PS02--rep-theory--solutions.md

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title: |
 ProblemSet 2 -- Representations and characters
author: George McNinch
date: 2024-01-29

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\newcommand{\trivial}{\mathbf{1}}

In these exercises, $G$ always denotes a finite group and all vector spaces are assumed to be finite dimensional over the field $F = \mathbb{C}$.

In these exercises, you may use results stated but not yet proved in class about characters of representations of $G$.

1. In this problem, we identify the character $\chi_\Omega$ of the permutation representation $(\rho,F[\Omega])$ of a group $G$.

   a. Let $V$ be a vector space and let $\Phi:V \to V$ a linear mapping If $\mathcal{B}$ is a basis for $V$, recall that the trace of $\Phi$ is defined by $$\operatorname{tr}(\Phi) = \operatorname{tr}([\Phi]_{\mathcal{B}}).$$

   apologies -- this is just explanatory; it isn't actually a question

   b. Recall that the dual of $V$ is the vector space $V^\vee = \operatorname{Hom}_F(V,F)$ of linear functionals on $V$.

   If $b_1,\dots,b_n$ is a basis for $V$, let ${b_j}^\vee:V \to F$ be defined by ${b_j}^\vee(b_i) = \delta_{i,j}$. Show that ${b_1}^\vee,\dots,{b_n}^\vee$ is a basis for $V^\vee$; it is known as the dual basis to $b_1,\dots,b_n$.

   ::: {.solution} We must show that the vectors $\{ {b_i}^\vee \}$ are linearly independent and span $V^\vee$.

   First, linear independence:

   Suppose that $\alpha_1,\cdots,\alpha_n \in F$ and that $$0 = \sum_{i=1}^n \alpha_i {b_i}^\vee$$ (note that this equality "takes place in the vector space $V^\vee$").

   We must argue that all coefficients $\alpha_j$ are zero. Well, fix $j$ and consider the vector $b_j$. We apply the functional $\sum_{i=1}^n \alpha_i {b_i}$ to $b_j$: $$\left(\sum_{i=1}^n \alpha_i {b_i}^\vee\right)(b_j) = \sum_{i=1}^n \alpha_i {b_i}^\vee)(b_j) = \alpha_j.$$ Since the functional $\sum_{i=1}^n \alpha_i {b_i}^\vee$ is equal to 0, we know that $\left(\sum_{i=1}^n \alpha_i {b_i}^\vee\right)(v) =0$ for every $v \in V$. In particular, this holds when $v=b_j$ and we now conclude that $\alpha_j = 0$.

   This proves linear independence.

   Finally, we prove the vectors span $V^\vee$.

   Let $\phi \in V^\vee$, and for $1 \le i \le n$ write $\alpha_i = \phi(b_i)$. We claim that $$(\clubsuit) \quad \phi = \sum_{i=1}^n \alpha_i {b_i}^\vee.$$

   TO prove this equality of functions ("functionals") we must argue that $$\phi(v) = \left(\sum_{i=1}^n \alpha_i {b_i}^\vee\right)(v)$$ for every $v \in V$.

   And it suffices to prove the latter equality for vectors $v$ taken from the basis $\{b_i\}$.

   But now by construction, for each $j$ we have $$\phi(b_j) = \alpha_j = \left(\sum_{i=1}^n \alpha_i {b_i}^\vee\right)(\alpha_j)$$

   This proves $(\clubsuit)$ so that the ${b_i}^\vee$ indeed span $V^\vee$. :::

   c. Prove that the trace of the linear mapping $\Phi:V \to V$ is given by the expression $$\operatorname{tr}(\Phi) = \sum_{i=1}^n {b_i}^\vee(\Phi(b_i)).$$

   ::: {.solution} Recall that $\operatorname{tr}(\Phi)$ is defined to be the trace of the matrix $[\Phi]_{\mathcal{B}}$ where $\mathcal{B}$ is a basis of $V$. It is a fact that this definition is independent of the choice of basis.

