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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
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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

1 Answer 1

This exercise makes a lot more sense in the context of the book. In fact it's probably one of the few problems about representations of finite groups for which the approach in that chapter is more suitable than character theory :-). (The authors themselves state that the approach is "superfluous" for finite groups and they're only using it to introduce an idea that's useful for Lie groups.)

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 $\text{Sym}^{k+6}V \cong \text{Sym}^kV \oplus R$.

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