@ZhenLin Hello. Do you have any Linear Algebra book you'd recommend?

@skullpatrol Come again?

:6456594 "Come again?" means something like "paraphrase" =D Why "high and mighty"?

@PeterTamaroff Wait and find out...

@skullpatrol You made him leave!

@PeterTamaroff I'm not the one who ignored you.

@PeterTamaroff Gilbert Strang from MIT writes great Linear Algebra books.

Wow, elliptic integrals are so weird!!

@PeterTamaroff He is also a Rhodes Scholar.

@skullpatrol Nice.

Why isn't there a math class or tutorials for made specifically for programmers.

Where you could learn math applications and extend off them. I want to learn math for games development but I only want to learn the math I needs... @JasperLoy

I've searched on Google for this a million times and haven't found anything relevant.

I'm surprised they don't offer something of that matter. :\ I do not know what I need therefore I'm unable to learn. I don't want to go on khanacademy.org and learn it all because I don't need all of it.

Because lol when I look at all these calculations in open-source games, I'm like how in the **** did he do that. I can't look it up because I don't know what kind of formula is being applied.

Exactly... It just upsets me because I am able to apply like 15 calculations to my code... The rest is just solid trial and error. I've been programming for like 6 years now. I have came a long way... I guess its just experience thing.

Hmm I don't have any particular questions. Half the resources on the web I look at confuse me because I don't understand the symbols they give.

@MarianoSuárez-Alvarez Hola, Mariano.

hola )

@JasperLoy I had pinged anon.

@MarianoSuárez-Alvarez Como va?

todo tranquilo

cansado! :-0

vos

@MarianoSuárez-Alvarez Bien. Hoy fue un buen dia.

bien :-)

@MarianoSuárez-Alvarez Para mi te estresó subir tantas algerias sobre el pais al FB =P

jajajaj

Creo que a veces es sano olvidarlas, por lo menos por un rato!

recién subí una hermosa canción

para variar :-)

@MarianoSuárez-Alvarez Ahh! A ver. Yo estoy con esto

me encanta el primer comentario :D

@MarianoSuárez-Alvarez Jeje, la gente se explaya muuuuucho en YouTube.

«Are you high?»

@MarianoSuárez-Alvarez Jaja =) si, pero digo, ese contesta a "guys.... I thought I was listening to the movie of life itself (...)"

sip, los comentarios de youtube son un espectáculo en sí mismo

@MarianoSuárez-Alvarez JAJAJJAJA

@MarianoSuárez-Alvarez Entre otros temas, tengo un agujero en una demostración. Y lo que sigue de esa demostracion es sobre Bolzano Weierstrass, continudiad de una $f$, continuidad uniforme de $f$, y otras cosas interestantes.

Es mas, es algo que me habias contado vos.

cual es el problema

@MarianoSuárez-Alvarez Demostrar que dado $A\subset \mathbb R$, entonces $x$ es un piunto de acum de $A$ sii para cada $\epsilon >0$ existen infintos $a\in A$ con $|x-a|<\epsilon$

Tengo algo hecho, pero no se si es totalmente correcto.

cual es tu definición de punto de acumulación?

(...) si para todo $\epsilon >0$ existe $a\in A$ tal que $0<|x-a|<\epsilon$.

ok

supongamos que $x$ de acumulación de $A$

@MarianoSuárez-Alvarez Puedo yo? =)

claro

Sea $x$ un p. de acuml. Entonces para cada $\epsilon>0$ y en part para cada $n\in\Bbb N$ existe un $a_n\in A$ tal que $|a_n-x|<1/n$. Luego, dado $\epsilon>0$, sea $N\in \Bbb N$ tal que $1/n<\epsilon$ si $n>N$. Entonces todo $a\in \{a_n:n>N\}$ cumple que $|a_n-x|<\epsilon$

Lo que no puedo asegurar es $\{ a_n:n>N\}$ no tenga repeticiones que lo conviertan en finito.

Al menos, tendria que probarlo.

If one of those points were repeated infinitely many times, say $a$, then you'd have $|x-a|<1/n$ for infinitely many different values of $n$

@MarianoSuárez-Alvarez Pero la idea no es mostrar que existe, para cada $\epsilon$, un (sub)conjunto infinto de puntos de $A$ con $|x-a|<\epsilon$?