   Also recall that the trace of the $n\times n$ matrix $M = (M_{i,j})$ is given by $$\operatorname{tr}(M) = \sum_{i=1}^n M_{i,i}.$$

   We consider the basis $\mathcal{B}$ of $V$, and the dual basis $\mathcal{B}^\vee$ of $V^\vee$, as above.

   Recall that the matrix $M = [M_{j,i}] = [\Phi]_{\mathcal{B}}$ of $\Phi$ in the basis $\mathcal{B}$ is defined by the condition $$\Phi(b_i) = \sum_{j=1}^n M_{j,i} b_j$$ (for $1 \le i \le n$).

   Thus, $${b_i}^\vee(\Phi(b_i)) = {b_i}^\vee\left(\sum_{j=1}^n M_{j,i} b_j\right) = M_{i,i}$$.

   Summing over $i$ we find that $$\sum_{i=1}^n {b_i}^\vee(\Phi(b_i)) = \sum_{i=1}^n M_{i,i} = \operatorname{tr}(M) = \operatorname{tr}([\Phi]_{\mathcal{B}}) = \operatorname{tr}(\Phi),$$ as required.

   :::

   d. Suppose that the finite group $G$ acts on the finite set $\Omega$, and consider the corresponding permutation representation $(\rho,F[\Omega])$ of $G$. Recall that $F[\Omega]$ is the vector space of all $F$-values functions on $\Omega$, and that for $f \in F[\Omega]$ and $g \in G$, we have $$\rho(g)f(\omega) = f(g^{-1}\omega).$$ In particular, we saw in the lecture that $$\rho(g)\delta_\omega) = \delta_{g\omega},$$ where $\delta_\omega$ denotes the Dirac function at $\omega \in \Omega$.

   Show that $$\operatorname{tr}(\rho(g)) = \#\{\omega \in \Omega \mid g\omega = \omega\};$$ i.e. the trace of $\rho(g)$ is the number of fixed points of the action of $g$ on $\Omega$.

   ::: {.solution} Recall that the vector space $F[\Omega]$ has a basis consisting of the vectors $\delta_\omega$ for $\omega \in \Omega$.

   We write $\delta_\omega^\vee \in F[\Omega]^\vee$ for vectors of the dual basis. The linear functional $$\delta_\omega^\vee : F[\Omega] \to F$$ is defined by $$\delta_\omega^\vee(\delta_\tau) = \left \{ \begin{matrix} 1 & \tau = \omega \\ 0 & \tau \ne \omega \end{matrix} \right .$$

   Fix $g \in G$. According to our earlier work, we know that

   $$\operatorname{tr}(\rho(g)) = \sum_{\omega \in \Omega} \delta_\omega^\vee ( \rho(g) \delta_\omega ) = \sum_{\omega \in \Omega} \delta_\omega^\vee(\delta_{g \omega}).$$

   Now, $\delta_\omega^\vee (\delta_g\omega)$ is 1 when $\omega = g\omega$ and is 0 otherwise. This shows that $\operatorname{tr}(\rho(g))$ is given by the number of $\omega \in \Omega$ for which $g\omega = \omega$, as required.

   :::

2. Let $V$ be a representation of $G$, suppose that $W_1,W_2$ are invariant subspaces, and that $V$ is the internal direct sum $$V = W_1 \oplus W_2.$$

   Show that the character $\chi_V$ of $V$ satisfies $$\chi_V = \chi_{W_1} + \chi_{W_2}$$ i.e. for $g \in G$ that $$\chi_V(g) = \chi_{W_1}(g) + \chi_{W_2}(g).$$

   ::: {.solution}

   Let $\mathcal{B} = \{b_1,\cdots,b_n\}$ be a basis of $W_1$ and let $\mathcal{C} = \{c_1, \cdots, c_m\}$ be a basis of $W_2$.

   Since $V = W_1 \oplus W_2$, we know that $\mathcal{B} \cup \mathcal{C} = \{b_1,\cdots,b_n,c_1,\cdots,c_m\}$ is a basis for

   We consider the dual basis $b_1^\vee, b_2^\vee, \cdots, b_n^\vee,c_1^\vee,\cdots,c_m^\vee$ of the dual vector space $V^\vee$.

