User y macdisi - MathOverflow

Answer by Y Macdisi for How does hyperbolicity of space time affect our lives?

2013-03-15T18:33:02Z

For a creative (and more digestible) twist on this try http://gregegan.customer.netspace.net.au/ORTHOGONAL/00/PM.html

Answer by Y Macdisi for A basis for Schur functors

2013-02-04T21:50:37Z

You can construct "canonical" basis for the decomposition of $V^{\otimes n}$ under the symmetric group (and simultaneously under $GL(V)$) through "Young symmetrizers";

see http://en.wikipedia.org/wiki/Young_symmetrizer for starters..

I'm sure the map $S^\lambda(M)$ can also be constructed "canonically". You might want to look at this software by Brian Wybourne for details on the calculations...

http://schur.sourceforge.net/

Answer by Y Macdisi for What are the symmetric and anti-symmetric representations of $6\times6$ of $SU(6)$ in $SU(3)\times SU(2)$?

2013-01-30T22:00:28Z

$6 \otimes 6 \to 18 \oplus 6 \oplus 9 \oplus 3$

$Sym^2(6) \to 18 \oplus 3$

$Ext^2(6) \to 9 \oplus 6$

here $18=(6 \otimes 3)$,$6=(6 \otimes 1)$,$9=(3 \otimes 3)$,$3=(3 \otimes 1)$ of $SU(3) \times SU(2)$.

I was looking at something similar

http://mathoverflow.net/questions/117308/su6-su3-branching-rule

Clifford Lie Algebras

2013-01-16T21:07:47Z

I'm studying the "Clifford Lie Algebra" (see http://arxiv.org/pdf/1007.2481.pdf page 30). It's basically a way to look at Clifford algebras and their properties in a Lie algebraic setting (which I find appealing). I'm looking for a reference that looks at this in more detail; one that discusses the lie subalgebras even better. Thanks.

SU(6) -> SU(3) branching rule

2012-12-27T07:50:59Z

I read in at least one paper and in the wiki below

http://en.wikipedia.org/wiki/Quark_model

that the 56 symmetric irrep of SU(6) breaks down into 10^{3/2} + 8^{1/2} irreps of SU(3)xSU(2). Here the first is 40 dimensional (10 of SU(3) x 4 of SU(2)) and the second is 16 dimensional (8 of SU(3) x 2 of SU(2)).

The problem is that this is simply not true. I checked the branching rules and this just doesn't show up. Can someone please double check if the above decomposition is correct or not.

Thanks.

stabilizer always abelian?

2012-01-03T20:05:05Z

Let G be a finite group acting on a finite dimensional vector space V. Let C be a nontrivial subspace of V. Let H be the subgroup of G that fixes C pointwise (the stabilizer of C). I'm fairly sure that H has to be abelian. The argument going along the lines that C contains simultaneous eigenvectors of all the elements of H so these must commute...

I have two questions :

(1)Is the above correct? Is there a reference with a derivation.

(2)Is there a more general statement or setting for this?

Thanks.

Comment by Y Macdisi

2013-03-11T21:21:10Z

...potential is also harmonic in Lorentz gauge

Comment by Y Macdisi

2013-02-05T03:26:29Z

The Young symmetrizers are just a starting point. They are used to construct special elements in the group algebra of $S_n$. There are still several steps after that to get a basis for the invariant suspaces of $V^{\otimes n}$. The calculations are not for the faint of heart...a lot of manipulations with Young tableaux. I'll try to dig up more reference later.

Comment by Y Macdisi

2013-01-24T02:34:43Z

Thanks for the response. It is actually the other elements (non-quadratic) that make things more interesting. Altogether these give a $2^n$ dimensional lie algebra which can stand on its own as an abstract lie algebra that includes $so(n)$ as a lie subalgebra. The multivectors correspond to invariant subspaces of this subalgebra (adjoint action). Also $so(n+1)$ and I believe $so(n+2)$ also occur as lie subalgebras...all this motivated the question.

Comment by Y Macdisi

2013-01-23T06:03:32Z

The 7 numbers 2,6,8,10,12,14,18 are the degrees of polynomial invariants of the Weyl reflection group of $E_7$. These generate (freely) the full ring of polynomial invariants of the group in the reflection rep. These can be calculated many ways; look at invariant theory of reflection or Coxeter groups. See also en.wikipedia.org/wiki/Coxeter_element

Comment by Y Macdisi

2013-01-17T21:06:58Z

I agree with you on the dimension, it should be $2^n$; there is an element in the lie algebra that commutes with everything; my guess is that the authors did not want to include it. Clifford algebras are a large field and my guess is that angle is just not very popular. One motivation for me is the ability to look at the adjoint rep of $cl(n)$ as a rep of $so(n)$; for example $cl(4)$ $\to 1 \oplus 4 \oplus 6 \oplus 4 \oplus 1$ as an $so(4)$ rep...

Comment by Y Macdisi

2013-01-17T06:54:03Z

Thanks I'll try to track this down. Finding $so(n)$ in $cl(n)$ is, as expected, easy to do; here $cl(n)$ is the Clifford lie algebra; it has dimension $2^n-1$ (I think this can also be extended to be $2^n$). $so(n+1)$ also shows up in $cl(n)$ if I did my calculations correctly.

Comment by Y Macdisi

2012-12-28T22:03:48Z

As it turned out the sla package can already find all such subalgebras. It found the two we've been talking about...but it also found a third! $L_1=A_2A_1$,$6 \to (3 \otimes 2)$,$S^3(6)\to (10\otimes 4) \oplus (8 \otimes 2)$; $L_2=A_2A_1$,$6 \to (3 \otimes 1) \oplus (1 \otimes 2) \oplus (1 \otimes 1)$; $L_3=A_2A_1$,$6 \to (3 \otimes 1) \oplus (1 \otimes 3)$.

Comment by Y Macdisi

2012-12-28T19:06:22Z

Your argument looks right. I'll accept that what's happening here is that there's more than one way to embed $SU(3)\times SU(2)$ in $SU(6)$ and the branching rules are different for each. I did a quick search for a subalgebra of $A_5$ with semisimple type $A_2A_1$ that keeps the $SU(6)$-6 irrep irreducible...so far I haven't found it.

Comment by Y Macdisi

2012-12-28T00:15:24Z

I use GAP (groups algorithms and programming) with sla package for quick branching rules (it works with algebras...) I also have a collection of code I've written over the years that is somewhat mature and reliable...but it's always good to check these calculations from different angles...

Comment by Y Macdisi

2012-12-28T00:06:56Z

But the wiki article is really in an SU(6) setting. So the "6" is referring to an irrep of SU(6). The subgroup I'm using is the one you identified : SU(6)-6 branches to 3 + 2 + 1. I can't think of a $SU(3)\otimes SU(2)$ subgroup where the SU(6)-6 would remain irreducible; (is that possible?). So the article talks about the decomposition : 6x6x6 = 56 + 70 + 70 + 20 where all these are SU(6) reps. The 56 here is $S^3(6)$ and as an $SU(3) \otimes SU(2)$ module it does not decompose into 40 + 16.

Comment by Y Macdisi

2012-12-28T00:06:31Z

In a pure $SU(3) \otimes SU(2)$ setting, you're right : $S^3(3 \otimes 2) \cong (10 \otimes 4) \oplus (8 \otimes 2)$ or in terms of dimensions $S^3(6)=40+16$; the 6,40,and 16 are all $SU(3) \otimes SU(2)$ irreps.