[proofplan] For a torus $G = S^1$, every finite-dimensional continuous representation diagonalises into one-dimensional weight spaces under the action of $z \in S^1$. Choosing a weight basis $\{e_i\}$ for $V$ and $\{f_j\}$ for $W$ with $\rho_V(z)e_i = z^{n_i}e_i$ and $\rho_W(z)f_j = z^{m_j}f_j$, the [tensor product](/theorems/2441) acts diagonally on the basis $\{e_i \otimes f_j\}$ with eigenvalue $z^{n_i + m_j}$. Summing the diagonal entries factors as a product, giving $\chi_{V \otimes W} = \chi_V \cdot \chi_W$ on $S^1$. For $G = \operatorname{SU}(2)$, the identity reduces to the torus case because every conjugacy class of $\operatorname{SU}(2)$ meets $T \cong S^1$ ([Conjugacy in $\operatorname{SU}(2)$](/theorems/2475)) and characters are determined by their restriction to $T$. [/proofplan] [step:Reduce $G = \operatorname{SU}(2)$ to $G = S^1$ via the torus restriction] We treat both cases. For $G = \operatorname{SU}(2)$, both $\chi_{V \otimes W}$ and $\chi_V \cdot \chi_W$ are continuous class functions on $G$. By [Conjugacy in $\operatorname{SU}(2)$](/theorems/2475) (part 4), every conjugacy class of $G = \operatorname{SU}(2)$ meets the maximal torus \begin{align*} T = \{ t_z := \mathrm{diag}(z, z^{-1}) : z \in S^1 \}, \end{align*} so two continuous class functions on $G$ agree iff their restrictions to $T$ agree. Hence it suffices to prove $\chi_{V \otimes W}(t_z) = \chi_V(t_z) \chi_W(t_z)$ for $z \in S^1$. This is the assertion of the theorem for the subtorus $T \cong S^1$ acting via the restriction representation $V|_T$ and $W|_T$. For $G = S^1$, no reduction is needed: we work directly. In both cases, the remaining problem is: given continuous finite-dimensional representations $V, W$ of $S^1$, prove $\chi_{V \otimes W} = \chi_V \cdot \chi_W$ as functions on $S^1$. [/step] [step:Diagonalise $V$ and $W$ into one-dimensional weight spaces under $S^1$] Let $V, W$ be finite-dimensional continuous representations of $G = S^1$. By the spectral theory of compact abelian groups (the [Peter–Weyl Theorem](/theorems/???) specialised to abelian compact groups, or the elementary argument that every continuous unitary representation of $S^1$ decomposes via Fourier analysis), every continuous finite-dimensional representation of $S^1$ decomposes as a direct sum of one-dimensional representations. The one-dimensional continuous representations of $S^1$ are exactly the characters $z \mapsto z^k$ for $k \in \mathbb{Z}$. Concretely: choose an inner product on $V$ that is $G$-invariant (such an inner product exists by the [Maschke's Theorem for Compact Groups](/theorems/2473) averaging construction with the Haar measure). With respect to this inner product, $\rho_V(z): V \to V$ is unitary for every $z \in S^1$. The family $\{\rho_V(z) : z \in S^1\}$ is commuting (since $S^1$ is abelian), so by the [Spectral Theorem for Commuting Unitaries](/theorems/???), there is an orthonormal basis $\{e_1, \ldots, e_n\}$ of $V$ simultaneously diagonalising every $\rho_V(z)$. Each common eigenvalue is a continuous group homomorphism $S^1 \to S^1$, hence of the form $z \mapsto z^{n_i}$ for some $n_i \in \mathbb{Z}$: \begin{align*} \rho_V(z) e_i = z^{n_i} e_i, \quad i = 1, \ldots, n = \dim V. \end{align*} Similarly, choose an orthonormal weight basis $\{f_1, \ldots, f_m\}$ of $W$ with \begin{align*} \rho_W(z) f_j = z^{m_j} f_j, \quad j = 1, \ldots, m = \dim W. \end{align*} The characters in this basis read off as the diagonal traces: \begin{align*} \chi_V(z) = \operatorname{tr}\, \rho_V(z) = \sum_{i=1}^n z^{n_i}, \qquad \chi_W(z) = \sum_{j=1}^m z^{m_j}. \end{align*} [guided] The key fact is the simultaneous diagonalisation of the family $\{\rho_V(z) : z \in S^1\}$. Why does this hold? Because (i) $S^1$ is compact abelian, so by [Maschke's Theorem for Compact Groups](/theorems/2473) there is a $G$-invariant inner product on $V$ — average any inner product over $G$ using the [Haar measure](/theorems/2472); (ii) under this inner product each $\rho_V(z)$ is unitary; (iii) unitary operators that commute are simultaneously diagonalisable by the spectral theorem for normal operators on a finite-dimensional Hilbert space. The eigenvalues $z \mapsto z^{n_i}$ are integers because the only