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

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.

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 $$\mathrm{triv}\oplus\mathrm{std}$$.

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.

• Thank you very much! Jul 7 at 17:08

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: $$\langle\mathrm{triv},Sym^3V\rangle=\frac16(1.4+3.0+2.1)=1.$$

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*}

• I'm sorry! This exercise is just before character theory, probably to show how you'd do it for other fields. I don't really want to move on before understanding if this is correct, would you be able to confirm if my final result for $Sym^3 V$ is correct (no need for a proof)? That is, is $Sym^3 V \cong \text{trivial} \oplus \text{alternating} \oplus V$? In either case thank you, I'll check out your proof after reading the section on character theory. Jul 7 at 8:13
• Yeah, that looks right. I can try to type up a non-character theoretic proof when I have time. Jul 7 at 9:21
• I'm sorry, but shouldn't the denominator on std be 6 instead of 3? Apologies if this is correct, but Fulton Harris says the regular representation, R (space of functions from $S_3$ to $\mathbb C$ indexed by elements of $S_3$ with $f_g(x) = 0$ if $x \neq g$ and $f_g(x) = 1$ if $x = g$), should satisfy the property that $Sym^{n+6} V= Sym^n V \oplus R$. However I got $R \cong triv \oplus sgn \oplus 2 std$. Which dimension-wise makes sense as it should be a 6-dimensional vector space. While through your formula I'd need $R \cong triv \oplus sgn \oplus 4 std$ which has dimension 10.Edit: mistake Jul 10 at 14:43
• Oh yes, the denominator should probably be $6$! Will fix later. Jul 10 at 23:58