Mathematics - 2017-11-06 (page 1 of 3)

Mary Star

00:20

I am looking at the exercise:
At a tango course $12$ married couples participate. Find an appropriate measurable space and calculate the probability that exactly nine couples dance together.

Is the measurable space a 2-tuple $(\Omega, \mathcal{F})$, where $\Omega=\{1,2,3\ldots,12\}^2$, since we have 12 men and 12 women? And $\mathcal{F}$ is a $\sigma$-Algebra on $\Omega$, or not ?

Ted Shifrin

01:51

@MeowMix Yes.

Meow Mix

coolio

Ted Shifrin

I answered your email.

Meow Mix

just read

god, balancing all my subjects is gonna kill me

i have 3 tests tomorrow, cool

then a 2 tests and a project due the next day

Ted Shifrin

Well, stop Spivaking and put priority on your schoolwork.

Meow Mix

I've been schoolworking the past few hours

Ted Shifrin

02:00

well, make sure to get plenty of sleep!

anakhronizein

Anyone fond of representation theory of groups?

Daminark

I like it but am a noob

Like I barely know shit

anakhronizein

Anything about branching graphs?

And subreps?

Ted Shifrin

Hi noob Demonark

anakhronizein

I am having troubles wrapping my head around subreps of cyclic groups of the order 2^n

There are the irreducible reps that correspond to the nth roots of unit

Meow Mix

02:15

btw Ted

Ted Shifrin

Oh, that makes sense, @anakhronizein.

anakhronizein

So then suppose I take the restriction of these subreps to the cyclic group of order $2^{n-1}$:

what are the subreps of these restrictions?

Meow Mix

It's the average of all the $a_i$ right?

Ted Shifrin

Yes, but what about b.? It's so cool.

(If I remember right.)

Meow Mix

??? how did you know what b is

anyways it's like

sum of $|x - a_i|$

Ted Shifrin

02:18

Right.

Meow Mix

wow, this is interesting

Ted Shifrin

@anakhronizein: Doesn't it depend on which $k$ divide $n$ or $n-1$?

Meow Mix

well, looking at the derivative doesnt prove to be very meaningful

anakhronizein

Yeah that is what I am thinking, something along those lines.

But the subreps will correspond to subspaces of C

which there is only 2

So maybe I am missing something

Meow Mix

since it isn't differentiable at any of the $a_i$ you dont have any calculus to do

but i guess the problem does mention that

WAIT

Ted Shifrin

02:23

I've actually forgotten the definition of subrep.

Do you have to restrict the action to a proper subspace? I don't think so.

anakhronizein

I guess not.

Ted Shifrin

What's the careful definition?

Meow Mix

ok so Ted hear me out

anakhronizein

I thought it was the restriction to a $\sigma$-invariant subspace.

Daminark

@Ted yo

Meow Mix

02:25

suppose $x < a_1$ (and thus less than all the other $a_i$)

Ted Shifrin

I want a precise, correct definition, @anakhronizein.

Meow Mix

in that case, $f(x)$ looks like $nx - \sum a_i$

anakhronizein

If $(\sigma,V)$ is a representation of $G$, then a subrepresentation of $V$ is a $\sigma$-invariant subspace $W$ of $V$. The subrepresentation is given by $(\sigma|_W,W)$.

Meow Mix

then, when $a_1 < x < a_2$, you have $f(x) = a_1 + (n-2)x - \sum a_i$ for $i > 1$

Ted Shifrin

@anakhronizein: $\sigma|_W$ doesn't make sense.

Meow Mix

02:29

oops, these are supposed to be negative

Ted Shifrin

@Meow: Maybe a graphical approach would be helpful.

Meow Mix

What im trying to say here is that

anakhronizein

$\sigma|_W\colon G\to W\colon g\mapsto \sigma(g)$.

It makes sense to me.

Meow Mix

if $n$ is odd, the minimum occurs at $a_{(n + 1 )/ 2}$

Ted Shifrin

that $|_W$ notation is for restriction of domain, not of range.

anakhronizein

02:30

Sorry, that's just the notation I use.

Ted Shifrin

It's bad.

anakhronizein

Depends on the book you refer to.

