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finitely come up with one?
@anon you mean finally?
@anon ALl the previous proofs were wrong
you mean finally
do a ctrl+f for "finitely come" on your question
07:10
@anon Oh god I have used that word too many times today.......
too
oh god
Via visual intuition I expect it should generalize to manifolds that are dim > 2.
@anon I am uploading a pic now for the visualisation in $\BBb{R} 3$
maybe given two base points we can consider projections onto a dim 3 affine space containing them
(which might reduce to dim 3 anyway)
or just a lower dim affine space, doesn't matter. that might be what you're talking about anyway, but I don't feel like juggling coordinates
07:30
@anon What is zhen saying in his comment to my question?
apparently zhen just misunderstood what you were saying. seemed clear to me though.
lol ranlge
yeah :D :D
I'm having a brainfart. Why is every irrep a summand of the reg rep? Notes say it's cause the former will be a homomorphic image of the latter, but I fail to justify this.
@ZhenLin did you see my comment?
07:38
you are the pingiest guy in all of mse chat Ben
@ZhenLin Well even the unit cube in $\Bbb{R}^3$ not all the z - coordinates of the vertices are the same
@anon yes
@BenjaLim I agree with anon. It's irritating.
sorry
I am just worked up over this problem....
I think it's entertaining
what is?
07:40
@anon The regular representation has a universal property. Use it!
@ZhenLin Actually I think I'm getting bogged down in details.....
You seem to be almost there.
there has been zero mention of a UP
bah
@ZhenLin you mean my proof is nearly there? But isn't that it what I posted?
Yeah, just satisfy yourself that $A$ and $B$ are homeomorphic to $\mathbb{R}^n$ minus finitely many points and you're done.
07:42
and given the regular representation has a default construction, I expect one would have to prove the universal property rather than take it as a given, so I'd be back to where I was
@ZhenLin hooray!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
@ZhenLin Finally a rigorous proof of this fact without handwaving!!!
handwaving is good exercise
for the arms, and stuff
@anon I find in AT there is a lot of handwaving
I hear that a lot
Let $G$ be a group. The regular representation $k G$ has the universal property of a free $k G$-module on one generator.
07:44
@anon In the case of a finite group like say $S_3$
don't you have that the regular representation is a direct sum of the other irreducible reps?
the regular representation now is 6 dimensional
I recall that every rep decomposes to a sum of irreps. I don't immediately see why the reg rep contains every irrep.
yes that's true
(assuming algebraically closed field or semisimple group algebra or whatever)
Well, that is a non-obvious fact. But it is easy to show that every irreducible representation is the homomorphic image of the regular representation, assuming that we are in the situation where we have complete reducibility.
Sigh, I always seem to be getting bogged down in a ton of details
07:45
(so $G$ finite and $\operatorname{char} k = 0$, for example)
zhen where can I find a proof of the fact that anon wants?
not sure if serre has it
how would one show this
Use the universal property of the regular representation! :p
@ZhenLin where can I find a reference for the universal prop
wiki doesn
't have it
I dunno. It's an entirely obvious fact to me that $k G$ is a free $k G$-module on one generator.
07:48
what is the generator
$1$, obviously. :p
ok.....
what is this universal property - that for every kG module M there is a homomorphism of kG-modules kG->M?
I get the distinct feeling this universal property doesn't come for free
It's the universal property of a free module. There's nothing special about the fact that we have a group algebra $k G$.
I don't know commutative algebra / module theory, dammit! :O
I just wanted to learn representation theory :S
to wikipedia I go
07:52
@ZhenLin I find using the van kampen theorem for my problem is like black magic
somehow my loop in $\Bbb{R}^n -\{x_1,\ldots,x_m\}$ I can always homotope it such that when deforming it to zero I miss $x_m$.....
Let $R$ be a ring (with unit, not necessarily commutative, blah blah). $M$ is a free $R$-module generated by $S \subseteq M$ if and only if for every $R$-module $N$ and every map of sets $S \to N$ there is a unique $R$-module homomorphism $M \to N$ extending $S \to N$.
@BenjaLim Yeah. But why is that so surprising? $\mathbb{R}^n$ has lots of wiggle-room, and you have deleted only finitely many points.
okay, for every mapping $\{1\}\to M$ with $M$ a $kG$-module, we can extend by linearity to get a mapping $kG\to M$. the space of mappings $\{1\}\to M$ is iso to $M$ itself, and by extension we get a mapping $kG\to M$ that is surjective. right?
not sober enough to do this
yes the result that it is simply connected is not surprising. It's the way to prove it that is
somehow van kampen "encodes" information about deforming the loop.....
