Discussions between Mercio and Rudi Brinbaum

Rudi_Birnbaum

07:04

Hello world

mercio

hi

Rudi_Birnbaum

08:00

Hi

Now I am available.

but its possible that I have to leave now and then!

mercio

08:18

okay I was rethinking about the quaternioninc representations

they still are the sum of two complex conjugate irreducible representations

except that this time the complex irrep is isomorphic to its complex conjugate

I remember a little

so we still have decomposition of real representations into real irreps

but some of them are a bit less rigid than what we have in the complex case

and anyway it shouldn't interfere with what we do

Rudi_Birnbaum

08:38

left and back.

I quickly paste the Referees point:

btw. have you got access to the complete referee statements?

mercio

I'm not sure

I have a version from june 12 with some referee comment

Rudi_Birnbaum

here the one statement6) The Authors wrote on page 3 of the text that they use only the real representations instead of complex. It can be understood as a ban on the use of Schur's lemmas for some axial groups.
(The right constructions were used in SM). This peculiar point is important in the analysis of the transformations for gradients, which still has to be done correctly for point groups with complex one-dimensional representations (the note at the top of page S2 is incorrect in this case and should not be used).

mercio

"the note at the top of page S2" is that something in our paper ?

also what is SM ?

Rudi_Birnbaum

do you want me to send you the comlpete statements or would you prefer that "I walk us thorough"?

Supplementary Material

S2 is the second page of that.

mercio

that page is about Schur functors

oh wait

"Finally, if Γ is a real irreducible representation by a group G, then
Sym 2 (Γ) always has exactly one totally symmetric Γ 0 component in its decomposition, so that
for any two dimensional real irreducible representation E"

that is wrong

but

wait no it is right

Rudi_Birnbaum

08:45

:-)

mercio

Sym²

it is Alt² that gets all the extra Gamma0s

in case of a sum of conjugate pairs

Rudi_Birnbaum

So you have the current version?

But wait I'll send you the real current submission version by email

mercio

I don't think my version has the same referer comments

the note on S2 is true but I might have to explain it a bit more then

Rudi_Birnbaum

I sent it, please go through that I'll be away for 20 min

mercio

I think that's the one I have

I still can't find the comments you are quoting

mercio

09:13

the classical result is that an occurence of the trivial rep in Sym²(rho) corresponds to a G-linear automorphism V -> V, and then by Schur's lemma, there is only one of them

but that's over C

over R, in case rho = r + r'

actually they might be right

I have to check

Sym²(rho) = Sym²(r) + Sym²(r') + r * r'

:(

I was so sure they would be split ?

I do remember checking this

let's check on the cyclic group of order 3

ahaha

I think there is a dual thing I forgot

but then the Schur's lemma argument doesn't work as is

it's not Sym²(r) that corresponds to an automorphism

it's r' * r

Sym²(r) is more about symmetric bilinear forms

probably

so in Sym²(r+r') the Gamma0 component comes from r * r'

alright, the dual representation

is given by rho*(g) = transpose of rho(inverse of g)

and so an automorphism of V

corresponds to a Gamma0 component in the tensor product of V and V*

... that still doesn't give the "easy" proof that there is exactly one Gamma0 component in rho tensor rho when rho is a complex irreducible representation

well it does for unitary representations

and every finite group representation is unitarizable if I'm reading wikipedia correctly

Rudi_Birnbaum

09:43

@mercio Sure these are the referee comments. I will send them to you as well. But mind that then you still miss the comments from the first round of reviewing. We might ome to that later in case of need. I anyway suggest we two now and here focus on the specific numbered issie list from rev 4 for now.

The problem is that the whole thing is already very confusing with two rounds of reviewing ...

So lets try to stay focused.

@mercio Hm why can't we just go back to reps over C for the proof, and check that we have all groups included that matter?

mercio

there are also some elementary proofs that if A is a matrix of finite order then tr(inverse of A) = conjugate of tr(A) so I guess that works

I think they are talking about the (S1) statement

where I say that Sym²(rho) has exactly one Gamma0 component when rho is a real irreducible representation

Rudi_Birnbaum

OK

mercio

I think the usual argument, over C, goes like this

Rudi_Birnbaum

btw. does comment 7 help here?

mercio

I still have not found comment 6

which page are they on ?

Rudi_Birnbaum

09:51

I have submitted another email

mercio

ah

Rudi_Birnbaum

with the two complete referee statements

mercio

I got it

Rudi_Birnbaum

good

mercio

7) Many constructions would be much shorter and clearer if a known (and easily inferred)
expression were used: the symmetrized square of the representation Г ( dimГ = n ) has
dimension n(n+1)/2, and n(n-1)/2 corresponds to the antisymmetrized square

yes but the dimension doesn't tell everything there is to know about a representation ?

