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15:00
I think now you've understood?
@AbhasKumarSinha good :) I want to study the variation of $g_{\mu,u}$ and show that $g_{\mu, u}=\eta$
Take example, $6! \geq 4!$, so, $6!$ is divisible by $4!$
@geocalc33 bery long calculation, (I'm studying that part), so, I'd skip that :P
@AbhasKumarSinha What should I start with then??
@AbhasKumarSinha Yes, but $(n!)^{(n-1)!}$ is not of form as 4! or any other thing
@Knight Do you know induction method?
@geocalc33 Lagrangian in 4D Space, then, minimize it ;)
15:02
@AbhasKumarSinha I tried that, you cannot do (I mean I couldn’t do)
By induction
@Knight can you show that for $n=1$?
@AbhasKumarSinha Yeah
@AbhasKumarSinha I work in 2D spacetime for now haha :) but I'll try that out :)
@Knight $$ (n!)! > (n!)^{n-1}!$$
awkei?
@AbhasKumarSinha Please put that factorial on RHS at top
15:05
@Knight see his question carefully, he has factorial on the whole expression, not on power.
@Knight okay, I see his brackets, my fault, sorry,.....
Okay, to prove $$ (n!)! \, | \, (n!)^{(n-1)!} $$
Is that right?^
@Knight If that is the right question, then gimme a minute,
Take a minute, well you told me that you’re studying GR then how can you be so mundane with time, :-)
@AbhasKumarSinha Do you got something?
There's an obvious copy of $\underbrace{S_n\times...\times S_n}_{(n-1)!\text{ times}}$ in $S_{n!}$, so $(n!)^{(n-1)!}\vert(n!)!$ by Lagrange.
Trying to get something out of nothing.
@Thorgott What is Lagrange?
Lagrange's theorem, in the mathematics of group theory, states that for any finite group G, the order (number of elements) of every subgroup H of G divides the order of G. The theorem is named after Joseph-Louis Lagrange. == Proof == This can be shown using the concept of left cosets of H in G. The left cosets are the equivalence classes of a certain equivalence relation on G and therefore form a partition of G. Specifically, x and y in G are related if and only if there exists h in H such that x = yh. If we can show that all cosets of H have the same number of elements, then each coset of H has...
15:18
I thought it was a normal high school problem, but it came out to be a Group Theory problem
Wow!
there's a group theory solution
doesn't mean there isn't an elementary one
@Knight If it can be solved with LT, then it's also a high school problem
@Semiclassical It can be,
i'm also not sure how 'obvious' it is that $S_n^{\times (n-1)!}$ is a subgroup of $S_{n!}$
@AbhasKumarSinha But I think if it can be solved by elementary methods then it must involve some tricks
15:20
@Knight that's what I think
@AbhasKumarSinha Please add “also” coz I first thought of that LOL
Yeah, I'm pretty sure there's an elementary solution, though I don't immediately see it and I don't feel like thinking about it since I'm very content with thinking about combinatorics in terms of gropu theory :p
@Thorgott I never find combinatorics intuitional
please don't
15:22
@Semiclassical $\{1,...,n!\}$ can be partitioned in $(n-1)!$ sets with $n$ elements each, let a copy of $S_n$ operate on each of these via permutation.
@AbhasKumarSinha Where? Which message?
no nothing
(tempted to star the "please don't")
43 secs ago, by Semiclassical
please don't
15:23
please don't
hahahahah XD
@Thorgott nice
Should I follow it or star it?
No, just delete back all messages related to it
A similar argument shows that, if $m_1+...+m_k=n$, then $m_1!\cdot...\cdot m_k!\vert n!$
And this is just saying that multinomial coefficients are integers
So you should be able to infer an elementary solution from that angle
right
probably one could convert the group theory argument into some kind of counting solution
e.g. the old "count a committee in two different ways" notion
15:26
@Semiclassical Do you come here before going to your university or do you help me from the university?
from the university
@Knight He's a very smart teacher of Classical Physics.
