[proofplan] We prove the decomposition by establishing the corresponding identity of characters and then invoking unique decomposition into irreducibles. Assume WLOG $n \geq m$. Using the closed form $\chi_n(z) = (z^{n+1} - z^{-(n+1)})/(z - z^{-1})$ and the expansion $\chi_m(z) = z^m + z^{m-2} + \cdots + z^{-m}$, the product $\chi_n(z) \chi_m(z)$ telescopes term-by-term into the sum $\sum_{j=0}^m \chi_{n + m - 2j}(z)$. The hypothesis $n \geq m$ ensures all $m + 1$ summands have non-negative index $n + m - 2j \geq n - m \geq 0$, so the sum is exactly $\chi_{n+m} + \chi_{n+m-2} + \cdots + \chi_{n - m}$. By [Multiplicativity of Characters](/theorems/2478), the left-hand side equals $\chi_{V_n \otimes V_m}$, so $V_n \otimes V_m$ has the same character as $\bigoplus_{j=0}^m V_{n+m-2j}$. Since characters of distinct irreducibles are linearly independent (orthogonality on the compact group $\operatorname{SU}(2)$), the two representations are isomorphic. [/proofplan]

[step:Reduce to a character identity using multiplicativity and unique decomposition] By [Multiplicativity of Characters under Tensor Product](/theorems/2478), the character of $V_n \otimes V_m$ is the pointwise product of characters: \begin{align*} \chi_{V_n \otimes V_m}(g) = \chi_{V_n}(g) \chi_{V_m}(g) \quad \text{for all } g \in \operatorname{SU}(2). \end{align*} By [Maschke's Theorem for Compact Groups](/theorems/2473), $V_n \otimes V_m$ is completely reducible, i.e., a direct sum of irreducibles. By [Completeness of the $V_n$](/theorems/2477), every irreducible summand is isomorphic to some $V_k$. So \begin{align*} V_n \otimes V_m \cong \bigoplus_{k \geq 0} V_k^{\oplus c_k(n, m)} \end{align*} for unique multiplicities $c_k(n, m) \in \mathbb{Z}_{\geq 0}$. The corresponding character identity is \begin{align*} \chi_n \chi_m = \sum_{k \geq 0} c_k(n, m) \chi_k. \end{align*} By the linear independence of characters of distinct irreducibles (orthogonality with respect to the [Haar measure](/theorems/2472), which is a consequence of [Schur's Lemma](/theorems/2414)), the $c_k(n, m)$ are determined by the character identity above. To prove the theorem, it suffices to show that for $n \geq m \geq 0$, \begin{align*} \chi_n(z) \chi_m(z) = \chi_{n+m}(z) + \chi_{n + m - 2}(z) + \cdots + \chi_{n - m}(z) \quad \text{for all } z \in S^1 \setminus \{\pm 1\}. \end{align*} (We restrict to $z \neq \pm 1$ to avoid the indeterminate form in the closed-form expression for $\chi_n$. By continuity of both sides in $z$, the identity then extends to all $z \in S^1$.) The case $n < m$ follows by the symmetry $V_n \otimes V_m \cong V_m \otimes V_n$ of the tensor product. [/step]