No, it's not necessarily surjective. To get surjectivity you need to know $M$ is irreducible.
@ZhenLin how do we go from irreducible to surjective?
07:55
Well, its image is a sub $kG$-module. Complete reducibility means it's a summand. Irreducibility means it's the whole module.
or trivial - how do we rule that out?
Because the only way you get a trivial image is if you chose the zero homomorphism...
@ZhenLin thanks
okay. obviously not every set map {1}->M is trivial, so kG->M won't be trivial either.
that clears it up
The tricky part is to get $M$ as a summand of $k G$. I've forgotten how to do that.
08:00
I know reps decompose to sums of irreps, and hom distributes through sums, and schur's, so it all adds up.
at least when kG is semisimple (or less stringently k is alg. closed)
Ah, Schur's lemma!
There's some nasty finiteness assumption lurking in the background here, but I'm too lazy to check everything...
I'll tackle that sort of thing later in the game
Introductory representation theory is usually done with $G$ finite, $\operatorname{char} k = 0$, and $k$ algebraically closed. But then you may as well take $k = \mathbb{C}$...
I think, to make Schur's lemma and Maschke's theorem go through, it is enough to take $G$ finite, $k$ algebraically closed with $\operatorname{char} k$ not dividing the order of $G$.
|G| needs to be invertible in k more precisely, IIRC
wait that's the same
facepalm
Hmmm, I wonder if there's a $p$-adic analogue of Haar measure. Since $\mathbb{Q}_p$ and friends are one of the few other topological fields...
08:08
you mean where measure takes p-adic values?
yeah
Then we can talk about $p$-adic representations of compact groups.
does nonnegativity of measures make sense in the p-adics?
and would monotonicity have an anologue?
No, I guess we'd lose those.
I don't know enough rep theory to see if those aspects of measures are actually relevant though
Would you happen to know of any interesting examples of "exponentiation" other than by numbers?
08:15
right actions in general behave like exponentiation. (in fact I have a group theory text that writes $g^\alpha$ for $g\in G$ and $\alpha\in\mathrm{Aut}(G)$.)
(it writes automorphisms from the right for some reason)
No, conjugation has the wrong "distributive" law.
I mean right actions. the Aut(G) thing was just a tangent.
Well, right actions in general.
real exponentiation is kind of like making something a right $\Bbb R$-module
I think of exponentiation as being characterised by these equations: $$m^{a_1 + a_2} = m^{a_1} \cdot m^{a_2}$$ $$m^{a_1 a_2} = (m^{a_1})^{a_2}$$ $$m^1 = m$$ $$m^0 = 1$$
08:18
but there's more to it in that there's the obvious $\Bbb Z$-action
The first one is a kind of distributivity law, but a right action satisfies $(m_1 \cdot m_2)^a = {m_1}^a {m_2}^a$ instead.
we could assume a commutative ring $S$ has the property that $S^\times$ is a right $R$-module, and speak of an extension of this to the rest of $S$ (so like a module acting on a monoid instead of a group, and the additive structure takes the sidelines)
Yeah, but are there any examples?
If $K$ is a field we can speak of $K^\times$ as a right $\Bbb Z[\mathrm{Aut}(K^\times)]$-module and extend to the remaining element $0\in K$
or something like that
or replace Aut with Aut^op technically, maybe
for it the be on the right
Hmmm... I'm not sure if the distributivity law is satisfied there.
08:25
Hmm, maybe $\mathrm{End}(K^\times)$
oh, well, it's trivially satisfied since every element of $\mathbb{Z}[\textrm{Aut}(K^\times)]$ is uniquely a sum of automorphisms of $K^\times$...
endomorphisms work too, I guess
doesn't it make sense to speak of eg Z[R] when R is a ring?
formally it should work out
I guess it'd be functions R->Z with convolution
Sure, but you probably have to forget something about $R$. I don't think it's feasible to remember both the addition and the multiplication.
yeah, we could forget the additive structure and make it a multiplicative monoid
Ultimately though, what I'm looking for is a non-commutative version of exponentiation. The only one I know of is the ordinal numbers, but that's still "numbers"...