Wow! Educational institutions always require us to reach there early in morning
@AbhasKumarSinha Of mathematics too!
It's 10:30 am here
@Semiclassical China or Japan?
15:28
USA
@Semiclassical here too
lol
I calculated it backwards
hehehe XD
@Semiclassical Hello people, I've come from the future to warn you
16:29 here
@Ante May I ask something about your name, if you don’t mind?
15:29
I guess a counting approach would start with (n!)!, which you could do as follows: Start with n students, and consider all possible ways to enumerate them. Then consider all the ways to prepare a list of those enumerations.
@Knight You can try.
hmm. i guess the real issue is to find a counting interpretation to $(n!)^{(n-1)!}$
@Ante How to pronounce your name? Is it same as Aunty or is it like Aunt ?
@Semiclassical Yes! Bcoz it’s not a factorial
@Semiclassical (I've not checked my approach), see can I say that way $(n!)^{(n-1)!}$ is divisible by $(n!)^{n!}$ which is divisible my $(n!)^n!$ which is divisible by $(n!)!$
15:32
@AbhasKumarSinha Explain your first chain
@Knight for any $a^{(b-n)}$ is divisble for $a^b$
along thorgott's approach: you can classify each enumeration of students according to which student is listed first. that gives n classes of enumerations, each containing (n-1)! lists
@Knight It is almost the same as is "ante" pronounced in the word "antenna". Practically, the same.
@AbhasKumarSinha Explain it further
@Ante OKAY! Now, I got it! Is it your real name? Coz it’s a great name
15:34
meh. i don't have the brainpower right now to think more about it
@Knight Yes, real name.
@Ante Wow! Are you a student or a instructor or a researcher ?
@Knight $$ \dfrac{a^b}{a^{(b-n)}} = \dfrac{a \times a \times a \dots \times a}{a \times a \times a \dots \times a} $$ (numerator has more $a$ than denominator)
@Semiclassical What's wrong with my approach?
@AbhasKumarSinha Now see the wordings of what you have written above
@Knight Amateur. Not on a college and not affiliated to some of them. Doing research in much of my free time.
15:36
@Knight ok
@AbhasKumarSinha hey ! But what about the factorials on the exponents
@Ante That’s really nice!
@Knight one has more value than other, so, the more one would divide the lesser one.
@Knight one factorial in the exponent is less than the second fatorial in the exponent
@AbhasKumarSinha Yes got ya there
now Explain your arguments
Please remember $a/b$ means b divides a
@Knight no, it's $a | b$
What are you trying to prove here, some divisibility of factorials with factorials?
15:40
$$ (n!)^{(n)!} | (n!)^{((n-1)!)}$$
It's strange. Take a line and slightly tilt it and suddenly it has a totally different meaning
47 mins ago, by Knight
7 hours ago, by Amresh Prasad Sinha
this was my question :- Prove that (n!)! is divisible by (n!)^{(n-1)!}
9 mins ago, by Abhas Kumar Sinha
@Semiclassical (I've not checked my approach), see can I say that way $(n!)^{(n-1)!}$ is divisible by $(n!)^{n!}$ which is divisible my $(n!)^n!$ which is divisible by $(n!)!$
@AbhasKumarSinha Abhas my brother, $4/2$ means 2 divides 4
@Knight sorry, IK
15:42
@AbhasKumarSinha All right Whats after that
@Knight $$ (n!)^n! | (n!)^{(n!)}$$
In algebra, number theory, and other fields, people will often write $a\mid b$ to say "a divides b." But you appear to be writing it as a fraction.
@AbhasKumarSinha How?
