something like that?

right, but we can just write these as $\tau \sigma$

As in composition?

yes, we compose automorphisms, we only multiply field elements

I see.

I think I have quite a few details to work out, but I think I get it now.

so we have (n/d)*d of these $\tau \sigma$

remember the $\tau$ fix H', so we get the $\sigma$-conjugates repeated n/d times, see?

OK! Huge aha moment! Yes, now I see. Right $\tau$ fixes $H'$. Yes.

that's why it's helpful to do a "double subdivision"

now $m_\alpha(x) = \prod_{\sigma} (x - \sigma(\alpha))$ right?

Yes.

so when we carry out the multiplication (in the algebraic closure) we get $a_0 = (-1)^d\prod_{\sigma} \sigma(\alpha)$

and the norm just repeats this factor (n/d) times, so there it is

There it is.

I got it.

Thank you so much!

the notation is really the worst part about it, the ideas aren't so bad

Yes. I've been struggling with the notation quite a bit.

the general idea is this: fields are complicated, but finite groups are a lot simpler.

so we try to push our reasoning on to "the group side" where the possibilities are more constrained.

so the galois correspondence is very strong....

I think I'm finally beginning to really see that.

a simple thing like an integer only having a finite number of divisors....that lets us simplfy which fields are even POSSIBLE

we can't get to a splitting field of $x^3 - 2$ with square roots. why? because 6 isn't a power of 2.

(the 6 comes from $S_3$ which is the galois group of $x^3 - 2$)

Rightrightright

@Mariano Are you around for a doubt about nilpotent elements and ideals? (Looking forward to you...)

shoot

I'm writing a complex Makefile

anything else is fun :D

in the simplest case, when we consider $\Bbb{Q}(i)$, the norm turns out to be $a^2+b^2$

$N = \{ 0, 1 + i, 1 + j, 1 + k, i + j, i + k, j + k, 1 + i + j + k\}$ is the set of nilpotent elements in the quaternion over $\Bbb F_2$.

ok

Now, when Matt and I discussed this, we concluded $N$ is not an ideal.... But, it looks to me $N$ is in fact an ideal.

hi

(The reason we gave was $1+i+j$ is not in this...but I fail to see this ...)

the quaternions over F_2 are a commutative ring, no?

Just a second, let me parse....

@KannappanSampath you can't have an odd number of terms

Yes, it looks like it. @Mariano

ok

Anyone remember what question it was (on the main site) on the limit as $x \rightarrow 0$ of $1/x$?

in a commutative ring, the set of nilpotent elements is an ideal

@MarianoSuárezAlvarez Yes, we have proved this.

(there is a question on the main site asking how to prove that, in fact :) )

@DavidWheeler did you ever figure out what the real subgroup of degree 4 over Q was?

and then why are you asking if the set of nilpotent elements is an ideal? :P

Ignorance that $Q(\Bbb F_2)$ is actually a commutative ring!?!!!

@EricGregor I answered, but sadly, I was ignored....

You were not ignored

@DavidWheeler i didn't see your answer. you just said it wasn't cyclic

@EricGregor that i did. i even gave a reason.

yes David, but then what's the real subfield of degree 4?

Who saw the answer on the main site on the injectivity of some map using a nuclear weapon??

@azarel I got your answer, thanks.

oh i see that Mariano was just asking whether it was cyclic or not

@BenjaminLim No problem

$\Bbb{Q}(\sqrt{2+\sqrt{2}})$ is the field i believe

@DavidWheeler, i'm actually slightly confused about your reasoning. you said the degree 4 real extension can't be cyclic because conjugation is (0,1) in the group

could you elucidate that for me slightly?

i would also love to know how you guessed that field

if conjugation fixes a field (that is, it is a subfield of the real numbers), and conjugation is an element of our given galois group, that the subgroup of the galois group that fixes our field must contain the conjugation automorphism

@Mariano Can you help me understand what some comments in the question you mention are--Jyrki and Matt E have left their comments.

but the only elements of $\mathrm{Gal}(\Bbb{Q}(\zeta))$ that are of order 4 are $\sigma_3,\sigma_{11}$

hey eric

are you good with number theory?

and they square to $\sigma_9$

ask your question, John. as mariano said no one is going to say yes to such a question

@EricGregor it was the real part of $\zeta$, multplied by a suitable rational number

@EricGregor I am looking at problem 379 of PE

tinyurl.com/89vh67c

much more mathematical

i wrote a function that will quickly handle f(n) but g(n) is horribly slow

i will think about it. i have my hands kind of full atm

@David, i understand, thanks

@DavidWheeler, oh and how did you find the field?

