[proofplan] We first compare the two characters on the fixed maximal torus $T$. The Weyl character formula shows that the restriction of an irreducible character to $T$ is determined entirely by the common highest weight $\lambda$ and the fixed positive root data. Since characters are invariant under conjugation and every element of the compact connected group $G$ is conjugate to an element of $T$, equality on $T$ extends to equality on all of $G$. Finally, compact-group character orthogonality implies that two irreducible representations with the same character are equivalent. [/proofplan] [step:Restrict both characters to the maximal torus and apply the Weyl character formula] Let $\chi_V|_T:T\to\mathbb C$ and $\chi_{V'}|_T:T\to\mathbb C$ denote the restrictions of the two characters to $T$. By the Weyl character formula for compact connected Lie groups, applied to the fixed maximal torus $T$, the fixed positive root system $R^+$, and an irreducible representation with highest weight $\lambda$, the restricted character is the unique function on $T$ given by the Weyl character expression attached to $\lambda$ and the root datum of $(G,T)$. We use this named external prerequisite in the following precise form: the Weyl character formula for compact connected Lie groups determines the restricted irreducible character on $T$ from $\lambda$, $T$, and $R^+$. Since $V$ and $V'$ have the same highest weight $\lambda$ with respect to the same choice of $T$ and $R^+$, the Weyl character formula gives the same restricted character for both representations. Hence \begin{align*} \chi_V(t)=\chi_{V'}(t) \end{align*} for every $t\in T$. [guided] We compare the characters first on $T$ because highest weights are defined by decomposing the representation after restriction to the maximal torus. The fixed choice of positive roots $R^+$ determines which weights are called dominant and which weight is the highest weight. The Weyl character formula for compact connected Lie groups states that, once $T$ and $R^+$ are fixed, the character of an irreducible representation restricted to $T$ is determined by its highest weight. More explicitly, the formula expresses $\chi_V|_T$ as a Weyl-invariant expression depending on the highest weight $\lambda$ and on the root datum of $(G,T)$. We use this named result as an external prerequisite. The hypotheses needed for this application are exactly in force: $G$ is compact and connected, $T$ is a maximal torus, $R^+$ is a fixed positive root system, and $V$ and $V'$ are continuous irreducible finite-dimensional complex representations whose highest weights are both $\lambda$. Therefore the Weyl character formula assigns the same function on $T$ to $V$ and to $V'$. Thus, for every $t\in T$, \begin{align*} \chi_V(t)=\chi_{V'}(t). \end{align*} [/guided] [/step] [step:Use conjugation invariance and the maximal torus theorem to extend equality to all of $G$] Let $g\in G$. By the [[Maximal Torus Theorem](/theorems/9713)][citetheorem:9713], applied to the compact connected Lie group $G$ and the maximal torus $T$, there exist $h\in G$ and $t\in T$ such that \begin{align*} hgh^{-1}=t. \end{align*} Equivalently, \begin{align*} g=h^{-1}th. \end{align*} Characters are invariant under conjugation. Indeed, let $E$ be a finite-dimensional complex [vector space](/page/Vector%20Space), let $\tau:G\to GL(E)$ be a finite-dimensional representation, and define the character $\chi_E:G\to\mathbb C$ by $\chi_E(x):=\operatorname{tr}(\tau(x))$ for $x\in G$. The cyclic invariance of trace gives \begin{align*} \chi_E(h^{-1}th)=\operatorname{tr}(\tau(h)^{-1}\tau(t)\tau(h))=\operatorname{tr}(\tau(t)). \end{align*} Applying this to $\rho$ and $\rho'$, and using the equality already proved on $T$, gives \begin{align*} \chi_V(g)=\chi_V(h^{-1}th)=\chi_V(t)=\chi_{V'}(t)=\chi_{V'}(h^{-1}th)=\chi_{V'}(g). \end{align*} Since $g\in G$ was arbitrary, \begin{align*} \chi_V=\chi_{V'} \end{align*} as functions $G\to\mathbb C$. [/step] [step:Apply irreducible character orthogonality to conclude the representations are isomorphic] Since $G$ is compact, the representations may be equipped with $G$-invariant Hermitian inner products by [[Averaging Inner Products](/theorems/9712)][citetheorem:9712], without changing their characters or their underlying isomorphism classes. Thus the compact-group irreducible character theory applies. By the [orthogonality and completeness of irreducible characters for compact Lie groups](/theorems/9742), as in [Orthogonality and [Completeness of Irreducible Characters](/theorems/2432) for Compact Lie Groups][citetheorem:9742], inequivalent irreducible unitary representations have orthogonal characters, while each irreducible character has non-zero norm in the Haar [inner product](/page/Inner%20Product). If $V$ and $V'$ were not isomorphic, their irreducible characters would be orthogonal. But $\chi_V=\chi_{V'}$, so the Haar inner product of $\chi_V$ with $\chi_{V'}$ is the norm square of $\chi_V$, which is non-zero. This contradiction shows that $V\cong V'$ as complex representations of $G$. [guided] The last step converts equality of class functions into an isomorphism of representations. Because the statement did not assume the representations are unitary, we first justify using unitary character theory. The group $G$ is compact, and the representations are finite-dimensional and continuous, so [Averaging Inner Products][citetheorem:9712] gives $G$-invariant Hermitian inner products on $V$ and on $V'$. Replacing the original vector spaces by the same vector spaces with these invariant inner products does not alter the linear maps $\rho(g)$ and $\rho'(g)$, so it does not alter the characters or the question whether the representations are isomorphic. Now use compact-group character orthogonality. The theorem [Orthogonality and Completeness of Irreducible Characters for Compact Lie Groups][citetheorem:9742] applies because $G$ is compact and the representations have been made finite-dimensional, continuous, irreducible, and unitary. It says, in particular, that irreducible characters are orthogonal unless the representations are equivalent, and that the inner product of an irreducible character with itself is non-zero. Suppose, for contradiction, that $V$ and $V'$ are not isomorphic as representations of $G$. Then their irreducible characters must be orthogonal. With $\mu_G$ denoting the normalized Haar probability measure on $G$, this means \begin{align*} \int_G \chi_V(g)\overline{\chi_{V'}(g)}\,d\mu_G(g)=0. \end{align*} But we have already proved $\chi_V=\chi_{V'}$ on $G$, so the same integral is \begin{align*} \int_G |\chi_V(g)|^2\,d\mu_G(g). \end{align*} By irreducible character orthogonality, this norm square is non-zero. The contradiction proves that $V$ and $V'$ must be isomorphic as complex representations of $G$. [/guided] [/step]