Rudi_Birnbaum hmm, no idea

no one is perfect Leaky

it includes $f^1/lnx $

it includes f^(1/lnx)

?

OK, then I go on

@LeakyNun you can take $\overline{F}$ to be a separable closure of course, but actually we will have $\operatorname{Gal}(F^{sep}/F) = \operatorname{Aut}_F(F^{alg})$

What's up chat?

so it doesn't really matter

yes, because $z = 1/\ln x$ is a complex number for $x \ne 0$

@MatheinBoulomenos :o

Take a set of points $S$ in $\Bbb R^3$. OK?

and $\ln$ is a particular branch of the general concept of logarithm

@Rudi_Birnbaum ok

so, my answer is:

study logarithms

my suggestion, rather

logarithms are used to define powers to complex numbers

I'll take a look at ()^z

Good grief, @Mathein is back to unsleeping just like a @Balarka

@LeakyNun about your question on the Krull topology from a topological perspective: I think the way that the Stacks project does it is quite topological: take the coarsest topology on $\operatorname{Ga}(E/F)$ such that $\operatorname{Gal}(E/F) \times E \to E$ is continuous

@Fargle !!!

Hi @Ted @MikeMiller

$f^z = \exp( z \log f)$ where the equality is an equality of sets geocalc

@Ted ?!?!?!

@MatheinBoulomenos does that work?

that's stupidly simple

@LeakyNun yes

Just surprizled to see you, Fargle.

compared to other definitions

@LeakyNun: The group $G$ is then a set of unitary transformations permuting these points. OK?

@Rudi_Birnbaum ok

so by studying logarithms, you'll be studying complex powers

unitary... orthogonal?

Seems like we've got all the nerds in one place

You'll need some field theory to show anything about that, though, e.g. that it is profinite

Now what do we do?

You're the chief nerd, Demonark.

h

Deminark, lead us, give us your guidance

@TedShifrin Haha, I'm here a lot, I just somehow always dodge you. :P

@LeakyNun of course, in the definition $E$ has the discrete topology

which?

Probably divine inspiration, Fargle.

even if it might have another topology, depending on the situation

I see from the other room that you're enjoying Hatcher, @Fargle. That's awesome.

since $E$ has the discrete topology, we want $\{\sigma \mid \sigma(x) = y\}$ to be open for each $x$ and $y$, right

It's a heck of a book, @Ted. I always figured I'd like alg top, but Hatcher is the first text that's confirmed that suspicion.

@MatheinBoulomenos is that right?

@LeakyNun here's a fun fact from which this follows more or less (you'll need some extension lemma, too): purely inseparable extensions are epimorphisms in the category of fields

@MatheinBoulomenos :o

yes orthogonal.

I'm also glad that he's email-able, because what few questions I've had he's been able to answer pretty effectively.

@Rudi_Birnbaum ok

He's more geometric and has wonderful exercises, @Fargle. Some people hate it, but it is demanding and, I think, worth it.

@Fargle So you mean to confirm you like geometric topology! :)

lol, I guess that's fair

Hatcher is a geometric topologist by trade, which is a profession that requires a darn good chunk of algebraic topology

@LeakyNun I don't see how that step follows

That's amazing that he's willing to be accessible like that, @Fargle. I'm super proud of him. I had heard he was ill ... :(

I hadn't heard anything about that. That's rough.

His claim-to-fame theorem is really beautiful (known as the Smale conjecture). I never really got through the paper, since it's pretty difficult.

@TedShifrin :(

@MatheinBoulomenos hmm

hi @loch

I hope I'm misremembering, @MikeM.

Hi @loch

Well, this correspondence has been over the past few days. I hope I haven't bothered him if that is the case.

@MatheinBoulomenos how would I describe the topology on Gal(E/F) then

@LeakyNun: Now we represent our set of points in $S$ as a vectors in Euklidian space.

At any rate, the exercises are wonderful. I hope to eventually do at least a majority of them.

@Fargle: You bugged him instead of bugging us. I'm impressed.

...I may have done both.

@Rudi_Birnbaum ok

@Fargle You have not.

@LeakyNun the details are in the stacks project, this definition is quite non-standard, despite its simplicity

I always claim (even though Jasper vehemently disagreed — I wonder what's happened to him) that exercises are what makes a great math book.

it's nontrivial to show that this is profinite

Thanks @Mike

Believe it or not, @Fargle, I've even helped people with alg top on occasion. Of course, I'm old and forgetful.

lol

which is perhaps why it isn't used more, I think it's easier if you just realize that $Gal(E/F)$ is naturally an inverse limit of finite groups, so you use the topology you get from that

@TedShifrin I think that's a fair statement, the books I always remember the best are the ones with exercises that are both challenging and elucidating.

I'm very proud of the exercises in my books, for the most part.

We interpret adding some vector $v$ to $S$ like transforming the set of points, in the sense that we shift them to other positions.

My exposition may or may not be good.

exercises are fun

I like them, FWIW

I wish I had catalogued cute things I figured out

I kept folders of that, @MikeM ... but they're gone.

I already gave you that advice.

@LeakyNun: OK?

@Rudi_Birnbaum yes ok

I never take good advice.

@MatheinBoulomenos If I'm interpreting stacks correctly, it's the compact open topology?

@MikeMiller I should start doing that. And now I regret not having done something like that earlier in my college career.

