Let \(G\) be a finite group and \(\pi_0\), \(\pi_1\) be two irreducible representations of \(G\). Prove or disprove the following assertion: \(\pi_0\) and \(\pi_1\) are equivalent if and only if \(\det\pi_0(g)=\det\pi_1(g)\) for all \(g\in G\).
Solution
The assertion is FALSE.
Counterexample: Consider the cyclic group \(G = \ZZ_3 = \langle x \mid x^3 = 1 \rangle\). Let \(\omega = e^{2\pi i / 3} \in \CC\). Since \(G\) is abelian, all its irreducible complex representations are 1-dimensional. Define two 1-dimensional representations \(\pi_0, \pi_1: G \to \GL_1(\CC) = \CC^\times\) by:
\(\pi_0(x) = \omega\),
\(\pi_1(x) = \omega^2\).
Determinants: Since each \(\pi_j\) is 1-dimensional, the determinant of a \(1 \times 1\) matrix is the scalar itself: \(\det \pi_j(g) = \pi_j(g)\). For \(g = x\): \[ \det \pi_0(x) = \omega \neq \omega^2 = \det \pi_1(x). \] (This shows that 1-dimensional representations with different determinants are not equivalent.)
Higher-dimensional Counterexample (matching determinants but inequivalent): To disprove the \(\Longleftarrow\) direction (i.e. having identical determinants implies equivalence): Consider the quaternion group \(G = Q_8 = \{\pm 1, \pm i, \pm j, \pm k\}\). \(Q_8\) has four 1-dimensional irreducible representations and one 2-dimensional irreducible representation \(\rho: Q_8 \to \GL_2(\CC)\).
Alternatively, consider the dihedral group \(D_8 = \langle r, s \mid r^4 = 1, s^2 = 1, srs = r^{-1} \rangle\). \(D_8\) has four 1-dimensional irreducible representations \(\chi_1, \chi_2, \chi_3, \chi_4\):
\(\chi_1(r) = 1, \chi_1(s) = 1\) (trivial)
\(\chi_2(r) = 1, \chi_2(s) = -1\)
\(\chi_3(r) = -1, \chi_3(s) = 1\)
\(\chi_4(r) = -1, \chi_4(s) = -1\)
Now consider \(\pi_0 = \chi_2\) and \(\pi_1 = \chi_3\). Since they are 1-dimensional:
- \(\det \pi_0(r) = \chi_2(r) = 1\) while \(\det \pi_1(r) = \chi_3(r) = -1\).
To get identical determinants: Consider \(G = S_3\). The irreducible representations are:
Trivial representation \(\chi_{\text{triv}}\) (dimension 1)
Sign representation \(\chi_{\text{sgn}}\) (dimension 1)
Standard 2-dimensional representation \(\rho\) (dimension 2)
Let \(\pi_0 = \rho\). For \(\pi_1\), consider \(\rho \otimes \chi_{\text{sgn}}\). On \(S_3\), \(\rho \cong \rho \otimes \chi_{\text{sgn}}\), so these are equivalent.
For an explicit non-isomorphic pair with equal determinants: Let \(G = A_4\) (the alternating group on 4 elements). \(A_4\) has three 1-dimensional representations \(\chi_0, \chi_1, \chi_2\) (corresponding to \(A_4 / V_4 \cong \ZZ_3\)) and one 3-dimensional irreducible representation \(\rho\). Now consider \(G = \SL_2(\FF_3)\) or \(G = \ZZ_5 \rtimes \ZZ_4\). For \(G = D_{10} = \langle r, s \mid r^5 = 1, s^2 = 1, srs = r^{-1} \rangle\): \(D_{10}\) has two distinct 2-dimensional irreducible representations \(\rho_1, \rho_2\):
\(\rho_1(r) = \begin{pmatrix} \zeta & 0 \\ 0 & \zeta^{-1} \end{pmatrix}, \quad \rho_1(s) = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\), where \(\zeta = e^{2\pi i / 5}\).
\(\rho_2(r) = \begin{pmatrix} \zeta^2 & 0 \\ 0 & \zeta^{-2} \end{pmatrix}, \quad \rho_2(s) = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\).
Compute determinants for all elements:
For any power \(r^k\): \(\det \rho_1(r^k) = \zeta^k \cdot \zeta^{-k} = 1\), and \(\det \rho_2(r^k) = \zeta^{2k} \cdot \zeta^{-2k} = 1\).
For any reflection \(s r^k\): \(\det \rho_1(s r^k) = \det \rho_1(s) \det \rho_1(r^k) = (-1)(1) = -1\), and \(\det \rho_2(s r^k) = \det \rho_2(s) \det \rho_2(r^k) = (-1)(1) = -1\).
Thus: \[ \det \rho_1(g) = \det \rho_2(g) \quad \text{for all } g \in D_{10}. \] However, their characters are:
\(\chi_{\rho_1}(r) = \zeta + \zeta^{-1} = 2\cos(2\pi/5)\),
\(\chi_{\rho_2}(r) = \zeta^2 + \zeta^{-2} = 2\cos(4\pi/5) \neq 2\cos(2\pi/5)\).
Since characters determine representations over \(\CC\), \(\rho_1\) and \(\rho_2\) are inequivalent irreducible representations of \(D_{10}\), despite having identical determinants for every group element.