(and $\textbf{Ord}$ is a proper class)
08:34
well
are you willing to allow the set of potential exponents to be a semiring?
I'm willing to allow it to be a rig, since $\mathbb{N}$ is a rig.
$\textbf{Ord}$ is not even a rig, because addition is not commutative...
actually wait I'm still thinking abelian, nevermind.
I was thinking tensor powers in the representation ring :P
Sure, but that's powering by $\mathbb{N}$, no?
yeah
maybe schur functors. I don't quite get what those are yet though.
Every monoid automatically has a "$\mathbb{N}$-powering", to coin a term.
08:38
is it possible to get unbanned from arxiv for scraping or do i need to get my ip changed?
@DanBrumleve I had no idea one could get banned from arXiv...
you could email them asking them to unban you. I've been banned from mathworld three times now for mass downloading
i downloaded about 10K papers a couple months ago but now i'm perma-ip-banned
@MattN. Na, I don't think that is true. However, if you are able to work for 8 hours it will only do more harm than good. It can be very useful if you say can only do that for 10 minutes. Don't take Erdös as an example, he's a crazy man.
10K papers...?
Why?
i feel like i should bother comcast instead of arxiv since comcast is paid for that, unless there is some easy process?
08:41
Downloading lots of papers sounds more like hoarding...
@DanBrumleve that's the sort of thing I would user stranger's wi-fi for. highly unlikely a random person would use arxiv :P
I would have banned you too.
well i won't do it again.
If you're banned that would be hard.
I wonder why you would want that... they get updated sometimes, and it is a permanent archive.
i wanted to put math.LO etc. on my kindle but it was a dumb idea since it can't handle that many files.
08:43
Doesn't Kindle have a web browser, powered by mysterious free internet?
airplanes.
magnets
magic
I just said that
@DanBrumleve How long are your flights...? 8-).
08:45
I want a blog with MathJax, but WordPress.com doesn't allow that. Hm.
@ZhenLin Hm... but they support just conversion to .png right?
doesn't allow it?
What is the "support" they give? You just need to add the javascript (perhaps through a link) and that's it 8-).
@Jonas, twice a month. :)
@DanBrumleve Right, but I usually could not read more than half a paper on a flight...
08:47
@JonasTeuwen Hideous...
Perhaps $2$ if I go to Canberra.
@ZhenLin "Install MathJax by robjohn!".
Well, the problem is that they don't allow adding arbitrary Javascript...
unless you pay I presume
08:47
yeah neither can i since my file browser is too slow.
Bloody monkey. Raining! AGAIN. 8-).
Oh, I'm not sure they'd let me do that even if I paid.
also need a better search to find anything cool to read.
... like pie?
If I were running a hosting platform, I certainly wouldn't allow it. Too many XSS worries...
08:48
samsung phone actually works a lot better than the kindle.
@ZhenLin If you paid you can just install it on your own webhost 8-).
Indeed.
Shouldn't be too expensive, by the way.
I have access to free webhosting anyway.
Grad student proof pricing.
Mm, yes.
08:50
The trouble, as I'm sure you're familiar with, is maintenance. :p
The software? Wordpress is quite easy to maintain.
The nice thing about hosted blogs is that you can just forget about it for a few years and it will still be there when you come back. Without having been hacked and defaced...
Just don't take very exotic templates and plugins.
Yes, but wordpress is pretty safe (if the server is).
There was once a time when WordPress was more of a system intrusion vector than a blogging platform...
Yep.
Joomla still is 8-).
08:51
i've been using these guys openhosting.com for 10 years and never had any trouble.
never use dreamhost.
@ZhenLin We have not been defaced after about a year (PhD organization) with wordpress. So if you check once in a year you should probably be fine...
Or you will have to go to the LaTeX support the Tao guy and such have.
I don't think my server ever got hacked in the last 6 years.
But it must be luck.
maybe it just wasn't interesting
I got a rootkit installed which was for a different architecture. "Why does my ls segfault?"
ahahaha
what was it?
08:54
how the dick
I don't know, but I just reinstalled the whole vm.
eh, I'm going to zzz
But what architecture was it? :-|
There were some malicious looking comments in the hex.
@anon Goodnight
08:55
@ZhenLin It was a virtual machine which emulated ARM.
And they tried to bullshit me with some x86 thing 8-).
ah... very unusual, certainly
but why ARM :-|
I would compile stuff on there for smartphones and such.
