Exploration 4: Character theory

February 27, 2024.

Questions:

• TODO

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There are two problems when it comes to working with matrix representations:

• You have to work with (possibly huge) matrices
• They implicitly require a basis, which seems arbitrary

Let’s start by defining what it means for two (general) representations $\rho:G\to\Aut(M)$ and $\rho':G\to\Aut(M')$ to be the same representation. Say that these two representations are equivalent if there is an isomorphism $\phi$ between $M$ and $M'$ that preserves the group action, i.e. $\phi\circ\rho(g)=\rho'(g)\circ\phi$ for all $g\in G$, i.e. the following diagram commutes:

$$\begin{aligned} M&\xrightarrow{\rho(g)}&M \\ \downarrow{\phi}&&\downarrow{\phi}&\quad\text{ where }\phi:M\to M'\text{ is an isomorphism}\\ M'&\xrightarrow{\rho'(g)}&M' \end{aligned}$$

Let’s explore what happens in the case of matrix representations. When $M,M'$ are free modules $R^n$, then $\rho,\rho'$ become matrix representations, and thus $Aut(R^n)$ is composed of $n\times n$ matrices. Then the problem of checking whether $M,M'$ are isomorphic comes down to comparing properties of the representative matrices.

Corollary: Conjugacy classes in $G$ are represented by similarity classes in $\Aut(R^n)$.

Corollary: Equivalent matrix representations differ only in which similarity class they assign to each conjugacy class.

In this section, we explore some facts about class functions.

Since representations of elements of $G$ in the same conjugacy class are similar matrices, we’d like to classify the similarity classes of matrices by defining a function on matrices that is unchanged under conjugation. So we’re interested in class functions on $G$, functions $G\to R$ that are invariant under conjugation in $G$.

Ideally we can decompose every class function $G\to R$ into a matrix representation $G\to\Aut(R^n)$ followed by a linear class function $\Aut(R^n)\to R$. This works out nicely because of the following theorem:

That is, the only linear class function is the trace, and therefore we can decompose every class function $G\to R$ into a matrix representation $\rho:G\to\Aut(R^n)$ followed by the trace (multiplied by some scalar).

In this section, we discover how to classify equivalent matrix representations.

Given a matrix representation $\rho:G\to\Aut(R^n)$, define the character of $\rho$ as $\chi_\rho=\tr\circ\rho$, essentially taking the trace of each representative matrix for $g\in G$. By construction, characters are one-dimensional representations $G\to R$ that are invariant under conjugation in $G$. Because equivalent matrix representations only differ by conjugation in $G$, representations that are invariant under conjugation are perfect for classifying matrix representations. In particular, each distinct value that $\chi_\rho(g)$ takes on is representative of a distinct conjugacy class in $G$.

Assume we’re in an integral domain $R$ whose characteristic is $0$, so that $\char R\nmid |G|$ is true for all finite $G$. Then Maschke’s theorem lets us describe any group representation in terms of its irreps, and so we shift our focus of study towards irreps. Characters of irreps are known as irreducible characters.

Characters have a number of other properties in an integral domain, and much more in a field. Let’s go over the integral domain properties first.

• $\chi(e)$ is the degree of $\chi$, which turns out to be equal to the dimension of the underlying representation.
  Theorem: $\chi(e)$ gives the dimension $n$ of its underlying representation in $\Aut(R^n)$.

  Since the identity element in $G$ is always represented by the identity matrix in $\Aut(R^n)$, taking the trace simply adds the $n$ ones along the diagonal.

• TODO

In this section, we view a strategy for classifying group representations over fields.

Recall Schur’s lemma:

Theorem: Any homomorphism $\sigma:M\to N$ between simple modules $M$ and $N$ is either trivial or an isomorphism.

A corollary (often also called Schur’s lemma) appears when $R$ is actually an algebraically closed field $F$:

Character theory is most often used over the field of complex numbers $\CC$. This is mostly because $\CC$ is an algebraically closed field, which gives us the above.

