# Representation Decomposition of $\operatorname{Sym}^{k+6} V$

In Fulton and Harris's Text Representation Theory: A First Course, exercise 1.12(b) asks to show that $\operatorname{Sym}^{k+6}V \cong \operatorname{Sym}^kV \oplus R$ as representations of $\frak S_3$. $V$ is the 2-dimensional standard representation of $\frak S_3$ and $R$ is the regular representation. Note that this is supposed to be done without using character theory.

The hint given by my professor was to show that $\operatorname{Sym}^6 V \cong U \oplus R \cong U^{\oplus 2} \oplus U' \oplus V^{\oplus 2}$, where $U$ is the trivial representation and $U'$ is the alternating representation. Then, we're supposed to use that to find copies of $\operatorname{Sym}^k V$ and $R$ in $\operatorname{Sym}^{k+6} V$ that intersect only at $0$, which will then prove the original isomorphism.

Proving the congruence was relatively straightforward, but I have no idea how we're supposed to find copies of $\operatorname{Sym}^k V$ and $R$ in $\operatorname{Sym}^{k+6} V$.

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In case I'm not the only one who hadn't seen this notation before: $\frak S_3$ is the symmetric group on a set of three elements, otherwise known as $S_3$. –  joriki Sep 7 '11 at 7:16
Just some thoughts which may or may not lead to a solution: There is a natural map $Sym^k V \otimes Sym^6 V \to Sym^{k+6}V$. Since $Sym^6 V \cong U \oplus R$, the left side of the map contains a copy of $Sym^k V \otimes U \cong Sym^k V$. I guess (but don't see quite how to prove) that it's isomorphic to its image in $Sym^{k+6}V$ so that gives you the copy of $Sym^k V$ in $Sym^{k+6} V$. To find a copy of $R$, at least if $k>1$, you can show that there is always a copy of the trivial representation in $Sym^k V$. Needless to say I haven't checked the details but this may work. –  Ted Sep 7 '11 at 8:23
@Ted: I was thinking along similar lines. The mapping $Sym^kV\otimes U\rightarrow Sym^{k+6}V$ should be injective. After all, the symmetric algebra is like the polynomial algebra with basis vectors of $V$ as unknowns. So multiplication by a homogeneous polynomial of degree six (=a basis element of $U$) is surely injective. I don't see a natural way of identifying the cokernel with $R$ though. –  Jyrki Lahtonen Sep 7 '11 at 8:34

So, using the notation and the basis $\alpha,\beta$ of $V$ from that chapter, the elements of $\text{Sym}^kV$ can be paired up as $\alpha^j\beta^{k-j}$ and $\alpha^{k-j}\beta^j$ (except for $\alpha^{k/2}\beta^{k/2}$ for even $k$). The transposition $\sigma$ transforms these into each other, and their eigenvalues under $\tau$ are $\omega^j\omega^{2(k-j)}=\omega^{2k-j}=\omega^{-(k+j)}$ and $\omega^{k-j}\omega^{2j}=\omega^{k+j}$, respectively. Thus, if $k+j\equiv0\pmod3$, they form a trivial and an alternating representation, and if $k+j\not\equiv0\pmod3$ they form the standard representation. Thus, if $k\gt6$, there are three successive values of $j$ for which the three pairs together form two standard, one alternating and one trivial representation, i.e. the regular representation. The remaining values of $j$ have the same remainders mod $3$ as the values for $k-6$, which implies $\text{Sym}^{k+6}V \cong \text{Sym}^kV \oplus R$.