@JohnSmith one thing i could just throw out there is that we can compute the lcm in terms of the gcd

if that's of any help at all

well

f(n) can be computed oeis.org/A018892

@Mariano When you are around, please do ping me.

i guess the hope is that this function might be multiplicative

or weakly multiplicative or something

Taking the exponents of the prime factors of n (call them e), and finding the product of all (2e+1), then at the end, (product+1)/2 will give you f(n)

i don't have time to try these ideas out

f(n) is fast, but g(n) still horrifically slow

i was talking about the function g(n)

is it possible that g(n) is multiplicative or obeys some nice identity?

like if you know g(n), can you find g(n^2)?

oeis.org/A182082 is g(n)

hmm, a recurrence relationship?

then you would be done, right?

try it for small numbers

i don't know, it's just an idea

it might not be too simple

@EricGregor $\mathrm{Re}(\zeta) = \frac{1}{2}(\zeta - \zeta^7)$ which shows it is a subfield, hmm?

but the givens imply that something like this is going to happen

?

how do you figure?

if there is a recurrence I am not even remotely sure how one would go about finding it

@JohnSmith, you are given that g(10^6), you want g(10^12)

well problems tend to give you smaller outputs to help you check that you're on the right track

they dont necessarily have relevance to the full problem directly

ok, i don't know

@DavidWheeler i agree with that equality. and ok it's a real subfield. how did you get what you got explicitly

forgive my obtuseness

Thanks again @DavidWheeler

can some one explain this answer to me - "A cheap and cheerful way to get a (special) unitary matrix close to the identity is to use the product of two Householder reflections, where the unit vectors defining them are close. I reckon the Frobenius norm of the difference between the product and the identity is proportional to the absolute value of the sine of the angle between the vectors."?

I wanted to get SU(3) matrices close to the identity for a numerical calculatin

@Didier: hi

A: Cogwheel-Preferences (on the top of the drop-down)-Languages (middle tab). Choose any other language, apply. Repeat until you meet familiar alphabet and finally choose English

Hi Ilya!

@Ilya survival tips for a long distance traveler?

Oh, I think I got it.

Maybe. Still workin on it..

Hi @Anon

hey

I am on a moving train !

cool. I've never ridden on a real train, believe it or not.

Oh !

train has become obsolete......all through my life i had travelled on train rarely on Bus and a handful of times on plane

Ok I am off......my battery is discharging..bye

@tb exactly :) also, sms here are screwed up as well as hotel reservation procedures

@Ilya heh :) hope you enjoy your stay anyway.

There's always the point where software becomes too "intelligent" to be useful. The Google engine is beyond that point for quite a while now...

i hate google....the internet has gotten so big that it no longer cuts through the detritus

@tb: I do :) also, Florence and Milan dissatisfied my quite a bit - so no high hopes anymore

the more i learn about math...the more despondent i become as to learning what anything "is", everything comes in so many flavors, it gets harder to tell when two things are just "wearing different clothes"

i've been thinking about the splitting field of $x^3 - 2$ a lot lately. if we call this field K, only one element of Gal(K/Q) is complex conjugation. why is this one transposition of S3 "special"?

the only "heuristic" explanation i can think of...is that 3 is odd, so someone has to be "the odd man out"

that is, only two of the cube roots are unity are primitive, which sort of breaks the "triune symmetry"

Is it okay to post questions that simply ask for references?

I have a 2D function (think f(x,y)) that I need to plot. The function is slow to compute on a computer, so I want to use an adaptive sampling method.

I am looking for books / sources that describe adaptive sampling methods that can be used in this situation.

Here's what I had a couple of months ago: mathematica.stackexchange.com/questions/216/…

I fixed some serious problems with that method since then.

I realized that this is not a trivial thing to get right, and I do not want to reinvent the wheel here ... so I'm looking for some sources on such methods. Googling for "adaptive sampling" doesn't give many useful things (it gives unrelated things).

Perhaps I need to search for another keyword.

@Szabolcs I believe so. Just use reference-request.

@Szabolcs It certainly is okay to ask for references and not much more but the more specific the question, the better. A word of warning, though: there are relatively few people here who appear to know their way around numerical methods, so I wouldn't expect too much.

and what @robjohn said :)

@tb :-) Good morning. I am just about to take Lilly for her morning constitutional.