   (Be careful! $W_1^\vee$ is not a subspace of $V^\vee$! Instead, it is a quotient of $V^\vee$...)

   Observe that the functional $b_i^\vee \in V^\vee$ is determined by the rules $$b_i^\vee(b_j) = \delta_{i,j} \quad \text{and}\quad b_i^\vee(c_j) = 0.$$

   Similarly, the functional $c_j^\vee \in V^\vee$ is determined by the rules $$c_j^\vee(b_i) = 0 \quad \text{and}\quad c_j^\vee(c_i) = \delta_{j,i}.$$

   Observe that we can restrict $b_i^\vee$ to $W_1$, and these restrictions $\{b_i^\vee\vert_{W_1}\}$ give the basis of $W_1^\vee$ dual to the basis $\{b_i\}$ of $W_1$.

   Similarly the restrictions $\{c_j^\vee \vert_{W_2}\}$ give the basis of $W_2^\vee$ dual to the basis $\{c_j\}$ of $W_2$.

   Now using the results of the previous problem applied to the mapping $g:W_1 \to W_1$, $g:W_2 \to W_2$ and $g:V \to V$, we see that

   $$\chi_{W_1}(g) = \operatorname{tr}(g:W_1 \to W_1) = \sum_{i=1}^n b_i^\vee(g\cdot b_i)$$$$\chi_{W_2}(g) = \operatorname{tr}(g:W_2 \to W_2) = \sum_{j=1}^m c_j^\vee(g\cdot c_j)$$$$\chi_V(g) = \operatorname{tr}(g:V \to V) = \sum_{i=1}^n b_i^\vee(g\cdot b_i) + \sum_{j=1}^m c_j^\vee(g \cdot c_j)$$

   Thus indeed $\chi_V(g) = \chi_{W_1}(g) + \chi_{W_2}(g)$ for each $g$, as required.

   :::

3. Let $G = A_4$ be the alternating group of order $\dfrac{4!}{2} = 12$.

   We are going to find the character table of this group.

   a. Confirm that the following list gives a representative for each of the conjugacy classes of $G$:

   $$1, (12)(34), (123), (124)$$

   (Note that $(123)$ and $(124)$ are conjugate in $S_4$, but not in $A_4$).

   What are the sizes of the corresponding conjugacy classes?

   ::: {.solution}

   Note that the centralizer $C_{A_4}( (12)(34) )$ contains the group $\langle (12), (34) \rangle$, which has 4 elements. On the other hand, $(12)(34)$ is not central in $A_4$ (e.g. $(23)$ doesn't commute with $(12)(34)$). Since $[A_4:\langle (12)(34) \rangle] = 3$ (a prime number), conclude that $C_{A_4}( (12)(34) ) = \langle (12), (13) \rangle$. We conclude that $(12)(34)$ has exactly $12/4 = 3$ conjugates in $A_4$.

   Next note that the centralizer $C_{A_4}( (123) )$ contains the subgroup $\langle (123) \rangle$ of order $3$. On the other hand, suppose that $\sigma \in C_{A_4}((123))$. Then $\sigma (123) \sigma^{-1} = (123)$. But we know $\sigma(123)\sigma^{-1} = (\sigma(1) \sigma(2) \sigma(3))$, and now the condition $$(123) = (\sigma(1) \sigma(2) \sigma(3))$$ implies that $\sigma \in \langle (123) \rangle$. Thus $C_{A_4}((123)) = \langle (123) \rangle$ has order 3, and the conjugacy class of $(123)$ has $12/3 = 4$ elements.

   Similarly, the centralizer of $(124)$ has order 3, and its conjugacy class has $4$ elements.

