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12:45 AM
In this video in the beginning, he describes the action on a square like rotation that leaves it unchanged / identical, and he puts a picture of pi with some eyes on it on the square. But when he rotates it 90 degrees, the pi with eyes is flipped on its side, and so the square isn't "identical" after the rotation.
So, in what sense of the word "identical" is he using?
 
the squares are identical, the squares with the pi figures attached are not identical
the pi figures only get attached so that you can keep track of what which of these actions does in the first place
 
ah, right -- which leads to his talk about compositions
thanks @Thorgott :)
@Thorgott do you plan to attend math grad school?
 
fun fact: what he discusses there is not "just" the dihedral group of order 8, it is the natural representation of the dihedral group of order 8
yes, I do
 
@Thorgott so for the dihedral group, the 8 actions -- rotating, flipping, staying the same -- are the elements of the group. What is the binary operation that combines two elements and forming a third element then?
 
1:00 AM
performing these actions after another
 
I see
 
most groups that one encounters naturally are groups of symmetries of some given object, so an important perspective often adapted in group theory is trying to understand a group by how it can act on various objects (which is aptly known as a group action); this is the viewpoint from which the 3b1b video approaches this
a representation of a group is simply a group action on a vector space in a way such that the group acts linearly and since the theory of vector spaces is well understood, these actions behave particularly well
 
@Thorgott I see; so for the dihedral group with 8 actions, a representation of this group would then fix / choose one of its 8 actions and let it act on some vector space?
 
yeah, though you wouldn't really fix an element, but let all of them act on the space
 
@Thorgott I see -- thanks :)
 
1:14 AM
the natural representation is the one in the video: the dihedral group of order 8 acts on the plane, via 4 rotations and 4 rotations + reflection
these are, in fact, the only rigid motions of the plane that leave the square fixed
so that's why it's called the symmetry group of a square
 
group of symmetries is just shorthand for group of symmetric actions, right?
actions that leave the square unchanged ...
be back later, thanks @Thorgott
 
 
2 hours later…
3:47 AM
hi @TedShifrin
 
 
3 hours later…
6:26 AM
In mathematical articles, is there any difference of expressions one ... and we ... ?
For example:
"To apply this formula, one needs the property of the function to be good enough" and "To apply this formula, we need the property of the function to be good enough"
Or it's just a choice of the author?
(And personally, which one do you prefer?)
 
 
2 hours later…
8:49 AM
My book is trying to teach me how to find the integral solutions of a first degree equation in two variables. He says:
> Any Eqaution of the first degree involving two unknowns $x$ and $y$ can be reduced to the form $ax \pm by = \pm c$, Where $a, b, c$ Are positive integers.
Then he writes:
> It is clear that the equation $ax + by =-c$ has no positive integral solution; and that the equation $ax-by=-c$ is equaibalent to $by-ax =c$; hence it will be sufficient to consider the eqautions $ax \pm by =c $
But I cannot see how can $ax \pm by =c$ give us $ax - by =-c$? From $ax \pm by =c$ we can get just two equations $$ax + by = c \\ \color{magenta}{ax-by =c}$$
The magenta one can be transformed into $$\color{violet}{ by - ax =-c}$$
But we can never get $ax -by =-c$, so why the book considered only two cases? What the book tried to say?
 
9:47 AM
@Knight $(a,x,b,y) \mapsto (b,y,a,x)$
 
10:05 AM
@LeakyNun Yes.
But I don’t what does $\mapsto$ mean
 
$x \mapsto f(x)$ is the same as $f$, it denotes the function
 
@Astyx Okay, but I’m unable to think how it resolves my issue? Maybe I need a little more hint
By the way, Astyx today I learned that in French “we” means “yes”
 
What he means is that you can just switch the role of a and b, and x and y, WLOG
It's not written "we" though
 
10:24 AM
How $ax -by =-c$ is equivalent to $by-ax=c$? I think multiplying my $-1$ and re-arranging will give us the desired result. But how $ax-by=c$ can ever become $ax-by=-c$?
 
10:34 AM
By switching a and b, and x and y
 
Still didn’t get you, I’m sorry
 
a,b,x,and y are just variables
to go from by-ax = c to ax - by =-c you can just rename your variables
If I want to solve 3y - 7x =25, (so with a = 7, b=3) is the same as solving 7x - 3y = -25 , which is the same as solving 7y - 3x = -25 (so with a = 3, b=7) up to switching x and y
 
10:59 AM
Okay
Thank you so much
 
@Astyx parlay voo from say?
 
 
1 hour later…
12:29 PM
Recall that a group $G$ has the Infinite Conjugacy Class (ICC) property if the conjugacy class of any nontrivial element is infinite. If $\{G_i\}$ is a directed sequence of ICC groups, will the direct limit of the $G_i$ also be ICC?
I was thinking in general no; e.g., if the maps between the groups are not injective...But I can't find a counterexample.
Wait...is the following a counterexample? Let $G$ be any ICC group, and define $G_i := G$ for every $i \in \Bbb{N}$. For every $i \in \Bbb{N}$, define $f_{ii} : G_i \to G_i$ by $f_{ii}(g) = g$ and if $i < j$, define $f_{ij} : G_i \to G_j$ by $f_{ij}(g) = 1$.
 
yes
 
The direct limit is the trivial group, right?
 
exactly
 
12:44 PM
If they are injective, will the direct limit be ICC? I think so.
 