Some statements are well known from the theory of vibrations. For
example, the matrices of the (x,y,z) coordinates representation and the corresponding
infinitesimal rotations ( i.e., the projections of the momentum) differ only in the factor – the
determinant of the matrix – and therefore are simultaneously reducible or irreducible.

well I don't know anything about what is well known or not in physics

and I don't know what they mean with "differ only in the factor"

(the note at the top of page S2 is incorrect in this case and
should not be used).

the note is correct but is phrased in a way that might look wrong

I should think about improving that part

also those tables need some vertical bars

Rudi_Birnbaum

10:16

here is a problem. at this stage we need to carefully protocol and comment all changes

do you know git?

mercio

I haven't used it much

Rudi_Birnbaum

me not either I just start learning it. But I think it would be the right tool now.

mercio

possibly

Rudi_Birnbaum

Well on the other hand if we only do punctual changes one after the other we can do without it as well .

For git in any case we'd need to set up a server. I had the remote hope you would be an experienced git user. But in this case maybe its easier without

I am wondering what background that referee has. I am certain he is no chemist

Y and Yh for what we call I and Ih is very uncommon.

mercio

don't worry, I and Ih is very uncommon too

Rudi_Birnbaum

10:26

:-)

mercio

from where I come from

Rudi_Birnbaum

sure

mercio

... does he want us to include definitions of vector spaces and representations and tensor products ?

Rudi_Birnbaum

no idea. I have already complaint about the very unspecific comments, its a really tough one, that paper ...

Well we both should try to fix 6. now

mercio

I have to admit I chuckled inside when he asked for a "definition statement proof" since I sometimes feel that physicists do the opposite in their writings

"experimental result - computations - therefore we should use this definition because it works"

Rudi_Birnbaum

10:29

there are so many aspects, this work is the worst in trouble making I ever had

mercio

I'm sorry it sounds hard

Rudi_Birnbaum

not at all your faut

fault

mercio

I see people wanting math stuff to be moved all to the supplementary section

then the next one wanting the opposite

Rudi_Birnbaum

exaclty

one wants it to be shortend to 2 pages

the other one extended to 3 publications ...

that all depends on soft criteria

mercio

the fact that it relies on some math stuff does sound like it makes the procedure a bit wonky

we are all accustommed to our own things and ways of describing things

Rudi_Birnbaum

10:31

yes

and in a sense its heavily interdisciplinary

Its very hard to focus on certain readership

I tried to and the the referess imply different readerships

but mostly ones that are different from themselves as well

its a terrible mess, to be honest :(

But then I think its mostly questions of style

I am aware that it is not greatly presented as it is, but I don't know how to do it substantially better.

In no case on two pages like ref 3 wanted

rev 4 has a very good understadning of the physics and maths but none for the field on which it should have the largest impact

mercio

ah yes there was this other referee

Rudi_Birnbaum

rev3 got some things quite wrong to begin with and that caused also lots of misunderstanding

mercio

apparently we said that D2d was abelian ?

Rudi_Birnbaum

Yeah I am so sorry

that was huge stupidity

mercio

I don't even remember what D2d is

Rudi_Birnbaum

10:38

He suggested to call then abelian and non-abelian

which is wrong (as we know)

so I wanted to give a counter example

mercio

ah

Rudi_Birnbaum

and then I got it wrong which caused troubles in this thrid round

mercio

I do remember one of the D2x ones being tricky

Rudi_Birnbaum

but they both understand now that its not non-abelian

mercio

and not being in the same family as his D3x D4x ... friends

Rudi_Birnbaum

10:39

well all the S2n are ones

all with pairs of imaginary 1D irreps

its actually the 1D imaginary that make the troubles

yes.

mercio

I'm going to read his group theory book

a little

Rudi_Birnbaum

but now he understood it and suggests stil not to use Kaaa, Kae and Kt but G1, G2, G3,

and call them 2-D represented which i think is again less precise and G_1 for me is one certain group

instead a family of groups ....

mercio

well I think G1 G2 G3 would be fine

Rudi_Birnbaum

the referee is less of a problem and has less clear insight.

@mercio Well, than kwhy not

one just has to define them properly

"1D, 2D and 3D represented" as he suggests is simply not good enough I think

anyway we should try to fix the most severe thing and that is 6. of rev4

because there are 2D irreps in G3 as well

mercio

yeah but they're not faithful

Rudi_Birnbaum

10:47

and there are 5D irreps in G3

mercio

and yeah I don't want to suggest there are families G4 G5 G6...

anyway

Rudi_Birnbaum

So the full formulated defintion is G1= 1D over R, G2= 2D but no higher over R, and G3= the rest.

