12:18 AM
Good Evening

1:15 AM
@TedShifrin Yeah I just bashed everything out and it popped up

1 hour later…
2:34 AM
Hi

Hi Bob

Any goods news related to the virus?

not sure, i am thinking about causes, it seems globally the standard of good living is collapsing

Yes, we are getting poorer as a planet and this is especially true in the US

in every sense and meaning of that word, yes, globally and locally that is true

2:48 AM
what we need to do is to live lower on the hog
work harder and/or spend less
I also think that we are not all that honest with our students at times
one thing I would like to see is lower standard to teach in college
that way, I might be able to get a teaching job
for example, if you are 20 years of age
and you have not had trig
you should not plan on getting a PhD in Physics

it depends, some experimentalists, like Tesla and Faraday, made much breakthroughs, and weren˙t specialzed at math at all, they knew only basic math, it depends on your views of what you consider as physics

COVID is probably going to change the course of history forever. It's hard to imagine the society pre-COVID will be getting any "good news" regarding the society post-COVID, because as far as I see everything pre-COVID will be eradicated after the event, be it standard of life, society on the whole, economy or dare I say a percentage of the generation prior to us.
In short, I very much doubt you will be getting any optimistic news any time soon.
Better get used to that

@Ante You have a good point but as I understand it to get a PhD in physics today you need a lot of math
I am thinking that we will beat COVID
we will have a vacinne
but it is going to take time

Not before 18 months.

and there are going to be a lot of deaths

2:57 AM
And the slow burn before that will be enough to change history, is what I think. Not just large scale death, but also the impact on global economy, politics, ...

I am thinking we will have a vaccine in 12 to 18 months

@BalarkaSen SARS changed our lives as Hongkongers
so probably COVID will do the same to the world
but the leaders will still be incompetent

when do you think the number of COVID cases will peak?

@LeakyNun can you shed more light on which ways? i am really ignorant about this, but shouldn't be
@Bob Peak where? We all live in different continents and countries.

either world wide or in the US

3:04 AM
I am from India, it's still growing, but we haven't hit the avalanche. In all probability we will, that's what the models say

Who knows

The more pressing issue here is that we are in a month's worth of lockdown, which may flatten the curve at the cost of the economic gamble equivalent to selling the whole country

I am looking at some US Data

Too much incomplete data

it looks horrible

3:05 AM
Oh yeah US is definitely having an avalanche alright
It crossed China and Italy by margins

The CDC is not telling the full story

And you should probably multiply the available number by 10 to get the actual number lol
so

i think the number of cases can be leisurely multiplied by, for example, 10, to have some real numbers

Lmao

@BalarkaSen we wear masks to protect ourselves / others amnesty.org/en/latest/news/2020/03/…
meanwhile Asians wearing masks in USA are attacked
2

3:07 AM
Oh right I saw people wearing masks during the protests

@BalarkaSen almost at the same time written and by the same factor multiplied, what to think

Dang, I didn't know that grew into a habit because of SARS

np pal

3:11 AM
@LeakyNun Yes; idiots. Something similar happened in Israel where the beat up an Asian decent who tested positive (in fact, he was not an immigrant/traveler, but an Israelite Jew! He was an Indian but to put it bluntly they thought he was a Chinese). That's what, I think post-COVID everyone will be extremely racist towards Asians and the political powers will gang up on China.

I do not think there will be much racist against the Asian people

It's already happening in my country and people are blaming the Chinese culture for this. It's hard to imagine this won't be tenfold the case in US! Full of white people with complete ignorance about other cultures.

it is increasing

seeing that there is racism against Mainland people in Hong Kong, I can say that there will definitely be racism against Chinese

After 911 they shot a turban wearing gas attendant

3:14 AM
I agree it is increasing but from what I have seen there is little to start and it is not increasing all that much

@BalarkaSen did you know that Sard's theorem implies a generic map has some good properties?