Thanks, I'll ask the question :)

@robjohn Good morning! Enjoy your walk :)

@Szabolcs I used to do some adaptive subdivision of curves for rendering in QuickDraw GX. I will see if anything can be adapted.

How is lim sup 1 < 1/2?

How can unpublished work pave the way for Riemann's work?...

Also, the name Bolyai is not mentioned in Riemann's collected works. The direct influence certainly was Gauss.

@MattN The comment you refer to seems to have disappeared.

I asked the question here:

@tb Some historians accuse Gauss of keeping quiet about the work of Bolyai on purpose.

@tb Bolyai was corresponding with Gauss.

@robjohn I'm very close to you :) just Pacific in between

@Szabolcs Accuse? That must be the influence of the Bolyais then... I'm aware that both Bolyais corresponded with Gauss, but mostly the father. In any case I don't think that paving the way for Riemann comes close to being accurate no matter what Gauss did or did not do.

@Ilya Could've told you as much : ) Italy is boring : )

As for using "Google" in China: are you even sure that what you think is Google is Google and not just a spoof site setup by the Chinese government to get your login data for gmail?

@MattN I would take a look on the Southern part now

@Ilya Ok : )

Holy monkey.

I have teh hunger like a horse.

@Ilya Where are you?

@JonasTeuwen You want oats and hay?

Ello : )

@JonasTeuwen Or a carrot perhaps?

hey folks! what's up?

@robjohn 8-).

@JonasTeuwen how are things?

@robjohn I've once tried that. I'm okay. You? Working on some homogeneous inequality 8-).

You know what I hate? BibTeX and inconsistent naming schemes across publishers.

</rant>

@JonasTeuwen taking care of animals and trying to get some work done and questions answered.

Hi guys

@JacopoNotarstefano word! :D

Sounds good :-).

Sleepy.

Was writing an answer and someone else posted the same seconds before me :,(

Call for votes.

I'm using dwm now :-).

What's that?

dwm.suckless.org

"suckless" : )

Yes.

Yay : ) +10 rep.

:).

I still haven't made a decision about whether it's better to stick to algebraic topology or try to find something else.

@MattN Something else.

: D

But badgering the teddy with that feels wrong.

And telling the supervisor that I want something else without having a good idea of what I want feels wrong, too.

Maybe I should ask it on main.

I should go to sleep.

Good night!

Good night!

@Jonas: I think this answer is okay. would you look it over, since I have some small worries about the second part. It just seems too easy; though, it might be.

@robjohn Hmm, looks solid!

@JonasTeuwen okay. I am just too tired to have real certainty :-) Thanks.

(I'm also tired! 8-))

I'm very satisfied with my new window manager.

The one you linked me to earlier?

Yes.

@JonasTeuwen cool. Do you run Linux?

@robjohn Yes, Arch.

@JonasTeuwen Ah, that's right. I remember that now.

@tb good evening

I'd prefer BSD but that doesn't support displayport 8-(.

Good evening!

Hi!

@JonasTeuwen I use Mac OS. I was very glad when they built it on Unix

same here :)

Yes, but it disturbs me that it isn't built on X11 :-).

So I cannot install a better window manager 8-(.

Well, we actually can, but then you need to run X11 apps.

@JonasTeuwen I think there would have been serious licensing problems going with Linux

And that sucks monkeys.

Yes, X11 is GPL right? Darwin was BSD-like.

I'd prefer a tiling window manager.

@JonasTeuwen I believe so

Oh no! MIT.

@JonasTeuwen what about MIT?

It has a MIT license.

Hey

hi

@tb Anyone know how much of direct limits I need to know in order to continue to chap 3 and the rest of AM?

Today there was a talk about discrete harmonic analysis on a Weyl alcove. Not sure what it was about really...

@JonasTeuwen beats me, too

@BenjaminLim I only looked quickly, but I think they will appear only in exercises (but in a lot of them). Chapters 10 and 11 depend on projective limits, though. But what's the matter, they're not that bad anyway :)

Hmmm because like there is this one exercise on proving that tensor products commute with direct limits

and one user on this site answered it with the Yoneda Lemma :( :(

well, but that's nothing but the adjoint relation with Hom

That's what Yoneda is good for: proving this sort of things; of course you can do it by hand, which is a good exercise.

@tb Well because I am just only starting out with this stuff

I mean for me to even see that every element in $\varinjlim M_i$ can be written as $\mu_i (x_i)$

that took a while!

The second instance of Jyrki, but sanity even this time...feels bad

@BenjaminLim what do you mean by that?

