@FrancescoS (1) You never know which flavor state produced the jets - in other words, it's impossible to say whether a jet originated from an up quark or a down quark or so on. In practice, when doing calculations we sum over all possible flavors.

(2) You certainly would take those processes into account if you are making a prediction using a model in which they are not suppressed.

@ACuriousMind I can't believe you forgot about @DavidZ :P

Wouldn't different flavor jets have different probabilities?

Can't you at least guess at the original

Sure, e.g. you're more likely to get an up quark producing a jet than a charm quark because there are more up quarks (to put it classically). Gluons are the most likely progenitors in many cases simply because there are so many of them.

@DavidZ thank you. @DavidZ, @ACuriousMind Another question: If I impose a flavour symmetry $G_F = U(3)_Q\times U(3)_d \times U(2)_u$ for the three-handed quarks $q_L$ the three RH-quarks $d_R$, the authors of the paper I cited above say that, for example, the operator $(\bar{q}_L\gamma^\mu q_L)^2$ doesn't contribute to $uu\rightarrow ss$. Why? The operator $\bar{q}_L\gamma^\mu q_L$ include $\bar{u}_L\gamma^\mu u_L$ and $\bar{s}_L\gamma^\mu s_L$, so, squaring I expect to get this process

But the jets don't retain any sign of what type of particle produced them.

@FrancescoS maybe the amplitude is zero?

@DavidZ Is it? Why? :O

I dunno, but you should probably do the calculation and find out :-P

I don't think I should do the calculation...

haha

I've said that before

Well, I'm not going to do it, I'm busy. It's up to you. There are various reasons a particular operator might not contribute to the amplitude of an interaction, e.g. it could be spin or some color thing, and things like this are generally not obvious.

@DavidZ Ok, but I don't see why the flavour symmetry should forbid this process. I mean... the operator $(\bar{u}_L\gamma^\mu u_L)(\bar{s}_L\gamma^\mu s_L) \subset (\bar{q}_L\gamma^\mu q_L)^2 $ and there is no reason because the amplitude of this vertex is zero. I think I am not understanding the notation and I am doing a mistake ;)

oh lordie that $\subset$ is a delicious abuse of notation

@bolbteppa currently failing at linear algebra :(

@FrancescoS in that case you should look up (or ask) what the notation means so that you can stop misunderstanding it ;-)

@0celo7 Read Axler

@DavidZ Yes, that's the point... In fact, I am asking ;) ahh

@bolbteppa No I just need one theorem

If a matrix $M$ is bijective can I change the basis to make it the identity matrix?

@0celo7 No.

If you're talking about the $GL$ thing above then you're doing Lie groups since you're mixing topology and algebra in a continuous fashion

@bolbteppa I'm not

I've accepted that as a black box and referenced the appropriate page in Lee in my report

@ACuriousMind :(

If $L:\Bbb R^k\to\Bbb R^l$ ($k<l$) is injective, can I change basis in $\Bbb R^l$ so that it has the matrix $(\frac{I_k}{0})$?

Where $I_k$ is the $k\times k$ identity matrix

Have you considered simple examples? How would you do that for the 1x1 matrix "2"?

be nice

erm, I would multiply the basis vector in $\Bbb R$ by $2$

That doesn't change anything - it's still the map that's "multiplication by 2", scaling the basis vector doesn't change the matrix at all.

yes it does

@0celo7 Let's see the transformation, then

@ACuriousMind Hmm, well, I'm not considering a matrix. I'm first considering a linear transformation and then the matrix of the transformation

That doesn't change anything

The map "multiplication by 2" is multiplication by 2 in every basis, you can make its matrix the identity matrix.

@ACuriousMind let me walk home and think on this

That's not a pathology of the one dimension - you can't make the matrix $n I_k$ in $k$ dimensions anything else because it commutes with all other matrices so no basis change has any effect on it.

Is there more than one way to permute a set $\{1,2,3\}$ by $(1~2~3)$ ?

not taking order into consideration

@Obliv what? $(1 2 3)$ is cycle notation for a specific permutation, what's your question?

what?

Well I'm observing the group action $S_n$ onto a subset of elements of $n$ with size $k \leq |n|$ and am trying to determine for what $k$ is this action faithful

I feel like it's faithful for all k

faithful means injective, right?

Yes. So how do you think that action is faithful for $k=1$?

It isn't, because there are multiple 1-tuples that permute k=1 subsets. so for all $1<k\leq n$?

in the solution manual it says if $|n|>1$ then $k<|n|$ not $\leq$

@ACuriousMind do you know the answer to the my question? :)

@Obliv Wait, what exactly do you mean by the "group action onto a subset of elements of size $k$"?

@FrancescoS No, else I would have told you

i have to switch to my phone brb

Ok I have walked home

Lost about 5 lbs in sweat @Slereah

@bolbteppa What's the standard graduate level linear algebra book

@0celo7 Axler is the first adult LA book, after that do LA as part of an abstract algebra book

Ok @ACuriousMind we will come to the bottom of this linear algebra trickery

Lang is a good 2nd resource

You're a crafty bastard but I've got God on my side

@acuriousmind the group $ S_n$ acts on the subset of n that is size $k \leq |n|$ so it permutes it

@Obliv wait why $|n|$

are you defining it for $n$ negative

0celo7 how hot is it in Tennessee for you to sweat in a 10 min walk

@Obliv I sweat sitting in my room

Are you in Weinberg's uni

No that's the size of the set , no?

