Exercise 1.11 Fulton Harris. Symmetric powers of standard representation of S_3

Question

I'm currently self studying Fulton Harris, unfortunately I don't really know where to check my answers, and I don't know if this logic is correct:

Notation: Let $U$, $U'$, $V$ be the trivial, alternating, standard representations of $S_3$ respectively, and $\omega = e^{2\pi i / 3}$. We already have a basis for $V$ that being $\alpha = (\omega, 1, \omega^2)$ and $\beta = (1, \omega, \omega ^2)$. We also have two transpositions $\sigma = (1 2)$ and $\tau = (1 2 3)$ which generate $S_3$, and these properties were already established:

\begin{align*} \tau \alpha =& \omega \alpha\\\\ \tau \beta =& \omega^2 \beta\\\\ \sigma \alpha =& \beta\\\\ \sigma \beta =& \alpha \end{align*}

And right before this exercise it's also established that: $$V \otimes V \cong U \oplus U' \oplus V$$ By using the basis $\\{\alpha \otimes \alpha,\alpha \otimes \beta,\beta \otimes \alpha,\beta \otimes \beta\\}$ And noticing that $\\{\alpha \otimes \alpha, \beta \otimes \beta\\}$ spans $V$ and $\alpha \otimes \beta + \beta \otimes \alpha$ spans $U$, and $\alpha \otimes \beta - \beta \otimes \alpha$ spans $U'$.

Exercise 1.11: Find a decomposition of the representations $Sym^2 V$ and $Sym^3 V$.

My answer: Since $Sym^2 V \leq V \otimes V$, the previous basis spans $Sym^2 V$, in particular since the action of $S_2$ on $\langle \alpha\otimes \alpha, \beta \otimes \beta\rangle$ is trivial so it's isomorphic to $V$. The span of $\langle \alpha \otimes \beta, \beta \otimes \alpha \rangle$ is one dimensional as $\alpha \otimes \beta = \beta \otimes \alpha$, and is thus equal to $\langle \alpha \otimes \beta + \beta \otimes \alpha\rangle = U$, so $Sym^2 V \cong U \oplus V$.

On $Sym^3 V$ we have a spanning set: $\\{\alpha \otimes \alpha \otimes \alpha,\alpha \otimes \alpha \otimes \beta, \alpha \otimes \beta \otimes \beta, \beta\otimes \beta \otimes \beta \\}$, we notice that by setting $v = \alpha \otimes \alpha \otimes \alpha$, we have $\sigma(v) = \beta \otimes \beta \otimes \beta$. And that both $v$ and $\sigma(v)$ are eigenvectors of $\tau$ with eigenvalue 1, and that $\langle \sigma(v) + v\rangle \cong U$ and $\langle \sigma(v) - v \rangle \cong U'$. By setting $w = \alpha \otimes \alpha \otimes \beta$, then $\sigma(w) = \alpha \otimes \beta \otimes \beta$, and $\tau(w) = \omega w$ and $\tau(\sigma(w)) = \omega^2 \sigma(w)$, and $\langle \sigma(v), v \rangle \cong V$, and therefore $Sym^3 V \cong U \oplus U' \oplus V$.

I'd be very grateful if anyone could check if this is correct, as this is my first exposure to symmetric powers.

Answer 1

We know $\mathrm{std}\otimes\mathrm{std}=\mathrm{triv}\oplus\mathrm{sgn}\oplus\mathrm{std}$. Now, $\wedge^2(\mathrm{std})$ is the determinant of $\mathrm{std}$, which is $\mathrm{sgn}$ (since $\mathrm{std}\oplus\mathrm{triv}=\mathbb C^3$ with the standard action, and determinant is multiplicative.)

Thus $\mathrm{Sym}^2(\mathrm{std})$, the complement of $\wedge^2(\mathrm{std})$, is simply $\mathrm{triv}\oplus\mathrm{std}$.

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Similarly, we have: $$ \begin{align*}\mathrm{std}\otimes\mathrm{std}\otimes \mathrm{std}&=(\mathrm{triv}\oplus \mathrm{sgn}\oplus\mathrm{std})\otimes \mathrm{std}\\ &=\mathrm{std}^{\oplus 2}\oplus(\mathrm{std}\otimes\mathrm{std})\\ &=\mathrm{triv}\oplus \mathrm{sgn}\oplus\mathrm{std}^{\oplus 3}. \end{align*}$$ Since $\mathrm{Sym}^3(\mathrm{std})$ is a $4$-dimensional sub-representation, by the combinatorics of the dimensions, it must be $\mathrm{triv}\oplus\mathrm{sgn}\oplus\mathrm{std}$ or $\mathrm{std}^{\oplus 2}$. To determine which scenario we are in, we can use the following argument motivated by Schur-Weyl duality:

Permuting the three copies of $\mathrm{std}$'s is a commuting $S_3$-action. It is the representation coming from the $S_3$-action on the $8$-element set $\{1,2\}^3$ by permuting the elements. This decomposes into the $S_3$-orbits:

• $(1,1,1)$
• $(2,2,2)$
• $(1,1,2),(1,2,1),(2,1,1)$
• $(2,2,1),(2,1,2),(1,2,2)$.