[step:Compute $\chi_n(z) \chi_m(z)$ in closed form via the Weyl denominator]We use the [closed-form formula for characters of $V_n$](/theorems/???): \begin{align*} \chi_n(z) = z^n + z^{n-2} + \cdots + z^{-n} = \frac{z^{n+1} - z^{-(n+1)}}{z - z^{-1}}, \quad z \in S^1, \, z \neq \pm 1. \end{align*} The closed form is the standard geometric-series identity: $\sum_{j=0}^n z^{n - 2j}$ is a geometric series in the ratio $z^{-2}$ with first term $z^n$, summing to $z^n \cdot (1 - z^{-2(n+1)})/(1 - z^{-2}) = (z^{n+1} - z^{-(n+1)})/(z - z^{-1})$ after simplification (multiply numerator and denominator by $z$). The strategy is to compute $(z - z^{-1}) \chi_n(z) \chi_m(z)$ by treating one factor in closed form and the other as an explicit sum, recognise the result as a sum of Weyl numerators $z^{k+1} - z^{-(k+1)}$, and divide back through by the Weyl denominator $z - z^{-1}$. Multiply the closed form for $\chi_n$ by the explicit sum $\chi_m(z) = \sum_{i=0}^m z^{m - 2i}$: \begin{align*} (z - z^{-1}) \chi_n(z) \chi_m(z) &= \bigl(z^{n+1} - z^{-(n+1)}\bigr) \sum_{i=0}^m z^{m - 2i} \\ &= \sum_{i=0}^m z^{n + m + 1 - 2i} - \sum_{i=0}^m z^{-(n+1) + m - 2i}. \end{align*} Reindex the second sum by setting $i' = m - i$, so $-(n+1) + m - 2i = -(n+1) - m + 2i' = -(n + m + 1) + 2i'$ and $i'$ ranges over $\{0, 1, \ldots, m\}$: \begin{align*} \sum_{i=0}^m z^{-(n+1) + m - 2i} = \sum_{i'=0}^m z^{-(n + m + 1) + 2i'} = \sum_{i'=0}^m z^{-(n + m + 1 - 2i')}. \end{align*} Combining the two sums under a single index $j := i = i'$: \begin{align*} (z - z^{-1}) \chi_n(z) \chi_m(z) = \sum_{j=0}^m \bigl( z^{n + m + 1 - 2j} - z^{-(n + m + 1 - 2j)} \bigr). \end{align*} The crucial point is that the reindexing pairs the $i$-th term of the positive-exponent sum with the $(m - i)$-th term of the negative-exponent sum, so that the exponents match in absolute value to give the Weyl numerator $z^{k_j + 1} - z^{-(k_j + 1)}$ with $k_j := n + m - 2j$. (This is what was missing from a naïve term-by-term identification: the negative-exponent term in the original product, before reindexing, has exponent $-(n+1) + m - 2j$, which is NOT $-(k_j + 1) = -(n + m - 2j + 1)$ in general; the matching only works after reindexing $i \mapsto m - i$.) Dividing both sides by $z - z^{-1}$ and recognising each summand as a character via the closed form: \begin{align*} \chi_n(z) \chi_m(z) = \sum_{j=0}^m \frac{z^{k_j + 1} - z^{-(k_j + 1)}}{z - z^{-1}} = \sum_{j=0}^m \chi_{k_j}(z), \quad k_j = n + m - 2j. \end{align*} The application of the closed form requires $k_j \geq 0$ (otherwise $\chi_{k_j}$ is not a character of an irreducible). We verify: $k_j = n + m - 2j$ for $0 \leq j \leq m$, so $k_j \geq n + m - 2m = n - m \geq 0$ by the hypothesis $n \geq m$. The indices range over $k_0 = n + m, k_1 = n + m - 2, \ldots, k_m = n - m$, an arithmetic progression of $m + 1$ terms with common difference $-2$, all non-negative.[/step]

[guided]The key algebraic step is the reindexing $i \mapsto m - i$ in the negative-exponent sum. To see why this is needed, observe that after distributing the product we have a "positive piece" $\sum_i z^{n + m + 1 - 2i}$ and a "negative piece" $\sum_i z^{-(n+1) + m - 2i}$. The exponents in the positive piece run $n + m + 1, n + m - 1, \ldots, n - m + 1$ (decreasing); the exponents in the negative piece run $-(n+1) + m, -(n+1) + m - 2, \ldots, -(n+1) - m$, i.e., $m - n - 1, m - n - 3, \ldots, -n - m - 1$ (also decreasing). To pair these into Weyl numerators of the form $z^{k+1} - z^{-(k+1)}$, we need to match a term $z^{k+1}$ in the positive piece with $z^{-(k+1)}$ in the negative piece; since the positive piece is decreasing and the negative piece (when read as $-(k+1)$) requires $k$ also decreasing in the same order, the natural pairing reverses the index of one of the two sums. Reversing the negative piece's index (i.e., reading it from the bottom up) produces exponents $-n - m - 1, -n - m + 1, \ldots, m - n - 1$, which are exactly $-(k_j + 1)$ for $k_j = n + m - 2j$ as $j = 0, 1, \ldots, m$. This reindexing — not a direct exponent identification — is the content of the telescoping. Equivalently: one can derive the identity by writing $\chi_n \chi_m = (z^{n+1} - z^{-(n+1)})(z^{m+1} - z^{-(m+1)}) / (z - z^{-1})^2$, expanding the numerator to $z^{n+m+2} - z^{n-m} - z^{-(n-m)} + z^{-(n+m+2)}$, and observing that $\sum_{j=0}^m (z^{n+m-2j+1} - z^{-(n+m-2j+1)})(z - z^{-1})$ telescopes to the same expression. The reindexing argument used above is the cleaner version of this telescoping. The hypothesis $n \geq m$ enters in exactly one place: it guarantees $k_j = n + m - 2j \geq 0$ for all $j \in \{0, 1, \ldots, m\}$. If $n < m$, some $k_j$ would be negative, and the same algebraic identity would still hold formally, but the summands $\chi_{k_j}$ with $k_j < 0$ would need to be reinterpreted via $\chi_{-k-2}(z) = -\chi_k(z)$ (which gives the cancellations that recover $\chi_{|n-m|} + \cdots + \chi_{n+m}$). The reduction to $n \geq m$ via tensor symmetry sidesteps this. The arithmetic progression $\{n + m, n + m - 2, \ldots, n - m\}$ of length $m + 1$ encodes the Clebsch–Gordan rule combinatorially: tensoring $V_n$ with $V_m$ produces the irreducibles whose indices range from the maximum $n + m$ down to the minimum $|n - m|$ in steps of $2$.[/guided]

