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 \(G\).
In this problem, we identify the character \(\chi_\Omega\) of the permutation representation \((\rho,F[\Omega])\) of a group \(G\).
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
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 \(b_1,\dots,b_n\).
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 \(V^\vee\).
Prove that the trace of the linear mapping \(\Phi:V \to V\) is given by the expression \[\operatorname{tr}(\Phi) = \sum_{i=1}^n {b_i}^\vee(\Phi(b_i)).\]
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.
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 \(\Omega\).
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.
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 \[\chi_V(g) = \chi_{W_1}(g) + \chi_{W_2}(g).\]
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.
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.
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?
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 \(A_4\).
Let \(K = \langle (12)(34), (14)(23)\rangle\). Show that \(K\) is a normal subgroup of index \(3\), so that \(G/K \simeq \mathbb{Z}/3\mathbb{Z}\).
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:
- \(\rho_i\left( (123) \right) = \zeta^i\)
- \(K \subseteq \ker \rho_i\).
Explain why \(\rho_0 = \mathbf{1},\rho_1,\rho_2\) determine distinct irreducible (1-dimensional) representations of \(G\).
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 \(G\).
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).
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 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 \(\chi_W = \chi_\Omega - \mathbf{1}\).
Problem 2 shows that \[\chi_{\Omega}= \mathbf{1}+ \chi_W.\]
This is an identity of \(F\)-valued functions on \(G\), and it immediately implies that \(\chi_W = \chi_\Omega - \mathbf{1}\) as required.
Now prove that \(\langle \chi_W, \chi_W \rangle = 1\) and conclude that \(W\) is an irreducible representation.
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 \(\chi_W = \chi_\Omega - \mathbf{1}\) 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.
Explain why \[\mathbf{1},\rho_1,\rho_2,W\] is a complete set of non-isomorphic irreducible representations of \(G\).
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.
Display the character table of \(G = A_4\).
\[\begin{array}{l|llll} & 1 & (12)(34) & (123) & (124) \\ & 12 & 4 & 3 & 3 \\ \hline \mathbf{1}& 1 & 1 & 1 & 1 \\ \rho_1 & 1 & 1 & \zeta & \zeta^2 \\ \rho_2 & 1 & 1 & \zeta^2 & \zeta \\ \chi_W & 3 & -1 & 0 & 0 \end{array}\]