@Rithaniel He's probably on mobile
@AbhasKumarSinha He is saying to you not to me
15:44
A cheap approach:
$$\frac{(n!)!}{(n!)^{(n-1)!}}=\frac{(n!)!(n!-n)!\cdot...\cdot(n!-((n-1)!-1)n)!}{n!(n!-n)!n!(n!-2n)!\cdot...\cdot n!(n!-(n-1)!\cdot n)}={n!\choose n}{n!-n\choose n}\cdot...\cdot{n!-((n-1)!-1)n\choose n}$$
Nah, the $b/a$ is strange to me after seeing $a\mid b$ so often. Hence why I mention it
in terms of multinomial coefficients: $$\binom{n!}{\underbrace{n,n,\cdots,n}_{n-1}}$$ is an integer
@Knight I'm out of my brain power now, need recharging
15:46
@Semiclassical Okay
@AbhasKumarSinha Drink something
@Knight koay
because $n(n-1)\leq n!$
Okay
Looks clickbaity to me
@Astyx confirm karne do
15:50
Let’s do something else
HAHA XDXDXD
Hey everyone. I'm wondering if there's a name for a certain probability distribution.
I'm modeling the stock market, and I'm assuming that each day, the stock price has a p_1 chance of going up g_1, a p_2 chance of going up g_2, ..., a p_n chance of going up g_n.
First, what's this distribution called? It's a real-valued categorical distribution... so is it just called exactly that?
Second, what do you call the distribution of the sum of t i.i.d. random variables with that distribution?
have to go to save the world....
bye :)
We can also make it work like this. Fix naturals $m_1,...,m_k$ such that $m_1+...+m_k=n$ and let $M$ be a set with $n$ elements. Let $\Omega=\{(A_1,...,A_k)\colon\text{partition of $M$},|A_i|=m_i,\,i=1,...,n\}$. Then $|\Omega|$ is the integer we are looking for and we need to show it's the multinomial coefficient ${n\choose m_1,...,m_k}$. Now let $\Omega^{\prime}=\{(w_1,...,w_n)\in M^n\colon w_i\neq w_j\text{ for }i\neq j\}$. Clearly, $|\Omega^{\prime}|=n!$.
Each element of $\Omega^{\prime}$, i.e. each ordering of $M$, induces an element of $\Omega$, i.e. a partition of $M$, by taking $A_1$
The first distribution is similar to a Bernoulli distribution, and the second is similar to a binomial distribution.
15:59
I am reading a book that says "In classical Euclidean geometry all points are the same. There is no distinguished point. The whole of the space is homogeneous.".
Can someone explain this statement? What do the authors mean by "all points are the same"?
In mathematics, a period is a number that can be expressed as an integral of an algebraic function over an algebraic domain. Sums and products of periods remain periods, so the periods form a ring. Maxim Kontsevich and Don Zagier (2001) gave a survey of periods and introduced some conjectures about them. == Definition == A real number is called a period if it is the difference of volumes of regions of Euclidean space given by polynomial inequalities with rational coefficients. More generally a complex number is called a period if its real and imaginary parts are periods. Periods are numbers that...
Hmm... so ring of periods is like an integral version of numbers that has a finite expansion under some base b
@nbro Well, geometers nowadays often assume that 2-dimensional space has one particular point which is the "middle", and one particular direction that is "up".
where now instead of summing over countably many terms, one integrate over intervals and that the base b can take funnier forms such as algebraic functions
Some neighborhood of some point is the same (locally) as exactly the same neighborhood of every other point, I think that is the meaning. @nbro
The author there is saying that in classical Euclidean geometry, you don't assume that there's a "middle" or an "up".
16:02
To add some context, this is a computer vision book and they have just introduced homogenous coordinates...
I am familiar with homogenous coordinates and that all points that differ by a multiplicative factor in HCs are the same point in the corresponding Euclidean space, but that doesn't seem to explain the statement above
16:29
@Knight and @AbhasKumarSinha my question was a IIT JEE level question.... I got doubt in it so I asked
16:44
Is there anyone with a PHD can give a personal third alternative solution on 2019 IMO problem 6 under 25 minutes imo2019.uk/wp-content/uploads/2018/07/solutions-r856.pdf
Can someone solve this?
0
Q: Can a nowhere differentiable Jordan curve pass only through points with both irrational coordinates?