Do you take bad advice?

I feel like my algebra and such wouldn't be as weak as it is if I had kept better track of realizations.

Fargle, you've developed and evolved a lot in the time I've "known" you here

@TedShifrin Probably.

well, I suppose it's a net plus that you take advice. Or maybe not.

lol, I'd be interested to know how. It's incredibly difficult to have an objective view of your own growth.

Complicating things is that I don't remember how long I've been here.

It's hard for me to be specific without doing research.

lol, don't worry about it

@LeakyNun: Great, I guess that was the toughest part :-). So now we choose some $v$ that with the following property: $v$ is transformed upon action of elements of $g$ on it in the same way as a certain irrep $\Gamma$ of $G$.

Just musing

I've commented to Balarka and Mathein about how their teaching skills (and patience) have improved over time. I think you are more confident in what you know even if you ask questions.

@LeakyNun indeed

cool

My teaching skills might have gotten better but I dunno about my patience

that's a little known-fact, I think

@MatheinBoulomenos but in the discrete topology, it's just the product topology

little-known fact?

@LeakyNun yes

I meant that the Krull topology is a compact-open topology

and hence a product topology

I always thought you were pretty good, @MikeM, so I didn't remark on your improvement. I think you are frustrated with life and occasionally are impatient with grad students you've TAed, but I haven't noticed it as a big issue. ... Hell, I lose patience too (but I'm old).

@LeakyNun products of discrete topologies are not necessarily discrete, that's why we can get something interesting

@Rudi_Birnbaum you lost me

@MatheinBoulomenos ok

@MikeM: Of course, in the old days we both lost patience with Balarka :)

Was about to mention that

@LeakyNun: To be more specific: Acting $g\in G$ on $v$: $g\circ v$ gives you $\lambda v$, with lambda being the character of $g$ in $\Lambda$.

LOL

well, he has grown up.

@LeakyNun: Back again?

more strongly, products of (non-empty) discrete topologies are never discrete, unless only finitely many factors have more than one element

And I think we both deserve partial credit for his evolution.

@MatheinBoulomenos sure

Even Leaky has improved a little bit as a teacher.

:)

That's my main role here — teacher training :P

I miss @Pedro and @DanielF.

@Rudi_Birnbaum so v is an eigenvector?

@TedShifrin I suppose so. I do feel less lost wading in all of it now.

@LeakyNun: Of what?

I'm TAing now for the second time. (The first time was LA, now number theory) I noticed a big improvement. I think it's partially due to my activity here

@Fargle: No question you've progressed. Of course, I still hope to chat diff geo, but now that you're hooked on Hatcher, I shan't complain.

@Rudi_Birnbaum g

Hi guys

I'll be curious to see if your evaluations are different, @Mathein, but I think you've changed — I've told you that before.

I think there was a long time--both in and out of this chat--where I was frustrated that the level of intuition I had developed for math up to college wasn't sufficient for the math I was about to tackle.

Your helping Kasmir was very impressive, in particular.

@TedShifrin That's still on my short list.

Though I'm sure that gets less and less believable years after I started.

It's OK, @Fargle. No, I'm not carping.

You should send Ken an update, btw.

Ah, you're right.

@LeakyNun: $g$ is a group element, how can group elements have eigenvectors?

Especially now that I'm doing Hatcher.

@Fargle Can I give you an exercise?

@Rudi_Birnbaum you represent $g$ as a matrix

@MikeMiller Go for it!

@Fargle: Well, he's a diff geometer. But I'm so proud of my former students who've become great teachers themselves. It makes me so happy.

Worst that happens is I give up

@LeakyNun: Yes. then you're right.

@LeakyNun: Its an Eigenvector simultaneously to all $g\in G$

OK?

ok

I really like symmetry. Of course, the kinds of symmetries that an object has depends on what kind of object you think of it as. Let's think about some highly symmetric metric spaces: a great example is $S^1$. There are other kinds of symmetry nearer and dearer to a topologist's heart, though: the space of all homeomorphisms, or diffeomorphisms.

As a devotee of homogeneous spaces, of course I adore symmetry, too.

@LeakyNun: Now you take the space $L$ of all such $v$.

Hmm, is devotee feminine? Maybe devote. With accents.

I also like symmetry. That's why I love groups

Exercise: Prove that the map $\text{Isom}(S^1) \to \text{Diffeo}(S^1)$ is a homotopy equivalence. The 'shape of the smooth symmetries' is the same as the 'shape of the rigid symmetries'

Symmetry is in my top 5 things.

@MikeMiller Disgusting. I'll get right to work.

That's pretty tough, @MikeM, but a great thing to think about.

You may replace the codomain with Homeo if desired. The result is the same in both cases.

@MatheinBoulomenos they don't call them symmetric groups for no reason :P

@Rudi_Birnbaum ok

@MikeMiller The inclusion, I take it?

Actually, if only @Fargle had done G&P, he'd have a main idea.

symmetric groups

I do have G&P around here.

@Ted It's definitely something a topology grad student can do given some thought. And my favorite introduction to a beautiful topic.

Of course, we are not all topology grad students.

In particular, I'm not, @Mikem :D

@Daminark no, it's more like groups are all possible symmetries that can exists for any object at all. symmetric groups are just symmetries of a naked set with no additional structure

You shift $L$ by $S$, and thats your deformation space.