## G10(a). Character Table of S4

Let’s construct the character table for $G = S_4$. First, we have the trivial and alternating representations (see examples 1 and 2 in G1), both of which are clearly irreducible.

Next, the action of G on {1, 2, 3, 4} induces a linear action of G on a space of dimension 4, by permuting the coordinates. The character for this last action is easy: since each $g\in G$ maps to a permutation matrix, the trace is precisely the number of fixed points. This gives:

Note that $\left<\chi_0, \chi_{\text{triv}}\right> = \frac 1 {24}(4+12+8) = 1$ so $\chi_0$ contains 1 copy of the trivial representation. And since $\left<\chi_0, \chi_{\text{alt}}\right> = 0$ it doesn’t include the alternating representation. Subtracting gives: $\chi_1 := \chi_0 - \chi_{\text{triv}} = (3, 1, 0, -1, -1).$

It’s easy to check that $\left<\chi_1, \chi_1\right> = 1$ so we’ve found another irreducible representation. Tensor product gives $\chi_1\chi_{\text{alt}} = (3, -1, 0, 1, -1)$ which is also easily checked to be irreducible. Since we’ve found 4 out of 5 irreducible representations, the remaining one is easy. First, the degree must be $\sqrt{24 - 1^2 - 1^2 - 3^2 - 3^2} = 2$ so the regular representation must contain two copies of it. This gives:

$\chi_2 := \frac 1 2 (\chi_{\text{reg}} - \chi_{\text{triv}} - \chi_{\text{alt}} - 3\chi_1 - 3\chi_1\chi_{\text{alt}}) = (2, 0, -1, 0, 2).$

We thus obtain the full character table:

## G10(b). Character Table of S5

With a bit more effort, we can do $G = S_5$ as well. As before, we have the trivial and alternating representations. Again, the natural action of G on {1, 2, 3, 4, 5} gives:

Now, $\left<\chi_0, \chi_{\text{triv}}\right> = \frac 1 {120}(5 + 30 + 40 + 30 +15)=1$ and $\left<\chi_0, \chi_{\text{alt}}\right> = 0$ so we take the difference $\chi_1 := \chi_0 - \chi_{\text{triv}} = (4, 2, 1, 0, -1, 0, -1).$ One easily checks that $\left<\chi_1, \chi_1\right> = 1$ so we’ve found an irreducible representation. With it comes $\chi_1\chi_{\text{alt}},$ which is easily checked to be irreducible as well.

Thus we’ve found four. For the remaining three, let’s consider tensor products. Now $\chi_1^2$ is a rather huge representation (of degree 16), so let’s take a subspace instead.

Interlude : Symmetric and Alternating Tensors

Let V be a vector space and consider $V\otimes V$. The swapping map $V\times V\to V\otimes V$ which takes $(x,y)\mapsto y\otimes x$ is clearly bilinear, so it induces a linear map $f:V\otimes V\to V\otimes V$ which takes $x\otimes y\mapsto y\otimes x.$ Clearly f2 is the identity, since it maps all elements of the form $x\otimes y$ back to themselves and the set of all such elements spans $V\otimes V.$ Since a power of f is the identity, f is diagonalisable. Its eigenvalues must satisfy $\lambda^2 = 1$ so:

$V\otimes V = \{w\in V\otimes V: f(w)=w\} \oplus \{w\in V\otimes V: f(w)=-w\}.$

We’ll denote the two spaces by $\text{Sym}^2 V$ and $\text{Alt}^2 V$ respectively. Clearly, if $\{e_1, \ldots, e_n\}$ is a basis of V, then a basis of $\text{Sym}^2 V$ (resp. $\text{Alt}^2 V$) is given by:

$\{e_i \otimes e_j + e_j\otimes e_i\}_{i\le j}\$ (resp. $\ \{e_i \otimes e_j - e_j\otimes e_i\}_{i).