AnteSuppose that $J$ is Jordan curve with the following properties: 1) $J$ is nowhere differentiable, that is, the injective continuous map $\phi$ mapping $[0,1]$ into $\mathbb R^2$ such that $\phi(0)=\phi(1)$ and the restriction of $\phi$ to $[0,1)$ is injective is nowhere differentiable. 2) $J$ p...

https://www.universiteitleiden.nl/binaries/content/assets/science/mi/scripties/bachelor/2017-2018/barinaga-bsc-scriptie.pdf
https://en.wikipedia.org/wiki/Vitali_set
@nbro The group of isometries (translations + rotations + reflections, if you like) takes any point to any other point. This is what it means for the space to be homogeneous. I have no idea what Terran's middle and up are. But I'm a differential geometer, so I suppose I shouldn't.
@Ante the only Jordan curves with property 2. are constant, so they're necessarily differentiable
16:57
Ok finally figured out the intuition: A nonmeasurable set is nonmeasurable (and why the inner and outer measures always mismatched in such cases) is because when trying to cover these sets with a sigma algebra, the set's extremely complicated and uncountably numerous point like structure lead to e.g. way too many sets to be used to cover it, leading to an overestimation, and e.g. way not enough sets to be used to lower bound them
@Thorgott Why they need to be constant?
It's like a set is nonmeasurable when trying to place it into a shrink wrap, and there are always gaps in between
whereas a measurable set, the shrink wrap always encase the set completely, so there are no gaps
The continuous image of a connected set is connected
The intuition came from this last step of the proof of the skerepiski construction: Lemma 1 showed exactly how the interval that tries to measure it blow up without a bound, and conclude with that A is measured to be 1
and that is because of those extra points distributed in complicated ways in A
And how to prove that $\mathbb I \times \mathbb I$ is disconnected?
17:02
@Ante It's missing entire lines of the form $(x,q)$ for a fixed $q\in\Bbb Q$ and $x\in\Bbb I$ varying
Hi, demonic @Alessandro
It's totally disconnected even, hence the image must be a singleton and the map constant
@AlessandroCodenotti Hm, right, that´s true.
Is $\mathbb R \times \mathbb R \setminus \mathbb Q \times \mathbb Q$ disconnected?
No, I did not mean to ask that, I mean if we take some point $A \in \mathbb R \times \mathbb R \setminus \mathbb Q \times \mathbb Q$ and a point $B \in \mathbb R \times \mathbb R \setminus \mathbb Q \times \mathbb Q$ can we continuously arrive from A to B?
yes
17:16
@Ante No, it's connected
Fun fact, there's circles contained in it
@AmreshPrasadSinha Where did you find it?
if $A = (a_1, a_2)$ and $B = (b_1, b_2)$, you have the maps $t\mapsto (ta_1 + b_1(1-t), a_2)$ and $t\mapsto (b_1, ta_2 + (1-t)b_2)$
for $t\in [0,1]$
@AmreshPrasadSinha Is it in some book or paper? Give me screen shot of the question
Is $\mathbb R \times \mathbb R \setminus \mathbb A \times \mathbb A$ also connected?
What's $\Bbb A$ ?
17:20
The set of all real algebraic numbers.
Yes
With the exact same maps I used above
:D
Salut, @Astyx
Wait no those maps don't always work
I'm saying nonsense
Salut
I think that if $X$ is everywhere disconnected and countably infinite then $\mathbb R^2 \setminus X^2$ has the property that for every pair of points $(A,B) \in \mathbb R^2 \times \mathbb R^2$ there is a continuous path from $A$ to $B$.
We could, if true, lift this up to $\mathbb R^n \setminus X^n$ somehow, I think.
17:29
But my maps almost work no ?
Sometimes, yes.
@Ante Given two countable dense subsets $A$ and $B$ of $\Bbb R^n$ there is always an homeomorphism of $\Bbb R^n$ mapping $A$ to $B$ bijectively, so you only need to consider $X=\Bbb Q^n$
With them you can connect any point from $X\times (\Bbb R\setminus X)$ and $ (\Bbb R\setminus X) \times X$ to a point $(\Bbb R\setminus X)\times (\Bbb R\setminus X)$
and with my argument you can connect any two points $(\Bbb R\setminus X)\times (\Bbb R\setminus X)$
@AlessandroCodenotti Does that theorem work also for $\mathbb R^{+ \infty} \setminus \mathbb Q^{+ \infty}$?