Digo: si el conjunto $\{a_n:n\in\mathbb N\}$ fueese finito, existiria uno de sus elementos que es igual a $a_n$ para infinitos valores de $n$

@MarianoSuárez-Alvarez Claro.

y entonces, si $a$ es ese elemento, tendríamos $|x-a|<1/n$ para infinitos valores de $n$

Esto último solo es posible si $a=x$, y eso es imposible

@MarianoSuárez-Alvarez Ah, claro!

@MarianoSuárez-Alvarez BTW, encontre lo de sucesiones equidistribuidas que te dije.

Esta en el Spivak.

mathese is always easy, @JasperLoy :-)

@JasperLoy Super slow? Why so?

@MarianoSuárez-Alvarez Ahi estalo que estoy haciendo ahora. 30. 31 y 32 en un rato

@JasperLoy I didn't find it in English, but even so, I prefer Spanish.

@JasperLoy And does that mean I have to read math in English? I mean, I do read most math in english...

@JasperLoy Can you read Spanish?

Darn. Gotta go.

I'll be back in a while.

Oh, nevermind. I'm back.

@PeterTamaroff so hi then!

@leo Knew you'd log in sooner or later

@leo Español?

@PeterTamaroff si

@leo Mira la pagina arriba del Spivak

Voy a demostrar la $(c)$ ahora.

@leo Que hora es ahi?

@PeterTamaroff 9:13pm

:-)

tengo clases mañana a las 7

@leo de que?

@PeterTamaroff de la mañana. Que hora es allá?

@leo Jaja pero digo la materia!

@leo Son las 12:15 pm

@PeterTamaroff ya la estoy viendo

@PeterTamaroff je je, disculpa, de Geo. Diferencial

@PeterTamaroff a cual teorema de Bolzano-Weierstrass se refiere en la 31.?

sucesión acotada y monótona $\implies$ convergente?

@leo "La forma usual..."

@leo No no, pero "casi" continua

@leo Ese es el tma de la convergencia monotona.

B-W es que toda sucesion acotada tiene una subsucesion convergente.

Que segun leí se traduce a nociones de compacidad.

@PeterTamaroff ah cierto! muy importante ese

@PeterTamaroff y cuál de ésos ejercicios estas haciendo?

@leo Ahora el $(c)$

@PeterTamaroff y que es el lim con la raya arriba?

@leo Es una definicion medio engorrosa. Dado un conjunto infinito $A$ de los reales, decimos que $x$ is un casi cota superior de $A$ si existen solo finitos $y$ de $A$ con $y\geq x$.

Entonces si $A^*$ es el conjunto de todas las casi cotas superiores, $\overline{\lim}\;A=\inf A^*$

El otro es el analogo para casi cotas inferiores.

Es el analogo para conjuntos de $\limsup \; a_n$ para sucesiones.

Es mas, si es $A=\{a_n:n\in \Bbb N\}$, Spivak pide probar que estos numeros coinciden.

(siempre que $a_n\neq a_m$ para $m\neq n$

@PeterTamaroff ya veo

@leo Es medio molesta la definicion, pero bueno.

@leo Creo que me voy a dormir. Este $(c)$ me va a tomar un tiempillo y mañana no tengo mucho tiempolibre para siestas!

@PeterTamaroff ya te acostumbraras :-)

@leo Jejeje si, me imagino. =)

@PeterTamaroff entonces buenas noches!

yo tambien estoy por dormir

@leo Buenas noches. Que estés bien.@MarianoSuárez-Alvarez IDEM!

@skullpatrol cool :-)

say gnite to Peter

@leo Ya, was I surprised :-)

@skullpatrol I suppose. sup?

@leo Not much, just chillin'... you?

@skullpatrol where are you? are you about to sleep as well?

@skullpatrol about to sleep

:-)

No, I just woke up :-)

@leo With Google translate I was able to read the entire transcipt :-D

@skullpatrol is it morning there?

@leo Yup.

@skullpatrol I see. I think math can be more transparent because the names of the things are similar

@skullpatrol so you are in some place in the other side

@leo Now I have google translate turned on for this room :-) when I mouse over anything I get the translations.

@skullpatrol cool

hi @peoplepower

@skullpatrol Why hello there.

hi @AlexanderGruber

hi there @skullpatrol

If I had a dollar for every dollar sign I typed...

Would you be a millionaire?

Just the owners and I are left, after everybody else left the math chat room :-D

@skullpatrol

@BenjaLim Whatz up?

hi @Chris'ssister

Hi all. Hi @skullpatrol!

@Chris'ssister How do like my double meaning "left" sentence above and below?

