Y Macdisi
|
Registered User
|
|
|
Mar 15 |
answered | How does hyperbolicity of space time affect our lives? |
|
Mar 11 |
comment |
maxwell’s equations and hodge theory ...potential is also harmonic in Lorentz gauge |
|
Feb 5 |
awarded | ● Commentator |
|
Feb 5 |
comment |
A basis for Schur functors 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. |
|
Feb 4 |
awarded | ● Teacher |
|
Feb 4 |
answered | A basis for Schur functors |
|
Jan 30 |
answered | What are the symmetric and anti-symmetric representations of $6\times6$ of $SU(6)$ in $SU(3)\times SU(2)$? |
|
Jan 24 |
comment |
Clifford Lie Algebras 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. |
|
Jan 23 |
comment |
degrees of the invariants for the action of $SL(V)$ on $\wedge^4V$ 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 |
|
Jan 17 |
comment |
Clifford Lie Algebras 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... |
|
Jan 17 |
comment |
Clifford Lie Algebras 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. |
|
Jan 16 |
asked | Clifford Lie Algebras |
|
Dec 28 |
comment |
SU(6) -> SU(3) branching rule 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! $L_1=A_2A_1$,$6 \to (3 \otimes 2)$,$S^3(6)\to (10\otimes 4) \oplus (8 \otimes 2)$; $L_2=A_2A_1$,$6 \to (3 \otimes 1) \oplus (1 \otimes 2) \oplus (1 \otimes 1)$; $L_3=A_2A_1$,$6 \to (3 \otimes 1) \oplus (1 \otimes 3)$. |
|
Dec 28 |
awarded | ● Scholar |
|
Dec 28 |
comment |
SU(6) -> SU(3) branching rule 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. |
|
Dec 28 |
awarded | ● Supporter |
|
Dec 28 |
comment |
SU(6) -> SU(3) branching rule 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... |
|
Dec 28 |
comment |
SU(6) -> SU(3) branching rule 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. |
|
Dec 28 |
comment |
SU(6) -> SU(3) branching rule 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. |
|
Dec 27 |
awarded | ● Student |
|
Dec 27 |
asked | SU(6) -> SU(3) branching rule |
