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Given $V = \Bbb C^2$ the standard representation of $\mathfrak{sl}_2\Bbb C$, on page 157 of Fulton and Harris's Representation Theory they state

Since $U = \mathrm{Sym}^3 V$ has eigenvalues $-3, -1, 1, 3$, the symmetric square of $U$ i.e., $\mathrm{Sym}^2(\mathrm{Sym}^3 V)$ has eigenvalues $-6, -4, -2 \text{ (twice)}, 0 \text{ (twice)}, 2 \text{ (twice)}, 4$ and $6$.

I'm not seeing how they computed these eigenvalues. I've tried taking sums (of $\mathrm{Sym}^3$ eigenvalues with themselves) and products (of $\mathrm{Sym}^2$ eigenvalues with those of $\mathrm{Sym}^3$) but I'm not seeing it. Is there a formula involving symmetric polynomials?

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  • $\begingroup$ It should just be sums. (-3)+(-3), (-3)+(-1), (-3)+(1), (3)+(3), (-1)+(-1), (-1)+(1), (-1)+3, (1)+(1), (1)+(3), (3)+(3). Page 152 says this a little more explicitly. $\endgroup$ – Jack Schmidt Apr 27 '13 at 18:26
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The Lie algebra $\mathfrak{sl}_2$ has basis $H=\begin{bmatrix} 1&0\\0&-1\end{bmatrix}$, $X=\begin{bmatrix} 0&1\\0&0\end{bmatrix}$, $Y=\begin{bmatrix}0&0\\1&0\end{bmatrix}$, and the Lie bracket $[A,B] =AB-BA$ obeys $[H,X]=2X$ and $[H,Y]=-2Y$ and $[X,Y]=H$.

The "eigenvalues" of a representation $U$ of $\mathfrak{sl}_2$ are defined to be the eigenvalues of $H$.

Adjoint

For instance, $U=\mathfrak{sl}_2$ is itself a vector space of dimension 3, and on the basis $\{ Y,H,X\}$, $H$ acts as the diagonal matrix with entries -2,0,2. This representation is called the adjoint representation and is irreducible, named $V^{(2)}$ after the largest eigenvalue of $H$.

Natural

The standard representation $V$ is two dimensional, and the matrix representation is already diagonal so its eigenvalues are clear: -1,1. This is called the standard or natural representation and is irreducible, named $V^{(1)}$ after the largest eigenvalue of $H$.

Tensor product

The tensor product of $U$ with $W$ has a basis consisting of $u \otimes w$ for $u,w$ in bases of $U,W$. Elements $G$ of $\mathfrak{sl}_2$ act as $G(u \otimes w) = G(u) \otimes w + u \otimes G(w)$. In particular, if $u,w$ are eigenvectors of $H$ with eigenvalues $\alpha,\beta$, then $$H(u\otimes w) = \alpha u \otimes w + u \otimes \beta w = \alpha (u\otimes w) + \beta (u\otimes w) = (\alpha+\beta) (u\otimes w).$$ This means that $H$ acts diagonally on $U,W$ too, and its eigenvalues are the sums of the eigenvalues on $U$ and $W$.

In particular, $V \otimes V$ would have eigenvalues $$\{ (-1)+(-1), (-1)+(1), (1)+(-1), (1)+(1) \} = \{ -2,0,0,2 \} = \{-2,0,2\} \cup \{0\}$$ so that $V\otimes V = V^{(2)} \oplus V^{(0)}$.

Symmetric Power

The $n$th symmetric power of $U$ has basis all multisets of size $n$ whose elements come from a basis of $U$. It is nice to write these as monomials, so that $u^3 v^2$ means the multiset with basis element $u$ repeated 3 times, and basis element $v$ repeated twice. An element $G \in \mathfrak{sl}_2$ acts similarly on a product $uv$: $H(uv) = H(u)v + uH(v)$. This works a lot like a derivative, so that $H(u^3v^2) = 3u^2H(u) v^2 + 2u^3vH(v)$. If $u,v$ are eigenvectors of $H$ with eigenvalues $\alpha,\beta$, then $H(u^3v^2) = 3\alpha u^3v^2 + 2\beta u^3 v^2 = (3\alpha+2\beta) u^3 v^2$. In particular, the eigenvalues of $H$ on the $n$th symmetric power of $U$ are just the sums of size $n$ multisets of eigenvalues of $H$ on $U$.

$\newcommand{\Sym}{\operatorname{Sym}}$ For example, the symmetric cube of $V$ has eigenvalues $\{ (-1)+(-1)+(-1), (-1)+(-1)+(1), (-1)+(1)+(1), (1)+(1)+(1) \} = \{ -3,-1,1,3 \}$ and so $\Sym^3(V) = V^{(3)}$ named after the highest eigenvalue $3$. Similarly, $\Sym^2(\Sym^3(V))$ has eigenvalues sums of any pairs of eigenvalues of $\Sym^3(V)$, namely $$\{ (-3)+(-3), (-3)+(-1), (-3)+(1), (-3)+(3), (-1)+(-1), (-1)+(1), (-1)+(3), (1)+(1), (1)+(3), (3)+(3) \} = \{ -6, -4, -2, 0, -2, 0, 2, 2, 4, 6 \} = \{ -6,4,-2,0,2,4,6 \} \cup \{-2,0,2\}$$ so that $\Sym^2(\Sym^3(V)) \cong V^{(6)} \oplus V^{(2)}$.

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  • $\begingroup$ Thanks Jack, this works perfectly. I too noticed that the elements of $\mathfrak{sl}_2$ were acting like derivatives in terms of reduction of powers (it was clear that they would act via the typical derivation action on tensor products). I appreciate you taking the time to work this out for me in a very motivated manner. $\endgroup$ – Moderat Apr 27 '13 at 19:48
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What you call eigenvalues are more properly called weights: the eigenvalues of the diagonalisable elements conjugate to $$\begin{pmatrix}t&0\\0&t^{-1}\end{pmatrix}$$ in $SL(2,\Bbb C)$, represented as (integer) powers of $t$. The multiset of weights is the character .The standard representation has character $\{\!\{+1,-1\}\!\}$, and taking $n$-th symmetric powers correspond to taking all different $n$-multisets of weights, forming their sums, and taking the multiset of those. For the $3$-rd symmetric power of the standard representation you get character $$ \{\!\{+1+1+1,+1+1-1,+1-1-1,-1-1-1\}\!\}= \{\!\{+3,+1,-1,-3\}\!\}. $$ Taking the $2$-nd symmetric power of that, you get character $$ \{\!\{+3+3,+3+1,+3-1,+3-3,+1+1,+1-1,+1-3,-1-1,-1-3,-3-3\}\!\} = \{\!\{+6,+4,+2,0,+2,0,-2,-2,-4,-6\}\!\}, $$ which after reordering is the same multiset as mentioned in the question.

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