Decomposing into irreducibles

1.40 Consider the standard representation C 2 tensored with itself: C 2 tensor C 2. This will turn out to have the following weight diagram:

In other words, there are 1-dimensional weight spaces with weights 2 and minus 2 and a 2-dimensional weight space in weight 0.

2.10 Inside this diagram, you can see the weight diagram of Sym 2 C 2 as a subdiagram. When you remove this subdiagram what's left is the weight diagram of the trivial 1-dimensional representation.

This will allow us to deduce that C 2 tensor C 2 is isomorphic to Sym 2 C 2 direct sum the trivial representation C.

2.45 A basis for C 2 tensor C 2 is given by e_1 tensor e_1, e_1 tensor e_2, e_2 tensor e_1 and e_2 tensor e_2 (where e_1, e_2 is the standard basis for C 2. We have H e_1 equals e_1, H e_2 equals minus e_2 Because H lives in the Lie algebra, we evaluate H of x tensor y using the product rule: H of (e_1 tensor e_1) equals (H of e_1) tensor e_1 plus e_1 tensor (H of e_1), which equals 2 e_1 tensor e_1

4.20 Therefore e_1 tensor e_1 is an eigenvector for H with eigenvalue 2, so it lives in the weight space W_2. Similarly we get H of e_1 tensor e_2) equals e_1 tensor e_2 minus e_1 tensor e_2 equals 0, and similarly for H of e_2 tensor e_1 which give us the 2-dimensional weight space W_0 and H of e_2 tensor e_2 equals minus 2 e_2 tensor e_2 which gives the weight space W_(minus 2). So our weight spaces are: W_(minus 2) spanned by e_2 tensor e_2, W_0 spanned by e_1 tensor e_2 and e_2 tensor e_1, and W_2 spanned by e_2 tensor e_2.

5.50 Following the procedure we used in proving the classification theorem, we pick a highest weight vector v equals e_1 tensor e_1 and use Y equals 0, 0, 1, 0 to generate a subrepresentation spanned by: v, Y v, and Y squared v These three vectors live in the weight spaces W_2, W_0 and W_(minus 2) respectively, and give a subrepresentation U isomorphic to Sym 2 C 2 (because it has the right weight diagram).

7.05 There is an invariant Hermitian inner product on the representation, and we take the orthogonal complement U perp to get a complementary subrepresentation. Since this complement is 1-dimensional, it must be isomorphic to the trivial 1-dimensional representation (the only irrep of this dimension).

8.20 We deduce that C 2 tensor C 2 is isomorphic to Sym 2 C 2 direct sum C. This should not be a surprise, because Sym 2 C 2 was defined to be a subrepresentation of C 2 tensor C 2, in fact it was defined to be the subrepresentation spanned by the symmetric tensors e_1 tensor e_1, e_1 tensor e_2 plus e_2 tensor e_1 and e_2 tensor e_2.

9.05 Let's check that this is what we are getting. Since Y equals 0, 0, 1, 0 we have Y e_1 = e_2 and Y e_2 = 0. If v equals e_1 tensor e_1 then Y v equals (Y e_1) tensor e_1 plus e_1 tensor Y e_1, which equals e_2 tensor e_1 plus e_1 tensor e_2, and Y squared v equals (Y e_2) tensor e_1 plus e_1 tensor (Y e_1) plus Y e_1 tensor e_2 plus e_2 tensor (Y e_1), which equals 2 e_2 tensor e_2 so we get precisely that U equals Sym 2 of C 2 on the nose.

9.35 The complementary subrepresentation turns out to be spanned by the antisymmetric tensor e_1 tensor e_2 minus e_2 tensor e_2, which spans the exterior square Lambda 2 C 2. It's an exercise to check that this antisymmetric tensor does span a subrepresentation (i.e. it is fixed by the action of SU(2)).

10.35 Let's figure out the decomposition of Sym 2 Sym 2 C 2. We know that Sym 2 C 2 is spanned by alpha equals e_1 squared, beta equals e_1 e_2 and gamma equals e_2 squared. In the same way we can think of Sym 2 C 2 as quadratic polynomials in e_1, e_2, we can think of Sym 2 Sym 2 C 2 as quadratic polynomials in \alpha, \beta, and \gamma. It therefore has a basis alpha squared, alpha beta, alpha gamma, beta squared, beta gamma, and gamma squared.

11.50 Let's calculate the weight space decomposition. We know that H alpha equals 2 alpha, H beta equals 0 and H gamma equals minus 2 gamma, so (using the product rule) we have H of alpha squared equals (H alpha) times alpha plus alpha times (H alpha), which equals 4 alpha squared H (alpha beta) equals (H alpha) times beta plus alpha times (H beta) equals 2 alpha beta H alpha gamma equals H beta squared equals 0, H beta gamma equals minus 2 beta gamma, and H gamma squared equals minus 4 gamma squared. The weight space decomposition is therefore: W_(minus 4) spanned by gamma squared, W_(minus 2) spanned by beta gamma, W_0 spanned by beta squared and alpha gamma, W_2 spanned by alpha beta, and W_4 spanned by alpha squared. with weight diagram

14.15 From this you can see that Sym 2 Sym 2 C 2 equals Sym 4 C 2 direct sum C The argument is exactly as before: you pick a highest weight vector like v equals alpha squared and you generate a highest weight subrepresentation U spanned by v, Y v, Y squared v, Y cubed v, Y to the 4 v, which is isomorphic to Sym 4 C 2 by the classification theorem. Then you take the orthogonal complement of U with respect to an invariant Hermitian inner product and you get a complementary subrepresentation whose weight diagram is obtained from the weight diagram of Sym 2 Sym 2 C 2 by stripping off the weight diagram of U.

15.15 The trivial complementary subrepresentation has to be spanned by some combination of beta squared and alpha gamma as these span the 0-weight space. In this case, I claim that the trivial complementary subrepresentation is spanned by beta squared minus alpha gamma (it will be an exercise to check this).