17:35
Doesn't my argument work ?
@Astyx I didn´t check all your argumentation but if your maps are good for $X^n$ then they are good also for $X^{+ \infty}$, is that what you mean?
Yes kinda
You only need two coordinates
17:53
@Knight it from Cengage algebra mathematics book
@nbro Yeah, I think the two meanings of "homogeneous" are unrelated.
18:50
@Semiclassical yea, the reason I was asking is because I'm trying to understand the Lorentz group better
Hi Sha, long time no see
wews, hello Astyx!
How have you been ?
19:12
Howdy @Sha
sorrysorry, my mum called, and she was freaking about about the Corona virus, and so I completely forgot what I was doing haha
but hello @Ted!
@Astyx I'm doing fine
though I'm a bit worried about the near future, as there's a realistic chance our uni is going to go full online :(
Yes, I am freaking, too. I'm rather vulnerable and have a cold/sinus mess now.
oh no:0 are you staying inside?
Mostly, yes.
rights, okay
19:14
Lots of universities in the US are going to on-line now. Faculty are freaking out (as I would be).
I'm hoping you have some friends/familiy members in the vicinity who can support you, in case that's needed
Yeah, I'll be OK, thanks.
yea, our computer science courses have gone full online as well, I'm wondering how long it will take when the same will happen to the math and physics department
Lots of my old friends are wrestling with lecture delivery on-line and — ha — testing.
I can image:p I'm somewhat lucky that next week is my last TA session for this year, so I won't miss on TA-ing (due to corona)
19:17
@TedShifrin yeah, we're really scratching our heads about the latter
right off the bat, the groupwork portion of our quizzes are out
but you also really can't do online short answer because it's so easy to cheat
right now we're mulling over online multiple choice?
I'm not a fan of that, but at least it's something
@ShaVuklia gotcha. in the Lorentz group case, the four components come down to whether a given O(3,1) element is symmetric or antisymmetric with respect to time reversal and to space inversion
I'm going to miss exams
this part of the lorentz transformation wiki page may be useful: en.wikipedia.org/wiki/…
@Semiclassical ah, I think I might see it for the determinant, as we can at least split O(3,1) in two parts using the determinant function
which I think corresponds with space inversion or not
see in particular that 2-by-2 table
19:22
@Astyx that's really a shame btw
are they going to reschedule them?
@Semiclassical yes, I know that table. I think I only just now realised that I could use this homomorphism to make topological remarks as well
They have to
Probably in September
Or we'll have to take the exams online
I don't know what that would look like
hm I see
main point is that there's no way to go smoothly from the identity transformation to the parity inversion transformation
whereas you can go smoothly from the identity transformation to a boost
to a particular boost, right?
19:25
right
an infinitesimal boost makes sense, whereas an "infinitesimal parity inversion" is nonsense
the proper and orthochrone ones
@TedShifrin lol, imagine the students with classes that are 100% final exam
You mean the European system, @topologicalorientablesurface?
now i'm forgetting, tho, whether there's such a thing as an improper boost. comes down to a matter of definition i suppose
19:27
@TedShifrin yeah
hm right. also, I was wondering about one thing
they talk about two homomorphisms
the determinant function
i guess improper boosts should be a thing, if only because proper/improper rotations are
"They" just deleted my answer to an elementary word problem.
and the one which sends equivalence classes of spacelike events to the same equivalente classes (or not)
would there be an explicit form of this one as well? which doesn't depend on coordinates
When someone posts a correct solution and one only has suggestions to improve it, I guess we can't say "Yes, this is correct. Here are some suggestions to improve it." What asses.
