Nailing down the Wedderburn decomposition

Wildon's Weblog 2024-09-02

The purpose of this post is to give a very explicit Wedderburn decomposition for the group algebra $\mathbb{Q}\mathrm{Sym}_3$ of the symmetric group of degree $3$ over the rational field. Even this small example is rich enough to show some interesting features, and I want to write down the calculation, rather than carry on doing it approximately once a year for the rest of my (mathematical) life.

One feature that is new (for me) this time around, is that the Wedderburn isomorphism is isometric with respect to the canonical inner product of $\mathbb{Q}\mathrm{Sym}_3$ , in which the group elements are an orthonormal basis. Despite this, there is no sense in which the chosen isomorphism is entirely canonical. I’ve tried to bring this out below by showing that, while $2$ -dimensional left ideals in $\mathrm{Mat}_2(\mathbb{Q})$ (corresponding to $2$ -dimensional left ideals in $\mathbb{Q}\mathrm{Sym}_3$ isomorphic to the irreducible Specht module $S^{(2,1)}$ ) are indeed parameterised by $\mathbb{P}^1(\mathbb{Q})$ , as one would expect, the parameterisation of the full set of idempotents is more subtle, and it needs two parameters to get the whole lot. As a deliberately slightly tricky exercise you might like to guess the value of

$\left\langle \left( \begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix} \right) , \left( \begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix} \right) \right\rangle$

where the inner product is defined from the canonical bilinear form on $\mathbb{Q}[\mathrm{Sym}_3]$ in which the group elements are orthonormal by taking the component of $2 \times 2$ matrices in the image of the isometric Wedderburn isomorphism

$\mathbb{Q}[\mathrm{Sym}_3] \cong \mathbb{Q} \oplus \mathbb{Q} \oplus \mathrm{Mat}_2(\mathbb{Q}).$

The answer is not dependent on the chosen isometric isomorphism, and can be found, without extensive calculation, using only the dimensions of the irreducible characters of $\mathrm{Sym}_3$ . As a small hint, even though this matrix is clearly an idempotent (it’s even one of the special class of ‘perpendicular’ idempotents defined below), the answer is not $1$ .

The general Wedderburn decomposition

We start with the observation that if $G$ is any finite group and $V$ is an irreducible $\mathbb{C}[G]$ -module with character $\chi$ then the multiplicity of $V$ in any direct sum decomposition of $\mathbb{C}[G]$ into irreducible is $\chi(1)$ . For instance, this follows from the fact that the character of $\mathbb{C}[G]$ as a left module for itself is the regular character $\pi$ , defined by $\pi(\mathrm{id}_G) = |G|$ and $\pi(g) = 0$ for $g \not= \mathrm{id}_G$ , and so $\langle \pi, \chi \rangle = \frac{1}{|G|} |G| \chi(1) = \chi(1)$ . Therefore, as a left $\mathbb{C}[G]$ -module we have

$\mathbb{C}[G] = \bigoplus_{V \in \mathrm{Irr}(G)} \bigl( V \oplus \cdots \oplus V \bigr)$

where there are $\mathrm{dim} V$ summands in each parenthesised direct sum. By Schur’s Lemma $\mathrm{Hom}_{\mathbb{C}[G]}(U, V) = 0$ if $U \not\cong V$ and $\mathrm{Hom}_{\mathbb{C}[G]}(U, U) \cong \mathbb{C}$ . It follows routinely that the endomorphism algebra of $\mathbb{C}[G]$ as a left $\mathbb{C}[G]$ -module satisfies

$\mathrm{End}_{\mathbb{C}[G]}(\mathbb{C}[G]) \cong \bigoplus_{V \in \mathrm{Irr}(G)} \mathrm{Mat}_{\mathrm{dim} V}(\mathbb{C}).$

For instance, to see this in the case relevant below where $\dim V = 2$ , fix a `reference copy’ of $V$ and choose a non-zero $v \in V$ , and take another copy $V'$ isomorphic to $V$ under a unique up to scalars (again by Schur’s Lemma) isomorphism sending $v \in V$ to $v' \in V'$ . Then $V \oplus V'$ is generated as a left $\mathbb{C}[G]$ -module by $v$ and $v'$ , and any $\mathbb{C}[G]$ -endomorphism $\theta$ of $V \oplus V'$ is determined by its effect on $v$ and $v'$ , which must be $\theta(v) = \alpha v + \beta v'$ and $\theta(v') = \gamma v + \delta v'$ for some scalars $\alpha,\beta,\gamma, \delta \in \mathbb{C}$ , since, composing with the projection to $V'$ , the map $V \rightarrow V'$ defined on the chosen generator by $v \mapsto \beta v'$ is a $\mathbb{C}[G]$ -homomorphism.

It might seem that we have landed on the Wedderburn decomposition, but there is one subtlety, which shows that a proof as above using Schur’s Lemma is required, rather than any more naive argument that says, in effect, “well I know the simple modules, and ‘obviously’ the group algebra is the sum of their endomorphism rings over the field”. (This argument works to decompose $\mathrm{C}[G]$ as a left-module for itself, but not as an algebra.) To explain this subtlety, recall that if $A$ is an associative algebra then the opposite algebra to $A$ is the algebra with the same underlying vector space as $A$ , but with multiplication defined by $x \cdot_\mathrm{op} y = yx$ .