Thus, the two spaces are of dimensions $\frac {n(n+1)}2$ and $\frac{n(n-1)}2$ respectively.

Computing Characters for Symmetric/Alternating Tensors

Now suppose V is a C[G]-module. We claim that the subspace $\text{Sym}^2 V\subseteq V\otimes V$ is invariant under every $g\in G.$ To prove this, it suffices to show g commutes with the swapping map above $f:V\otimes V\to V\otimes V$ which takes $x\otimes y\to y\otimes x.$ This fact isn’t hard to prove:

$f(g(x\otimes y)) = f(gx\otimes gy) = gy\otimes gx = g(y\otimes x) = g(f(x\otimes y))$

for any $x,y\in V$. Since f commutes with g, a standard result from linear algebra tells us g is invariant on the eigenspaces of f. Thus $g(\text{Sym}^2 V)\subseteq \text{Sym}^2 V$ and $g(\text{Alt}^2 V)\subseteq \text{Alt}^2 V.$

Our next task is to express $\chi_{\text{Sym}^2 V}(g)$ in terms of $\chi_V$.

If $\{e_1, \ldots, e_n\}$ is a basis of eigenvectors of g, with eigenvalues $\{\lambda_1, \ldots, \lambda_n\},$ then we have:

$g(e_i \otimes e_j + e_j\otimes e_i) = \lambda_i \lambda_j (e_i\otimes e_j + e_j\otimes e_i),$

so the trace of g on $\text{Sym}^2 V$ is $\sum_{i=1}^n \lambda_i^2 + \sum_{i. Since $\chi_V(g) = \sum_i \lambda_i$ and $\chi_V(g^2) = \sum_i \lambda_i^2,$ it follows that:

$\chi_{\text{Sym}^2 V}(g) = \frac 1 2 (\chi_V(g)^2 + \chi_V(g^2)) \implies \chi_{\text{Alt}^2 V}(g) = \frac 1 2 (\chi_V(g)^2 - \chi_V(g^2)).$

If we take $\chi = \chi_1 = (4, 2, 1, 0, -1, 0, -1)$ above, then:

$\chi_2 := \chi_{\text{Sym}^2 V} = (10, 4, 1, 0, 0, 2, 1)$ and $\chi_3 :=\chi_{\text{Alt}^2 V} = (6, 0, 0, 0, 1, -2, 0).$

Now $\chi_3$ is irreducible. On the other hand $\left<\chi_2, \chi_{\text{triv}}\right> = 1$ and $\left<\chi_2, \chi_1\right> = 1$ so taking the difference gives the sixth irreducible character:

$\chi_4 = \chi_2 -\chi_{\text{triv}} - \chi_1 = (5, 1, -1, -1, 0, 1, 1).$

The last one is obviously $\chi_4 \chi_{\text{alt}}$. Conclusion:

## G10(c). Character Table of A5

Now we take the alternating group $G = A_5$. Since this is a subgroup of S5, let’s take the above 7 characters and restrict them to $A_5 \subset S_5.$ However, we’re now left with only 4 since 3 of the characters are paired via $\{\chi, \chi'\}$ such that $\chi' = \chi\cdot\chi_{\text{alt}}$ (we say that the two characters are twists of each other).

A bit of effort tells us A5 has 5 conjugancy classes: e, (1, 2, 3), (1, 2, 3, 4, 5), (1, 2, 3, 4, 5)2, (1, 2)(3, 4). Restriction then gives:

The first three are irreducible, while the last one satisfies $\left<\chi_3, \chi_3\right> = 2$ so it’s a direct sum of two non-isomorphic irreducible representations. Since $\chi_3$ is orthogonal to the other three characters, it is the direct sum of the two remaining irreducible characters. Their dimensions satisfy $d_1 + d_2 = 6$ and $d_1^2 + d_2^2 = 18$ so $d_1 = d_2 = 3.$