   Finally, we should argue that $(123)$ and $(124)$ are not in fact conjugate in $A_4$. Of course, they are conjugate in $S_4$ by the transposition $(34)$. Arguing as above, the centralizer of $(123)$ in $S_4$ is still just equal to $\langle (123) \rangle$. So any element $\sigma$ of $S_4$ for which $\sigma(123)\sigma^{-1} = (124)$ has the form $(123)^i(12)$ for some $i$, and none of those elements is in $A_4$.

   We have

   class rep $g$	$C_{A_4}(g)$	size of conjugacy class of $g$
   1	12	1
   $(12)(34)$	4	3
   $(123)$	3	4
   $(124)$	3	4

   Since $$1 + 3 + 4 + 4 = 12$$ we have found all of the conjugacy classes in $A_4$.

   :::

   b. Let $K = \langle (12)(34), (14)(23)\rangle$. Show that $K$ is a normal subgroup of index $3$, so that $G/K \simeq \mathbb{Z}/3\mathbb{Z}$.

   ::: {.solution} One checks directly that $$K = \{ 1, (12)(34), (14)(23), (13)(24) \}$$ so that $K$ has order 4 and index 3 as asserted.

   Notice that - as a set - $K$ is the union of $\{1\}$ and the 3-element conjugacy class of $(12)(34)$. This makes clear that $\sigma \tau \sigma^{-1} \in K$ for all $\sigma \in A_4$ and $\tau \in K$, so that $K$ is a normal subgroup.

   Since $|G/K|=3$, of course $G/K \simeq \mathbb{Z}/3\mathbb{Z}$ ("groups of prime order are cyclic").

   :::

   Let $\zeta_3$ be a primitive $3$rd root of unity in $F^\times$ and for $i=0,1,2$ let $\rho_i:G \to F^\times$ be the unique homomorphism with the following properties:

   i. $\rho_i\left( (123) \right) = \zeta^i$ ii. $K \subseteq \ker \rho_i$.

   Explain why ParseError: KaTeX parse error: Undefined control sequence: \trivial at position 10: \rho_0 = \̲t̲r̲i̲v̲i̲a̲l̲,\rho_1,\rho_2 determine distinct irreducible (1-dimensional) representations of $G$.

   ::: {.solution}

   In fact, let $(\rho_1,V_1)$ and $(\rho_2,V_2)$ be representations of $G$ for which $V_1$ and $V_2$ are 1 dimensional. In this case, $\operatorname{GL}(V_i) = \operatorname{GL}_1(F) = F^\times$ is a commutative group.

   Since $V_1$ and $V_2$ have dimension 1, any isomorphism $\Phi:V_1 \to V_2$ is just given by multiplication with a scalar $\alpha \in F^\times$. So if the representations are isomorphic, we have for each $g\in G$ and $v \in V_1$: $$\rho_2(g)\Phi(v) = \Phi(\rho_1(g)v) \implies \alpha \rho_2(g)v = \alpha \rho_1(g)v$$

   Since $\alpha \ne 0$ and since this holds for all $v \in V_1$, we conclude that $\rho_1(g) = \rho_2(g)$ for each $g \in G$.

   In other words, two 1 dimensional representations are isomorphic iff they are equal (as functions $G \to F^\times)$.

   Now, the three homomorphisms $\rho_i$ $(i=0,1,2)$ are clearly distinct, because each maps the element $(123)$ to a different element of $F^\times$. Thus they constitute distinct irreducible 1 dimensional representations of $G$.

   :::

   c. Let $\Omega = \{1,2,3,4\}$ on which $G$ acts by the embedding $A_4 \subset S_4$.

   Compute the character $\chi_\Omega$ of the representation $F[\Omega]$. (This means: compute and list the values of $\chi_\Omega$ at the conjugacy class representatives given in a.)

   (Use the result of problem 1 above).

   ::: {.solution}

   According to problem 1, the trace of the action of an element $\sigma \in A_4$ on the permutation representation $F[\Omega]$ is equal to the number of fixed points of $\sigma$ on $\Omega$.

   Let's write $\chi_\Omega$ for the character of the representation $F[\Omega]$.