@skullpatrol: hmmm, since I'm not so good at English it's a bit hard to give you a good answer.

:D

It sounds interesting!

@Chris'ssister That's OK neither am I :)

Has anyone else just experience a weird spat of down voting behavior?

yeah.

I'm downvoted almost every day by DonAntonio.

Four of my questions were just down-voted in the span of less than a minute. They were chosen seemingly at random.

Not suer what to make of this.

@Potato: just be careless about that.

@Chris'ssister Indeed, I don't care, but it seems strange to me.

I was just wondering if I was being specifically targeted or if others have notice something similar.

Or some users were removed and then you lose your points.

(the points they gave you)

@Chris'ssister These show up as down votes on my profile. It's not users being removed.

@Potato: I never ever thought of downvoting a question just because the question seemed bad. There is no bad question. All of them are absolutely precious!

But people think differently.

@Chris'ssister Are you a math teacher?

@skullpatrol: No. But I suppose my brother will be. He's so brilliant.

@Chris'ssister With this kind of attitude: "There is no bad question. All of them are absolutely precious!" you would make a great teacher.

@skullpatrol: The attitude is very important. Any question is in fact a good opportunity to open the gates for great answers.

@anon Hey do you think my answer here is sufficient?

@JonasTeuwen Okay. Thanks. :-) Will get it up and running. Even I trust Mendeley more.

@JayeshBadwaik wassup

@BenjaLim I was trying to decipher it before you came in the room. The chronology of the invocations seems a bit strange...

@BenjaLim Good.I.Am.

Wassup with you?

@JayeshBadwaik Been addicted to representation theory

really addictive

@anon I can improve it?

For instance, you have to relate $T$ to $z,f$ somehow so the reader knows what's going on, and should you set $f$ to be a linear map on the alternating power from the get-go or should you show that any map out of the alternating power will happen to be of that form?

@BenjaLim Haha. Okay. I haven't got there yet. Right now, still doing problems on Todd-Coceter Algorithm.

@anon I just said that if we have any alternating map $f : V \times \ldots \times V \to \Bbb{C} \cong \wedge^n V$

then it must be multiplication by a scalar

that scalar given by the determinant.

@JayeshBadwaik It's not so bad

I'd say "here is an arbitrary alternating multilinear map $f$, then by universal property there is a unique endomorphism $L$ of the alternating power such that $f:V^n\to \Bbb C\cong \Lambda^n V$ and $V^n\to \Lambda^n V\xrightarrow{T}\Lambda^nV$ are the same, since the alternating power is dim=1 the map must be scalar multiplication, and then said scalar is uniquely determined by where the identity matrix gets sent," in that chronological order

@BenjaLim Hmm. Probably. Anyway, gotta go. Lunch. BBL.

@anon Feel free to edit my answer

@anon I think you want that $T$ to be $L$, and then conclude later that

$T = L$

you talk about the universal property after you forced $f$ to be a scalar multiplication, but I think it should go the other way around to make sense

I agree.

Do you know about the Mackey decomposition theorem? I'm trying to digest a proof of it, but I can't visualize the isomorphism $$k(HgK)\otimes_{kK}M\cong kH\otimes_{k(H\cap K^g)}M^g|_{H\cap K^g},$$ as given in (ii) of page 2 here

In any case, I'll still upvote.

@anon I am editing my answer. I

will get back to you. Hold on.

Also, if you know how to make tall up and down arrows in $\LaTeX$ that mark induction and restriction of modules (representations in particular), I'd love to know.

Then Mackey decomposition would look good like $$M\big\uparrow^G\big\downarrow_H\cong\bigoplus_{HgK}M^g\big\downarrow_{H\cap K^g}\big\uparrow^H.$$ (Checking to see how that looks.) Edit: Aha! \big seems to work nicely enough.

@anon I am trying to read the paper now

what is $M^g$?

hi @JonasTeuwen

M^g is defined right before the statement of the theorem on pg1

@anon I am confused as to what is $kx$

When they say $k(HgK) = \sum kx$

the vector subspace of k[G] generated by x is kx, and the sum of these subspaces over x in HgK will be k(HgK). essentially it means the subspace of k[G] spanned linearly by the elements of the double coset HgK.

@anon Ok so $kx$ is one dimensional

yes

@anon Are you confused as to why $\varphi_g$ is well - defined?

let's go with that

But once you show $\varphi_g$ is surjective and count dimension you're done

how did the subscript of the tensor product go from $k(H\cap K^g)$ to $kK$?