TODO either expand on character theory theorems like orthogonality in C OR go back and highlight the importance of

TODO: three proofs that link together to get schur’s lemma

• In an algebraically closed field, every matrix has eigenvalue
• if matrix has eigenvalue, its image is an eigenspace (by defn)
• if image of linear map is an eigenspace, the map is a multiplication by scalar

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Let ϕ : ρ_1 → ρ_2 be a nonzero homomorphism.

prove it in 2 parts:

If ρ_2 is irreducible then ϕ is surjective.
If ρ_1 is irreducible then ϕ is injective. 

The inner product $\langle,\rangle$ is a function of two arguments $R^n\times R^n\to R$ that satisfy the following:

Conjugate Symmetry Linearity in the first argument Positive-definiteness

Irreducible characters are orthogonal

Orthogonal means ⟨φ,ψ⟩=1 if they’re the same representation up to isomorphism, and =0 otherwise.

Irreducible characters are orthogonal, proof is in lecture 22

We call this property character orthogonality, as in “by character orthogonality…”

Computing character tables over ℂ

A character table has irreducible characters χ as rows and conjugacy classes of G as columns. Then each entry is χ((g)) where g is in that conjugacy class of G. For example, let ζ = a primitive third root of unity. Then use the degree 1 irreducible χ⁽ᵏ⁾ = g ↦ ζᵏ. Then the cyclic group C₃ = ⟨g⟩ has the character table:

       1   1   1      ← size of the conjugacy class of C₃
      (1) (g) (g²)    ← the conjugacy classes of C₃
     ------------
χ⁽⁰⁾ | 1   1   1      ← trivial representation g ↦ [1]
χ⁽¹⁾ | 1   ζ   ζ²     ← dim 1 representation g ↦ [ζ]
χ⁽²⁾ | 1   ζ²  ζ      ← dim 1 representation g ↦ [ζ²]

If we define χ⁽ᵏ⁾ = g ↦ iᵏ (a dim 1 representation), then the cyclic group C₄ = ⟨g⟩ has the character table:

       1   1   1   1      ← size of the conjugacy class of C₃
      (1) (g) (g²)(g³)    ← the conjugacy classes of C₃
     ----------------
χ⁽⁰⁾ | 1   1   1   1      ← trivial representation g ↦ [1]
χ⁽¹⁾ | 1   i  -1  -i      ← dim 1 representation g ↦ [ζ]
χ⁽²⁾ | 1  -1   1  -1      ← dim 1 representation g ↦ [ζ²]
χ⁽³⁾ | 1  -i  -1   i      ← dim 1 representation g ↦ [ζ³]

The symmetric group S₃ = ⟨(1 2),(1 2 3)⟩ has the character table:

       1    1    1      ← size of the conjugacy class of C₃
      (1) (12) (123)    ← the conjugacy classes of C₃
     ---------------
1    | 1    1    1      ← trivial representation π ↦ [1]
χˢ   | 1   -1    1      ← dim 1 sign representation π ↦ sign(π)
χᵁ   | 2    0   -1      ← dim 2 representation π ↦ (number of fixed points of π) - 1

The dihedral group D₄ = ⟨r,s | r⁴=s²=e, srs = r⁻¹⟩ has the character table:

       1    1    1       1       1       ← size of the conjugacy class of C₃
      {e} {r²} {r,r³} {s,sr²} {sr,sr³}   ← the conjugacy classes of C₃
     ---------------------------------
χ₁   | 1    1    1       1       1       ← trivial representation g ↦ [1]
χ₂   | 1    1   -1       1      -1       ← dim 1 representation g ↦ ?
χ₃   | 1    1    1      -1      -1       ← dim 1 representation g ↦ -1 if reflection
χ₄   | 1    1   -1      -1       1       ← dim 1 representation g ↦ ?
χ₅   | 2   -2    0       0       0       ← dim 2 representation g ↦ ?

