if two representations have the same character, they are isomorphic

hbghlyj

Let $G$ be a compact Lie group, and $\rho_1, \rho_2$ be finite-dimensional complex representations with $\chi_{\rho_1} = \chi_{\rho_2}$. By the complete reducibility theorem, every finite-dimensional representation of $G$ decomposes into a direct sum of irreducible representations. Write:
$$\rho_1 \cong \bigoplus_{i} n_i \sigma_i, \quad \rho_2 \cong \bigoplus_{i} m_i \sigma_i,
$$where $\sigma_i$ are distinct irreducible representations and $n_i, m_i \in \mathbb{N}$ are multiplicities. The characters satisfy:
$$\chi_{\rho_1} = \sum_{i} n_i \chi_{\sigma_i}, \quad \chi_{\rho_2} = \sum_{i} m_i \chi_{\sigma_i}.
$$Since $\chi_{\rho_1} = \chi_{\rho_2}$, we have $\sum_{i} (n_i - m_i) \chi_{\sigma_i} = 0$. By the orthogonality relations for irreducible characters on compact Lie groups, the $\chi_{\sigma_i}$ are linearly independent. Thus, $n_i = m_i$ for all $i$, implying $\rho_1 \cong \rho_2$.

hbghlyj

Define two 2-dimensional representations of $G = (\mathbb{R}, +)$:
$\rho_1(t) = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ (trivial representation),
$\rho_2(t) = \begin{pmatrix} 1 & t \\ 0 & 1 \end{pmatrix}$ (unipotent representation).
Characters: For all $t \in \mathbb{R}$,
$$\chi_{\rho_1}(t) = \text{Tr}(\rho_1(t)) = 2, \quad \chi_{\rho_2}(t) = \text{Tr}(\rho_2(t)) = 2.
$$Thus, $\chi_{\rho_1} = \chi_{\rho_2}$.
Non-isomorphism: Suppose there exists an invertible matrix $S$ such that $S \rho_1(t) S^{-1} = \rho_2(t)$ for all $t$. Since $\rho_1(t) = I$, this implies $S I S^{-1} = I = \rho_2(t)$, contradicting $\rho_2(t) \neq I$ for $t \neq 0$. Hence, $\rho_1 \not\cong \rho_2$.
$\rho_1$ and $\rho_2$ have equal characters but are non-isomorphic, demonstrating the failure of the statement for non-compact groups.