# Representation of the Symmetric Group $S_3$

I am reading Representation Theory A First Course by William Fulton and Joe Harris. In Section 3, Lecture 1, they gave a method to find all the irreducible representations of the symmetric group $$\mathfrak{S}_3$$.

Suppose $$W$$ is an representation of $$\mathfrak{S}_3$$. ($$W$$ is a complex vector space). Let $$\tau$$ be a generator of the subgroup $$\mathfrak{A}_3\subset \mathfrak{S}_3$$, for example, $$(123)$$. Then the space $$W$$ is spanned by eigenvectors $$v_i$$ for the action of $$\tau$$.

I don't understand why $$W$$ is spanned by eigenvectors of $$\tau$$. Even in complex field, not all invertible matrix is diagonalizable. I guess it may have something to do with the fact that $$\tau^3 = 1$$, and so the eigenvalues of $$\tau$$ are all roots of unit.

• A nondiagonalizable complex matrix cannot satisfy $\tau^3 = {\bf 1}$: To see this, observe that the $3$rd power (for that matter, the $n$th power for all $n \geq 1$) of any Jordan block of size $> 1$ is not the identity matrix. Sep 22, 2023 at 12:59
• Thank you. :) So every action by a group element on a finite-dimensional vector space is diagonlizable? That is a little surprising to me. Sep 22, 2023 at 13:04
• If the group is finite, yes, for that reason. But that's not always the case for infinite groups---just consider the standard action of a nondiagonalizable element of $\operatorname{GL}(\Bbb V)$ (where the underlying field has characteristic $0$). Sep 22, 2023 at 13:16

You can check this by hand, and then you will notice a pattern:

Say $$\tau$$ has order $$3$$, and acts on $$W$$. Let $$\omega = \exp\frac{2 \pi i}{3}$$. We can write every element $$w$$ in $$W$$ as

$$w = w_0 + w_1 + w_2$$

where $$w_k = \frac{1}{3} \cdot\sum_{l=0}^2 \omega^{-k l} \tau^l w$$

Also

$$\tau w_k = \frac{1}{3}\sum_l\omega^{-k l} \tau^{l+1} w= \omega^k w_k$$

Similar to " every function is the sum of an even function and an odd function".

Every action $$\rho(g)$$ by an element $$g$$ of a complex representation $$G \hookrightarrow GL(n, \Bbb C)$$ of a finite group $$G$$ is diagonalizable. Suppose not, that is, that for some $$g \in G$$ the Jordan normal form of $$\rho(g)$$ contains a Jordan block of size $$k > 1$$, say, $$J_k(\lambda)$$. Now, for $$n \geq 1$$, $$J_k(\lambda)^n \sim J_k(\lambda^n)$$, so $$\rho(g^n) = \rho(g)^n$$ also contains a Jordan block of size $$k > 1$$. On the other hand, if we set $$n$$ to be the order of $$g$$, then $$\rho(g^n) = \rho(1_G) = {\bf 1}$$, whose Jordan blocks all have size $$1$$, a contradiction.

(In fact, computing gives that the superdiagonal entries of $$J_k(\lambda)^n$$ are $$n \lambda^{n - 1} = n \lambda^{-1} \neq 0$$, so if $$k > 1$$ then $$J_k(\lambda)^n \not\sim {\bf 1}_k$$.)

The general result is the following. Let $$V$$ be a finite-dimensional vector space over a field $$k$$ and $$T : V \to V$$ a linear map.

Proposition: Suppose $$f(t) \in k[t]$$ is a polynomial which splits over $$k$$ and has no repeated roots such that $$f(T) = 0$$. Then $$T$$ is diagonalizable.

Note that $$f$$ doesn't have to be either the minimal or the characteristic polynomial. If $$G$$ is a finite group, $$\rho : G \to GL(V)$$ is a representation of $$G$$ over an algebraically closed field $$k$$, and $$g \in G$$ has order $$n$$, then applying the above proposition, $$\rho(g)^n = 1$$ implies that $$\rho(g)$$ is diagonalizable as long as $$\text{char}(k) \nmid n$$.

Proof. Write $$f(t) = \prod_{i=1}^n (t - \lambda_i)$$. We are actually going to write down, explicitly, the projections onto each eigenspace of $$T$$. The idea is that the polynomial $$\frac{f(t)}{t - \lambda_i} = \prod_{j \neq i} (t - \lambda_j)$$ vanishes everywhere except at $$t = \lambda_i$$, so the polynomial

$$f_i(t) = \frac{f(t)}{(t - \lambda_i) \prod_{j \neq i} (\lambda_i - \lambda_j)}$$

vanishes everywhere except at $$t = \lambda_i$$ where it takes the value $$1$$. Now if $$v \in V$$ we compute that

$$(T - \lambda_i) f_i(T) v = \frac{f(T)}{\prod_{j \neq i} (\lambda_i - \lambda_j)} v = 0$$

so $$f_i(T) v$$ is an eigenvector with eigenvalue $$\lambda_i$$. Moreover, if $$v_i \in V_i = \text{ker}(T - \lambda_i)$$ is already such an eigenvector then

$$f_i(T) v_i = f_i(\lambda_i) v_i = v_i$$

so $$f_i(T)$$ fixes every eigenvector with eigenvalue $$\lambda_i$$. It follows that $$f_i(T)$$ is a projection onto the eigenspace $$V_i$$ as desired (which may be zero since $$f$$ is not necessarily the minimal polynomial).

From here we want to show that every $$v \in V$$ is a sum of eigenvectors. This would follow if we could show that $$\sum_{i=1}^n f_i(T) = I$$. Now, the polynomial $$\sum_{i=1}^n f_i(t) - 1$$, by construction, takes the value $$0$$ at each $$\lambda_i$$. But it also has degree $$n-1$$, and has $$n$$ roots. So it must be identically zero, and we have $$\sum_{i=1}^n f_i(T) = I$$ and hence every vector $$v \in V$$ can be written as a sum of eigenvectors

$$v = \sum_{i=1}^n f_i(T) v$$

as desired. (We can also show that $$f_i(T) f_j(T) = 0$$ for $$i \neq j$$ so the $$f_i(T)$$ are a system of orthogonal idempotents corresponding to the direct sum decomposition $$V \cong \bigoplus_i V_i$$ into eigenspaces but it's not totally necessary.) $$\Box$$

This can also be proven more abstractly by observing that $$V$$ has the structure of a module over $$k[t]/f(t) \cong k^n$$ but this argument can feel a little bloodless sometimes and it's nice to know that it can be made a lot more explicit. In any case it requires knowing a little commutative algebra whereas the above argument requires only a solid grasp of how polynomials behave. It's overkill bordering on circular to cite the Jordan normal form theorem here, because a generalization of the above result is used as a lemma in some proofs of that theorem.

In the special case that $$f(t) = t^n - 1$$ relevant to representations of finite groups the explicit direct sum decomposition above is essentially the discrete Fourier transform, or what is known sometimes as the "roots of unity filter."