PS03--rep-theory--solutions.md

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title: |
 ProblemSet 3 -- Solutions
author: George McNinch
date: 2024-02-16

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

In these exercises, $G$ always denotes a finite group. Unless indicated otherwise, all vector spaces are assumed to be finite dimensional over the field $F = \mathbb{C}$. The representation space $V$ of a representation of $G$ is always assumed to be finite dimensional over $\mathbb{C}$.

1. Let $\phi:G \to F^\times$ be a group homomorphism; since $F^\times = \operatorname{GL}_1(F)$, we can think of $\phi$ as a 1-dimensional representation $(\phi,F)$ of $G$.

   If $V$ is any representation of $G$, we can form a new representation $\phi \otimes V$. The underlying vector space for this representation is just $V$, and the "new" action of an element $g \in G$ on a vector $v$ is given by the rule $$g \star v = \phi(g) gv.$$

   a. Prove that if $V$ is irreducible, then $\phi \otimes V$ is also irreducible.

   ::: {.solution}

   We prove the following statement: $(*)$ if $W \subset V$ is a subspace, then $W$ is invariant for the original action of $G$ if and only if it is invariant for the $\star$ action of $G$.

   First note that $(*)$ immediately implies the assertion of (a).

   To test invariance, let $w \in W$ and let $g \in G$. Since $W$ is a linear subspace and since $\phi(g)$ is a non-zero scalar, it is immediate that $$gw \in W \iff g \star w = \phi(g) gw \in W$$ Since this holds for all $w$ and all $g$, $(*)$ follows.

   :::

   b. Prove that if $\chi$ denotes the character of $V$, then the character of $\phi \otimes V$ is given by $\phi \cdot \chi$; in other words, the trace of the action of $g \in G$ on $\phi \otimes V$ is given by $$\chi_{\phi \otimes V}(g) = \operatorname{tr}( v\mapsto g \star v) = \phi(g) \chi(g).$$

   ::: {.solution} We just need to compute the trace of the linear mapping $V \to V$ given by $v \mapsto g \star v$.

   If the action of $g$ on $V$ is given by the linear mapping $\rho(g)$, then $$\chi_V(g) = \operatorname{tr}(\rho(g)).$$

   Now, the $\star$-action of $g$ is given by the linear mapping $v \mapsto g \star v = \phi(g) \rho(g)v$.

   So $\chi_{\phi \otimes V}(g) = \operatorname{tr}(\phi(g)\rho(g))$. For any scalar $s \in k$, trace of the linear mapping $s \rho(g)$ is given by $$\operatorname{tr}(s \rho(g)) = s \operatorname{tr}(\rho(g)) = s \chi_V(g)$$ ("linearity of the trace").

   Thus $$\chi_{\phi\otimes V}(g) = \operatorname{tr}(\phi(g)\rho(g)) = \phi(g) \chi_V(g).$$

   :::

   c. Recall that in class we saw that $S_3$ has an irreducible representation $V_2$ of dimension 2 whose character $\psi_2$ is given by

   $$\begin{array}{l|lll} g & 1 & (12) & (123) \\ \hline \psi_2 & 2 & 0 & -1 \end{array}$$

   Observe that $\operatorname{sgn} \psi = \psi$ and conclude that $V_2 \simeq \operatorname{sgn} \otimes V_2$, where $\operatorname{sgn}:S_n \to \{\pm 1\} \subset \mathbb{C}^\times$ is the sign homomorphism.

   On the other hand, $S_4$ has an irreducible representation $V_3$ of dimension 3 whose character $\psi_3$ is given by

   $$\begin{array}{l|lllll} g & 1 & (12) & (123) & (1234) & (12)(34) \\ \hline \psi_3 & 3 & 1 & 0 & -1 & -1 \end{array}$$

   (I'm not asking you to confirm that $\psi_3$ is irreducible, though it would be straightforward to check that $\langle \psi_3,\psi_3 \rangle = 1$).

   Prove that $V_3 \not \simeq \operatorname{sgn} \otimes V_3$ as $S_4$-representations.

   (In particular, $S_4$ has at least two irreducible representations of dimension 3.)

   ::: {.solution}

   We first consider the representation $V_2$ of $S_3$. Write $\chi_2$ of the character of this irreducible representation. The character of $\operatorname{sgn}\chi_2$ is then given by the product $\operatorname{sgn}\chi_2$.

   $$\begin{array}{l|lll} g & 1 & (12) & (123) \\ \hline \psi_2 & 2 & 0 & -1 \\ \operatorname{sgn} & 1 & -1 & 1 \\ \operatorname{sgn} \psi_2 & 2 & 0 & -1 \end{array}$$

   Inspecting the table we see that $\psi_2 = \operatorname{sgn}\psi_2$. This shows that $V_2$ is isomorphic to $\operatorname{sgn} \otimes V_2$ as representations for $S_3$.