Apparently we just don't share the notation.

Ted Shifrin

I've never seen that in my life.

OK, so you do need proper invariant subspaces.

Meow Mix

if $n$ is even, everything in $[a_{n/2}, a_{n/2 + 1}]$ is a local minimum

does that solution sound good to you?

Ted Shifrin

No.

Oh, I guess that's right.

So what's the interpretation of the minimum point(s)?

Meow Mix

02:32

it's sort of like the "middle" $a_i$

Ted Shifrin

There's a stat word for that.

anakhronizein

books.google.ca/…

Meow Mix

uhh

median?

anakhronizein

Here's a first for you @TedShifrin, look on page 2

Ted Shifrin

Yes, Meow.

Meow Mix

02:33

now that makes sense why he chose to make $a_1 < a_2 < \dots$

Ted Shifrin

@anakhronizein: Read more carefully. What they said agrees with me, not with you.

Daminark

Riperoni pepperoni

Meow Mix

cool problem.

Ted Shifrin

The multivariable generalization is also quite interesting, @Meow. Save that for later :P

Meow Mix

is there any multivariable stuff in spivaks?

Ted Shifrin

02:36

Nope. That's what my book is for :)

Meow Mix

approaches Ted's book geometrically

Ted Shifrin

You'll be back to computers soon enough :)

Go worry about your exams for tomorrow. Enough math here.

anakhronizein

Well I would link you to my professors notes where it is used, but then that discloses information about my identity.

Meow Mix

cya

anakhronizein

But that's the book he uses and I personally think that it is not bad notation.

Ted Shifrin

02:38

I don't really care, @anakhronizein, but did you see that they wrote $\sigma(g)|_W$? So they're restricting the domain of $\sigma(g)$, which is correct.

Where did they write what you said?

Daminark

@Meow it seems like when one of us gets really into math, the other starts to like compsci

:P

Catch you around

Meow Mix

well one reason my compsci shit is on hold right now is because

Ted Shifrin

You said you were waiting for chips or somethin'.

Meow Mix

i have no funding for my cpu project at the moment

Daminark

:(

gian

02:39

Why is it true that for an $n$-manifold, the intersection of a generally positioned $i$-cycle and $j$-cycle is an $(i+j−n)$- cycle?

Meow Mix

plus one of my projects got very hard very quickly

Ted Shifrin

Think about linear algebra, @gian. Take $i$, $j$ dimensional subspaces.

Meow Mix

and id like to return to it at a later date because it's the top on the "stack" of projects, where each of the lower project depends on the above projects

Daminark

I see. Hopefully that won't get stonewalled for too much longer

Ted Shifrin

"very hard" is exciting to you, Meow.

Daminark

02:40

I'm starting to shift toward compsci because in discrete we're starting graph theory and I remember how fun the stuff is

Meow Mix

oh, the project has to do with 3d graphics ;)

Ted Shifrin

That's why you liked projective geometry after all, Meow. :)

Meow Mix

TWO LINES ARE CONICS TOO

i mean, yeah

Daminark

Ted: What's projective geometry about exactly?

Meow Mix

projecting

your geometries

Ted Shifrin

02:42

Things that are $PGL$ invariant instead of Euclidean-group invariant?

anakhronizein

Take your normal euclidean geometry and instead of having parallel lines, have all the lines intersect.

Meow Mix

in reality you start by constructing real projective space

Ted Shifrin

For example, two shapes that appear the same when viewed on different viewing planes from different spots are projectively equivalent.

Meow Mix

iirc dont you define it by making equivalence classes of $\Bbb R^2$?

at least $R\mathbb P^1$

Ted Shifrin

That's defining projective space ... lines through the origin in $\Bbb R^{n+1}$.

Meow Mix

02:44

yeah, real projective space

then you do some theorems and stuff

anakhronizein

The Fano plane is the best image I think that sums up projective geometry.

Daminark

So does geometry of $\mathbb{RP}^n$ match nicely with PGL business? Or do the names just overlap?

anakhronizein

At least, naively.

Ted Shifrin

Yes, @Demonark. $PGL$ is the group of motions.

