2
$\begingroup$

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 $\operatorname{Sym}^{k+6} V$.

  • 0
    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 $S_3$.2011-09-07
  • 1
    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.2011-09-07
  • 0
    @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.2011-09-07

1 Answers 1