   :::

2. Let $V$ be a representation of $G$.

   For an irreducible representation $L$, consider the set $$\mathcal{S}=\{ S \subseteq V \mid S \simeq L\}$$ of all invariant subspaces that are isomorphic to $L$ as $G$-representations.

   Put $$V_{(L)} = \sum_{S \in \mathcal{S}} S.$$

   a. Prove that $V_{(L)}$ is an invariant subspace, and show that $V_{(L)}$ is isomorphic to a direct sum $$V_{(L)} \simeq L \oplus \cdots \oplus L$$ as $G$-representations.

   ::: {.solution}

   First of all, we note more generally that if $I$ is an index set and if $W_i \subset V$ is a $G$-invariant subspace for each $i \in I$, then $\displaystyle \sum_{i \in I} W_i$ is again an invariant subspace. (The proof is straightforward from the definitions). This confirms that $V_{(L)}$ is an invariant subspace.

   To prove the remaining assertion, we proceed as follows.

   Let us say that a $G$-representation $W$ is $L$-isotypic if every irreducible invariant subspace of $W$ is isomorphic to $L$.

   It is immediate that $V_{(L)}$ is $L$-isotypic. We are going to prove:

   If $W$ is an $L$-isotypic $G$-representation, then $W$ is isomorphic to a direct sum $$W \simeq L \oplus \cdots \oplus L.$$

   Proceed by induction on $\dim W$. If $\dim W = 0$ then $W = \{0\}$ and the result is immediate ($W$ is the direct sum of zero copies of $L$).

   Now observe that if $\dim W > 0$ then $W$ contains an invariant subspace isomorphic to $L$, so that $\dim W \ge \dim L$.

   Now if $\dim W = \dim L$, then $W \simeq L$ and the result holds in this case.

   Finally, suppose that $\dim W > \dim L$ and let $S \subset W$ be an invariant subspace with $S \simeq L$.

   By complete reducibility we may find an invariant subspace $U \subset W$ such that $W$ is the internal direct sum $$W = S \oplus U.$$

   Since $\dim W = \dim S + \dim U$, we have $\dim U < \dim W$. Moreover, $U$ is also $L$-isotypic. So by induction on dimension, we know that $$U \simeq L \oplus \cdots \oplus L,$$ (say, a direct sum of $d$ copies of $L$).

   But then $$W = S \oplus U \simeq L \oplus ( L \oplus \cdots \oplus L) = L \oplus L \oplus \cdots \oplus L$$ is isomorphic to a direct sum of $d+1$ copies of $L$.

   :::

   b. Prove that the quotient representation $V/V_{(L)}$ has no invariant subspaces isomorphic to $L$ as $G$-representations.

   ::: {.solution}

   Write $\pi:V \to V/V_{(L)}$ for the quotient map $v \mapsto v + V_{(L)}$; thus $\pi$ is a surjective homomorphism of $G$-representations.

   Suppose by way of contradiction that $S \subset V/V_{(L)}$ is an invariant subspace isomorphic to $L$, and let $S' \subset V$ be the inverse image under $\pi$ of $S$: $$S' = \pi^{-1}(S).$$

   Then $S'$ is an invariant subspace of $V$ containing $V_{(L)}$, and the restriction of $\pi$ to $S'$ defines a surjective mapping $$\pi_{\mid S'}:S' \to S \simeq L.$$

   If $K$ denotes the kernel of $\pi_{\mid S'}$, then complete reducibility implies that there is an invariant subspace $M$ of $V$ such that $S'$ is the internal direct sum $$(*) \quad S' = K \oplus M.$$

   In particular, the invariant subspace $M$ is isomorphic to $L$ as $G$-representations. But then by definition we have $M \subset V_{(L)}$ contradicting the condition $M \cap K = \{0\}$ which must hold by $(*)$. This contradiction proves the result.

   :::

   c. If $L_1,L_2,\cdots,L_m$ is a complete set of non-isomorphic irreducible representations for $G$, prove that $V$ is the internal direct sum $$V = \bigoplus_{i=1}^m V_{(L_i)}.$$

   ::: {.solution}

   We first note that $V$ is equal to the sum $$\sum_{i=1}^m V_{(L_i)};$$ indeed, if $W = \sum_{i=1}^m V_{(L_i)}$, then by complete reducibility $V = W \oplus W'$ for an invariant subspace $W'$. But if $W' \ne 0$ then $W'$ contains an irreducible invariant subspace, so that $W' \cap V_{(L_i)} \ne 0$ for some $i$ and hence $W' \cap W \ne 0$; this is impossible since the internal sum $V = W \oplus W'$ is direct. This argument shows that $W' = 0$ and hence that $V = W$.