@Obliv how could it be negative then

you only put $|n|$ if you're worried about it being negative

@bolbteppa no

That denotes the size of the set not abs value Bars

...that's shit notation

@Obliv I don't know what that's supposed to mean. Give me an example, let's say n=3,k=2. What is the action and what is the set it acts upon?

You should have $S_X$ as the permutation group of the set $X$

and this is isomorphic to $S_n$

where $S_n$ means the permutation group of $\{1,\dotsc,n\}$.

So you don't put $|n|$

So it seems like in the inverse function theorem you start from an open set and find a smaller open set which maps diffeomorphically onto the image, tubular neighborhoods arise by starting from the smaller open set and reconstruct that larger open set, if you believe Do Carmo

Because $n$ is not a set

@bolbteppa erm do Carmo does not talk about tubular neighborhoods

Page 109 on he does, also an exercise about a tubular surface using the Frenet frame on page 89

So {1,2,3} has permutations (12) (13)(23) and those permutes all subsets of {1,2,3} with 2 elements so like (12) , {1,2} = {1,2} and (12) , {2,3} ={ 1,3}

More linked to the implicit function theorem

@ACuriousMind Ok I'm trying to unravel a black box theorem from bundle theory. For this I need the theorem on page 15 of Guillemin-Pollack

But you're telling me they made a mistake

Or am I missing something crucial?

@bolbteppa wait which do Carmo

Diff Geom

Oh

What is dotsc? I can't read tex on my phone

I don't have that one

@Obliv ...

@Obliv oh I forgot to say that $X$ is finite and $|X|=n$

but $n$ is not a set

Wait my example is wrong if the subset doesn't contain numbers in the cycle perm. It can't be permuted

@0celo7 Ahhh, I didn't get that we are only choosing the basis of the target, but not of the source.

Yes it is. $S_n$is the permutation group of the set n

@0celo7

Then you get that basis rather easily by just choosing the image of the basis you chose for $\mathbb{R}^k$.

@Obliv you have shitty notating and are wrong

Symmetry group***

Take it up with Mr dummit and foote

no they were probably bullied enough as children

@ACuriousMind ...wat

@0celo7 Pick any basis of $\mathbb{R}^k$ and then complete its image to a basis of $\mathbb{R}^l$ however you like. In these bases, the map has the form they claim.

I don't understand "and then complete its image to a basis of R^l however you like."

@0celo7 it's a set of vertices that's why it's called a symmetry group. I think

Eh actually it would make more sense if it weren't a set since order doesn't matter in a set but it does when ur permuting elements

@FrancescoS well but I don't know what part of it you're confused about. In any case, I'm not so familiar with this fundamental QFT stuff so I'm not sure I could give you an answer. You could ask on the site!

@0celo7 It means "add arbitrary linearly independent vectors until you have a basis".

But why does this have that matrix

are these new vectors in the kernel of $dg_0$?

No, they are in its cokernel.

ok right they're not in the kernel, whoops

but I dunno what a coker is

I used to know

Back when I was smart

Oh I messed this up $S_n $ acts on a set of subsets of n that are of size less than or equal to n

@ACuriousMind What is the coker :(

In this case I should've better said that they are simply "not in the image of the map"

Yes, obviously

But I don't understand why the matrix has that form now

@0celo7 What are the columns of the matrix of a linear map?

images of the basis vectors IIRC

Exactly.

That doesn't help...

@DavidZ yes, i just posted a question physics.stackexchange.com/questions/262435/… ;)

@ACuriousMind Oh am I picking each basis vector in $\Bbb R^l$ to have the last $(l-k)$ components be zero?

Why can I do that

Is the question clear? or do you suggest me to add more details?

@0celo7 Uh...what do "components" mean before you have picked a basis?

@ACuriousMind I don't know. I have no clue what's going on :(

@ACuriousMind Let $\{e_i\}$ be a basis of $\Bbb R^k$. Then because $dg_0$ is injective, $\{dg_0(e_i)\}$ spans some $k$-dimensional subspace of $\Bbb R^l$. Define $f_i=dg_0(e_i)$, then these vectors form a basis of this subspace. We give each $f_i$ the representation $(0,\dotsc,1,\dotsc,0)^t$ where there are $l$ entries in total and the $1$ is in the $i$th spot. Then let $\{g_i\}$ be a basis of the complement of the subspace. Then $\{f_i,g_i\}$ is a basis with the desired properties.

I'm not sure what you mean by "We give each $f_i$ the representation". By definition of how one writes the components in a given basis, that's how you write the $i$-th basis vector.

o.O

wtf

I'm so confused

@ACuriousMind I'm afraid I have no clue what $(1,\dotsc,0)$, etc. means

what the hell are matrices

I'm so confused

I'm afraid I have no time right now to clear up another of your existential confusions

seriously, what is a matrix

what does $v=(v^1,\dotsc, v^k)$ even mean

"another"

be nice, @ACuriousMind

Hmm, why are rotations a bunch of reflections

What's the proof

are they

Yeah they are

What's a rotation by 45°

Not sure

But Math SE people think it's true

A pox on them I say