Meow Mix

oh, you have that cross thing that is invariant under projective transformations

cross-something

Ted Shifrin

02:45

cross ratio, yup.

It's the one invariant.

Daminark

Oh wait I guess that makes sense, they're both defined by scalar multiples being identified with each other

Ted Shifrin

right, Demonark

Projective differential geometry gets really cool ...

Daminark

Huh, might check that out at some point

Ted Shifrin

Projective connections instead of Levi-Civita :P

Good luck with your exams, Meow.

Meow Mix

before I go to bed

I saw one of the problems you gave to me one day in this book

"Find two convex functions that intersect only at the integers"

Daminark

02:48

We didn't quite do Levi-Civita in our class, or at least I don't remember our referring to it as such, what is that exactly?

Ted Shifrin

That is the general version of the $\nabla_X$ you learned for surfaces.

But there's no metric when you do projective differential geometry.

Meow Mix

anyways, good night and thanks

im going to do well on at least 1 of the tests

Ted Shifrin

Oh, right, @Meow ... that's in Spivak Chapter 11 appendix.

Do well. :) Night!

anakhronizein

What classifies as doing projective differential geometry? I have worked with quotients of projective varieties, but I get metrics there.

Ted Shifrin

Sure, projective varieties get the induced Kähler metric (so it's hermitian). But a projective connection comes from working with the projectivized tangent bundle, not the tangent bundle.

anakhronizein

02:52

So is the former not under "projective differential geometry"?

Ted Shifrin

No, it's Kähler geometry.

anakhronizein

I see.

Ted Shifrin

Terminology is admittedly confusing.

anakhronizein

I have been looking to do more complex geometry.

It would complement the co-adjoint orbit method stuff I want to look into as well.

Ted Shifrin

Well, more symplectic than complex, but there's a lot of overlap.

So you still don't concede the difference between the book's notation and yours?

My final word on it ... since I'm going to cook dinner.

anakhronizein

02:56

@TedShifrin in reference to the book's notation, my professor used the notation I used above (which no one had an issue with during class, and still 2 months later), and he heavily used the book. It doesn't clash with it, nor is it poor, IMHO.

Ted Shifrin

You just don't get it. Never mind.

anakhronizein

Indeed, I don't get the issue at hand with the notation.

In any case, goodnight. I think I solved my problem.

Secret

03:13

[Random]

Some rambles about the transfinite Veblen

\begin{align}
\varphi \binom{1\\1} \\
\varphi \binom{1\\2} \\
\varphi \binom{1\\3} \\
\varphi \binom{1\\4} \\
\varphi \binom{1\\5} \\
\cdots \\
\varphi \binom{1\\ \omega} \\
\end{align}

Then diagonal is clearly:

$\varphi (1,\cdots 1,1,1,1,1)$

yet there are only countably many Veblen expressions?

Leaky Nun

04:08

that eye-opening moment when my prof asked me what the trace and determinant of the empty matrix is

it got me thinking

user76284

@LeakyNun What is it?

Semiclassical

"what is the sound of one hand clapping"

Leaky Nun

@Semiclassical lol

user76284

Is it trace = 0 and determinant = 1?

Leaky Nun

@user76284 yes

user76284

04:14

Nice

I assumed so since trace is 'additive' and determinant is 'multiplicative'

Semiclassical

I think I have used something like that, actually, in the context of writing recursion relations for determinants of n-by-n matrices

Leaky Nun

determinant is a homomorphism to the units, so it must be the multiplicative identity

trace is a homomorphism to the additive group, so it must be additive identity

Semiclassical

though in my case it's really just a convenient way of expressing the recurrence rather than anything insightful

user76284

What is the exponential of the empty matrix?

Just empty again?

Leaky Nun

@user76284 how is exponential defined? the taylor series?

$\exp(T) := \displaystyle \sum_{n=0}^\infty \frac1{n!} T^n$

Semiclassical

04:17

there's also the formula $\det e^A = e^{\tr A}$

Leaky Nun

well it must be the empty matrix

user76284

@Semiclassical Yeah I was thinking of that

Semiclassical

i will never understand why \det is a thing but not \tr

Leaky Nun

I don't really think of it as "empty matrix", but rather the unique endomorphism of the trivial vector space

then $\exp(T)$ is an endomorphism, so it must be the empty matrix

Semiclassical

anyways. if A is the empty matrix, then tr(A)=0 as argued above. so for the formula to hold you'd need det(e^A)=e^0=1

in which case e^A has determinant 1

which is consistent with e^A being the empty matrix

user76284

04:18

And there is only one 'empty matrix'.