   Finally, we show that the sum $$\sum_{i=1}^m V_{(L_i)}$$ is direct, i.e. that for each $j$ we have $$(\clubsuit) \quad V_{(L_j)} \cap \left ( \sum_{i\ne j} V_{(L_i)} \right ) = 0.$$

   Wrote $I$ for the intersection appearing in $(\clubsuit)$; thus, $I$ is an invariant subspace of $V$. If $I$ is non-zero, it has an irreducible invariant subspace $S$. Since $I \subset V_{(L_j)}$ and since $V_{(L_j)}$ is $L_j$-isotypic, we conclude that $$S \simeq L_j.$$

   But then $S \cap V_{(L_i)} = 0$ for every $i \ne j$ so that $$S \cap \left( \sum_{i \ne j} V_{(L_i)} \right ) = 0.$$ Since $I \subset \displaystyle \left( \sum_{i \ne j} V_{(L_i)} \right )$, we conclude that $I = 0$.

   This completes the proof that $V$ is the direct sum of the $V_{(L_i)}$, as required. :::

3. Let $\chi$ be the character of a representation $V$ of $G$. For $g\in G$ prove that $\overline{\chi(g)} = \chi(g^{-1})$.

   Is it true for any arbitrary class function $f:G \to \mathbb{C}$ that $\overline{f(g)} = f(g^{-1})$ for every $g$? (Give a proof or a counterexample...)

   ::: {.solution}

   Let $\rho(g):V \to V$ denote the linear automorphism of $V$ determined by the action of $g \in G$. Then $\chi(g) = \operatorname{tr}(\rho(g))$.

   Now, since $\rho(g)$ has finite order, say $n$, its minimal polynomial divides $T^n - 1 \in \mathbb{C}[T]$, and hence every eigenvalue of $\rho(g)$ is an $n$-th root of unity.

   For any $n$-th root of unity $\zeta$, note that $\overline{\zeta} = \zeta^{-1}$.

   Write $\alpha_1,\cdots,\alpha_d$ for the eigenvalues of $\rho(g)$, with multiplicity (so that $d = \dim V$). Notice that $\rho(g^{-1})$ has eigenvalues $\alpha_1^{-1},\cdots,\alpha_d^{-1}.$

   Thus

   $$\chi(g) = \sum_{i=1}^d \alpha_i\quad \text{and} \quad \chi(g^{-1}) = \sum_{i=1}^d \alpha_i^{-1}.$$

   Now, we see that $$\overline{\chi(g)} = \sum_{i=1}^d \overline{\alpha_i} = \sum_{i=1}^d \alpha_i^{-1} = \chi(g^{-1})$$ as required.

   It is not true that $\overline{f(g)} = f(g^{-1})$ for every class function $f$ and every $g \in G$. Indeed, let $f = \alpha\delta_1$ be a multiple of the characterisitic function $\delta_1$ of the trivial conjugacy class $\{1\}$.

   Then $\overline{f(1)} = \overline{\alpha}$ while $f(1^{-1}) = f(1) = \alpha$, so that if $\alpha \not \in \mathbb{R}$, we have $\overline{f(1)} \ne f(1^{-1})$.

   :::

4. For a prime number $p$, let $k=\mathbb{F}_p = \mathbb{Z}/p\mathbb{Z}$ be the field with $p$ elements. Let $V$ be an $n$-dimensional vector space over $\mathbb{F}_p$ for some natural number $n$, and let $$\langle \cdot,\cdot \rangle: V \times V \to k$$ be a non-degenerate bilinear form on $V$.

   (A common example would be to take $V = \mathbb{F}_{p^n}$ the field of order $p^n$, and $\langle \alpha,\beta\rangle = \operatorname{tr}_{\mathbb{F}_{p^n}/\mathbb{F}_p}(\alpha \beta)$ the trace pairing).