@tb I mean when I look back at this I will say it is all really trivial

LIke now when I look back at tensor product it was not that hard

So: Step 1) prove that the tensor product commutes with direct sums

(can you do that?)

@tb Yes

The proof is rather long

Yoneda does that in one line :)

pffffffffff????????????????????????????????????????

@tb Hell that proof was like so long man

@KannappanSampath People make mistakes all the time. You only need to ask theo about the stupid things I make!!

Maybe you should get maggot to forget about things? 8-).

@JonasTeuwen I have not had alcohol in a while now

8-)!

@tb Which exercises would you recommend to do for direct limits, to get familiar for the rest of the book?

I think the tensor product commutes with direct limits one is good

That I can't tell.

ok

@JonasTeuwen If you want to drive here you need to know the alcohol restrictions

Restrictions...?

L license - 0.00 BAC P1 license, P2 license 0.00 BAC

@tb I can't believe how in the last week I have gone from like tensor products to direct limits to man....!!!!

I can't drive, so no problem.

awwww

not rendering theo

huhuh?

lemme debug on main :)

$$\DeclareMathOperator{\Hom}{Hom}\begin{align*}\Hom{\left(\left(\bigoplus M_i \right) \otimes M, N\right)} &=\Hom{\left(\bigoplus M_i, \Hom{(M,N)}\right)} = \prod \Hom{(M_i,\Hom{(M,N)})} \\ & = \prod \Hom{(M_i \otimes M, N))} = \Hom{\left(\bigoplus (M_i \otimes M),N\right)}\end{align*}$$

ok

so

and replacing $\bigoplus$ with $\varinjlim$ and $\prod$ with $\varprojlim$ you get the result that tensor products commute with direct limits from Yoneda.

@tb I don't know about how hom relates with the direct product, as well as inverse limits

you see theo because your mathematics is at a much higher level than me you can see all these connections

Well, $\Hom{\left(\bigoplus M_i, N\right)} = \prod \Hom{(M_i, N)}$. Let us think about what that really says.

Giving an element of the left hand side, is just giving a morphism $f: \bigoplus M_i \to N$.

ok

yes

ah ok

Now giving an element on the right hand side ...

a morphism like that is just a direct sum of morphisms from $M_i$ to $N$?

Well, $\bigoplus M_i$ is a module and you have the inclusions $j_i: M_i \to \bigoplus M_i$.

that just says

yes

do by the universal property of the direct product

Now $\left (f \circ j_i\right)_{i \in I} \in \prod \Hom{(M_i,N)}$, right?

yeah

Conversely (now comes the universal property):

Given $(g_i)_{i \in I} \in \prod \Hom{(M_i,N)}$

wait wait

hold on

sure

what is the map that sends an element on the left to one on the right

let me figure that out

(I gave that in the two messages before "Conversely" :))

yes

wait let me digest

I am not so fast

The map is $f \mapsto \left( f \circ j_i \right)_{i \in I}$ where $j_i : M_i \to \bigoplus M_i$ are the inclusions.

ok

well

hhhhhh

oh ok

we get the element $(fj_1,fj_2,\ldots fj_n.....)$ in the direct product

possibly all terms non-zero

exactly.

right

Ready for the other direction?

yes

So, given an element $(g_i)_{i \in I} \in \prod \Hom{(M_i,N)}$

That, is given $g_i: M_i \to N$ for all $i \in I$.

then $g_i : M_i \rightarrow N$

(your turn)

By the universal property of the direct sum

there is precisely one linear map

$g : \bigoplus M \to N$

such that...

defined by $g( ( m_i)_{i \in I}) = \sum g_i(m_i)$

whoops

wait

Now the sum makes sense because all but finitely many terms are zero

maybe better to write $g \left( \sum_{i \in I} m_i\right) = \sum g_i(m_i)$

yeah ok

And even better to write: there's a unique $g \in \Hom{\left(\bigoplus M_i, N\right)}$ such that $g_i = g \circ j_i$ for all $i \in I$.

yeah

so you used $j_i$ for the canonical inclusions, ok

okokokokokokokokok

now that is good

@tb So in your chain of equalities before you applied yoneda

Now these maps: $\Phi: f \mapsto \left( f \circ j_i\right)_{i \in I}$ and $\Psi: (g_i)_{i \in I} \mapsto g$ such that $g_i = g \circ j_i$ are mutually inverse.

oh crap forgot to check that!!

That's what the universal property tells you!

oh ok

yes

So $f = \Psi(\Phi(f))$ comes from the uniqueness of the map $g$ such that blabla.