@Jonas, do you really spend 4+ hours reading half a paper? just curious about learning strategies which I've been meaning to ask a question about...
That's what crosscompilers are for!
@DanBrumleve Or much more.
@ZhenLin Yea, but some stuff really sucks in cross compiling... Good luck with R for example.
@DanBrumleve If I want to study it rather than read what it is about: yes.
08:57
You'd need a cross-compiler if your target runs a different OS anyway...
my way is very different, i'll spend half an hour reading, a week thinking, then maybe go back later to confirm understanding
Yes, but that is a bit easier because some configures bootstrap it.
I have been known to take several days to read a paper - but I am a retard :)
@DanBrumleve Mmm... that's scary. Half an hour? A paper is too much info for me to read and fully process on a flight.
... although I'm mostly learning introductory stuff not research papers
08:58
I've read a couple in about a week, my advisor was like wow, you're fast.
I fall asleep reading textbooks, nevermind papers!
And airplanes are full of idiots. Including myself, yes.
well i guess the difference is that i don't process while i read, i'll read/memorize, and then process very slowly later.
They don't even think about paying me business class. They just laugh 8-(.
@DanBrumleve Perhaps if you read some substantial papers I'm sure you will not be able to read more than 2-3 on any flight (before you can access the internet)...
well imma gonna keep tryin, i'd be so proud to say i've read all of math.LO.
09:00
Or perhaps just a textbook.
"I've read all of math.LO... forgot all too!"
Must be mind numbing... I'd rather read a phonebook.
i've read a lot of dover books but few modern textbooks
yeah i'd rather read a phonebook than skymall too.
i used to love skymall but then i realized it's mostly the same every month.
Well, there are quite some cool modern textbooks as you are aware. I used to take many books on a trip and read 0.
phonebook is different on every page.
if only i could turn them faster on the kindle, sheesh.
09:03
Yep.
I usually now only take one book. If I am back within two weeks.
unfortunately i'm sick of all my dover books because i'm interested in more modern topics.
@ZhenLin Actually a better question would be: why is it possible to deface a compile farm...? Because I am an idiot. 8-). Forgot about it.
Well, springerlink contains many nice .pdf of newer books.
(Other places too, 8-))
Of new and old books alike!
@JonasTeuwen Yeah - but we people outside academia can't access them :(
@OldJohn 8-(. But if you drop me an e-mail...
09:06
yeah i think i'm gonna turn a new leaf and put gutenberg on there instead. read some chaucer for once.
Read some Schöpenhauer. No need to have 10K papers...
@JonasTeuwen When I go on holiday I always take exactly one maths book - some holidays are more successful than others ...
Would be pretty unsuccesful if it was Federer.
Holidays should not feel like selftorture 8-).
what are some good holiday math books?
Some recreational math perhaps. Penrose stuff?
09:09
@JonasTeuwen I wouldn't : ) And I'm quite surprised that I can do maths for so many hours. I knew I could write code for say 10 hours (-ish) but that's much less intense. I guess with the right sort of motivation one can do lots of things one would not normally be capable of.
@JonasTeuwen I once took Rudin's Fourier analysis on groups - not a good holiday :(
@MattN. It is a very unusual skill. Most people certainly cannot do that (well, not good anyway). Amphetamines would totally fuq it up. You'd be jumping around instead.
@OldJohn Heh... 8-).
@MattN. They seem to work by fixing some understimulation instead of adding to already good stimulation... Then you will be very non-productive. Many ideas, write nothing down would be the best option.
Usually, after 2-3 hours of coding, I'd like to vomit. So no 10 hours for me.
also they fuck up your teeth (even the oral pharmaceuticals). i haven't done speed in 10 years but i'm going to the dentist tomorrow.
Because of the dry mouth?
i dunno why exactly.
09:12
I believe it is due to the dry mouth. The meth mouth more has to do with very bad oral hygiene...
@JonasTeuwen It's like ETH: I'm full of energy and motivation and interest by nature and then ETH is like sticking a pipe down your throat and pumping amphetamines into you. --> zero productivity x_x --> death (eventually)
well u can do meth and brush twice a day but i guess you will still get lots of cavities. whatever is the cause it is a systemic effect.
@MattN. Haha... but a degree from there is very valuable. It is the price you'd have to pay.
They alternate between amphetamines (=deadlines) and valium (=lectures).
Very healthy.
