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 \(\mathbb{C}\).
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 \[g \star v = \phi(g) gv.\]
Prove that if \(V\) is irreducible, then \(\phi \otimes V\) is also irreducible.
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.
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 \[\chi_{\phi \otimes V}(g) = \operatorname{tr}( v\mapsto g \star v) = \phi(g) \chi(g).\]
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 \[\chi_{\phi\otimes V}(g) = \operatorname{tr}(\phi(g)\rho(g)) = \phi(g) \chi_V(g).\]
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.)
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 \(S_3\).
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 \[V_{(L)} = \sum_{S \in \mathcal{S}} S.\]
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.
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 \(L\).
Prove that the quotient representation \(V/V_{(L)}\) has no invariant subspaces isomorphic to \(L\) as \(G\)-representations.
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.
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 \[V = \bigoplus_{i=1}^m V_{(L_i)}.\]
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.
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…)
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 \(\overline{f(1)} \ne f(1^{-1})\).
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 \[\Psi_v(w) = \psi( \langle w,v \rangle ).\]
Show that \(\Psi_v\) is a group homomorphism \(V \to \mathbb{C}^\times\).
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.
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?)
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.
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 \(V \to \mathbb{C}^\times\).
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.