Lemma $\mathrm{End}_{\mathbb{C}[G]}(\mathbb{C}[G]) \cong \mathbb{C}[G]^{\mathrm{op}}$

Proof. Let $\theta \in \mathrm{End}_{\mathbb{C}[G]}(\mathbb{C}[G])$ . Observe that since $\theta$ is a homomorphism of left $\mathbb{C}[G]$ -modules, once we know $\theta(\mathrm{id}_G)$ , an arbitrary image $\theta(g)$ is determined by

$\theta(g) = g \theta(\mathrm{id}_G).$

Conversely, we may choose any $x \in \mathbb{C}[G]$ as $\theta_x(\mathrm{id}_G)$ , and then the displayed equation determines $\theta$ uniquely. Thus, as a vector space, there is an isomorphism $\mathrm{End}_{\mathbb{C}[G]}(\mathbb{C}[G])$ defined by $\theta \mapsto \theta(\mathrm{id}_G)$ . If under this isomorphism, $\theta \mapsto x \in \mathbb{C}[G]$ and $\phi \mapsto y \in \mathbb{C}[G]$ then

$(\theta \phi)(\mathrm{id}_G) = \theta(y) = y\theta(\mathrm{id}_G) = y x$

and so the linear isomorphism is an algebra isomorphism as required in the lemma. $\Box$

There is however a way to remove the `opposite’ above, using that $\mathrm{Mat}_d(\mathbb{C})$ is isomorphic to its opposite algebra, by the map $A \mapsto A^\mathrm{tr}$ transposing a matrix $A$ . Thus having paid due attention to this technical point, we may remove it and obtain Wedderburn’s Theorem in the more usual form:

$\mathbb{C}[G] \cong \bigoplus_{V \in \mathrm{Irr}(G)} \mathrm{Mat}_d (\mathbb{C}).$

We finish these generalities by remarking that if $\chi \in \mathrm{Irr}(G)$ then

$e_\chi \frac{1}{|G|} \sum_{g \in G} \chi(g^{-1}) g \qquad (\star)$

has constant coefficients on the conjugacy classes of $G$ and so lies in $Z(\mathbb{C}[G])$ . By the row orthogonality relations for irreducible characters, the $Ze_\chi$ are mutually orthogonal idempotents: $Ze_\chi^2 = e_\chi$ and $Ze_\chi e_\phi = e_\phi e_\chi = 0$ for $\phi \not= \chi$ . By the column orthogonality with the column for the identity element, $\sum_{\chi \in \mathrm{Irr}(G)} e_\chi = \mathrm{id}_G$ . Moreover the $e_\chi$ are primitive because the `character table is square’: we have $|\mathrm{Irr}(G)|$ idempotents and it is easily seen that the dimension of $Z(\mathbb{C}[G])$ is the number of conjugacy classes of the group. We therefore have

$\mathbb{C}[G] = \bigoplus_{\chi \in \mathrm{Irr}(G)} e_\chi \mathbb{C}[G] e_\chi$

and we can append the rider ‘where $\mathrm{Mat}_{\mathrm{dim} V}(\mathbb{C})$ is isomorphic to the opposite algebra of $e_\chi \mathbb{C}[G] e_\chi$ ‘ to $(\star)$ above.

Wedderburn blocks for $\mathrm{Sym}_3$

To apply this theory to $\mathrm{Sym}_3$ we recall that the irreducible characters of $\mathrm{Sym}_3$ are $1_G$ , the trivial character, $\mathrm{sgn}$ , the sign character and $\chi$ , the degree $2$ irreducible constituent of the natural permutation character, satisfying $\chi(g) = |\mathrm{Fix}(g)| - 1$ . To construct a left-module corresponding to $\chi$ , take $M = \langle e_1, e_2, e_3 \rangle$ , the natural permutation module, and then take the submodule $S = \langle e_1 - e_2, e_1 - e_3 \rangle$ . All these modules are defined over the rational numbers, and so we may work over $\mathbb{Q}$ , obtaining three orthogonal central idempotents

$\begin{aligned} e_{1_G} &= \textstyle \frac{1}{6} \bigl( 1 + (1,2,3) + (1,3,2) + (1,2) + (1,3) + (2,3) \bigr) \\ e_\mathrm{sgn} &= \textstyle\frac{1}{6} \bigl( 1 + (1,2,3) + (1,3,2) - (1,2) - (1,3) - (2,3) \bigr) \\ e_\chi &= \textstyle\frac{1}{3} \bigl(2 - (1,2,3) - (1,3,2) \bigr).\end{aligned}$

The Wedderburn decomposition $(\star)$ becomes

$\begin{aligned} & \mathbb{Q}[\mathrm{Sym}_3] \\ &\quad = e_1 \mathbb{Q}[\mathrm{Sym}_3] e_1 + e_\mathrm{sgn} \mathbb{Q}[\mathrm{Sym}_3]e_\mathrm{sgn} + e_\chi \mathbb{Q}[\mathrm{Sym}_3] e_\chi \\ &\quad\cong \mathbb{Q} \oplus \mathbb{Q} \oplus \mathrm{Mat}_2(\mathbb{Q}).\end{aligned}$

Making the $2 \times 2$ -block explicit

To be continued.

Nailing down the Wedderburn decomposition

Wildon's Weblog 2024-09-02

The general Wedderburn decomposition

Wedderburn blocks for

Making the -block explicit

Wedderburn blocks for $\mathrm{Sym}_3$

Making the $2 \times 2$ -block explicit