I am currently working on representation theory in my algebra class and we are asked the the following question. I was hoping I could get some verification of my proofs although I'm pretty sure that my solution for 3 is wrong as I have found the answer elsewhere using character theory, and it does not match up with mine. Part of the reason I am seeking help is that my linear background is rather weak, so I don't really understand what is going on in the background with the eigenvectors and what have you.
Let $V$ denote the standard representation of $S_3$. Using the ad hoc approach presented in Fulton and Harris, compute the decompositions into direct sums of irreducible representation of the following representations of $S_3$
- Sym$^2V$, Sym$^3V$, and Sym$^kV$ for all $k\in\textbf{N}$
- $\bigwedge^2(V),\bigwedge^3(V)$, and $\bigwedge^k(V)$ for all $k\in\textbf{N}$
- $V^{\otimes2},V^{\otimes3}$, and $V^{\otimes k}$ for all $k\in\textbf{N}$
For all of the problems, I am taking $\tau=(123)$ and $\sigma=(12)$. For 1, I am fairly certain I am correct so I will post only my answer and if that is wrong I will gladly put up my whole proof.
Sym$^2(V)$ decomposes as $V\oplus U$ where $V$ is the standard representation of $S_3$ and $U$ is the trivial representation, Sym$^3(V)$ decomposes as $V\oplus V$, and in general if $k$ is even, then Sym$^k(V)$ decomposes as $\bigoplus\limits^{\frac{k}{2}}V\oplus U$, and if $k$ is odd, then Sym$^k(V)$ decomposes as $\bigoplus\limits^{k-1} V$.
For 2, I am less sure.
Recall that $\bigwedge^2(V)=\textbf{C}$-span$\{u\wedge v|u,v\in V\}$ where $u\wedge v=\frac{1}{2}(u\otimes v-v\otimes y)$. Using the same setup as in problem 1, we see that a basis for $\bigwedge^2(V)$ is $\{\alpha\wedge\beta\}$ as $\alpha^2=\beta^2=0$, and $\beta\wedge\alpha=-\alpha\wedge\beta$ where concatenation is understood to be the wedge product. We compute that $\tau(\alpha\wedge\beta)=\frac{1}{2}(\alpha\otimes\beta-\beta\otimes\alpha)$ and likewise that $\sigma(\alpha\wedge\beta)=\frac{1}{2}(\beta\otimes\alpha-\alpha\otimes\beta)$. Since $\tau$ has sign 1 and $\sigma$ has sign -1, we conclude that $\bigwedge^2(V)$ is equivalent to the alternating representation.
For $\bigwedge^3(V)$ we follow the same procedure and get that our ``basis" is $\{\vec{0}\}$, meaning that our vector space is the zero vector space. Therefore, $\bigwedge^3(V)$ does not decompose into a sum of irreducible representations of $S_3$. Likewise, we have that $\bigwedge^k(V)$ does not decompose into a sum of irreducible representations of $S_3$ for all $k\geq 3$ where $k\in\textbf{N}$.
For 3,
Recall that $V^{\otimes2}=\textbf{C}$-span$\{u\otimes v|u,v\in V\}$ where $\otimes$ is not commutative. Using the same setup as in problem 1, we see that a basis for $V^{\otimes2}$ is $\{\alpha^2,\alpha\beta,\beta\alpha,\beta^2\}$ where we understand the operation to be the tensor product. Thus, $\tau(\alpha^2)=\omega^2\alpha^2,\tau(\alpha\beta)=\alpha\beta,\tau(\beta\alpha)=\beta\alpha$, and $\tau(\beta^2)=\omega\beta^2$. Likewise, $\sigma(\alpha^2)=\beta^2,\sigma(\alpha\beta)=\beta\alpha,\sigma(\beta\alpha)=\alpha\beta$, and $\sigma(\beta^2)=\alpha^2$. Thus, $\{\alpha^2,\beta^2\}$ is a subspace of $V^{\otimes2}$ with dimension 2 that is $S_3$ invariant, so it is equivalent to the standard representation. Likewise, $\{\alpha\beta,\beta\alpha\}$ is a subspace of $V^{\otimes2}$ with dimension 2 that is $S_3$ invariant, so it is equivalent to the standard representation. Therefore, $V^{\otimes2}$ decomposes as $V\oplus V$.
For $V^{\otimes3}$ we follow the same procedure using $\{\alpha^3,\alpha^2\beta,\alpha\beta\alpha,\beta\alpha^2,\alpha\beta^2,\beta\alpha\beta,\beta^2\alpha,\beta^3\}$ as our basis. Thus, we get that $\tau(\alpha^3)=\alpha^3,\tau(\alpha^2\beta)=\omega\alpha^2\beta,\tau(\alpha\beta\alpha)=\omega\alpha\beta\alpha,\tau(\beta\alpha^2)=\omega\beta\alpha^2,\tau(\alpha\beta^2)=\omega^2\alpha\beta^2,\tau(\beta\alpha\beta)=\omega^2\beta\alpha\beta,\tau(\beta^2\alpha)=\omega^2\beta^2\alpha$, and $\tau(\beta^3)=\beta^3$. Likewise, $\sigma(\alpha^3)=\beta^3,\sigma(\alpha^2\beta)=\beta^2\alpha,\sigma(\alpha\beta\alpha)=\beta\alpha\beta,\sigma(\beta\alpha^2)=\alpha\beta^2,\sigma(\alpha\beta^2)=\beta\alpha^2,\sigma(\beta\alpha\beta)=\alpha\beta\alpha,\sigma(\beta^2\alpha)=\alpha^2\beta$, and $\sigma(\beta^3)=\alpha^3$. Thus, $\{\alpha^3,\beta^3\},\{\alpha^2\beta,\beta^2\alpha\},\{\alpha\beta\alpha,\beta\alpha\beta\},\{\beta\alpha^2,\alpha\beta^2\}$ are all subspaces of $V^{\otimes3}$ with dimension 2 that are $S_3$ invariant, and thus are all equivalent to the standard representation. Therefore, $V^{\otimes3}$ decomposes as $V\oplus V\oplus V\oplus V$.
In general, if $k\in\textbf{N}$ then each pairing of linear combinations of $\{v_1\otimes...\otimes v_k\}$ where each $v_i\in\{\alpha,\beta\}$ forms $2^{k-1}$ subspaces of dimension two. Therefore, $V^{\otimes k}$ decomposes as $\bigoplus\limits^{2^{k-1}}V$.
Any help would be appreciated.