19:28
@Semiclassical I would say so yes
Guys, if anyone could read through my answer and perhaps give feedback on what I wrote here: math.stackexchange.com/questions/3576245/… Id much appreciate it
@ShaVuklia so a coordinate-independent map, in the same way that the determinant is coordinate independent?
yes
I have something that comes close to it (maybe even 100%), let me find it
For the lorentz group the matrices are caracterised by wether the determinant is 1 or -1 and wether the (0,0) component is greater than 1 or less than -1
oh right, well what Astyx says
19:30
right
I don't know why I was against that.. I think because my teacher's notes were written poorly
I guess that's an argument you can carry to O(p,q)
yea, I think so too
the determinant carries over, yes
I'm not sure how "the (0,0) component is positive/negative" carries over
maybe something like positive/negative definite?
but that's a guess
also, I can see how the (0,0) component being strictly positive means that causality is preserved, but do any of you happen to know what is preserved when det = 1? would it be space orientation?
19:33
Maybe the determinant of the top left $p\times p$ block ?
@TedShifrin is there an uprising?
I'm not sure
Sorry, @topologicalorientablesurface. I refuse to read something that long for what should be a one-line proof.
my guess is that it means that space inversion and time reversal have the same parity
@Sha: You mean space orientation assuming the $(0,0)$ component is positive?
19:34
i.e. it's testing whether doing both preserves orientation
Right. They can both reverse orientation.
@anakhro: What uprising?
@TedShifrin yes, I think under that condition indeed
Deleting your answers.
@Sha, then yes, that's right.
I shouldn't bother dealing with elementary questions, @anakhro, but I thought I would offer constructive suggestions on how to write up word problems rather than just a pile of formulas.
as an example, take a playing card. if you flip it horizontal or vertical, you don't get back what you started with. but if you do both, you do
19:35
Sometimes I write comments which are answers, but then questions sit around un-answered.
@Semiclassic: Indeed, any rotation of the plane is a composition of two reflections.
(I'm pretending for a moment that playing cards are unchanged if you rotate by 180 degrees. that's fine for face cards but not literally true otherwise)
right yea
the thing is, I am trying to motivate physically why we would only look at the proper orthochrone Lorentz group (along with translation of course). the idea in my mind is that we are firstly asking for two invariants: we want the spacetime interval to be invariant (which leads O(3,1) and translations), and we want causality to be preserved, which means we can only deal with the orthochrone Lorentzgroup (and dilatations and translations).
however, if we also ask for orientation of space, then we end up with just the proper orthochrone Lorentzgroup (and translations). and in some way that could than be considered a physical motivation to study this object
instead of the whole thing
and I'm speaking then in the context of studying irreducible representations
my teacher's answer was "we don't throw away anything, it's just that if we know how the proper orthochrone part works, then we know all the other things as well"
through the factor group (factoring O(3,1) through the proper orthochrone lorentzgroup), but I wasn't entiiiirely convinced that it was just a mathematical shortcut. but maybe it is.
@TedShifrin I thought maybe including different ways of thinking about a particular problem could be helpful, for both, myself and the reader.
(sorry for spam)
@topologicalorientablesurface: Nothing against you. I just don't have the patience to read 5 pages when one line is all that's needed to be clear.
19:42
@TedShifrin oh alright. haha
It's just definition of subspace topology. Nothing more.
@ShaVuklia i take the point there to be that, if you have an element of the improper orthochrone component, and act by the space inversion transformation, then what you get is in the proper orthochrone group
and similarly for the proper antichrone / improper antichrone
that is fair, but aren't transformations that for instance don't preserve causality in some sense undesirable?
Depends what your purpose is, I suppose
19:48
(a^b)%m = x, i know a, m and x. How to find b?
What do you mean by causality ? @Sha
For field theory purposes (which care a lot about discrete symmetries) I can see the argument for talking about the full Lorentz group
@Astyx you can have a partial order on space time, that is
@kayush: I don't understand as you typed it. You have some percentage of $m$ that is $x$?
each event that is time like (that is, the invariant interval is greater than zero)
19:49
Hi everyone
a quotient b, maybe?
can be compared to another timelike event, by looking at the time component
% = mod
Hi @Alessandro
LOL, that's not well known.