Critical observation: every element of G is conjugate to its inverse so any character satisfies:

$\chi(g) = \chi(g^{-1}) = \overline{\chi(g)}$

where the last equality follows from G7. Thus we can think of the characters of A5 as elements of a real vector space and obtain the last two irreducible characters via linear algebra. Indeed, let’s apply Gram-Schmidt to the orthonormal set $\{\chi_{\text{triv}}, \chi_1, \chi_4\}$ to obtain some orthonormal basis:

$\{\chi_{\text{triv}}, \chi_1, \chi_4, \phi_1, \phi_2\}\$, where $\ \phi_1 = \frac 1 {\sqrt 3}(3, 0, 3, -2, -1)$ and $\phi_2 = \sqrt{\frac 5 3}(3, 0, 0, 1, -1)$.

If $\chi$ is one of the last two irreducible characters, then we can write $\chi = a\phi_1 + b\phi_2$ for real values ab satisfying $a^2 + b^2 = 1.$ Since $\chi(1) = 3$ we also have $a\sqrt 3 + b\sqrt {15} = 3$ which gives:

$(a,b) = \left(\frac{\sqrt 3 \pm \sqrt{15}}6, \frac{5 \mp \sqrt 5}{2\sqrt {15}}\right) .$

Thus, $\chi = (3, 0, \frac{1+\sqrt 5}2, \frac{1-\sqrt 5}2, -1)$ or $(3, 0, \frac{1-\sqrt 5}2, \frac{1+\sqrt 5}2, -1)$ and our work is done.

## G11. Restricted and Induced Representations

If H ≤ G is a subgroup, then we can restrict any representation $\rho:G\to GL(V)$ to that of H. We’ll denote this restricted representation by $\text{Res}_G^H\rho$ and the corresponding character $\text{Res}_G^H\chi.$

As we saw in G10(c), even if $\chi:G\to \mathbf{C}$ is irreducible, its restriction to H may not be. Even worse, even if$\chi_1$ and $\chi_2$ are orthogonal (i.e. $\left<\chi_1, \chi_2\right>=0$), their restrictions to H may not be. We’ll see this in an example later on.

Conversely, given a representation $\phi:H\to GL(V)$, we’ll define an induced representation (denoted $\text{Ind}_H^G\rho$ and $\text{Ind}_H^G\chi$). Abstractly, if V is a C[H]-module, we take:

$W := \mathbf{C}[G]\otimes_{\mathbf{C}[H]} V$

which is now a C[G]-module. Clearly, both Ind and Res are functorial, in the sense that if V and W are C[H]-modules with a C[H]-linear map fV → W, then this induces a C[G]-linear map $\text{Ind}_H^G V \to \text{Ind}_H^G W$ – this follows from the more general fact that if $R\subseteq S$ are rings, then $S\otimes_R -$ is functorial. The case for Res is obvious. Furthermore, we have:

Proposition. We have natural isomorphisms:

$\text{Res}_H^K (\text{Res}_G^H V) \cong \text{Res}_G^K V$ and $\text{Ind}_H^G (\text{Ind}_K^H W)\cong \text{Ind}_K^G W$

for any C[G]-module V and C[H]-module W.

Proof.

The case for Res is obvious. For Ind, this follows from the more general fact that if $R\subseteq S\subseteq T$ are rings, then there’s a natural isomorphism $T\otimes_S (S\otimes_R M) \cong T\otimes_R M$ for any R-module M. ♦

Frobenius Reciprocity Law

Next, we have the following general result for modules.

