04:00 - 21:0021:00 - 23:00

4:30 AM
Cool! Thankss, btw any ideas on how to generalize the above for
$$\dfrac{{xn}\choose{(x-2)n}}{{(x-1)n}\choose{(x-3)n}}$$ ....For x=3, I got 27/16...and x=4 gives 64/27 ....kind of seeing a pattern

3 hours later…
7:23 AM
Representation theory is annoying. Why does the regular representation encode any information about the irreducible representations of $G$? To me both seem completely separate entities. (I do get proof-wise how you would show it, but I don't have any intuition whatsoever)

7:38 AM
@SayanChattopadhyay Well, it encodes the information because it contains all of the irreducibles

@TobiasKildetoft I get that, but why should I expect it to?

I am not sure there is a completely intuitive reason
In some sense it is because the fact that the regular representation contains the identity element of the group, you get to map to any representation you want to via a non-trivial map
Another is that the regular representation is induced from the trivial subgroup

@SayanChattopadhyay because any module is a quotient of $\Bbb CG^n$
so if you think of modules as numbers and irreducible modules as prime numbers

Ah okay, I see

then every number is a divisor of a big enough power of $\Bbb CG$
so $\Bbb CG$ has to contain all the primes
@TobiasKildetoft there's something special about $\Bbb CG$ that makes this work
for example it doesn't work with $\Bbb Z$ because $\Bbb Z/2\Bbb Z$ is a simple module that isn't a direct summand of $\Bbb Z$

7:49 AM
@LeakyNun Yes, we need to be careful to not mix up the left- and the right induced modules in the general case
and hence not mix up whether we get submodules or quotients

well $\Bbb Z$ is commutative

That does not make then coincide

Oh so if I am doing it with the algebraic closure of some other field this won't work?

what's the general statement, if any?
@SayanChattopadhyay no the same proof works with any algebraically closed field with characteristic 0

Not sure what the precise general statement would be for infinite groups

7:51 AM
(you need characteristic 0 to divide by the order of the group)

Usually you add some topological stuff to make things behave nicely in those cases

or more generally, with characteristic not dividing the order of the group

Ahh, so the question is what makes group algebras special in this case?

in general they're all quotients of the ring?

Well, given a simple $R$-module $L$, can we define a module homomorphism $R\to L$?
(non-zero one)

7:54 AM
pick a nonzero element of $L$ lol

anyway this seems to be saying that every simple $\Bbb CG$-module is projective

well yes, since the ring is semisimple

aha, this is saying that the $\Bbb CG$ is semisimple
so in general a ring is not semisimple

very true

7:56 AM

according to wiki, semisimple rings are all artinian
In algebra, the Artin–Wedderburn theorem is a classification theorem for semisimple rings and semisimple algebras. The theorem states that an (Artinian) semisimple ring R is isomorphic to a product of finitely many ni-by-ni matrix rings over division rings Di, for some integers ni, both of which are uniquely determined up to permutation of the index i. In particular, any simple left or right Artinian ring is isomorphic to an n-by-n matrix ring over a division ring D, where both n and D are uniquely determined.Also, the Artin–Wedderburn theorem says that a semisimple algebra that is finit...
and then it decomposes to matrix rings over division rings
so there's the classification
so $\Bbb Z$ already fails because $\dim_{\text{Krull}}(\Bbb Z) = 1$

@AnindyaPrithvi sure... $\frac{x^x\,(x-3)^{x-3}}{(x-1)^{x-1}\,(x-2)^{x-2}}$

8:11 AM
I am going to start to learn Linear Algebra. I came across Gilbert Strang's course but someone here said that it is focused more on computation and I don't want that. Can you guys suggest a book?

@LeakyNun Can you give me a brief review?