Seems like a Sard-ish result. Elaborate on good properties?
Morse? That'd be for maps to $\Bbb R$

nice chatting

Similar results can be stated for maps between arbitrary manifolds, but require effort to describe precisely. That's singularity theory, as far as I know.

I wish you the best of health
good night

3:17 AM
Take care, @Bob
You too

cya pal, take care

If $F: M \times P \to N$ is a submersion and $S \subseteq N$ is a submanifold then for a generic $p \in P$ the indexed map $f_p: M \to N$ is transverse to $S$ @BalarkaSen

Ah. Actually you only need $F$ to be transverse to $S$ for this result to hold, I believe.

really? the proof given to me used submersion

Let me quickly check mentally
Yeah, you only need that

3:22 AM

Suppose $F$ is transverse to $S$. $F^{-1}(S)$ is a submanifold of $M \times P$; consider the restriction of the projection $\pi : M \times P \to P$ to $F^{-1}(S)$. Whenever $a$ is a regular value of this, $\pi^{-1}(a) = M \times \{a\}$ cuts $F^{-1}(S)$ transversely. Then $F|M \times \{a\} : M \to N$ is then transverse to $S$

yeah why is $F^{-1}(S)$ a submanifold?
I guess you only need $F$ to be transverse to $S$? I don't know anything
what does implicit function theorem say

Whenever $f : X \to Y$ is a map transverse to a submanifold $Z$ of $Y$, $f^{-1}(Z)$ is a submanifold of $X$, yeah.
Yep, it follows from the implicit function theorem
For any point $z \in Z$ find chart $U$ around $z$ in $Y$ such that $U \cap Z$ is cut out of $U$ by a map $g : U \to \Bbb R^k$ as $g^{-1}(0)$, where $0$ is a regular value of $g$ - this is essentially implicit function theorem.
Then the composition $g \circ f : f^{-1}(U) \to U \to \Bbb R^k$ also has regular value $0$, so $f^{-1}(Z)$ is also locally cut out by preimage of some map to a Euclidean space transverse to $0$
The composite has regular value $0$ because of $f \pitchfork Z$

I'm sorry
I don't see how the implicit function theorem tells me that lol

3:37 AM
What is you version of the implicit function theorem
I can translate it to that if you tell me the statement

if $F: \Bbb R^m \times \Bbb R^n \to \Bbb R^n$ has invertible derivative in the second coordinate at a particular point then good stuff happens

math

@LeakyNun Okay, I will use inverse function theorem instead. Suppose $\iota : Z \to X$ is an embedding of manifolds such that $\iota(z) = x$. Then you can choose coordinate neighborhoods $(U, z_1, \cdots, z_k)$ around $z$ and $(V, x_1, \cdots, x_n)$ around $x$ such that $d\iota : T_z Z \to T_x X$ is conjugate to the inclusion $\Bbb R^k \to \Bbb R^k \times \Bbb R^{n-k} = \Bbb R^n$ as first $k$ coordinates.
Consider the map $\Phi : U \times \Bbb R^{n-k} \to \Bbb R^n$ given by $\Phi(z_1, \cdots, z_k, y_1, \cdots, y_{n-k}) = \Phi(z) + (0, \cdots, 0, y_1, \cdots, y_{n-k})$. $d\Phi_{(z, 0)}$ is the identity map now, so $\Phi$ has a local inverse by inverse function theorem.
Eh, you have to replace some thing now, and this is always hard for me to figure out. You want to replace $(V, x_1, \cdots, x_n)$ by a little neighborhood of $(z, 0)$ in $U \times \Bbb R^{n-k}$ on which $\Phi$ is invertible, and the coordinates on $U \times \Bbb R^{n-k}$ thereof.
Let's call that coordinate neighborhood $(V_1, z_1, \cdots, z_k, y_1, \cdots, y_{n-k})$, denoting the coordinates with deliberation because that's what they are on $U \times \Bbb R^{n-k}$ hence on $V_1 \subset U \times \Bbb R^{n-k}$ as well
Then $\iota : Z \to X$ is conjugate to the inclusion $(U_1, z_1, \cdots, z_k) \to (V_1, z_1, \cdots, z_k, y_1, \cdots, y_{n-k})$ where $U_1$ is small enough to fit inside $V_1$, say $U_1 = (U \times 0) \cap V$.
Aka any embedding - and we only used immersion in this proof - can be written in coordinates as the canonical embedding - immersion - of $\Bbb R^k$ in $\Bbb R^n$ in the first $k$ coordinates
Once that is done, let $Z \subset X$ be a submanifold. Choose coordinates around $z \in Z$ such that $Z$ is the subset of points of the form $(x_1, \cdots, x_k, 0, \cdots, 0)$ in the chart with coordinates $(x_1, \cdots, x_k, x_{k+1}, \cdots, x_n)$.
Then on that chart $Z$ is given by the equation $x_{k+1} = \cdots = x_n = 0$. Letting $g = (x_{k+1}, \cdots, x_n)$ with respect to those coordinates, $Z \cap U = g^{-1}(0)$, $U$ being your chart around $z$.