Zero-dimensional

Leaky Nun

right

user76284

04:47

If I'm training a neural network in an online setting with samples which are generated from the true distribution of data, overfitting shouldn't be a problem, right?

Leaky Nun

05:19

5

Q: Proving that $\frac{\phi^{400}+1}{\phi^{200}}$ is an integer.

How do we prove that $\dfrac{\phi^{400}+1}{\phi^{200}}$ is an integer, where $\phi$ is the golden ratio? This appeared in an answer to a question I asked previously, but I do not see how to prove this..

golden-ratio

hmm, I wonder if it can be solved more elegantly

Konformist Liberal

Can someone give a hint about the following: Take a point on a CW complex X and then define a new CW structure on X such that this point is a 0-cell?

Jan

06:06

How to call the coefficient that is used for helping to solve equation? Like in Jensen's inequality. Is it called just the coefficient or it has some special word for it?

Konformist Liberal

06:26

This is the second exercise from here math.cornell.edu/~hatcher/AT/ATapp.pdf

Daminark

07:27

Yo @Tasty!

TastyRomeo

Morning

Daminark

How's it going?

TastyRomeo

Okay, but busy (as always :P )

TA'ing graph theory to a blind student is taking up a lot of my time, though

And you?

Daminark

Anything interesting you've been up to recently?

Oh, I see

I'm TAing discrete math and it has recently reached graph theory

I'm only now remembering how much I like that subject

mick

Hi

0

Q: $i_a = (i_b + i_c + 1)/2$ and $f(x) = f(x - f(x-1)) /2$?

Consider the set $i_0,i_1,...$ defined as $i_0$ is the smallest element and $i_0 = 0$. If $ (i_a - i_b)^2 < 1$ then $i_c$ is Also in the set and given by $$ i_c = (i_a + i_b + 1)/2$$ [*] Let $T(x,y)$ be the cardinality of $i_j$ Numbers in the interval $[x,y]$. Notice $1/2,3/4,7/8,...$ are all...

sequences-and-series combinatorics recurrence-relations multilinear-algebra tetration

Daminark

07:33

Any fun problems you've given when TAing graph theory? @Tasty

TastyRomeo

Well, it's graph theory aimed towards computer science (sadly), so not really I'm afraid :P

Like, the hardest exercises I've given so far are like

Prove that a graph with 5 vertices and at least one vertex of degree 3 must be planar

and, eh

Prove that the following are equivalent for a connected graph: G is maximal, G is a triangulation, e = 3v-6

mick

Hmm

Daminark

I see

The problems these students got were pretty slick

One wasn't even graph theory, find $f$ and $g$ such that $\lim_{x\to\infty} f(x) = \lim_{x\to\infty} g(x) = \infty$, $g = \Theta(f)$, and $\lim_{x\to\infty} \frac{f(x)}{g(x)}$ does not exist

TastyRomeo

What's the big theta mean? :O

Daminark

Okay so, $f(x) = O(g(x))$ as $x\to L$ if there exists some constant $C$ such that for all $x$ sufficiently close to $L$, we have $|f(x)| \le C|g(x)|$

And we say $g(x) = \Theta(f(x))$

Now, if $f = O(g)$ and $f= \Theta(g)$, then $f = \Theta(g)$

mick

07:48

Anyone willing to try my question ??

Daminark

Hey @Mathei

MatheiBoulomenos

Hi

Daminark

08:20

Okay so I just want to make sure my argument for why groups of order 1001 are abelian isn't complete nonsense

So we have $H_1$ of order $7$, $H_2$ of order 11, and $H_3$ of order 13

Oh wait yup, it's wrong

Yeah I'm gonna have to resort to Sylow here, I've got no clue how else to proceed

MatheiBoulomenos

yeah, you have to use Sylow

Daminark

I dunno if we're gonna prove that in class by the time it's due, which is why I was hesitant, but I did a short expository paper a few years ago in which I proved Sylow, so I'll just cite that

Daminark

08:48

Is it true that an abelian group of order pqr is cyclic?

