Young tableaux and tensor product

[302021895]

I am working on all of the problems from Georgi's book in Lie algebras in particle physics (independent study), but I am stuck on one of them. The question is the following:

"Find (2,1)x(2,1) (in su(3) using Young tableaux). Can you determine which representations appear antisymmetrically in the tensor product, and which appear symmetrically?"

I understand the first part, and I get that (diagram - representation - dimension)

xxx (2,1) [15]
x

times

xxx (2,1) [15]
x

=

xxxxxx (4,2) [60]
xx

+

xxxxxx (5,0) [21]
x
x

+

xxxxx (2,3) [42]
xxx

+

xxxxx (3,1) [24]
xx
x

+

xxxx (0,4) [15]
xxxx

+

xxxx (1,2) [15]
xxx
x

+

xxx (0,1) [3]
xxx
xx

+

xxxxx (3,1) [24]
xx
x

+

xxxx (1,2) [15]
xxx
x

+

xxxx (2,0) [6]
xx
xx

I don't quite get the second part. How can one determine from this which representations appear symmetrically or antisymmetrically in the tensor product? Any suggestions?

 

[Bill_K]

To do part two you need to use the "standard" Young's tableaux, which are obtained by filling in the frames with integers in standard order, left to right and top to bottom. Use primed integers for the second factor. Thus the standard frames for (2,1) x (2,1) are:

1 2 3
4

1' 2' 3'
4'

You decomposed the direct product by starting with the first factor and attaching the squares of the second factor to it in all possible ways. So now in the tableau, consider the permutation 1 ↔ 1', 2 ↔ 2', etc. This is an involution, and therefore either an odd or even permutation. Even means that rep belongs to the symmetric part of the product, odd means it's antisymmetric.

For example, take [42], which was

xxxxx
xxx

The standard tableau is

1 2 3 1' 2'
4 3' 4'

Switching the 1's, 2's and 4's are all even since they only involve horizontal moves. But switching 3 with 3' is a vertical move, hence odd. We conclude therefore that [42] is antisymmetric.

 

[302021895]

Awesome! Thanks for the help!