3:56 AM
so up till here we're basically saying, WLOG $Y = \Bbb R^n$ and $Z = \Bbb R^k$?
29 mins ago, by Balarka Sen
Whenever $f : X \to Y$ is a map transverse to a submanifold $Z$ of $Y$, $f^{-1}(Z)$ is a submanifold of $X$, yeah.

Yes, and WLOG $Z \subset Y$ is $\Bbb R^k \subset \Bbb R^n$, the first k coordinates hyperplane

Since $\Bbb R^k \subset \Bbb R^n$ is describable as preimage of $0$ by the projection map $\Bbb R^n \to \Bbb R^k$ which is transverse to $0$, you're through.

aha

"Submanifolds are locally cut out transversely by equations" is the punchline. Not true globally.

3:58 AM
thanks

No problem. You should read Guillemin-Pollack, all of this is so lucidly written there

let's say the punchline for me is "locally you can quotient by the submanifold"

and get a Euclidean space, yeah

@BalarkaSen chess?

4:22 AM
@LeakyNun Nah

Hello people

Hey @feynhat

So, we define orientation for topological manifolds using the homology groups, $H_n(M, M-x)$, and for smooth manifolds using say a non-vanishing top form (or a frame satisfying the continuity condition or an orientable atlas etc.) How do I show that for a smooth manifold, both these notions agree?
(I am suspecting we may have to invoke the de Rham isomorphism)

Nah, it's direct. Here's one way to see it: If you have a frame $(e_1, \cdots, e_n)$ at a point $x \in M$, it gives rise to a class in $H_n(M, M - x)$ as follows; consider the simplex $\Delta$ in $T_x M$ spanned by $0, e_1, \cdots, e_n$, and map it to $T_n M$ in a way that the image of the interior contains $0$ (project from a point orthogonal to the face spanned by $e_1, \cdots, e_n$ in $\Delta)$.
Equip $M$ with a background Riemannian metric and compose with the exponential map $T_n M \to M$ to get a singular simplex $\sigma : \Delta \to M$ containing $x$ in the interior, so with $\sigma(\partial \Delta) \subset M - x$.
Then $\sigma$ gives rise to an element of $H_n(M, M - x)$ by considering it as a singular $n$-cycle on $M$ relative to $M - x$
This is the correspondence between a frame and a local orientation.
Uh, the map $\Delta \to T_x M$ should just be translation by $\varepsilon$ so that the interior of the translated simplex contains $0$. I wrote some nonsense, forget that.
You can go for an abstract proof using de Rham philosophy, by showing that, in case $M$ is compact, compatible local orientation at every point is equivalent to choice of an isomorphism $H_n(M) \cong \Bbb Z$ and in the other direction a choice of a non-vanishing top form is equivalent to an isomorphism $H^n_{dR}(M) \cong \Bbb R$, and one of these isomorphisms naturally give rise to the other because of (universal coefficient theorem and) de Rham pairing being nondegenerate.
If $M$ is not compact you have to fix this appropriately by using Borel-Moore homology on one hand and de Rham cohomology with compact support on the other, and that's a bit of a mess.
There are many, many other proofs. Orientability is equivalent to not having a particular degree 2 cover, or not having an embedded $n$-dimensional Mobius strip (the unique nontrivial $B^n$-bundle over $S^1$) sitting inside your manifold, etc etc etc
You can use any x number of those things to come up with a proof interpolating topological and smooth orientation