   Let us fix a non-trivial group homomorphism $\psi:k \to \mathbb{C}^\times$ (recall that $k = \mathbb{Z}/p\mathbb{Z}$ is an additive group, while $\mathbb{C}^\times$ is multiplicative). Thus $$\psi(\alpha + \beta) = \psi(\alpha)\psi(\beta) \quad \text{for all} \quad \alpha,\beta \in k.$$ If you want an explicit choice, set $\psi(j + p\mathbb{Z}) = \exp(j \cdot 2 \pi i/p) = \exp(2 \pi i /p)^j.$

   For a vector $v \in V$, consider the mapping $\Psi_v:V \to \mathbb{C}^\times$ given by the rule $$\Psi_v(w) = \psi( \langle w,v \rangle ).$$

   a. Show that $\Psi_v$ is a group homomorphism $V \to \mathbb{C}^\times$.

   ::: {.solution}

   For $w,w' \in V$ notice that $$\begin{align*} \Psi_v(w + w') &= \psi(\langle w + w',v \rangle) & \\ &= \psi(\langle w,v \rangle + \langle w',v \rangle) & \text{since the form is bilinear} \\ &= \psi(\langle w,v \rangle) \cdot \psi(\langle w',v \rangle) & \text{since $\psi$ is a group homom} \\ &= \Psi_v(w)\cdot \Psi_v(w') & \text{by definition.} \end{align*}$$ This confirms that $\Psi_v$ is a group homomorphism.

   :::

   b. Show that the assignment $v \mapsto \Psi_v$ is injective (one-to-one).

   (This assignment is a function $V \to \operatorname{Hom}(V,\mathbb{C}^\times)$. In fact, it is a group homomorphism. Do you see why? How do you make $\operatorname{Hom}(V,\mathbb{C}^\times)$ into a group?)

   ::: {.solution}

   One checks that $\operatorname{Hom}(V,\mathbb{C}^\times)$ is a multiplicitive group (this is the dual group $\widehat V$ of $V$, mentioned in the lectures); the product of $\phi,\psi \in \widehat V$ is given by the rule $g \mapsto \phi(g) \cdot \psi(g)$.

   We note that the assignment $v \mapsto \Psi_v$ is a group homomorphism. For $v,v' \in V$ we must argue that $\Psi_{v+v'} = \Psi_v \Psi_{v'}.$

   For $w \in W$ we have $$\begin{align*} \Psi_{v+v'}(w) =& \psi( \langle v + v',w \rangle ) & \\ =& \psi(\langle v,w \rangle + \langle v',w\rangle) & \text{since the form is bilinear}\\ =& \psi(\langle v,w \rangle )\cdot \psi(\langle v',w\rangle) & \text{since $\psi$ is a group homom}\\ =& \Psi_v(w)\cdot \Psi_{v'}(w) & \text{by defn} \\ \end{align*}$$

   Now to show that $v \mapsto \Psi_v$ is injective, it is enough to argue that the kernel of this mapping is $\{0\}$.

   So, suppose that $\Psi_v$ is the identity element of $\widehat{V}$. In other words, suppose that $\Psi_v(w) = 1$ for every $w \in V$. This shows that $\psi(\langle v,w \rangle) = 1$ for every $w \in V$. Since $\psi$ is a non-trivial homomorphism $\mathbb{F}_p \to \mathbb{C}^\times$, we know that $\ker \psi = \{0\}$ (remember that $k$ has prime order...) and we conclude that $\langle v,w \rangle = 0$ for every $w \in W$.

   (Note that $\langle v,w \rangle = 0$ is an equality in $k = \mathbb{F}_p = \mathbb{Z}/p\mathbb{Z}$).

   Since the form $\langle \cdot,\cdot \rangle$ is non-degenerate, so we may now conclude that $v = 0$.

   This proves that the kernel of the mapping $v \mapsto \Psi_v$ is $\{0\}$, hence the mapping is injective.

   :::

   b. Show that any group homomorphism $\Psi:V \to \mathbb{C}^\times$ has the form $\Psi = \Psi_v$ for some $v \in V$.

   Conclude that there are exactly $|V| = q^n$ group homomorphisms $V \to \mathbb{C}^\times$.

   ::: {.solution}

   We observed in class that for any finite abelian group $A$, there is an isomorphism $A \simeq \widehat{A}$.

   In particular, $|A| = |\widehat{A}|$.

   Applying this in the case $A = V$, we conclude that $$|V| = |\widehat{V}| = |\operatorname{Hom}(V,\mathbb{C}^\times)|.$$

   Now, we have define an injective mapping $$v \mapsto \Psi_v:V \to \widehat(V).$$ Since the domain and co-domain of this mapping are finite of the same order, the mapping $v \mapsto \Psi_v$ is also surjective.

   Thus the pigeonhole principal shows that every homomorphism $\Psi:V \to \mathbb{C}^\times$ has the form $\Psi = \Psi_v$ for some $v \in V$, as required.

   :::