I never read that book before lol

@LeakyNun pimping Ted's book? ;-)

@RobinSingh I mean, watch 3b1b's series
it completely changed what linear algebra is to me, if you want my review

8:14 AM
@LeakyNun I did watch the full playlist

@RobinSingh good
@RobinSingh then you can watch Ted's lecture series
(or his book that I linked to above)

@LeakyNun I never came across those lectures. Thanks! :-)

1 hour later…
9:26 AM
@robjohn Thank you so much, I'll write it out as a question soon and ping ; Thanks again 😊

@AnindyaPrithvi note that that does not agree with your value for $n=3$
that gives $\frac{27}4$

yes I have seen that
it goes 0^0

@AnindyaPrithvi but in this case, that is $1$
Numerically, $\frac{27}4$ looks right

@robjohn The key says 27/16
I am attaching the file
@robjohn numerically 27/16 fits...I checked on calc

I just took $\binom{3n}{n}^{1/n}$ numerically, and it tends to $\frac{27}4$ pretty closely
$\binom{2n}{0}=1$

9:34 AM
3ncn/2ncn....not the above...I think it should diverge

It is not $\binom{2n}{n}$ in the denominator

Ohh so the general representation was for the num and den seperate?
5 hours ago, by Anindya Prithvi
Cool! Thankss, btw any ideas on how to generalize the above for
$$\dfrac{{xn}\choose{(x-2)n}}{{(x-1)n}\choose{(x-3)n}}$$ ....For x=3, I got 27/16...and x=4 gives 64/27 ....kind of seeing a pattern
@robjohn so this is for $({xn} \choose {(x-2)n)})^{\frac{1}{n}}$ ?

$\left(\frac{\binom{3n}{n}}{\binom{2n}{0}}\right)^{1/n}$ for $x=3$

Ohhhh I see
@robjohn I interpreted it wrongly..Thanks...did you go by the attached method?
Or some known approximations?

I just simplifed it to $\frac{\binom{xn}{n}}{\binom{(x-2)n}{n}}$ and used Stirling

9:43 AM
ohh so, stirling is a sword...I thought stirling will not give me a rational answer in p/q form considering (n/e) in the expression
Thanks a lot

10:19 AM
If possible: could someone give me advice how to improve my question on SE? I get only some comments and no answers. Any help is welcome.
2

After analysis the following limit for prime gaps was found: $$\lim_{n \rightarrow \infty}\frac{3g_{n}^2}{p_{n}}=0$$ Is this limit known in literature? Is the presented method valid (see text below)? Closest found is: Oppermann's conjecture. Method. Two functions are defined: $\varepsilon_1$ and...

10:39 AM
If the limit is $0$ is the $3$ really necessary?

Good remark, the number 3 is what I found by correlation. The 3 and the 2 (from square) are the best fit. So according comment Erick Wong this limit is slightly stronger then existing ones (but not as strong as observed).

1 hour later…
12:04 PM
Are there any representation theorists out here

Tobias probably

I can neither confirm nor deny those allegations

hahah

@SayanChattopadhyay Oh fantastic, I came here to ask the same question. I think this is because if $V$ is an simple left $\Bbb CG$-module, then it must be a module over the various matrix algebras $\Bbb CG$ breaks up into. There aren't too many simple modules over a matrix algebra.
The only nontrivial simple left $M_n(\Bbb C)$-module is $\Bbb C^n$.

I really liked Leaky's prime number intuition

12:09 PM
Artin-Wedderburn is really a noncommutative version of primary factorization
be back later

You know that representations split as sums of irreducibles, and that this splitting is preserved by direct sum, and that quotients split. So since every finite dimensional G-module is finitely generated, it's a summand of some (CG)^n; and if it's irreducible it is a summand of CG itself, right?
Doesn't need AW I don't think

Right, as we noticed above, it works for any ring where all simple modules are projective

2 hours later…
1:57 PM
@MikeMiller Yeah that's a good way to see it
+1
I realized this after I left

The prime number intuition of course breaks down once the ring is not semisimple
That is when you break out the Lego block metaphor instead

Yeah, that $\Bbb CG$ is semisimple is literally Artin-Wedderburn

What? Artin-Wedderburn is a statement about semisimple algebras

Eh, it's Maschke

Yeah

2:02 PM
You can also arrive at the conclusion of Artin-Wedderburn explicitly for $\mathbb{C}G$, without invoking the general theorem

The general theorem is very natural so I don't care

Yeah, Maschke plus Schur gets you there nicely

I've never learned the general theorem lol

Actually, Maschke plus Schur is basically the general proof, isn't it?