4:54 AM
@BalarkaSen I've gotten so much better in bullet

Scary man

i'm now 1600

@BalarkaSen Thanks. Although I was expecting to work something along the lines of what you described in your second proof using de Rham philosophy (I was hoping to argue that by naturality of de Rham isomorphism choosing a generator for homology group will give me a generator for de Rham etc), I would rather go with the first one.

Question: Should I write a bit about constant curvature surfaces or not
There's a bunch of results which are worth collecting but also must be general knowledge
@feynhat Yeah, I just find the first one the cleanest.

@BalarkaSen U1700 bullet arena starting now lichess.org/tournament/AbbCRTTB

5:02 AM
Tight, @LeakyNun
You had to flag em lol

@BalarkaSen come join

too hard for me bruh

lol stalemate

Lmao nice stalemate
Rekt
Nice

yay

5:16 AM
Nasty, Leaky, nasty

dirty flags

its gonna hurt
I like how he got back his rook and blundered another one

could have done the endgame better

its all good
u played well
Ouch
Lol premove blunder

oops the firouzja technique didn't work for him

5:26 AM
lmao
so much pain in this tournament
i cant bear to watch this, im bailing out

I premoved everything because there was no time left
and I lost after the tournament was over
so my loss didn't count

lmfao

Performance 1625
Games played 11
Win rate 55%
Berserk rate 0%
Average opponent 1537
@BalarkaSen just some tournament strategies

cool

wanna play some regular chess?

5:29 AM
nah man i havent slept in like a day
gonna try now

rip
goodnight

Front for the arXiv seems to be down.

even though it's almost noon

should play eventually tho
gn

oh no 3b1b has a new video
gonna watch it later

5:30 AM
lol

so ODE

infection dynamics is the hype these days
3b1b is too much trendy math for me

its math nonetheless

maybe i should write some pop math in my blog
infection dynamics and machine learning
@LeakyNun true

sure why not

5:32 AM
currently its set to riemannian geometry mode
and i have like a draft of a spectral sequence thingy

cool

1 hour later…
6:55 AM
@LeakyNun They attack everyone, they have right to bear arms.

1 hour later…
8:12 AM

@BalarkaSen Why aren't you using MathJax or KaTeX?

2 hours later…
10:21 AM
hello

Hi

10:38 AM
how are you with confinement

11:21 AM
Why for a finite pole of f, the non zero $b_n$s are consecutive?

1 hour later…
12:46 PM
@Knight Everyone

1:03 PM
supoose i have an element of the form $(a_1, a_2, a_3)(a_4, a_5) /in S_n$ to find how many there are i can find how many 2- cycles exist how many 3 cycles and just take the product right?

only if you are sure that they all define different elements

supoose i have an element of the form $(a1,a2,a3)(a4,a5) \in$Sn to find how many there are i can find how many 2- cycles exist how many 3 cycles and just take the product right? In other words can 2 different 2 cycles and 2 different 3 cycles give the same product?

@feynhat You need plugins for that, no?
Those are not freely available

1:18 PM
if $(a_1,a_2) (b_1, b_2) (a_3 , a_4 , a_5) ( b_3 ,b_4 ,b_5)$ different then if $(a_1, a_2)(a_3, a_4, a_5)=(b_1,b_2)(b_3,b_4,b_5)$ is impossible since they are disjoint if $a3 -> a4$ then a3 and a4 as numbers must either be $b_1 , b_2$ impossible since a4 does not go to a3 etc..but apparently cant use the argument for general k cycles. consider 2x2 cycles (12)(34)=(34)(12) ..
so i ahve to count them directly
i cant break the counting into parts

1:30 PM
@BalarkaSen ah... I see. I have no experience with Wordpress.

