Decomposing into irreducibles

1.40 Consider the standard representation ${𝐂}^2$ tensored with itself: ${𝐂}^2\otimes {𝐂}^2$ . This will turn out to have the following weight diagram:

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

2.10 Inside this diagram, you can see the weight diagram of ${\mathrm{Sym}}^2({𝐂}^2)$ 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 ${𝐂}^2\otimes {𝐂}^2\cong {\mathrm{Sym}}^2({𝐂}^2)\oplus 𝐂$ .

2.45 A basis for ${𝐂}^2\otimes {𝐂}^2$ is given by $$e_1\otimes e_1,e_1\otimes e_2,e_2\otimes e_1,e_2\otimes e_2$$ (where $e_1,e_2$ is the standard basis for ${𝐂}^2$ . We have $$H{e}_1=e_1,H{e}_2=-e_2.$$ Because $H$ lives in the Lie algebra, we evaluate $H(x\otimes y)$ using the product rule: $$H(e_1\otimes e_1)=(H{e}_1)\otimes e_1+e_1\otimes (H{e}_1)=2e_1\otimes e_1.$$

4.20 Therefore $e_1\otimes e_1$ is an eigenvector for $H$ with eigenvalue $2$ , so it lives in the weight space $W_2$ . Similarly we get $$H(e_1\otimes e_2)=e_1\otimes e_2-e_1\otimes e_2=0=H(e_2\otimes e_1)$$ which give us the 2-dimensional weight space $W_0$ and $$H(e_2\otimes e_2)=-2e_2\otimes e_2$$ which gives the weight space $W_{-2}$ . So our weight spaces are: $$W_{-2}=𝐂\cdot (e_2\otimes e_2),W_0=𝐂\cdot (e_1\otimes e_2,e_2\otimes e_1),W_2=𝐂\cdot (e_1\otimes e_1).$$

5.50 Following the procedure we used in proving the classification theorem, we pick a highest weight vector $v=e_1\otimes e_1$ and use to generate a subrepresentation spanned by: $$v,Yv,Y^{2}v.$$ These three vectors live in the weight spaces $W_2$ , $W_0$ and $W_{-2}$ respectively, and give a subrepresentation $U$ isomorphic to ${\mathrm{Sym}}^2({𝐂}^2)$ (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 $U^{⟂}$ 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 ${𝐂}^2\otimes {𝐂}^2\cong {\mathrm{Sym}}^2({𝐂}^2)\oplus 𝐂$ . This should not be a surprise, because ${\mathrm{Sym}}^2({𝐂}^2)$ was defined to be a subrepresentation of ${𝐂}^2\otimes {𝐂}^2$ , in fact it was defined to be the subrepresentation spanned by the symmetric tensors $e_1\otimes e_1$ , $e_1\otimes e_2+e_2\otimes e_1$ and $e_2\otimes e_2$ .

9.05 Let's check that this is what we are getting. Since we have $Y{e}_1=e_2$ and $Y{e}_2=0$ . If $v=e_1\otimes e_1$ then $$Yv=Y{e}_1\otimes e_1+e_1\otimes Y{e}_1=e_2\otimes e_1+e_1\otimes e_2,$$ and $$Y^{2}v=Y{e}_2\otimes e_1+e_2\otimes Y{e}_1+Y{e}_1\otimes e_2+e_2\otimes Y{e}_1=2e_2\otimes e_2$$ so we get precisely that $U={\mathrm{Sym}}^2({𝐂}^2)$ on the nose.

9.35 The complementary subrepresentation turns out to be spanned by the antisymmetric tensor $e_1\otimes e_2-e_2\otimes e_1$ , which spans the exterior square ${\mathrm{\Lambda }}^2{𝐂}^2$ . 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 ${\mathrm{Sym}}^2{\mathrm{Sym}}^2({𝐂}^2)$ . We know that ${\mathrm{Sym}}^2({𝐂}^2)$ is spanned by $\alpha =e_1^2$ , $\beta =e_1{e}_2$ and $\gamma =e_2^2$ . In the same way we can think of ${\mathrm{Sym}}^2({𝐂}^2)$ as quadratic polynomials in $e_1,e_2$ , we can think of ${\mathrm{Sym}}^2{\mathrm{Sym}}^2({𝐂}^2)$ as quadratic polynomials in $\alpha$ , $\beta$ , and $\gamma$ . It therefore has a basis $${\alpha }^2,\alpha \beta ,\alpha \gamma ,{\beta }^2,\beta \gamma ,{\gamma }^2.$$

11.50 Let's calculate the weight space decomposition. We know that $H\alpha =2\alpha$ , $H\beta =0$ and $H\gamma =-2\gamma$ , so (using the product rule) we have $$H({\alpha }^2)=(H\alpha )\alpha +\alpha (H\alpha )=2{\alpha }^2+2{\alpha }^2=4{\alpha }^2,$$ $$H(\alpha \beta )=(H\alpha )\beta +\alpha (H\beta )=2\alpha \beta$$ $$H(\alpha \gamma )=0,H({\beta }^2)=0,H(\beta \gamma )=-2\beta \gamma ,H({\gamma }^2)=-4{\gamma }^2.$$ The weight space decomposition is therefore: $$W_{-4}=𝐂\cdot {\gamma }^2,W_{-2}=𝐂\cdot \beta \gamma ,W_0=𝐂\cdot ({\beta }^2,\alpha \gamma ),W_2=𝐂\cdot \alpha \beta ,W_4=𝐂\cdot {\alpha }^2$$ with weight diagram

14.15 From this you can see that $${\mathrm{Sym}}^2{\mathrm{Sym}}^2({𝐂}^2)={\mathrm{Sym}}^4({𝐂}^2)\oplus 𝐂.$$ The argument is exactly as before: you pick a highest weight vector like $v={\alpha }^2$ and you generate a highest weight subrepresentation $U$ spanned by $v,Yv,Y^{2}v,Y^{3}v,Y^{4}v$ , which is isomorphic to ${\mathrm{Sym}}^4({𝐂}^2)$ 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 ${\mathrm{Sym}}^2{\mathrm{Sym}}^2({𝐂}^2)$ by stripping off the weight diagram of $U$ .

15.15 The trivial complementary subrepresentation has to be spanned by some combination of ${\beta }^2$ and $\alpha \gamma$ as these span the 0-weight space. In this case, I claim that the trivial complementary subrepresentation is spanned by ${\beta }^2-\alpha \gamma$ (it will be an exercise to check this).