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Let $\mathfrak{g}$ be the Lie algebra $\mathrm{sl}_2(\mathbb{C})$. There is a classification of irreducible representations of $\mathfrak{g}$: each of them is defined by the only natural number $n$, which is the eigenvalue of $H=\begin{pmatrix}1 & 0\\0 & -1\end{pmatrix}$, namely $\mathrm{Sym}^n(V)$, where $V$ is a standard $2$-dimensional representation of $\mathfrak{g}$ (a good description of the construction is in Fulton-Harris). Now let $W$ be an arbitrary representation of $\mathfrak{g}$. Suppose we know the set $\sigma$ of eigenvalues of $H$ ($H$ now is an endomorphism of $W$). How to get the decomposition of $W$ into the sum of irreducible ones? I know that $W$ is irreducible if and only if $\sigma=\{n,n-2,n-4,\ldots-n\}$ for some $n\in \mathbb{N}$. What should I do if $\sigma$ is not as above?

Fulton and Harris in their "Representation theory..." give the following example (I describe it briefly,it is $\S 11.2$): let $W=\mathrm{Sym}^2(V)\otimes \mathrm{Sym}^3(V)$, where $V$ is a standard $2$-dimensional representation. The eigenvalues of $H$ can be easily evaluated, because $\{3,1,-1,-3\}$ and $\{2,0,-2\}$ are the sets of eigenvalues of $H$ on $\mathrm{Sym}^2(V)$ and $\mathrm{Sym}^3(V)$ respectively and it follows that the eigenvalues of $H$ on $W$ are all sums: $\{5$ and $-5,3$ and $-3 $ (taken twice)$, 1 $ and $-1 $ (taken three times)$\}$. After that they write some explanation which I ask you to help me to understand, which shows that $\mathrm{Sym}^2(V)\otimes \mathrm{Sym}^3(V)=\mathrm{Sym}^5(V)\bigoplus \mathrm{Sym}^3(V)\bigoplus V$.
So, could you explain this trick? Namely, why we notice that $\{5,3,1,-1,-3,-5\}$ are eigenvalues of $H$ and immediately conclude that $\mathrm{Sym}^5(V)$ is subrepresentation of $W$?!

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You have an error in the second sentence: $n$ is not “the eigenvalue of $H$”, $n$ is “the largest eigenvalue of $H$.” The other eigenvalues of $H$ on $\operatorname{Sym}^n(V)$ are $n-2, n-3, \ldots, -n$ as you said. The eigenvalues of a direct sum are the union (with multiplicity) of the eigenvalues of the summands. I think that should answer the rest of your question.


In case not, here is a more detailed write up:

Let $\sigma(W) : \mathbb{R} \to \{0,1,\ldots\}$ record the multiplicity of the eigenvalues of $H$ when restricted to the $H$-invariant subspace $W$. For example, $\sigma_n= \sigma( \operatorname{Sym}^n(V) )$ is the function:$$\sigma_n(\lambda) = \sigma( \operatorname{Sym}^n(V) )(\lambda) = \begin{cases} 1 & \text{if } \lambda \in \{ n,n-2,n-4,\ldots,4-n,2-n,-n\} \\ 0 & \text{otherwise} \end{cases}$$

Then $\sigma(W \oplus U) = \sigma(W) + \sigma(U)$ follows from the matrix of $H$ being block-diagonal on $W \oplus U$.

Now you need to know a very special thing about $\mathfrak{sl}_2$: every finite-dimensional representation is a direct sum of irreducible representations, and every irreducible sub-representation is a direct summand.

Now we have $$\sigma( \operatorname{Sym}^2(V) \otimes \operatorname{Sym}^3(V) )(\lambda) = \begin{cases} 2 & \text{if } \lambda \in \{3,1,-1,-3\} \\ 1 & \text{if } \lambda \in \{5,-5\} \\ 0 & \text{otherwise} \end{cases}$$

It is not too hard to see the only way $\sigma$ can be written as a sum of $\sigma$ of irreducible representations is as $\sigma_5+\sigma_3$. This is because the only way to get a $5$ with to use $\sigma_n$ for $n=5+2k$ ($k \geq 0$), but if $k > 0$, then we'd also get $7$. Hence $\sigma = \sigma_5 + \text{something}$ and we can handle $\sigma-\sigma_5$ by induction (in general) or (here) just noticing that it is exactly equal to $\sigma_3$.

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A mechanical way is to use characters: the character of a representation $W$ of $\mathfrak{sl}_2$ is $\chi_W(x)=\sum_i m_i\,x^{n_i}$, where $n_i$ runs over all the eigenvalues of $H$ in $W$ and $m_i$'s are their multiplicities. Characters satisfy $\chi_{W_1\oplus W_2}=\chi_{W_1}+\chi_{W_2}$ and $\chi_{W_1\otimes W_2}=\chi_{W_1}\chi_{W_2}$, and most importantly, $(x-x^{-1})\chi_W(x)=\sum_k d_k (x^{k+1}-x^{-k-1})$, where $d_k$ is the multiplicity of the irreducible representation $\mathrm{Sym}^k(V)$ in $W$. So to find $d_k$'s just multiply $\chi_W$ by $x-x^{-1}$ and read the coefficients for the positive powers of $x$. In particular, for $W=\mathrm{Sym}^2(V)\otimes \mathrm{Sym}^3(V)$ you have to compute $(x-x^{-1})(x^2+1+x^{-2})(x^3+x+x^{-1}+x^{-3})=(x^2+1+x^{-2})(x^4-x^{-4})=\dots$ (so I leave something for homework :)

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