@BalarkaSen wait then how am I seeing the rendered maths

Wordpress has a built in LaTeX system

cool

The rendering is not too great, is all. Also you have to type "dollar latex ... dollar" everytime which is annoying

1:48 PM
How one can use Markov chains to compute combinatorics problem?

2:32 PM
@BalarkaSen I am trying to grok Hatcher's notion of orientation
The one in terms of homology.
Wow those sent out of order.
So we have $H_n(R^n,R^n\setminus\{x\}) \cong \mathbb Z$ for all $x\in R^n$
So we have two equivalence classes for generators.
And a local orientation at $x$ is given by a choice of those two classes.

Mhm

The nice part is that if you have a sufficiently small ball $B$ around $x$, then we have induced isomorphisms $H_n(R^n,R^n\setminus\{x\}) \cong H_n(R^n,R^n\setminus B) \cong H_n(R^n,R^n\setminus\{y\})$ for any other $y\in B$.
And so this is why it is locally defined, right?

Well, you demand that the chosen orientation at $x$ maps to the chosen orientation at $y$ under those canonical isomorphisms.
But that's the compatibility criterion, yep

I think I can prove the following:
$$\text{lower bound} \left(\sum _{n=1}^{N} \left(\sum _{k|n} \mu (k) H_{\frac{n}{k}}-1\right)\right)<1-H_N$$
It only depends on a property of $$\varphi^{-1}(\gcd(n,k))$$ when $n=p_i-1$ where $p_i=$ $i$-th prime number, and where $\varphi^{-1}(n)=\sum_{d|n}\mu(d)*d$

Oh no, that's a painful way to do orientations

2:39 PM
I disagree, @Alessandro

Well all ways to do orientations are somewhat painful I guess

Haha that I agree with maybe.

I find orientations only "nice" in smooth manifolds. :(

That's why you don't do orientation, you just say, well, stuff flips if you loopy

But I guess this is an okay definition.

2:41 PM
Is it a known result?

This is my first time in a chat, is it for asking questions or what?

Aight, thanks

Ideally one would like to prove something like:
$$\text{lower bound} \left(\sum _{n=1}^{N} \left(\sum _{k|n} \mu (k) H_{\frac{n}{k}}-1\right)\right)= -\sqrt{N}*(H_N)^2$$

@BalarkaSen so Hatcher has this one exercise: "removing a point doesn't affect orientability". The orientable case seems "trivial" but I don't know if I am thinking about it the right way: any local orientation restricts to $M\setminus\{x\}$. But of course what if we extend this idea to something like removing a Mobius strip or something non-orientable...there should be argumentation that a single point is small enough not to obstruct orientability.

2:50 PM
@EnjoysMath Hello , did not notice you texted me sorry

@anakhro The direction that $M$ is orientable implies $M \setminus \{x\}$ is orientable is clear, yes. What about the other direction?
That's what you really need to show, and that'd answer your question since orientability of $M \setminus N$ doesn't imply orientability of $M$, if $N$ is an embedded submanifold-with-boundary.
(Like you said, take $\Bbb{RP}^2$ and take out a Mobius strip from there; you're left with a 2-ball, a very orientable object)

Suppose we have an orientation on $M-x$, when we put back x, neighborhood of x will induce an orientation at x by the excision maps, right?

Right.

Hmm, that makes sense.

why is the ordered pair (x,y) is defined as {{x} ,{x,y}}

3:01 PM
@anakhro Hi Hello Hola

Good knight @Knight

Hi knight.

@JackOhara I have just waken up

@JackOhara that's just a convenient way of doing it.

@anakhro Can you help me with some Physics Concepts ?

3:03 PM
Convenient for what purpose?

They like to define everything in terms of sets, so how would you define (x,y) as a set? Certainly not by {x,y} because this is also {y,x} and then you can't tell the difference between the two.