Oh wait yeah that should be CRT

Actually didn't even need that

Yasharyan Gaikwad

09:07

Is there a name given to numbers of form a + b√c

Daminark

Where $a,b\in\mathbb{Q}$? I think that's sometimes called $\mathbb{Q}[\sqrt{c}]$ but don't take me on faith

Yasharyan Gaikwad

Ok thanks. I had some notation ideas but this one looks like the best one.

Balarka Sen

09:48

@Daminark click on the sad face

@gian Think about an $i$-subspace and a $j$-subspace inside $\Bbb R^n$. What is their intersection?

abenthy

If $M$ is a connected, oriented manifold and I know that $H^k(M, M \setminus K) \neq 0$ for some compact set $K$, does it follow that $H^k(M) \neq 0$?

Leaky Nun

🤦🏻‍♂️

Balarka Sen

blow up is good

TastyRomeo

10:08

@BalarkaSen terrorist alert

Balarka Sen

I don't watch anime though

TastyRomeo

195

A: Mathematical "urban legends"

This happened just last year, but it certainly deserves to be included in the annals of mathematical legends: A graduate student (let's call him Saeed) is in the airport standing in a security line. He is coming back from a conference, where he presented some exciting results of his Ph.D. thesis...

Balarka Sen

yeah i know about that

i was referencing the fact that they found anime among bin laden's files

TastyRomeo

o.O

Didn't know that

Balarka Sen

it's going viral on the meme corners

bin laden be like "Omae Wa Mou Shindeiru"

TastyRomeo

10:12

Bush and Osama tsundere yaoi art when

Balarka Sen

lmao

avz2611

11:54

is anybody in here :P

anakhronizein

Hi

Mary Star

Hey!!

I am looking at the exercise:
At a tango course $12$ married couples participate. Find an appropriate measurable space and calculate the probability that exactly nine couples dance together.

Is the measurable space a 2-tuple $(\Omega, \mathcal{F})$, where $\Omega=\{1,2,3\ldots,12\}^2$, since we have 12 men and 12 women?

Or do we consider the $12$ men and the dance-couple assignment as a permutation, so $\Omega$ is the set of all possible permutations of the $12$ women, i.e., $S_{12}$ ? So, is $\Omega$ the set of all possible dance-couples?

avz2611

for probability think of 12 men lined up, and 12 women to be setup in line, and opposite ones dance, so 12! different combination possible, now for exactly 9 to be dancing you have 12C9 for couples dancing and for the rest 3 you apply derangement

can i ask optimizations questions on math.stack or should i ask in cross validation?

Mary Star

But by 12C9, we get the number of ways to get 9 couples, or not? We don't get the number of probability that 9 men dance with their wife, do we? @avz2611

avz2611

@MaryStar you divide that by total number of possibilities

basically (12C9*3!(1/2!-1/3!))/12!

Mary Star

12:06

Does it hold that
$P(\text{exactly 9 couples}) = \frac{\text{#permutations of 12 women with 9 fixed points}}{\text{#permutations of 12 women}}$ ?

It holds that $\text{#permutations of 12 women} = S_{12} = 12!$, right?
Does it hold that $\text{#permutations of 12 women with 9 fixed points}= 12C9*3!(1/2!-1/3!)$ ? Could you explain to me how we get that formula? @avz2611

avz2611

how do you convert the tex in chat?

its tough to read :P

Mary Star

@avz2611 math.ucla.edu/~robjohn/math/mathjax.html

avz2611

@MaryStar so my argument is this, i have arranged all men in order from 1-12, and now there are 1-12 vacant positions for women, so basically there are 12! arrangements possible. Now for exactly 9 couples, i am selecting 9 positions out of 12 that contain women with their respective other half, and other 3 not with their other half. so 12C9*derangement(3)=12C9*3!(1/2!-1/3!)

user193319

12:28

Quote: "Let $H,K,N$ be nontrivial normal subgroups of $G$ and suppose $G = H \times K$...."