@MattN. how do you pronounce Féjer?
09:15
@JonasTeuwen I won't pay for an MSC's, I think.
as in féjer's theorem?
@BenjaLim What nationality was he? French?
no hungarian
@DanBrumleve I believe the biggest cause really is not systemic. For one it is quite acidic, secondly dry mouth and third grinding teeth. Finally poor oral hygiene. You'd don't see it that bad in pharmacotherapeutic use.
Oh... I always pronounced it in the French way.
Then I have no idea lol, why do you think I can speak Hungarian? : D
09:16
@BenjaLim Ask Pierre! 8-).
@MattN. Good, doesn't matter.
i used to value the perceived productivity increase but now i tune into a similar benefit in my own adhd itself even though i can't control it as easily.
@JonasTeuwen he says it
fair - zhe
like
@DanBrumleve And you would read for 4+ hours on a plane with ADHD? Cool.
@BenjaLim Yes, French.
no féjer is hungarian
@JonasTeuwen he says it like it's in french
The pronounciation you write down is... French to me.
09:17
because in french
j is pronounced g
For me it is pronounced like a j, but maybe that is because I learnt French when I was very young 8-).
@JonasTeuwen Just need to complain from time to time in order to be happy. : ) You know me.
@MattN. Yes.
It's Fejér!
Anyway... need to bring out the thrash. Not enough money yet to pay someone to do that 8-(.
09:19
@JonasTeuwen : D Not that much work either. So not worth paying someone to do it for you.
Perhaps I can hire one of the Chinese PhD students. They have much less...
@MattN. I have like 1,5m of stacked papers.
@MattN. Why? Purely practical reasons.
@Jonas sure can, just couldn't sit still in school so i got put on drugs.
Sure, I should've written "sounds so ..."
09:20
Because it is deadly boring? I know that.
nothing to do with reading tolerance :)
You don't want to read boring retarded stuff and then they say you have ADHD.
I'm going to do some more diagram chasing. Hey, if I post a photo of my diagram chase in here, would you check it for me?
You don't want to talk to people because they are idiots and then you are asocial and autistic.
@Jonas, i guess i don't, but that's what happened.
09:21
@MattN. I'm not sure if I can, but perhaps someone else can.
Okay : )
haw haw well that is a nice theory
but i've never been called either of those.
I have, but perhaps that is because they are idiots!
actually for me it was more of a physical thing (hyperactivity) than not wanting to read boring shit (which i won't tolerate either although it's not as bad as staying in one place).
"It is not that I don't want to talk, but I feel bad after I talk to idiots... like you"
09:23
@DanBrumleve I would sue the asshole who put you on drugs.
And if it was my parents I'd delete them from my life and hope they die soon.
Come to think of it: aren't parents supposed to protect their children from stuff like this?
@JonasTeuwen Sorry, missed that.
Never mind : )
I have to go and do more stuff : )
wow two consecutive inappropriate comments i guess i won't respond.
Good luck!
See you in a bit!
Thanks : )
@DanBrumleve Well, they are written by Matt. Put it into context! 8-).
okay
my mom started me on ritalin and she died in 2005.
that's the context so matt can stfu.
09:26
Yes, I am sure he does not mean it like that as he does not know this.
Parents surely want the best for their children, but that doesn't necessarily mean it is the good thing. But anyway. Intentions are what matters here.
it's actually okay for you to call me an idiot. i think you are snide but also fun.
I also call myself an idiot.
Right, I was going to put out the trash and go to work. See you. Have a nice day!
cya @Jonas
@JonasTeuwen Later Jonas
Bye guys :-).
09:31
while i'm here i guess i should promote my latest question math.stackexchange.com/questions/179360/… which yesterday i was afraid of getting closed but i got some good feedback and information from @joriki
Ah cool, a question I could comment on. Will do later on perhaps.
Internet says: heavy showers. I see the sun. Which one should I believe?
i just left a comment on @joriki's answer where i attempt to classify heuristics into four categories
now i'm wondering to what extent is that classification correct or complete.
You could perhaps read the article in the Stanford philosophy encyclopedia: This.
hmm i guess i will thx.
fwiw, going on ritalin was really awesome at age 13... it was so much easier to tune out the class and work on software in my notebook. but it got problematic later and i didn't totally quit speed until my early 20s.
(switched from ritalin to dex at some point but it was pretty much the same thing.)

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