I see your point but I don't see why they would want to define everything in terms of sets but that might be a good thing to do

@JackOhara historically they had this idea of finding a "foundation" for mathematics.

well yeah that works I guess since (y,x) will be different

On which everything in mathematics is built.

3:07 PM
I see thanks @anakhro

Sets are a very fundamental object, so trying to build everything out of sets is basically how you can naively start doing mathematics.

@anakhro Now would you help me with Physics ?

@Knight I am not a physicist.

@anakhro But as you have more experience than me and my question would involve very much of maths I hope that you can do something for me
All it requires is will to do it

Well you can ask, but no guarantee.

3:10 PM
I’m reading Sommerfeld’s Lectures on Theoretical Physics, in the 2nd Vol (Fluid Mechanics) we have this:
At a given point the pressure acts on any surface element in the direction of its normal $i$
, it’s magnitude being the same for all directions $i$
; $$p_{11}= p_{22}= p_{33}$$

, for any three orthogonal directions. Pascal, it appears, was the first to perceive this law.

We can summarise this result in the statement that the hydrostatic pressure is a scalar quantity. It will be denoted by $p$
, A’s is usually down with omission of the subscripts that are now abundant. Note that there is no preferential loading of the horizontal cross sections if a vertical force such as gravity acts on th
I’m unable to understand things from “We can summarise this result..”

Yeah, sorry I am afraid I can't follow what it is saying either.

You need more context?

Probably not. I just don't have physical intuition.

Okay!
Anakhro I’m struggling with entrance exams, I don’t like exams but I like Maths and Physics. What should I do?

Entrance exams for what?
This passage seems to just be saying that it acts the same in any direction, thus it is a scalar (a vector would not act the same in any direction).

3:19 PM
University exams

But that's the best I got.
Entrance into university, and they are having you learn real analysis and fluid mechanics?
Fancy university.

@anakhro I’m unable to understand that vertical cross section and horizontal cross section thing

Yeah, not sure about that part.
Sorry.

@anakhro No, they want me to know trick not the concepts. But if I won’t follow them I won’t get any university
I’m about to go for UG

What is UG?

3:22 PM

What university?

Cambridge or Imperial College
Leave this, just take the example of Indian exam IIT

The IIT is horrible.
I don't know why they get students to do that.
@LeakyNun you go to Imperial, don't you?
Do they have entrance exams for Imperial?

@anakhro yeah

Do you need to know stuff like this, fluid mechanics and elementary real analysis?

3:26 PM
not really

Elementary real analysis is hard

I’m fine with deep thinking and pondering, what I hate is quickness and tricky questions

elementary real analysis is elementary and real

@Knight Some IISER's are pretty good right now, and the exams are not that competitive.

@feynhat What is IISER?
And what is that competitive ?

3:29 PM
@Knight the Imperial College mathematics admissions test seems to follow the following syllabus according to their website: maths.ox.ac.uk/system/files/attachments/syllabus_1.pdf

that would be IIT :)

Which happens to be the oxford one, too.

@Knight They are like IITs but for basic sciences.

What's that test that starts with J in India, is that JEE or something?
I had a friend who was obsessed with taking it.

@feynhat The IISERs are broadly better for math than IITs I think.

3:31 PM
@anakhro The matter is high merit and rot learning (IMO)

Unless you have IIT, Bombay in mind which is great at math but that's because TIFR faculties join there after retirement and some are adjunct, etc.
IISER, Pune has a good math faculty/environment.

@BalarkaSen Yes, some IISER have excellent math departments.

The BS-MS thing is confusing to me though
Why would you do that

This box of Nerds says share with 8 other people, but I ate it alone.

@anakhro WHAT?

3:33 PM
@BalarkaSen Its because for the first two year you study all the subjects.
So, by the end of third year, you actually haven't done many courses in your specialization.

Yeah that's terrifying haha high school was enough nightmares

This is one of the few things that NISER does better that IISER.
We only have to go through that for one year.

Ah that's much NISER

@BalarkaSen Hah!