(and why did the restriction on $M$ disappear)

ok let me see

Ok do we know that $kH$ is a right $k(H \cap K^g)$ module?

yes

$M_g$ a left $k(H \cap K^g)$ module?

yes

hi @warl0ck

Just one thing: $\sum_{x \in HgK} kx$ that is a direct sum yes?

it is in fact direct

yes because we want to say that our group algebra is now a direct sum of one dimensional subspaces

you mean the module induced from the double coset space that is the codomain of the map?

technically I don't think it's an algebra

cause HgK need not be closed under the group operation

hmmmm

yeah

anyway I need to go to bed, I will review it in the morning

Ok.

@anon good night

@anon

eh?

Hi.

We can define a $k(H \cap K^g)$ balanced map from $kH \times M^g|_{stuff} \to k(HgK) \otimes_{kK} M$

that sends $(h,m) \to hg\otimes m$

Now this map I just checked is $k(H \cap K^g)$ balanced

The reason is if we take $l \in k(H \cap K^g)$

by linearity assume that $l$ is just "in" $(H\cap K^g)$

then

$f(hl,m) = hlg \otimes m$

And now $f(h, l \cdot m) = hg \otimes l\cdot m = hg \otimes (g^{-1}lg) m = hg(g^{-1}lg) \otimes m = hlg \otimes m = f(hl,m)$

Where I can move $g^{-1}lg$ to the left of the elementary tensor because it is in $kK$

okay, I see.

Did you get this bit?

I see now.

Because now the universal property of tensor products tells me that I have a unique group homomorphism of abelian groups from $kH \otimes_{k(H \cap K^g)} M^g|_{k(H \cap K^g)} \to k(HgK) \otimes_{kK} M$

@anon But we're not done yet because this is just a group homomorphism.

Now for the tensor product on the right

It makes sense to say that $M$ is a left $kK$ module because a priori $M$ is a left $K$ module

and also I think it is clear that $k(HgK)$ is a right $kK$ module

so we can form their tensor product over $kK$

@anon Maybe I should have said this earlier...

But I think the point now is the map that we have is not just a group homomorphism

This may sound stupid but $kH$ is a ring yes?

We can view it as a left module over itself

so that now $kH$ is a $kH - k(H \cap K^g)$ bimodule

and I believe that the map now is not just a group homomorphism but also a homomorphism of left $kH$ - modules

Hmm but why should $k(HgK)$ be a left $kH$ module...

@anon But at least we know the existence of such a $\varphi$.

Oh maybe it actually makes sense

yeah.

@JonasTeuwen Why did you leave the room?

This^ is not really a "room." ;-)

Do you have a room?

@BenjaLim Leave it?

Could be worse, could be better. Kinda "okay". You?

Perfect.

@JasperLoy: Hi. I'm learning so much stuff. Moreover, I like to create new problems.

@JasperLoy: thanks for your question.

:D

$\Huge\text{:D}$

@JasperLoy Correction: Chris's sister.

Just Jonas, robjohn and I are left, after everybody left the math chat room.

@JasperLoy Good!

@JayeshBadwaik I am using scons to compile JT-thesis.pdf. Much... better.

At least it remembers checksums and such in a sqlite db.

I also use xetex.

Now my main dir is not polluted with auxiliary stuff.

Does anyone remember the question about upper bound of size of sigma algebra containing a set? I'm asking because of this. If anyone can, please help out and post a link. Thank you!

@JayeshBadwaik It has a sane font support (can use all fonts) ánd, it is not in Pascal.

@JasperLoy Hello there!

I got to go again.

See you all later!

Also, stuff like names can have é etc.

@MattN. Bye!

@JayeshBadwaik Also biber + biblatex.

@JonasTeuwen Hmm. Good! Yup, sane font support is nice.

@JasperLoy That's messy, xelatex has native support.

@JasperLoy Duh. But it is an ugly hack.

@JasperLoy It does now.

As of a year or two.

Also, it is much easier to handle and more "modern".

@JasperLoy It does support it.

The main reason is inertia, and the fact that the TeX Live people work on LuaTeX.

@JonasTeuwen Yeah.

Mmm, LuaTeX is also quite okay. But anyway, it is already much easier to handle certain issues with XeTeX. (protrusion works). The layout it at the end, eh.

But UTF-8 native support instead of a hack is much better.

@JasperLoy That is not the task of the typesetting engine right!

That is more like JFetchBrain.

@JasperLoy There you go.