Mathematics

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

Sayan Chattopadhyay

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)

Tobias Kildetoft

7:38 AM

@SayanChattopadhyay Well, it encodes the information because it contains all of the irreducibles

Sayan Chattopadhyay

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

Tobias Kildetoft

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

Leaky Nun

@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

Sayan Chattopadhyay

Ah okay, I see

Leaky Nun

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 $\Bbb Z$

Tobias Kildetoft

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

Leaky Nun

well $\Bbb Z$ is commutative

Tobias Kildetoft

That does not make then coincide

Sayan Chattopadhyay

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

Leaky Nun

what's the general statement, if any?

@SayanChattopadhyay no the same proof works with any algebraically closed field with characteristic 0

Tobias Kildetoft

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

Leaky Nun

7:51 AM

(you need characteristic 0 to divide by the order of the group)

Tobias Kildetoft

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

Leaky Nun

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

@TobiasKildetoft I'm talking about irreducible modules over a general ring

Tobias Kildetoft

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

Leaky Nun

in general they're all quotients of the ring?

Tobias Kildetoft

Well, given a simple $R$-module $L$, can we define a module homomorphism $R\to L$?

(non-zero one)

Leaky Nun

7:54 AM

pick a nonzero element of $L$ lol

Tobias Kildetoft

yep

Leaky Nun

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

Tobias Kildetoft

well yes, since the ring is semisimple

Leaky Nun

aha, this is saying that the $\Bbb CG$ is semisimple

so in general a ring is not semisimple

Tobias Kildetoft

very true

Leaky Nun

7:56 AM

maybe Artin and Wedderburn have something to say about this

Tobias Kildetoft

I can guarantee that both of those people had a lot to say about this subject in their time

Leaky Nun

according to wiki, semisimple rings are all artinian

Artin–Wedderburn theorem

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 $\dim_{\text{Krull}}(\Bbb Z) = 1$

robjohn

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

Robin Singh

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?

Leaky Nun

Linear Algebra: A Geometric Approach

Robin Singh

@LeakyNun Can you give me a brief review?

Leaky Nun

I never read that book before lol

robjohn

@LeakyNun pimping Ted's book? ;-)

Leaky Nun

@RobinSingh I mean, watch 3b1b's series

it completely changed what linear algebra is to me, if you want my review

Robin Singh

8:14 AM

@LeakyNun I did watch the full playlist

Leaky Nun

youtube.com/playlist?list=PLZHQObOWTQDPD3MizzM2xVFitgF8hE_ab

@RobinSingh good

@RobinSingh then you can watch Ted's lecture series

(or his book that I linked to above)

Robin Singh

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

Anindya Prithvi

9:26 AM

@robjohn Thank you so much, I'll write it out as a question soon and ping ; Thanks again 😊

robjohn

@AnindyaPrithvi note that that does not agree with your value for $n=3$

that gives $\frac{27}4$

Anindya Prithvi

yes I have seen that

it goes 0^0

robjohn

@AnindyaPrithvi but in this case, that is $1$

Numerically, $\frac{27}4$ looks right

Anindya Prithvi

@robjohn The key says 27/16

I am attaching the file

@robjohn numerically 27/16 fits...I checked on calc

robjohn

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

$\binom{2n}{0}=1$

Anindya Prithvi

9:34 AM

@robjohn desmos.com/calculator/bdauauw8va

3ncn/2ncn....not the above...I think it should diverge

robjohn

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

Anindya Prithvi

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 $({xn} \choose {(x-2)n)})^{\frac{1}{n}}$ ?

robjohn

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

Anindya Prithvi

Ohhhh I see

@robjohn I interpreted it wrongly..Thanks...did you go by the attached method?

Or some known approximations?

robjohn

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

Anindya Prithvi

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

OOOVincentOOO

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

Q: Is Primegap $\lim\limits_{n \to \infty} \frac{3g_{n}^2}{p_{n}}=0$ known in literature?

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...

robjohn

10:39 AM

If the limit is $0$ is the $3$ really necessary?

OOOVincentOOO

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).

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$Mathematics$ Mathematics