Things to note:

• The first row is all 1s (since [1] has trace 1)
• You can compute the last row using the rest of the rows
• The first column is always the dimension of the representation. This is because the representation is an identity matrix, which has (dim ) ones.
• A linear (degree 1) character describes a homomorphism : G → Cˣ. We can always find such homomorphisms by taking the abelianization Gᵃᵇ = G/[G,G] where [G,G] is the commutator subgroup of all xyx⁻¹y⁻¹.
• Row self-product: Self-inner product of a row with itself is always |G|, but you need to count differently. (If a conjugacy class has 3 elements, count that entry three times.)
• Column self-product: Self-inner product of a column with itself is always |C|² where |C| is the size of the conjugacy class.
• Row orthogonality: Inner product of two different rows is always zero, but you need to count differently. (If a conjugacy class has 3 elements, count that entry three times.)
• Column orthogonality: Inner product of two different columns is always zero. (Use this to compute the last row.)
• Magic formula: sum of squares of degrees of the irreducible characters equals |G|. (Use this to complete the first column.)

This website gives all the character tables for all groups with at most 10 irreps: https://people.maths.bris.ac.uk/~matyd/GroupNames/characters.html

TODO

Recall that a field $F$ is algebraically closed iff every nonconstant polynomial in $F[x]$ has a root in $F$.

Characters completely determine representations up to isomorphism

The original motivation for characters is to identify representations without regard to a basis or its matrix at all. There is also an amazing theorem that:

Theorem: ⟨χ,χ’⟩ = (1/N)∑_g (̅χ(g))χ’(g), which is just a normalized dot product in the complex vector space.

Corollary: In particular, ⟨χ,χ⟩=1 iff χ is irreducible. (This must be equal to ∑xᵢ², which can only be the dot product of a elementary basis vector with itself. Therefore irreducible)

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Characteristic polynomials

Every $n\times n$ matrix $A$ over $F$ has a characteristic polynomial $\chi_A$ equal to $\det(xI-A)$.

Motivation: To find the eigenvectors. $\chi_A(\lambda)=0$ iff $\lambda$ is an eigenvalue of $A$. So just factor $\chi_A$ to get the eigenvalues.

Minimal polynomials

Recall you can evaluate a polynomial $m$ at a matrix $A$: $m(A)$.

Every $n\times n$ matrix $A$ over $F$ has a unique minimal polynomial $m_A$ that is monic, lowest degree, and $m_A(A)=0$.

Motivation: the roots of $m_A$ are the eigenvalues. Proof: WTS $m_A(\lambda)=0$ iff $\lambda$ is an eigenvalue of $A.$ Since $m_A$ is lowest degree, there should be no other factors in $m_A$ other than the eigenvalues, if this is true. (->) $m_A\mid\chi_A$, so $\chi(\lambda)=0$, which means $\lambda$ is an eigenvalue of $A$. (<-) We know $m_A(A)=0$ so we know $m_A(\lambda)\cdot v=0$. If you factor $m_A$ in some field where it splits into $x-\mu_i$, then we must have $(x-\mu_i)\cdot v=0$ for some $\mu_i$.

Another motivation: to detect diagonalizability of $A$. If $m_A$ factors into distinct factors it’s diagonalizable. Proof involves finding the Jordan block matrix conjugate to $A$…

Given $f$ is the minimal polynomial of $\alpha$ over $F$, we can make an isomorphism: $\varphi:F[x]/(f)\iso F(\alpha)$ as $\varphi=g+(f)\mapsto g(\alpha)$

Conjugation doesn’t change these polynomials

Recall a conjugate matrix of $A$ is a matrix $PAP^{-1}$ where $P$ is any matrix of the same size in the same field as $A$. Note:

• $f(PAP^{-1})=f(A)$
• $\det(tI-PAP^{-1})=\det(xPIP^{-1}-PAP^{-1})=\det(P(xI-A)P^{-1})=\det(xI-A)$

So conjugation doesn’t change either of these polynomials.

Cayley-Hamilton theorem: the characteristic polynomial χ_A of T is det(tI-T)=f₁f₂…fᵣ ∈ F[t]. Then χ_T(T) = O.

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Character theory: - We care about the conjugacy classes of groups, since they help us find the irreducible characters of groups

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