« first day (3242 days earlier)      last day (1788 days later) » 

6:01 PM
What's a good book to attempt to learn this stuff, for future reference?
 
Gruson/Serganova "A Journey through Representation Theory" has lots of good stuff
 
Thanks!
 
"Let $K$ be the splitting field of $x^6+3$ over $\Bbb{Q}$." Does the fact that the definite article is used indicate that the sentence is talking about the smallest splitting field? Would that just be $\Bbb{Q}(r_1,...,r_6)$, where $r_i$ are roots of $f$?
 
yes
"the" splitting field is the smallest one
 
"Let $\omega$ be a primitive sixth root of unity over $\Bbb{Q}$." What exactly is $\omega$...I don't understand.
 
6:08 PM
it's a generator of the group of sixth roots of unity over $\Bbb Q$
 
So, $e^{i \frac{\pi}{6}}$?
 
if you embed everything into $\Bbb C$, that's either $e^{2\pi i/6}$ or $e^{5 \cdot 2 \pi i/6}$
you're off by a factor of two, but otherwise, yeah, that's a possible choice
 
"over $\Bbb{Q}$"? What does that mean? I know what what it means to say that a vector space is over something.
 
well, it is contained in some field extension of $\Bbb Q$. There's also a primitive sixth root of unity in $\Bbb F_7$, for example
 
Since $7$ is prime, I can take $\Bbb{F}_7 = \Bbb{Z}/7$?
 
6:12 PM
yes and the multiplicative group is cyclic
 
Ah, thanks!
 
Von Neumann correctly predicted how cells replicate before Watson and Crick did lol
sniped
 
 
1 hour later…
7:32 PM
What does this syntax bring to mind? $C_3^0$
 
Don't some people use that for binomial coefficients?
 
Well, it's meant to be a semigroup. Cyclic group of order 3 unioned with a "zero."
 
7:49 PM
@Rithaniel i'd jokingly call it carbon but that's ${^{12}}C$
but i'd suspect that "C with sub/superscripts" is hopelessly overloaded in general
 
8:06 PM
Fair. Though "$\mathbb{Z}/3\mathbb{Z}\bigcup\{ a\} $ where $ax=xa=a $ for all $x\in\mathbb{Z}/3\mathbb{Z} $" is a bit of a mouthful.
 
yeah
hmm, would a^2 be different than a? I mean, one has $a^2 x = a(ax)=a^2=xa^2$ as well
(if it's a zero, then I'd expect $a^2=a$)
 
Ah, yeah, that's an oversight in my expression of the semigroup.
It should be $ax=xa=a$ for all $x\in\mathbb{Z}/3\mathbb{Z}\bigcup\{ a\}$
 
Could also just do it as $ax=xa=a=a^2$ for all x in Z/3Z
 
That works too.
 
what i like about that example: 1 in Z/3Z still acts like an identity element for "most" of the semigroup
 
8:15 PM
Well, it actually is an identity, still. I suppose I should refer to it as a monoid, and not a semigroup.
 
ah, yeah
the issue isn't the identity, it's the lack of an inverse for a
 
Yes, that prevents it from being a group.
More explicitly it is the multiplicative structure on the field of four elements, though.
 
 
2 hours later…
10:02 PM
whats that theorem called listing a bunch of equivalent conditions for a riemannian metric to be complete? its not riemann roch
ok it was hopf-rinow
unfortunately doensn't contain the cases I was interested in :S
 
11:00 PM
@Rithaniel I wonder about the inverse problem
Given such a structure, is it the additive structure of some ring
or field
 
You mean the multiplicative structure, right?
Considering the additive structure it's pretty boring.
Well, not boring, but pretty well-explored, at the very least.
 
hi DogAteMy, @Rithaniel
 
Heya Ted.
Currently trying to look up the statement of the fundamental theorem of finitely generated abelian groups.
 
That's not so difficult.
 
@s.harp what are you looking for
 
11:05 PM
howdy @Ryan
 
Hi ted
@ÉricoMeloSilva Zhou gave a fantastic talk today
 
I wonder if the answer is manifolds with boundary.
 
Yeah, the only issue is that I've chosen to look it up in a book which lacks a table of contents. :P
 
Did you try an index instead?
Or is the "book" lacking both?
 
Just found it.
So yeah, given a particular structure, if it is the additive structure of some ring, then it is an abelian group. If it's finitely generated, then it's isomorphic to $(\oplus_{k=1}^n\mathbb{Z})\oplus F$ where $F$ is a finite abelian group. I don't know the case of an infinitely generated abelian group, though.
 
11:09 PM
@ÉricoMeloSilva I stumped him with a question
 
Also, yeah, it's not actually a book. It's the class notes from Abstract Algebra. I just have them bound up in a binder because there's so much of them.
 
Ah, sometimes books are more convenient.
 
Indeed, such as for an index.
 
Is there a rap stack exchange where I can post lyrics
 
I love exchanges with pompous undergraduates like this.
 
11:15 PM
I bet there's a reddit for that, Ultradark, but probably not a stack exchange. There might be a stack exchange specializing in music, though.
 
yeah I just found "Music fans stack exchange" I'll look on reddit also
 
I've been doing too much abstract stuff. I should get into more concrete calculations again.
 
Hey secret is in the chat there!
 
I love the typo that reads "hatred symbols are not vectors"
 
lol
 
11:27 PM
I guess I shouldn't correct wrong answers when the questions are themselves messed up. Answerers shouldn't understand what's correct.
@Thorgott: I think that was a Freudian hatred.
 
Ah, the classification of infinitely generated abelian groups is an open problem. I wasn't expecting that.
 
I'm not sure I even would know how to begin writing down any sort of definitive list.
Heya @Karl. Here's someone who might know :P
 
11:45 PM
Heh
 
lol
 
Looking to release my first album by next summer
 
hi Demonark
 
1
Q: Central Linear Functionals form a Closed Subspace

user193319Let $B$ be a Banach algebra and let $B^\ast$ denote its corresponding dual space, normed with the supremum norm over the unit ball. It is (I think?) easy to verify that $\phi_n \to \phi$ in $B^\ast$ in norm if and only if $\phi_n \to \phi$ uniformly on the unit ball in $B$. A linear functional $\...

 
I looked at that question
1 view = 1 respect
 
11:52 PM
Hey @Ted!
 
anyone here a terpischorean?
 
Gesundheit!
 
ugh I have logorrhea
 

« first day (3242 days earlier)      last day (1788 days later) »