   Thus, the trace is given by

   $\sigma$	$\chi_\Omega$
   $1$	4
   $(12)(34)$	0
   $(123)$	1
   $(124)$	1

   :::

   d. The span of the vector $\delta_1 + \delta_2 + \delta_3 + \delta_4 \in F[\Omega]$ is an invariant subspace isomorphic to the irreducible representation $\rho_0$ (the so-called trivial representation).

   Thus $F[\Omega] = \rho_0 \oplus W$ for a $3$-dimensional invariant subspace. Explain why problem 2 shows that the character of $W$ is given by ParseError: KaTeX parse error: Undefined control sequence: \trivial at position 24: … \chi_\Omega - \̲t̲r̲i̲v̲i̲a̲l̲.

   ::: {.solution}

   Problem 2 shows that ParseError: KaTeX parse error: Undefined control sequence: \trivial at position 16: \chi_{\Omega}= \̲t̲r̲i̲v̲i̲a̲l̲ ̲ ̲+ \chi_W.

   This is an identity of $F$-valued functions on $G$, and it immediately implies that ParseError: KaTeX parse error: Undefined control sequence: \trivial at position 24: … \chi_\Omega - \̲t̲r̲i̲v̲i̲a̲l̲ as required.

   :::

   Now prove that $\langle \chi_W, \chi_W \rangle = 1$ and conclude that $W$ is an irreducible representation.

   ::: {.solution} Write $\sigma_1,\sigma_2,\sigma_3,\sigma_4$ for class representatives $1,(12)(34), (123), (124)$. And write $c_i$ for the order of the centralizer of $\sigma_i$.

   Notice that the values of ParseError: KaTeX parse error: Undefined control sequence: \trivial at position 24: … \chi_\Omega - \̲t̲r̲i̲v̲i̲a̲l̲ are given in the following table:

   $\sigma_i$	$c_i$	$\chi_\Omega(\sigma_i)$	$\chi_W(\sigma_i)$
   $1 = \sigma_1$	12	4	3
   $(12)(34) = \sigma_2$	4	0	-1
   $(123) = \sigma_3$	3	1	0
   $(124) = \sigma_4$	3	1	0

   We calculate

   $$\begin{align*} \langle \chi_W,\chi_W \rangle =& \sum_{i=1}^4 \dfrac{1}{c_i} \chi_W(\sigma_i) \overline{\chi_W(\sigma_i)} = \dfrac{1}{12} 3 \cdot 3 + \dfrac{1}{4} (-1) \cdot (-1) + \dfrac{1}{3} 0 \cdot 0 + \dfrac{1}{3} 0 \cdot 0 \\ =& \dfrac{9}{12} + \dfrac{1}{4} = \dfrac{9 + 3}{12} = 1 \end{align*}$$

   It follows from the results described in lecture that a representation $V$ is irreducible if and only if $\langle \chi_V,\chi_V \rangle = 1$, so we conclude that $W$ is an irreducible representation.

   :::

   e. Explain why ParseError: KaTeX parse error: Undefined control sequence: \trivial at position 1: \̲t̲r̲i̲v̲i̲a̲l̲,\rho_1,\rho_2,… is a complete set of non-isomorphic irreducible representations of $G$.

   ::: {.solution}

   We know that $G$ has 4 conjugacy classes, so up to isomorphism there are exactly 4 irreducible representations of $G$.

   We've already pointed out that $1,\rho_1,\rho_2$ are non-isomorphic irreducible representations each of dimension 1. Now, we've seen that $W$ is an irreducible representation; since $W$ is 2 dimensional, it is not isomorphic to any of the representations $1,\rho_1,\rho_2$.

   Thus we have found 4 non-isomorphic irreducible representations, and we can conclude that any irreducible representation is isomorphic to once of these 4.

   :::

   f. Display the character table of $G = A_4$.

   ::: {.solution}

   ParseError: KaTeX parse error: Undefined control sequence: \trivial at position 96: …\ \hline \̲t̲r̲i̲v̲i̲a̲l̲& 1 & 1 & 1 & 1…

   :::