I don't see a way to make it simpler

mercio

yeah that would clash if somebody would read G3 = 3D but no higher

Rudi_Birnbaum

Its simply not a trivial partition.

mercio

the way I have them in my head is G1 = faithful 3d reps are all of the form a+a+a ; G2 = faithful 3d reps are all of the form a+e ; G3 = faithful 3d reps are all of the form t

and then it's a nontrivial result that there is no group having two kinds of faithful 3d rep

Rudi_Birnbaum

10:51

yes.

mercio

anyway, the usual equivalent of the note (S1) is that for a complex rep rho, there is exactly one Gamma0 in rho tensor conjugate(rho)

Rudi_Birnbaum

The most "cryspy" way is to call the 3 or higher represented ones a name. And say all non-NAME that are not 1D over R.

ok

mercio

because conjugate(rho) = dual of rho, then rho tensor dual of rho = maps from V to V, and invariant maps = G-automorphism, and Schur's lemma says there is only one

and I am saying that for a real irrep rho there is exactly one Gamma0 is Sym²(rho)

but in fact

depending on the kind of irrep rho is there can be 1,2, or 4 Gamma0s in rho²

which is I think what he is thinking about

because rho² would be the real equivalent of rho * conjugate of rho

Rudi_Birnbaum

but not in the point groups afaik?

mercio

some of the point groups have 2 in rho²

it's always exactly one in the symmetric part

Rudi_Birnbaum

10:55

oh yes!

mercio

and 0,1, or 3 in the alternate part

Rudi_Birnbaum

yes

mercio

and that's something I haven't expanded upon

Rudi_Birnbaum

is there an example of 3?

mercio

yes with quaternionic representations

the archetypal example being the quaternion group

Rudi_Birnbaum

10:56

ok but not in the point groups

mercio

I don't think so

but I would have to check

in G3

Rudi_Birnbaum

I am relatively sure, but checking is better

There is only very few e

mercio

I don't think e can be quaternionic

it can only happen on dimension at least 4

the quaternion group doesn't have a faithful representation in R³

Rudi_Birnbaum

I dont think so. It doesnt ring a bell ...

if its not T it cant be anything I guess

but T is A4

if i rem correctly ?

mercio

link me your friend's website again ?

Rudi_Birnbaum

11:00

Oh and there is S4

gernot-katzers-spice-pages.com/character_tables

did you notive btw its a html programm

?

mercio

yeah

Rudi_Birnbaum

Everything is computed ...

:-)

a pitty he hates complex numbers ... :-D

mercio

if rho = r1 + r2 where r1 and r2 are different (and conjugate of each other) then rho² = r1²+r2²+2r1r2, so they are 2 Gamma0 because there are 2r1r2

Rudi_Birnbaum

OK

mercio

if rho = 2r (with r complex and isomorphic to its conjugate) then rho² = 4r²

and that will be 4 Gamma0

but 3 of them are in the Alternate part

that's the quaternionic case

Rudi_Birnbaum

11:04

OK.

mercio

so for example for the representation of a cyclic group (n >= 3) on a plane with rotations

it has two Gamma0 in its square

and that corresponds to two automorphisms

well dimension 2 rather

while Schur's lemma would say there can only be homotheties

in this case you will also have rotations

Rudi_Birnbaum

yes

mercio

those are also endomorphisms that are compatible with the action of G

Rudi_Birnbaum

I mean its exactly those which spoil the fun just using reps over C

mercio

I'm kinda sad there is no quaternionic point group

Rudi_Birnbaum

11:12

sorry for that

S2?

mercio

well I looked at the tables and only I and Ih had irreps of dimension > 3 and apparently they weren't quaternionic

Rudi_Birnbaum

sorry S4

sure ..

not is only the 6 of them in Kt / G3

mercio

I don't think any of the symmetric groups have quaternionic stuff but I could be very very wrong

my intuition fails me

I know we have a thing to biject their C irreps with Young tableaux

Rudi_Birnbaum

Have you heared that soemthing like Jahn-Teller exists only in prime dimension spaces?

mercio

lol

no

Rudi_Birnbaum

11:15

Yeah

So in real 4D space there would be degeneracies that cannot be lifted by vibrational motions ....

mercio

don't break spacetime now

Rudi_Birnbaum

OK :-)

mercio

so I don't know if I can just add a remark

saying there can be one Gamma0 in the Alternate part

or if I have to explain more and prove it

Rudi_Birnbaum

Well it would be actually also not bad to change the whole prove, that would calm down the editor a bit

mercio

:'(

Rudi_Birnbaum

11:19

But as you like

the only point is to adress problem 6.

and that the paper is correct in the end

mercio

This peculiar point is important in the analysis of the
transformations for gradients, which still has to be done correctly for point groups with complex
one-dimensional representations

the transformations for gradients ?

Rudi_Birnbaum

the calculation you do, i think

−i (r′ ×∇′) = −i (Ur×U∇) = (U ×U)l

∇′ = U ∇ because U is orthogonal,

mercio

ah, so the gradient was for the nabla

Rudi_Birnbaum

yes, that is the same (for some)

Rudi_Birnbaum

15:39

So please feel free anytime to give a message if you have a good fix. I would like to try resubmission during the weekend if possible.

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$Mathematics$ Discussions between Mercio and Rudi B…