Presumably I am suppose to interpret the equal sign as an isomorphism between $G$ and $H \times K$.

Mary Star

12:43

@avz2611 By 12C9 we have the number of ways to select 9 positions out of 12. How do we know that these 9 have their correct wife? Do we know that because we multiply the derangement(3) ?

avz2611

13:00

@MaryStar the husbands are in their position so we just need to make sure the wives are in their position, lets just think of this as we are selecting 9 couples that are dancing together i.e 12C9

Mary Star

So, there are 12C9 ways to select 9 couples out of 12. The remaining 3 men have to choose the wrong woman. The number of ways is equal to #{(2,3,1), (3,1,2)} = 2, right?
Therefore, the number of of ways to get 9 couples and the remaining 3 people are with the wrong partner is equal to 12C9 * 2.
Have I understood that correctly? @avz2611

James

Hi everyone!

avz2611

@MaryStar yep!

James

For the third day, I still haven't gotten my question answered, may I ask for your help, please?

2

Q: Gaussian and Mean Curvatures for a Ruled Surface

We are asked to prove the following theorem found in page 88 of Differential Geometry: Curves, Surfaces, Manifolds by Wolfgang Kühnel. Using standard parameters, calculate the Gaussian curvature and the mean curvature of a ruled surface as follows: $K = -\dfrac {{\lambda}^2} {{{\lambda}^2 +v^2}...

differential-geometry curvature

abenthy

Can someone explain to me why the $\mathbb{Q}$-rank of the lattice $\Gamma = SL_2(\mathbb{Z}[\sqrt{3}])$ embedded into $G = SL_2(\mathbb{R}) \times SL_2(\mathbb{R})$ via $a+\sqrt{3}b \mapsto (a,b)$ on each of the four coordinates is one?

James

13:06

I wouldn't be asking if I knew...

Mary Star

I got it now!! Thank you!
Which is the measurable space?
The measurable space is a 2-tuple $(\Omega, \mathcal{F})$, where $\mathcal{F}$ is a $\sigma$-algebra on $\Omega$, or not? But what is $\Omega$ is this case?
Is it maybe the set of all the possible permutations of 12?

ÍgjøgnumMeg

13:27

Grüezi mitanond

anon

13:55

@user193319 or interpret $G$ as an internal direct product of $H$ and $K$

user193319

@anon Oh. I see. I am trying to prove that $N$ is in the center of $G$ or $N$ intersects one of $H,K$. Would the problem make more sense if $G$ were the internal direct product of $H$ and $K$?

anon

it's not a matter of "we have to assume it means this to make sense of the problem," it's, "that's what the author means"

@abenthy I don't understand how that embedding makes sense. For instance, $$ \begin{bmatrix} \sqrt{3} & 2 \\ 1 & \sqrt{3}\end{bmatrix}]\mapsto\left(\,\begin{bmatrix} 0 & 2 \\ 1 & 0\end{bmatrix},\begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}\,\right), $$ in which case the first coordinate is not in $SL_2(\Bbb R)$.

user193319

@anon Well. I am not sure what the author intends. What I mean "makes sense" is, is the theorem false if we take $G=H \times K$ to mean one thing over another, since, presumably, the author wouldn't want us to prove a false theorem. So I think my question is relevant, especially since I don't know what the other intends.

anon

@user193319 what you mean doesn't make sense, because internal and external direct products are isomorphic

it really doesn't matter. just do the problem.

user193319

@anon Okay. I didn't know that. But being isomorphic doesn't mean being equal.

anon

14:05

which is why I told you what the author means is internal

in any case people abuse the equals sign all the time when two things are conventionally identified

user193319

@anon So then $G=H \times K$ can be sensible or senseless depending upon how you interpret it.

anon

your hangup is senseless. do the problem, you're procrastinating.

abenthy

@anon Oh sorry, i mean $a + \sqrt{3}b \mapsto (a+b,a-b)$.

user193319

@anon Okay. so $G$ being the internal direct product of $G$ means that $G= \langle H \cup K \rangle$ with $H \cap K = \{e\}$; and since at least one of the two subgroups is normal, $\langle H \cup K \rangle = HK$? Is this also called the internal weak direct product?