Yeah
The general fact is that any simple left-Artinian ring is a matrix ring over a division ring

2:09 PM
I learned the proof for quals and forgot it the day after I passed

I am trying to prove representation theory can be done without character theory

Prove to whom?
Also, just because something can be done does not mean it should

I don't understand characters
I refuse to use them

Then fix that

I refuse them, in short

2:11 PM
They encode some nice stuff

Deformation of characters

let me know when you got a proof of the odd order theorem

Oh yeah shit that shit is bizarre
That's not a proof
It's magic
Characters take algebraic values or some absolute garbage

tell me a representation-theoretic, but not character-theoretic proof of Burnside for starters
would be helpful if you have one, cause the character-theoretic proof is magic and the group-theoretic proofs are too complicated

I think I did a decent job in my notes of making the idea of taking the trace of a representation appear natural

2:14 PM
Trace is the unnatural part
Trace of a matrix is meaningless
Sum of diagonal entries??? What

You don't believe that you like Ric

@BalarkaSen Here is one road to get there: Let's take an endomorphism of an irrep (but endomorphism just as a vector space)

something derivative of the determinant

Let's call it $\varphi$.

The Ricci curvature of a representation

2:15 PM
@MikeMiler Ric(X) measure how fast the volume of a geodesic ball bubbles in the X direction
No sum of diagonal entries
It's the 2nd order term in the Taylor expansion of vol

Now, let us turn this into an endomorphism of representations by the usual Maschke way, and call the result $\widetilde{\varphi}$

Yeah but that's because determinant has derivative = trace

Now, we know by Schur that this is in fact just a scalar, let's call it $\lambda_{\varphi}$
So we ask: What scalar is it, and the answer is $\frac{1}{m}\mathrm{Tr}(\varphi)$ where $m$ is the dimension of the original irrep

@MikeMiller This is neat

It's Leaky's idea
Or Maschke's
I'm just repeating and rephrasing

2:19 PM
Of course, the trace here is just a special case of the general concept of a symmetrizing trace on an algebra

Of course

So it is a way to turn the algebra of matrices into a symmetric algebra

@BalarkaSen Oh yeah this was my question as well, why trace?

@SayanChattopadhyay See what I wrote above for one reason to consider the trace

Hi all
According to Taylor's theorem (or just using L'Hopital), it should hold that $\lim_{x \to 0} (log(1+e^x) - log(2) - x/2)/x = 0$. However, when plotting $(log(1+e^x) - log(2) - x/2)/x$ or even just calculating it for small $x$, it looks like it goes to somewhere around $-0.28$. Does anyone see what's wrong here?

2:27 PM
W|A doesn't agree with you that the plot looks like it goes to -0.28
Are you sure you didn't make an arithmetic mistake

Weird. You're right, it doesn't. I had just typed the function in to google and that's what it gave me. I guess I can't trust google for plotting functions. Thanks.

The problem is that google interprets log as base 10. If you type ln you should get the correct plot.

loogle
log base googol
mwahahah

Thank you! That idea might have just not only solved that problem for me but also a bigger problem. I'm guessing that the log function built into R might also be base 10 then

Yeah just compute log(10) to find out

2:39 PM
@BalarkaSen hahah. pretty funny. i only just found about googol recently
@MikeMiller will do, thanks!

