# Decomposition on irreducible representations

Please check my solution of the following problem: Let $V_\lambda$ be an irreducible representation of Lie algebra $\mathfrak{sl}(2,\mathbb{C})$ with the highest weight $\lambda$. Decompose into a direct sum of irreducible representations $\mbox{Sym}^2(V_5)$.

Since $\lambda = 5$ is the highest weight, the character is given by $\chi_{V_5} = q^5 + q^3 + q^1 + q^{-1} + q^{-3} + q^{-5}$. $$(q^5 + q^3 + q^1 + q^{-1} + q^{-3} + q^{-5})^2 = q^{10} + 2q^8 + 3q^6 + 4q^4 + 5q^2 + 6 + 5q^{-2} + 4q^{-4} + 3q^{-6} + 2q^{-8} + q^{-10}$$

So the decomposition is $\mbox{Sym}^2(V_5) = V_{10} \oplus V_8 \oplus V_6 \oplus V_4 \oplus V_2 \oplus V_0$ and the dimensions of irreducible components are $\mbox{dim} V_{10} = 1$, $\mbox{dim} V_8 = 2$, $\mbox{dim} V_6 = 3$, $\mbox{dim} V_4 = 4$, $\mbox{dim} V_2 = 5$, $\mbox{dim} V_0 = 6$.

Is it right? It seems to me that I have done the task for $V_5 \otimes V_5$ but not for $\mbox{Sym}^2 V_5$.If so, please explain to me what to do with the symmetric power.

• Right, you have decomposed the tensor square, which gives you both the symmetric and the alternating part. Now you need to pick out the symmetric part. BTW, it is a poor idea to use the same notation for a simple module with some highest weight and for the weight spaces of that module. – Tobias Kildetoft Dec 21 '16 at 13:21
• Your notation change now makes your decomposition not actually the decomposition into irreducibles. – Tobias Kildetoft Dec 21 '16 at 13:38
• Thank you for the quick response. I have corrected the notation. But it is still unclear to me how to distinguish the symmetric part from it. It seems that I already use that $q^k q^l = q^{k+l} = q^l q^k$ when opened the brackets and collected the same summands. – Hasek Dec 21 '16 at 13:40
• Yes, the characters live in a commutative ring. What you need to use is that you know the dimension of the symmetric powers, the dimensions of the irreducibles and then to note that the highest weight vector (i.e. the one of weight $10$) of the tensor square is a symmetric element. – Tobias Kildetoft Dec 21 '16 at 13:43
• No, the total sum of dimensions of those irreducibles is more than $21$. – Tobias Kildetoft Dec 21 '16 at 14:32