Thus, as an abstract representation, it must decompose as $\mathrm{triv}\oplus\mathrm{triv}\oplus(\mathrm{triv}\oplus\mathrm{std})\oplus (\mathrm{triv}\oplus\mathrm{std})=\mathrm{triv}^{\oplus 4}\oplus\mathrm{std}^{\oplus 2}$. Now, if $\mathrm{Sym}^3(\mathrm{std})=\mathrm{std}^{\oplus 2}$ above then the other $S_3$-action is $\mathrm{triv}^{\oplus 4}$, and the complement is $\mathrm{triv}\oplus\mathrm{sgn}\oplus\mathrm{std}$, which cannot be given a commuting $\mathrm{std}^{\oplus 2}$-structure.

Answer 2

This can be done by character theory. First, recall the character values of the standard representation $V$: $$\chi_V(e)=2,\chi_V(12)=0,\chi_V(123)=-1.$$

In general, $\chi_{Sym^2V}(g)=\frac12(\chi_V(g)^2+\chi_V(g^2))$. Thus, when $V$ is the standard representation: $$\begin{align*} \chi_{Sym^2V}(e)&=\frac12(\chi_V(e)^2+\chi_V(e))=3\\ \chi_{Sym^2V}(12)&=\frac12(\chi_V(12)^2+\chi_V(e))=1\\ \chi_{Sym^2V}(123)&=\frac12(\chi_V(123)^2+\chi_V(123))=0. \end{align*}$$ It is easy to see this is just $\chi_{V}+1$, so $Sym^2V\cong V\oplus\mathrm{triv}$.

A similar computation applies for $Sym^3V$, where we know the character formula $$\chi_{Sym^3V}(g)=\frac16(\chi_V(g)^3+3\chi_V(g^2)\chi_V(g)+2\chi_V(g^3)),$$ so $$\begin{align*} \chi_{Sym^3V}(e)&=\frac16(\chi_V(e)^3+3\chi_V(e)\chi_V(e)+2\chi_V(e))=4\\ \chi_{Sym^3V}(12)&=\frac16(\chi_V(12)^3+3\chi_V(e)\chi_V(12)+2\chi_V(12))=0\\ \chi_{Sym^3V}(123)&=\frac16(\chi_V(123)^3+3\chi_V(123)\chi_V(123)+2\chi_V(e))=1. \end{align*}$$ You can figure out the multiplicity of each representation in $Sym^3V$ by taking inner products. I will just do this for $\mathrm{triv}$, and leave everything else as an exercise: $$\langle\mathrm{triv},Sym^3V\rangle=\frac16(1.4+3.0+2.1)=1.$$

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Let's generalize this, to the decomposition of $Sym^nV$. If a matrix $A$ acts on a vector space $V$ with eigenvalues $\alpha,\beta$, then the action on the symmetric power $Sym^nV$ has eigenvalues $\alpha^n,\alpha^{n-1}\beta,\dots,\beta^n$. Thus the trace is $\alpha^n+\cdots+\beta^n=(\alpha^{n+1}-\beta^{n+1})/(\alpha-\beta)$.

Since the eigenvalues of $\pi_V(e)$ are $\{1,1\}$, the eigenvalues of $\pi_V(12)$ are $\{1,-1\}$, and the eigenvalues of $\pi_V(123)$ are $\{\zeta_3,\zeta_3^{-1}\}$, we have: $$\begin{align*} \chi_{Sym^nV}(e)&:=n+1\\ \chi_{Sym^nV}(12)&=\frac12(1-(-1)^{n+1})\\ \chi_{Sym^nV}(123)&=(\zeta_3^{n+1}-\zeta_3^{-n-1})/(\zeta_3-\zeta_3^{-1}). \end{align*} $$ Thus this depends on $n$ modulo $6$: $$\begin{align*} \mathrm{Sym}^n(V)&=\begin{cases} \frac{n+6}6triv+\frac{n}6sgn+\frac{n}3std&n\equiv0\pmod 6\\ \frac{n-1}6triv+\frac{n-1}6sgn+\frac{n+2}3std&n\equiv1\pmod 6\\ \frac{n+4}6triv+\frac{n-2}6sgn+\frac{n+1}3std&n\equiv2\pmod 6\\ \frac{n+3}6triv+\frac{n+3}6sgn+\frac{n}3std&n\equiv3\pmod 6\\ \frac{n+2}6triv+\frac{n-4}6sgn+\frac{n+2}3std&n\equiv4\pmod 6\\ \frac{n+1}6triv+\frac{n+1}6sgn+\frac{n+1}3std&n\equiv5\pmod 6 \end{cases} \end{align*} $$