@MikeMiller what is this, actual numbers?

lolol

Let $N$ be normal in $G$ and $G_0$ normal in $G_1$. I want to verify that $G_0N$ is normal in $G_1N$. So I let $g_1n'g_0n(g_1n')^{-1} = g_1n'g_0nn'^{-1}g_1^{-1}$ but i m stuck at this stage.

well you want to prove that it's in $G_0 N$ right

yes

2:45 PM
so move $g_0$ all the way to the leftmost position

oh because N is normal

on the way you'll change $g_0$ but keep it in $G_0$

and because N is normal the rest of the stuff lives in N and that's it!
thanks

3:00 PM
If anyone is curious, R does use exp(1) as the base for its log function after all
Is it always true that $\lim_{x \to 0} f(x) g(x) = \lim_{x \to 0} f(x) \lim_{x \to 0} g(x)$ or are there exceptions? If so, when does it not? I thought it is always true but it seems like my above function multiplied by $1/x$ goes to $1/8$, not $0$ as it would if that rule applied here

what you wrote is always true, but I assume you wanted to write something else

oh yeah i did sorry
lol
$\lim_{x \to 0} f(x) g(x) = \lim_{x \to 0} f(x) \lim_{x \to 0} g(x)$

If the two expressions on the RHS exist (that is, f and g converge to some number as x -> 0; in particular neither goes off to infinity), then it is a theorem that the left side converges too, and to the expression on the RHS

or actually i guess just for any limit not necessarily $x \to 0$

If the expressions on the RHS don't exist then you can't say anything about the left side