 
2 hours later…
2:33 PM
@LeakyNun lmao
 
 
1 hour later…
3:45 PM
@Semiclassical Can you please come here for a moment?
 
hi, I had a question
suppose we have a polynomial form such as P(x) = x^2, or a multivariate one say P(x_1,x_2) = x_1 ^2 + x_1x_2 + x_2^4
so, how can we say something like : if we evaluate P over a ring R and also over a ring E, then in both of these we would have P is non-zero
or I mean, in general how can we say about simultaneous zero behaviour of a polynomial form over possibly different rings
any clue ?
 
in complete generality, probably not much
for any polynomial with integer coefficients, you can find a ring such that the associated polynomial function over it is the zero function
 
 
1 hour later…
5:14 PM
Hi all
When we consider the group $\mathbb{Z}_p \times \mathbb{Z}_p$, the elements $g_1=(1,0)$ and $g_2=(0,1)$ are generating elements. There $g_1$ can not be generated by $g_2$ and vice versa, right?
$p$ is an odd prime
 
Well, you can check for yourself, right?
 
Hmm, yes, I got as it can not be and wanted to clarify. Then can we say that a relationship like $g_1^m = g_2^{-n}$ is never possible since they can not be generated from each other?
Thanks a lot in advance. :)
 
No, such a relation is possible. The subgroups generated by $g_1$ and $g_2$ intersect in what?
 
The identity
 
Precisely. So what you said is wrong.
 
5:22 PM
It is because if $g_1^m=e$ and $g_2^{-n}=e$ then $g_1^m=g_2^{-n}$?
 
Yes, sure.
And you know precisely for which $m$ and $n$ that can happen.
 
For $g_1^m=e$ and $g_2^{-n}=e$ to be equal to the identity, $m$ and $n$ should be the orders of those two elements or multiples of them. Otherwise we can say that such equality is not possible :)
right?
Many many thanks @TedShifrin
 
Right. :)
 
:) :)
Have a nice day!
Bye
 
You too.
 
5:28 PM
:)
 
Does finding a gcf or lcm by multiplying bases raised to the min or max, respectively, of the inputs' prime factorization exponents require both inputs to have all the same non-trivial prime factorization bases?
 
no, there's no need to think about what you call "bases"
just think of the other primes as appearing in the factorization with exponent 0
 
Hmm, I was calling those factors trivial.
If you allow those factors to be included in min(), wouldn't the gcf of everything be 1?
 
no, but, of course, only the prime factors that appear in both numbers will matter for the gcd
 
5:44 PM
okay, thanks
 
5:55 PM
Ted can you please help me with continued fractions?
I’m having hard time in understanding what are the convergents
 
I'm not a continued fraction expert at all. Best to ask someone else.
 
Okay
 
guys, sanity check: an intertwiner for the same irreducible representation can only have one eigenvalue, right?
(an intertwiner being an linear map that commutes with the representation)
 
6:14 PM
any Eigenspace is stable under the group action
so you will either have no Eigenvalues or a homothety
the latter always happens over algebraically closed fields
 
ye, that was my reasoning too
alright, thx for the confirmation
 
6:35 PM
olas @Astyx
 
Heya
Howdy
 
Salut @Astyx, @Sha.
 
Salut
 
olasolas @Ted
@Astyx pretty good:0
I was wondering
have you ever studied a foreign language? (apart from high school stuff)
 
You asking Astyx?
It would seem his English is beyond proficient.
 
6:48 PM
I would assume that English is part of the high school curriculum
I was asking him, but I'd be interested to hear anyone's answer
 
I've been in the UK a few times, that helped a lot as well
 
Yes, but I've met plenty of Europeans whose English is mediocre, despite years in high school.
I studied lots of foreign languages.
 
In high school, I learned a little bit of german
 
@TedShifrin oooo, which ones:0
 
And Apart from that I studied a bit of chinese
 
6:49 PM
@Astyx :00000
 
I majored in French in college, took 5 semesters of German and 2 semesters of Russian.
2
I'm not counting high school Latin.
 
I've been wanting to learn Russian and Chinese forever. And I've finally taken serious steps
 
I'm not very good at all
 
You both mentioned my linguistic crushes<3
@Astyx Ye I'm at 0 atm:p but I'll be taking a summer course in Chinese in august
and Russian is close to my mother language, so I can learn snippets on my own without much distress
 
That's very cool !
 
6:56 PM
I've learned Latin in high school and have retained almost nothing
 
Ah yeah, I learned a bit of Ancient Greek and Latin as well if those count
 
why is it accusing me of being a bot or script and asking me to check a capitula box an infinite number of times
 
7:19 PM
@TedShifrin did you do a grade 13 sir?
 
That's only in Canada.
 
you must've done a double major, yes?
 
In college? Almost.
 
yup
 
7:58 PM
The section of 16th street in front of the White House is now officially “Black Lives Matter Plaza”.
 
I don’t whether I will be flagged for this but I cannot even think how monstrous a man can be who will not leave another man saying “Please leave me, I cannot breathe”
 
Trumpet is nothing, why should we even hark at a man like him?
 
because he barks racism
 
Whatever happened to George was a crime against God and Mother Nature
As far as Trumpet is concerned, don’t worry coming generations and history teachers will describe him properly.
 
 
3 hours later…
11:14 PM
@Knight it's hard to simply "not care" about him when he's the one who gets to order law enforcement to drive out peaceful protestors to make a photo op possible
unfortunately
 
11:26 PM
@Semiclassical have you seen this?
possibly coming soon to multiple states across the nation
 
ugh
not all states, thankfully
but
if there's a governor who wants to show they're firmly on Trump's side?
eesh
 
Either you are with me or against me.
 

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