Hi @Alessandro
19:50
and so preserving causality means that this partial order is preserved
Ha, ok
Anyone familiar with reduced and/or divisible modules?
orr... didn't I answer your question (edit; it seems I did)
is there some special cases where we can find b
Well, it won't ever be unique.
You have huge numbers, I assume?
19:51
i have one other relation as well
yes
Yeah, computational number theory people know the answer to this, but I don't.
no probs, thanks for helping
Have you googled modular logarithm?
I'm sure there are posts about it on MSE.
@astyx suppose you set up a flow like $dx/d\tau=Mx$ where $x$ is some 4-vector. If M is orthochrone, then the 0-component of x will increase with the flow parameter tau
didn't knew about those, i will check
19:54
If it’s antichrone, then it’ll decrease with tau instead
I am not crazy, and $Hom(\mathbb Z,\mathbb Z/6\mathbb Z) \cong \mathbb Z/6\mathbb Z$?
In more physics terms, M being orthochrone means that you only expect to see time dilation if you boost and not time-reversal
when you talk about a flow, do you mean that in the context of diffgeo?
It can probably be posed that way, yeah. All I really mean is some first-order system tho
@anakhro In which category is this Hom taken?
20:00
Groups. I have concluded I am not crazy.
It's true that Hom(Z,G) is G.
(Issues of causality are necessarily a big subtle, alas. In special relativity, you can very well have it be the case that a question like “did event A or B happen first” has different answers in different reference frames)
I was asking because in rings $\Bbb Z$ is the initial object
(No such thing as absolute simultaneity)
20:02
@AlessandroCodenotti unital rings. ;)
On the other hand, “event A is in the future light cone of B” is a frame-independent statement
@anakhro This is true by the universal property of free groups. A free group $F(X)$ on a set $X$ is such that for every group $G$ and every function $f\colon X\to G$ there is a group hom $F(X)\to G$ extending $f$. $\Bbb Z$ is free over a one element set
@anakhro The only rings you mean
Can’t talk about causal order for space-like pairs of events, but just fine to do so for time-like pairs
@AlessandroCodenotti only for the modern folk
To be fair in the first algebra course I took in my undergrad rings weren't assumed to be unital
20:06
@Semiclassical right ye. but didn't we just say that it's only the orthochrone Lorentz transformations that preserve causality?
wouldn't the boost with negative (0,0) component still reverse the order?
Yeah, I’m slipping back into my usual “boosts are proper” carelessness
ah okays
And, tbf, if you write down the boost you’d see in a physics book
It’s always proper
As are rotations, for that matter—ye olde ((cos,sin),(-sin,cos)) example
20:29
sorry, got distracted, as I just heard the news that there is not going to be any physical classes anymore throughout all faculties
how fragile we are as a society:(
21:13
@ShaVuklia same on our campus for at least the rest of the month
And that’s a very tentative “at least”
I'm quite grateful that I'm into theoretical things
at least I can continue doing what I want at home from behind my computer, or just using pen and paper
though times have become quite uncertain, but o well, that's life
ABC
ABC
21:51
I have $\int_C \frac{z}{(sin z)^2(1-cos z)}dz$ complex integral on Circle with radius 5

I know that $sin(z)=\frac{z-z^{-1}}{2i}$ and $cos(z)=\frac{z+z^{-1}}{2}$
Then I know that $z=5*e^{it}$

So I get $\int_0^{2\pi}{ i*e^{it} \frac{5e^{it}}{(\frac{e^{it}-e^{-it}}{2i})^2(1-\frac{e^{it}+e^{-it}}{2})}}$

Then I do this substitution $u=e^{it}$, $du=ie^{it}$ ... I get a complex integral around a circle with radius 1.

Now I use residue but I get a wrong result.


I know that I can appy Residue THM in the first integral, but I don't understand where is the error here.
Why if I use Residue THM in the newer integral I get a wrong result?

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