3:09 PM
this can be extended to some cases where the limit is infinite as well

Thanks, @MikeMiller. But why is it then, that it seems to hold that $\lim_{x \to 0} (log(1+e^x) - log(2) - x/2)/x = 0$ and $\lim_{x \to 0} 1/x = 0$ but \lim_{x \to 0} (log(1+e^x) - log(2) - x/2)/(x^2) = 1/8$(according to L'hopital and also visible when plotting it)? if one of the limits is finite and non-zero and the other is pm infty, then the limit of the product is pm infty$\lim_{x\rightarrow0}1/x$doesn't exist Oh man. AKA the answer to why it seems that way is I'm stupid. Thanks And sorry, idk what's wrong with my brain it's fine, we all make mistakes 3:37 PM People who learned analysis from Baby Rudin: did you cover Polish spaces? What's your goal with this stuff? pretty sure there's no mention of Polish spaces in Baby Rudin whatsoever I've never used Polish spaces in my life for what it's worth polish spaces is where a bunch of stochastic shit can be done i assume @MikeMiller They're coming up in my measure theory class. Baby Rudin is the textbook for the prereq course. 3:41 PM nailed it That'll do ya I suspect that this prof is teaching this class primarily with the perspective of preparing students to be probability researchers. I've not found a single textbook that covers this material the way he does. It's not what I expected, but I guess I'll either get pushed now or pushed later. A Polish space is just something homeomorphic to a complete metric space with a countable dense subspace. Most likely you won't need a lot of properties very specifically Polish The reason this is useful in probability is because to define continuous stochastic processes often you define em on the countable dense subspace Eg Levy's construction of the Brownian motion @BalarkaSen Yeah, my topology background is WEAK. I recognize all of those terms as topology terms, but I don't know what most of those words mean. And then take limits to extend it to all time (time parameter varies over your space now instead of [0, infty)) taking limits is where you need complete 3:44 PM I will have to do some really serious studying soon Just think of trying to define a sequence of random variables$\{W_t\}$where$t$varies over a space$X$, not$[0, \infty)$who said time is linear? Lol Mike Weak bit, deleted weak convergence of weak bits So basically, it sounds like you need this stuff to define more general stochastic processes. I'm only familiar with discrete-time and continuous-time processes. yeah who said time is 0, 1, 2, or [0, infty) time can be really fucked up thats what all you need to remember is what Levy did, which is passing to the rationals -- define it on rational times and then extend 3:49 PM Howdy, a @Balarka, @Clarinet, @MikeM rationals is a countable dense subset of R, a complete metric space I have a TON of topology to learn soon. That much is clear. This prof assumes we know about Polish spaces coming in Morning @Ted I know nothing about Polish spaces. But I'm 25% Polish. Lol 3:50 PM I'd do better with Russian spaces. All gibberish to me at this point minus "topological space" and "complete metric space," per Wikipedia: "In the mathematical discipline of general topology, a Polish space is a separable completely metrizable topological space; that is, a space homeomorphic to a complete metric space that has a countable dense subset." Polish is the right condition on groups to force measurable homomorphisms to be automatically continuous or something, right I think I'll have to take some days off and just dive into topology Nobody cares about Polish group man @TedShifrin Definition: A Polish space X is a topological space that will, over the course of three treaties and twenty years, become a disjoint union of a Prussian space, an Austrian space, and a Russian space. (communicated to me by Fargle) Count on @Fargle :P 3:56 PM huh, there's an open mapping theorem for Polish groups Sometimes I wonder why stats curriculum works the way it does. You basically have three phases of stats: the stuff that only requires college algebra, the stuff that requires calculus + linear algebra, and the stuff that requires measure theory. Why not just push the math prerequisites through from the beginning and learn the advanced stuff as it should be treated? I think it's important to develop intuition and learn lots of examples without all the fancy machinery. This applies in virtually all areas of mathematics. Yeah the fancy stuff obscures basic understanding; you then work to see why your basic understanding matches with what the fancy stuff is doing The main problem I'm having to deal with - as a data science faculty member - is people thinking they know everything there is to know about stats when they've only taken the college-algebra-level course. I get your points. What I find interesting is from talking to people who hold bachelor's degrees in Stats from Canada and Europe is that they're pushed to take measure theory in their bachelor's degree. Not the case in the US. I wonder if there's a difference in priorities with countries. In Europe everyone taking "calculus" learns analysis, seriously. 