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On page $65$ of Terry Gannon's "Moonshine beyond the Monster: The Bridge Connecting Algebra, Modular Forms and Physics" (via cambridge.org), it says that the result $$\frac{\sin(mx)\sin(nx)}{\sin x} = \sin((m+n)x) + \sin((m+n-2)x)+\dots+\sin((m-n)x)$$ is a special case of the general theory of representation theory of simple Lie algebras, but it doesn't explain what did they mean.

I think this has something to with any finite dimensional irreducible $\frak{sl}(2,\mathbb{C})$ module $V$'s obeying $$V = V_{m} \,\oplus V_{m-2}\,\oplus\dots\oplus V_{-m}$$ where $m = \text{dim}(V)-1$ and $V_k$ being the weight-spaces with eigenvalue $k$ (i.e locus of vectors with $h\cdot v = kv$) since leaps of $2$ happens in both of results, but I am not sure what did the author mean by that. Since it is not hard to show the first equation by other means, I wonder if anybody could provide a representation theoric proof of the trigonometric result.

Blue
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Rias Gremory
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1 Answers1

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For each integer $m\ge0$ there is a unique irreducible $(m+1)$-dimensional representation of $SU(2)$, called $V_m$.

For an element $g=\begin{pmatrix}e^{i\theta}\\&e^{-i\theta}\end{pmatrix}\in SU(2)$, we have $$tr_{V_m}(g)=\frac{\sin(m\theta)}{\sin(\theta)},$$ which follows for example from Weyl character formula. Now the Clebsch-Gordan rule says that for $m\le n$, $$V_m\otimes V_n\simeq V_{m+n}\oplus V_{m+n-2}\oplus\cdots \oplus V_{n-m}.$$ Taking the trace of $g$ on both sides gives $$\frac{\sin(m\theta)}{\sin(\theta)}\cdot\frac{\sin(n\theta)}{\sin(\theta)}=\frac{\sin((m+n)\theta)}{\sin(\theta)}+\cdots+\frac{\sin((n-m)\theta)}{\sin(\theta)}.$$ By clearing the denominator you get the desired formula.

Kenta S
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