Maxime

14:36

Should the mathematical questions be on math.se?

physics.se is waste of the time for this

i think they're answering theoretical questions because all questions I've seen had been put off-topic lmao

abenthy

15:25

If $F = \mathbb{Q}(\sqrt{d})$ is a real quadratic number field, why is $\text{SL}_2(F)$ a linear algebraic group? Whats is its embedding into $\text{GL}_n(\mathbb{C})$?

Faust

15:46

good morning

Sha Vuklia

16:29

@bala

you know what happened today

Balarka Sen

?

Sha Vuklia

I was talking to two girls, and at some point I made a joke about a meme or sth. and THEN

one of them started showing me illuminati videos...:P

and I just did not have the time to react appropriately to that

they were dying over illuminati jokes. but like, the most basic jokes ever:P

Balarka Sen

illuminati? what is this, 2005?

Sha Vuklia

I had to get that off my chest

yes........ I know

we were watching a vid (me reluctantly) and they were crying with laughter

and then I showed them a zucc meme, and they were like 'haha that's funny'

they probably need 10 more years to appreciate it :P

Balarka Sen

zucc is p old too

Sha Vuklia

16:32

ugh alright, some supreme meme then

Balarka Sen

~~supreme~~ sheep

Sha Vuklia

lol I didn't even mean it in that context

:P

user193319

Is the semidirect product different from the internal direct product?

Balarka Sen

@Sha old memes are only watchable if you mix it up with new ones

Sha Vuklia

yeaa i agree

Balarka Sen

16:39

that's how crash bandicoot came back in the game

Sha Vuklia

oh I never knew how to appreciate that one:P

but it doesn't matter,

because I have zucc memes now

abenthy

Another question, is the Weil restriction $R_{\mathbb{Q}(\sqrt{3})/\mathbb{Q}}(SL_2)$ isomorphic as a $\mathbb{Q}$-group to $SL_2 \times SL_2$?

Sha Vuklia

he gained my respect when he responded to "are you a lizard":P

like, I was excepting him to joke about it, but he literally just burned the person:P

Balarka Sen

the zucc is just a weird person

Sha Vuklia

he's not even a person..:l

Balarka Sen

16:42

he doesn't get humor other than fast and furious jokes

Sha Vuklia

hahahahhahaha

ah man

what a legend

Balarka Sen

truly

user193319

16:55

$Mathematics$ Linear & Abstract algebra

For any discussion concerning linear, abstract or even element...

Secret

17:14

[Random]

A random ordinal:

$$\psi_{LVO}({}^{LVO}\Omega)$$

Balarka Sen

17:33

Quiet day

user193319

Problem: Let $\{G_i \mid i \in I\}$ be a family of groups, then $\prod^w G_i$, the external weak direct product, is the internal weak direct product of the subgroups $\{i_k(G_k) \mid k \in I\}$, where $i_k : G_k \to \prod G_i$ is the canonical embedding.

I don't see how this is true. Specifically, isn't $\langle \bigcup i_k(G_k) \rangle$ bigger than $\prod^w G_i$.

I could see $\prod G_i$, the 'regular' direct product, being the internal weak direct product of these subgroups, but not $\prod^w G_i$. What am I missing?

Kasmir Khaan

17:53

@MatheiBoulomenos mathei :D

Alessandro Codenotti

Wouldn't $\prod G_i$ be the internal direct product of $i_k(G_k)$ rather than the weak one?

V.7

Hey all :)

I'm trying to find out what's wrong here ...

So ... as you see one person says it's an illusion, but I think he's wrong

0

Q: How to draw an object and rotate it in oblique frontal projection

How to draw an object and rotate it in oblique frontal (dimetric) projection properly ? An illustration of projection: I've already made a program (Pascal with Graph unit) which does it, but I think that it draws an object incorrectly. program p7test; uses PtcCrt, PtcGraph; type TPixel ...

math 3d rotation pascal projection-matrix

Daminark

@Mathei turns out our class won't likely get to Sylow this week

So I think they expect the group of order 1001 business to be done without

$Mathematics$ Linear & Abstract algebra

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$Mathematics$ Mathematics