continuous group homomorphisms $S^1 \to S^1$ are $z \mapsto z^k$ for $k \in \mathbb{Z}$ (the integer characters of the circle group). This is a basic fact from Fourier analysis: continuous functions on $S^1$ that are also multiplicative are exactly the $z^k$. [/guided] [/step] [step:Compute the action on $V \otimes W$ in the tensor basis and trace it] The tensor product representation $V \otimes W$ has the action $\rho_{V \otimes W}(z) = \rho_V(z) \otimes \rho_W(z)$ (by [Tensor Product of Representations](/theorems/2441)). The vectors $\{e_i \otimes f_j : 1 \leq i \leq n, \, 1 \leq j \leq m\}$ form a basis of $V \otimes W$, with \begin{align*} \rho_{V \otimes W}(z) (e_i \otimes f_j) = (\rho_V(z) e_i) \otimes (\rho_W(z) f_j) = z^{n_i} z^{m_j} (e_i \otimes f_j) = z^{n_i + m_j} (e_i \otimes f_j). \end{align*} So the basis $\{e_i \otimes f_j\}$ is an eigenbasis of $\rho_{V \otimes W}(z)$ with eigenvalues $z^{n_i + m_j}$. Tracing, \begin{align*} \chi_{V \otimes W}(z) = \operatorname{tr}\, \rho_{V \otimes W}(z) = \sum_{i, j} z^{n_i + m_j} = \sum_i z^{n_i} \sum_j z^{m_j} = \chi_V(z) \chi_W(z). \end{align*} The factorisation $\sum_{i, j} z^{n_i + m_j} = (\sum_i z^{n_i})(\sum_j z^{m_j})$ is the standard distributivity of finite sums. [/step] [step:Conclude via class-function equality] For $G = S^1$, the identity $\chi_{V \otimes W}(z) = \chi_V(z) \chi_W(z)$ for all $z \in S^1$ is precisely the statement of the theorem. For $G = \operatorname{SU}(2)$: by Step 1, two continuous class functions on $\operatorname{SU}(2)$ agree iff their restrictions to $T$ agree. Both $\chi_{V \otimes W}$ and $\chi_V \cdot \chi_W$ are continuous class functions on $\operatorname{SU}(2)$ — characters are continuous (composition of continuous representations and trace) and class (by cyclicity of trace), and the pointwise product of class functions is a class function. Their restrictions to $T$ are \begin{align*} \chi_{V \otimes W}(t_z) &= \chi_{(V \otimes W)|_T}(t_z) = \chi_{V|_T \otimes W|_T}(t_z) \\ &= \chi_{V|_T}(t_z) \chi_{W|_T}(t_z) \quad \text{(by the $S^1$ case applied to the restrictions $V|_T, W|_T$)} \\ &= \chi_V(t_z) \chi_W(t_z), \end{align*} where the second equality uses $(V \otimes W)|_T \cong V|_T \otimes W|_T$ as $T$-representations (the action of $T \subseteq G$ on $V \otimes W$ is by definition $\rho_V|_T \otimes \rho_W|_T$). Hence $\chi_{V \otimes W}|_T = (\chi_V \cdot \chi_W)|_T$ as functions on $T$, and by class-function equality from Step 1, $\chi_{V \otimes W} = \chi_V \cdot \chi_W$ as functions on $\operatorname{SU}(2)$. [guided] This step assembles the argument: Step 2 produces weight bases for $V$ and $W$ over $S^1$; Step 3 takes the tensor product of those bases and computes the trace as a product; Step 4 lifts the $S^1$ identity to $\operatorname{SU}(2)$ using the principle that class functions on $\operatorname{SU}(2)$ are determined by their torus restrictions. Why must the same identity for $S^1$ imply the same identity for $\operatorname{SU}(2)$? A class function on $\operatorname{SU}(2)$ is a function $G \to \mathbb{C}$ constant on conjugacy classes. By [Conjugacy in $\operatorname{SU}(2)$](/theorems/2475), every conjugacy class meets $T$, so a class function $f$ is determined by $f|_T$. If two continuous class functions on $G$ agree on $T$, they agree on all conjugacy classes meeting $T$ — i.e., on all of $G$. Both $\chi_{V \otimes W}$ and $\chi_V \cdot \chi_W$ are continuous class functions, so $\operatorname{SU}(2)$-equality reduces to $T$-equality, which is the $S^1$ case. The multiplicativity of characters under tensor product is the algebraic shadow of a fundamental geometric fact: the trace of a tensor product of operators is the product of traces, $\operatorname{tr}(A \otimes B) = \operatorname{tr}(A) \operatorname{tr}(B)$. The proof of this in the basis-free setting uses that, for $A: V \to V$ and $B: W \to W$, the map $A \otimes B: V \otimes W \to V \otimes W$ has matrix entries $(A \otimes B)_{(i,j), (k, \ell)} = A_{ik} B_{j\ell}$, so $\operatorname{tr}(A \otimes B) = \sum_{i, j} A_{ii} B_{jj} = \operatorname{tr}(A) \operatorname{tr}(B)$. The character identity is the special case where $A = \rho_V(g), B = \rho_W(g)$. [/guided] [/step]