[step:Conclude the isomorphism by linear independence of irreducible characters]By Step 2, for $z \in S^1 \setminus \{\pm 1\}$, \begin{align*} \chi_n(z) \chi_m(z) = \sum_{j = 0}^m \chi_{n + m - 2j}(z). \end{align*} Both sides are continuous functions on $S^1$ (in fact Laurent polynomials), and the identity holds on $S^1 \setminus \{\pm 1\}$. By continuity, the identity extends to all $z \in S^1$. By Step 1 and the linear independence of characters, \begin{align*} V_n \otimes V_m \cong \bigoplus_{j=0}^m V_{n+m-2j}. \end{align*} Setting $k = n + m - 2j$ and rewriting the index range: $j$ ranges over $\{0, 1, \ldots, m\}$, so $k$ ranges over $\{n + m, n + m - 2, \ldots, n - m\}$. With the assumption $n \geq m$, this is the same as the set $\{n + m, n + m - 2, \ldots, |n - m|\}$ (since $n - m = |n - m|$). For the case $n < m$ (handled by tensor symmetry), the same conclusion holds with the absolute value. In summary, \begin{align*} V_n \otimes V_m \cong V_{n + m} \oplus V_{n + m - 2} \oplus \cdots \oplus V_{|n - m| + 2} \oplus V_{|n - m|}. \end{align*}[/step]

[guided]The conclusion uses the same linear-independence machinery as in [Completeness of the $V_n$](/theorems/2477). The orthogonality relations for compact groups (Schur orthogonality) state that for irreducibles $V_k, V_{k'}$, \begin{align*} \langle \chi_k, \chi_{k'} \rangle = \int_G \chi_k(g) \overline{\chi_{k'}(g)} \, d\mu_G(g) = \delta_{k, k'}. \end{align*} In particular, the characters $\{\chi_k\}$ are linearly independent. The character identity from Step 2 expresses $\chi_{V_n \otimes V_m}$ as a non-negative integer combination of the $\chi_k$, and the multiplicities are uniquely determined by the inner products $\langle \chi_{V_n \otimes V_m}, \chi_k \rangle$. From the identity, this multiplicity is $1$ for $k \in \{n+m, n+m-2, \ldots, n - m\}$ and $0$ otherwise. The Clebsch–Gordan rule has a representation-theoretic interpretation in terms of weights: a tensor product representation of $\operatorname{SU}(2)$ has weights given by sums $n_i + m_j$ where $n_i, m_j$ are the weights of the factors. Taking the highest weight $n + m$ and "carving off" the irreducible component generated by it (which is $V_{n+m}$), the residual decomposes by induction with highest weight $n + m - 2$, and so on, down to $|n - m|$. The number of summands is $m + 1$ (assuming $n \geq m$), matching the dimension count $(n+1)(m+1) = \sum_{j=0}^m (n + m - 2j + 1)$, which is a standard combinatorial identity.[/guided]