@BalarkaSen @feynhat do any of these entrance exams deal with fluid mechanics and/or elementary real analysis?
For the Indian colleges

3:35 PM
So how do you people managed to develop a likeness for these entrance/competitive exams?

Elementary real analysis, yeah. Fluid mechanics, I dunno, I only gave math papers for the colleges I applied to
I never gave JEE or other usual entrances

@BalarkaSen Analysis, really? Did you mean calculus?

Oh maybe who knows
Like this random sequence converges and this random series diverges
Integrate this bullshit
Stuff like that

So calc II kind of stuff
Computational

Yeah right. Most of the stuff is taught without rigor in school.

3:38 PM
@anahro Listen to Caravan, "Nine Feet Underground" if you get the time

Yes and most of them requires trick not the concept’s depth

Will listen right now, @BalarkaSen !
@Knight from the questions you've been asking me, it seems you have been studying the wrong material if you are wanting to study for these exams.

its v good im relistening rn

@anakhro I’m studying those things on my own because I couldn’t develop an affinity for entrance exams.

@BalarkaSen is that a Hammond organ?

3:40 PM
@anakhro Most Indian schools do teach basic fluid mechanics in physics courses. So yeah, some entrance exams do deal with these topics. Mine did. I remember studying two or three things about hydrostatics, Bernoulli principle and viscosity.

Wikipedia says fuzztone organ, @anakhro

Old school prog, a very fine choice

the Canterbury prog scene lmao
I like their style
the album feels like im chewing strawberry bubblegum
ngl

Reminds me of Edgar Winters
@feynhat on the level of using multivariable calculus and tensor analysis?

No not tensors

3:47 PM
Someone could argue on the possibility of constructing a reducible spinor tensor field from tensor product of three irreducible tensors?

@anakhro No. Nothing like that. Just one-variable calculus. (atleast I don't remember doing those stuff)

The so-called Wigner-Eckart theorem is the playing card.

@feynhat the book Knight is reading takes tensor analysis and multivariable calculus as pre-reqs.

I face a confusion about that, since the lemma states

$F_{\kappa}^{(k)}=\int T_{\kappa}^{(k)}\left(\mathbf{r}_{1}, \mathbf{r}_{2}, \ldots\right) d \tau_{1} d \tau_{2} \ldots=0 \quad$ if $\quad k \neq 0$

I’m doing it because I really want to study Physics, I don’t want just the reputed university tag

3:50 PM

here $T$ is a physical scalar field of $k \neq 0$

@BalarkaSen 10:50 part is good.

Yeah that transition is great

Do you like Can, Balarka?

What I understood so far is the above lemma is telling us we cannot construct a reducible tensor if the condering scalar tensor is of degree different from zero
So, does the nature of integrand is independent of $F$? i.e., whether it is a scalar, spinor or higher rank tensor field.

3:53 PM
@anakhro Can you help me with entrance exams?

I can't guarantee my advice will be helpful on entrance exams, especially since I have never taken entrance exams.

@anakhro The Canterbury scene? Sorta, yeah

@BalarkaSen no, the krautrock band, Can.

Oh, never heard.

They are pretty good.

3:55 PM
I'll try it out
The closest I have gotten to krautrock is the Finnish band Circle

@anakhro we have to solve 30 questions in 1 hour. I just hate it

Speaking of krautmusik, what happened to Lindemann?
I heard he got the virus.

@Knight find something fun in it, then.

But now I am reading other articles about it being just a fever.

@anakhro I have failed in it thrice

3:58 PM
@Knight why did you fail the three times?

Off-topic: I forgot the hyperboloid model for $\Bbb H^2$. I think it's the positive sheet of the hyperboloid $x^2 - y^2 - z^2 = 1$ where the geodesics are what you get if you cut by planes passing through origin.

(...when math is off-topic)

Lol, you never know.

@anakhro Because I couldn’t qualify the cut off marks

@Knight so what was your issue? Did you do all the questions?

3:59 PM
Math classes for the moment youtube.com/watch?v=BSxM6_zEw2k
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