Proposition. Let $R\subseteq S$ be rings, M be an R-module and N be an S-module. Then there’s a canonical isomorphism:

$\text{Hom}_S(S\otimes_R M, N) \cong \text{Hom}_R(M, N).$

Let’s consider the case where $R = \mathbf{C}[H]$ and $S = \mathbf{C}[G]$ and let the characters for M and N be denoted by $\psi:H\to \mathbf{C}$ and $\chi:G\to \mathbf{C}$ respectively. We get an isomorphism of complex vector spaces in the above proposition, with dimensions:

\begin{aligned}LHS:& \dim_\mathbf{C} \text{Hom}_{\mathbf{C}[G]}(\mathbf{C}[G]\otimes_{\mathbf{C}[H]} M, N) = \left<\text{Ind}_H^G \psi, \chi\right>_G,\\ RHS:& \dim_\mathbf{C} \text{Hom}_{\mathbf{C}[H]} (M, N)= \left<\psi, \text{Res}_G^H \chi\right>_H.\end{aligned}

Conclusion:

Frobenius Reciprocity Theorem. If $\psi:H\to \mathbf{C}$ and $\chi:G\to\mathbf{C}$ are characters of representations, then:

$\left<\text{Ind}_H^G \psi, \chi\right>_G = \left<\psi, \text{Res}_G^H \chi\right>_H.$

Since the set of such characters spans the space of class functions, equality actually holds for any class functions $\psi$ and $\chi.$

[ Abstractly, one says that the Ind and Res maps are adjoint to each other. Roughly, this means that their underlying matrices are transposes of each other, as we’ll see below. ]

Example

Let’s consider groups $S_3 \le S_4.$ Their character tables are given below:

This gives us the following restrictions:

• $\text{Res}\chi_{\text{triv}} = \psi_{\text{triv}}$;
• $\text{Res}\chi_{\text{alt}} = \psi_{\text{alt}}$;
• $\text{Res}\chi_2 = \psi_1$;
• $\text{Res}\chi_1 = \psi_{\text{triv}}+\psi_1$;
• $\text{Res}\chi_1\chi_{\text{alt}} = \psi_{\text{alt}}+\psi_1$.

The corresponding matrix for Res is then:

$\begin{pmatrix} 1 & 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 & 1\\ 0 & 0 & 1 & 1 & 1\end{pmatrix}.$

To compute Ind ψ’s, we could use the Frobenius reciprocity theorem, but let’s do it the hard way via explicit computation since we’d like to show the formula for Ind ψ.

Theorem. We have:

$(\text{Ind}_H^G \psi)(g) = \frac 1 {|H|}\sum_{x\in G} \psi(xgx^{-1})$,

where ψ is extended to the whole of G by taking zero outside H.

Proof.

The ring C[G] is free as a C[H]-module, with basis given by $g_1, \ldots, g_k$ where the gi‘s is a set of left coset representatives for G/H. Hence if M is a C[H]-module, then $M' :=\mathbf{C}[G]\otimes_{\mathbf{C}[H]} M$ is the direct sum of $g_i M,$ where M is identified with $1\otimes M \subseteq M'.$

Hence $g\in G$ permutes the blocks $g_i M$ around; if $gg_i H \ne g_i H$ then there’s no contribution from $g_i M$ to the trace. Otherwise, $g_i^{-1} gg_i \in H$ so the contribution to the trace is exactly $\psi(g_i^{-1}g g_i),$ i.e.

\begin{aligned}(\text{Ind}\psi)(g) = \sum_{i=1}^k \psi(g_i^{-1}g g_i) = \sum_{i=1}^k \frac 1 {|H|}\sum_{h\in H} \psi(h^{-1} g_i^{-1} g g_i h) \stackrel{x=g_i h}{=} \frac 1 {|H|}\sum_{x\in G}\psi(x^{-1}g x),\end{aligned}

where the second equality follows from the fact that ψ is a class function on H. ♦

Going back to the example, let’s compute $\chi := \text{Ind} \psi$; it turns out to be much more convenient to apply the first equality above: $\chi = \sum_i \psi(g_i^{-1} g g_i).$ For coset representatives of $S_3 \le S_4,$ let us take $g_i = (1, 2, 3, 4)^i$ for i = 0, 1, 2, 3:

• Let g=e. Then $g_i^{-1} e g_i = e$ for each i. Thus $\chi(e) = 4\psi(e).$
• Let g=(1, 2). Then $g_i^{-1} g g_i \in S_3$ only for i=0, 3. Thus $\chi((1, 2)) = \psi((1, 2))+\psi((2,3)).$
• Let g=(1, 2, 3). Then $g_i^{-1} g g_i \in S_3$ only for i=0. So $\chi((1, 2, 3)) = \psi((1, 2, 3)).$
• Let g=(1, 2, 3, 4). Then $g_i^{-1}g g_i\not\in S_3$ so $\chi((1, 2, 3, 4)) = 0.$
• Similarly if g=(1, 2)(3, 4), then $\chi(g) = 0.$

This gives the following characters:

$\text{Ind}\psi_{\text{triv}} = (4, 2, 1, 0, 0) = \chi_{\text{triv}} + \chi_1, \\ \text{Ind}\psi_{\text{alt}} = (4, -2, 1, 0, 0) = \chi_{\text{alt}} + \chi_1\chi_{\text{alt}}, \\ \text{Ind}\psi_1 = (8, 0, -1, 0, 0) = \chi_2 + \chi_1 + \chi_1 \chi_{\text{alt}}.$

So the matrix for Ind is:

$\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 1\\ 0& 1 & 1\end{pmatrix}$

which is transpose to the matrix for Res, exactly as we’d expect it to!

Exercise.

Repeat the above computations for $S_4\subset S_5$ and $A_4\subset S_5$. [ Hint: to compute the character table of A4, pick the normal subgroup $N = \{e, (1,2)(3,4), (1,3)(2,4), (1,4)(2,3)\}$ and consider the character table of A4/N. ]

## G12. Omake

We have one final result to present:

Proposition. The degree of an irreducible representation of G divides |G|.

Proof.

The proof requires some elementary theory of algebraic integers. First, suppose we have a class function fG → C where every f(g) is an algebraic integer. Let $\alpha = \sum_{g\in G} f(g)g\in \mathbf{C}[G].$

• For each conjugancy class $C\subset G$, take the element $e_C := \sum_{g\in C} g\in \mathbf{C}[G].$
• Check that $e_C$ commutes with anything in C[G].
• Since $e_C\in \mathbf{Z}[G]$ and Z[G] is a ring which is a finite Z-module, $e_C$ is integral over Z.
• Now, $\alpha = \sum_g f(g)g$ is a linear combination of $\{e_C\}$ with coefficients which are algebraic integers. Thus, α is a sum of commuting elements, each of which is integral over Z. Conclusion: α satisfies a monic polynomial with integer coefficients.

Since f is a class function, $\alpha g = g\alpha\in \mathbf{C}[G]$ for any $g\in G.$ Hence, if V is a simple C[G]-module, then multiplication-by-α on V is a C[G]-linear map. Schur’s lemma says this map is a scalar multiple λ of the identity, which must be an algebraic integer. To compute λ, we take the trace of α on V:

\begin{aligned}\lambda\cdot\dim V = \sum_{g\in G} f(g)\text{tr}(g|_V) =\sum_{g\in G} f(g)\chi_V(g).\end{aligned}

In particular, let $f(g) = \overline{\chi_V(g)}$; each f(g) is an algebraic integer since the eigenvalues of g are roots of unity and thus algebraic integers. Now $\lambda\cdot\dim V = \sum_{g\in G} |f(g)|^2 = |G|$ by orthonormality of irreducible characters. Thus $\lambda = |G|/\dim V$ is an algebraic integer which lies in Q, so it’s an integer. ♦

In fact, one can even prove that the degree divides [GZ(G)] where Z(G) is the centre of G. Interested readers may refer to Serre’s book “Linear Representations on Finite Groups” (Graduate Texts in Mathematics vol. 45).

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