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How do I find the corresponding tensor component $v^{ij}$ of the six dimensional representation of $SU(3)$ with Dynkin label $(2,0)$?

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And this is when physics became confusing. –  BMS May 30 at 5:04

1 Answer 1

up vote 4 down vote accepted

The irreducible $SU(3)$ representations of Dynkin indices $(n,0)$ are the $n-$ symmetric tensor powers of the fundamental representation.

Therefore let $e_1$, $e_2$, $e_3$ be an orthonormal basis of the fundamental representation space, then

$ v_{ii} = e_i \otimes e_i $

$ v_{ij} = \frac{e_i \otimes e_j + e_j \otimes e_i }{\sqrt2}$, $i \ne j$

Identifying the fundamental representation basis with the weight vectors follows:

$ e_1 = (1,0)$

$e_2 = (-1,1)$

$e_3 = (0,-1)$,

The weights of the $(2,0)$ representation space can be obtained by inspection:

$ v_{11} = (2,0)$

$ v_{22} = (-2,2)$

$ v_{33} = (0,-2)$

$ v_{12} = (0,1)$

$ v_{13} = (1,-1)$

$ v_{23} = (-1,0)$

There exist a lot of algorithms for the construction of group representations. One excellent reference is Slansky's seminal article: Group Theory for Unified Model Building.

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I didn't understand how you got fundamental representation basis with the weight vectors. What does (1,0) for e.g. mean here? Does it mean the wavefunction or tensor |>? Shouldn't it be |1/2, sqrt(3)/6>, |-1/2, sqrt(3)/6> for the three roots. Sorry, if I am sounding silly, but I don't seem to be getting the hang of it at all. Help will be much appreciated. Sorry for not using LaTex, I still have to learn it. –  ramanujan_dirac May 16 '12 at 6:10
The two dimensional vectors are just the weights corresponding to the basis vectors, Sorry for the abuse notation. Please see first equation (5.4), on page 32 of Slansky, where the weight diagrams of the fundamental as well as the (2,0) representations are given. The algorithm for the construction of the weight diagram from the highest weight is described in words in the few lines preceding the equation. The principle is that in the weight basis, the primitive roots correspond to the rows of the Cartan matrix. –  David Bar Moshe May 16 '12 at 6:46
Continued, . If a weight has a a positive component of value $n$ at the $m-$th place, then there are weights in the representation obtained by $1, 2, ., ., ., n$ subtractions of the $m-th$ primitive root. Please, see a less trivial example (which you can use as an exercise if you wish) on page 84 (the representation 16 of SO(9)). –  David Bar Moshe May 16 '12 at 6:49
Hey, thanks for the explanation. I get it now. And thanks for the link to the pdf also. –  ramanujan_dirac May 16 '12 at 7:35

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