4:00 PM I always thought notions of sufficiency, completeness are nontrivial in statistics. It's a very different approach to education. A sufficient statistic is pretty much literally a measured foliation as Thurston thought of it Oh, that's interesting. So they dive in to things like Spivak and Apostol Lol I have no idea what a sufficient statistic is now @BalarkaSen Oh, I thought that's standard there's no such thing as calculus here 4:01 PM This was in my Stat-II course. @Clarinet: More like Rudin than Spivak/Apostol, in fact. @TedShifrin Wow, that's intense No idea what a measured foliation is. The standard text that I've seen people use for the master's-level sequence in statistics is Casella and Berger's Statistical Inference, which defines a sufficient statistic as given here: en.wikipedia.org/wiki/… Well, students select early on (like before high school?) there whether they're going to study science or other things. It's totally a different model of education than in the US. @Clarinetist Ah I don't expect you to know what a measured foliation is, no. I was confused that you had no idea what a sufficient statistic is. Lol 4:04 PM Sufficiency, minimality, completeness etc are introduced in my Bachelors' course Honestly, I was disappointed in that class. The whole discussion of UMVUEs is interesting, but once you realize how restrictive the assumptions are to discuss that topic, you start wondering about the practical implications I am very much not a stat guy, I just thought the math was interesting I confess that I never learned any statistics. I am not proud of this fact. It's totally nonobvious that a boundedly complete minimal sufficient statistic is independent of an ancillary statistic. Balarka is speaking gobbledygook. 4:06 PM It has weird implications like, if$X$and$Y$are Gamma distributions,$X + Y$and$X/(X + Y)$are independent(!!) I only got into stats because I didn't appreciate how much hand-waving was done in actuarial science, and I didn't see myself going to grad school for pure math @BalarkaSen Basu's Theorem is such a remarkable result I agree!! The proof is a mere 2-lines. The nontrivial part is to come up with those definitions. Completely agree Here is what I meant by a measured foliation, by the way. So$T$be your statistic, which is a map from the sample space$\Omega$to$\Bbb R$.$T^{-1}(t)$then "foliates" or partitions$\Omega$Oh, that's interesting 4:09 PM Ok, let's not say it that way. Let me see. So basically the point is you have a family of distributions$X_\theta$parametrized by$\theta$And distribution of$T(X)$is completely independent of$\theta$That means you have an evolution of a measure$\mu_\theta$on$\Omega$such that on each leaf$T^{-1}(t)$, the evolution is constant:$\mu_\theta$restricted to these leaves is constant. Interesting That is somehow exactly what a measured foliation is, but just with nice smooth spaces ("manifolds"). Some day, I'll get myself to open up Schervish's text lol You have some evolution of a measure which are time independent along leaves, but varies in the "transverse" direction Oh, I don't know that book It's an old book that people still use for PhD-stats qualifying exams in mathematical statistics Theory of Statistics, Schervish 4:12 PM Ah alright So, it looks like I have a ton of topology to learn on my to-do list I don't envy full-time teaching faculty these days who are online. It's gotta be overwhelming. @TedShifrin Suppose you have two separate radioactive materials, and$X$and$Y$represent the time that an$\alpha$-particle emanates out of one of them, respectively. These are random times, so random variables. One 3-credit class is enough for me already The theorem is that$(X + Y)/2$(average time) and$X/(X + Y)$(proportion) are independent. Which is algebraically totally nonsense ("of course those expressions have a relation, multiple one with the other, get X.. oh wait, X is random, but surely something works...") @BalarkaSen Yeah, it's hard for me to accept when someone claims two random variables are independent these days without a proof by definition. Too many people handwave that step. 4:17 PM Independence is such a nonobvious notion I remember I had a really nasty independence problem on MSE one time... let's see if I can find it Ahh yes classic You can say something intuitive about that though, multiply both with$1/\sqrt{2}$and note that you're just rotating the 2D normal distribution And Gaussians are rotationally symmetric My MS qualifying exam dealt with piecewise Jacobian transformations. THAT was something to study lol 2 hours later… 6:02 PM Example of a function for which contour inequality is equality 1 hour later… 7:29 PM$z^2-x^2-y^2 = 1 \implies x^TQx = 1$$where$$Q = \begin{pmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1\\ \end{pmatrix}$I'm stuck on how to rotate the original function so that the forward sheet lies in the first octant (+,+,+). I'm pretty sure I have to change the entries in the matrix Q is that right? nevermind 1 hour later… 8:41 PM If a sequence converges to$0$, does its rearrangement also converge to$0$? I'm planning to ask a question on the main site that basically asks, "given how Stack Exchange runs moderator elections, is it possible for a candidate to win if the election was being run with x seats, but lose if the election was being run with x+1 seats?". I'm not familiar with the site rules, so what sort of info am I supposed to put in it in order for it to be a good fit? Or rather if a sequence does not converge to$0$, does its rearrangement also not converge to$0$? what's the difference between a normal extension and the algebraic closure? @Hawk:$\Bbb Q[i]$is a normal extension. So is$\Bbb Q[\root4\of 2,i]$. The algebraic closure of$\Bbb Q\$ is gigantic.
Right. A normal extension is a splitting field of a single polynomial.
Every polynomial splits in the algebraic closure.

@Hawk why do you think they're the same?

8:55 PM
@LeakyNun i read the definitions and they look like they are saying the same thing

can you tell us what definitions you read?

it was just on wikipedia

can you copy the wikipedia definitions?

@Hawk: What we learned from our discussion the other day is that you often don't pay attention to little details that are super important in mathematics. You need to force yourself to pay attention to them.

On compact smooth manifolds, do vector fields always admit global flows?

8:57 PM
well i m reading them now and there is a requirement that we have to use irred polynomials to talk about normal extensions @LeakyNun

that's not a definition

Yes, @Sayan.

@LeakyNun F is alg closed if every non-constant poly in F[x] has a root in F. L/K if every irreducible poly over K that has at least 1 root in L splits over L.

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