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Preamble

$\DeclareMathOperator\SO{SO}$Let $\mathbb{S}^m\subset \mathbb{R}^{m+1}$ be the standard unit sphere. An observation of Do Carmo and Wallach states that

If $G$ is a subgroup of $\SO(m+1)$ acting transitively on $\mathbb{S}^m$, and $V^\lambda$ is a (finite dimensional vector) space of solutions to the eigenvalue equation $-\Delta_{\mathbb{S}^m} u = \lambda u$ such that $V^\lambda$ is invariant under the action of $G$, then letting $v_1, \dotsc, v_k$ be an orthonormal basis of $V^{\lambda}$, (a suitable rescaling of) the mapping $\mathbb{S}^m \to \mathbb{R}^k$ given by $p\mapsto (v_1(p),\dotsc, v_k(p))$ is a harmonic map to $\mathbb{S}^{k-1}$.

Since spherical harmonics are related to homogeneous harmonic polynomials, this procedure generates a special type of harmonic maps. In general, however, the mapping here are not mappings from $\mathbb{S}^m$ to itself. The simplest case is to take $G = \SO(m+1)$ itself, in which case the only choice of $V^\lambda$ is the full eigenspace of Laplacian. In this case $\lambda = \ell(\ell + m - 1)$ for $\ell$ a natural number, and we have $$ \dim(V^\lambda) = \binom{m+\ell - 1}{m-1} + \binom{m+\ell - 2}{m-1} $$ which is usually much larger than $m+1$.

Question

What do we know about those harmonic maps from $\mathbb{S}^m$ to itself that arise from homogeneous harmonic polynomials mappings of $\mathbb{R}^{m+1}$ to itself? Specifically, given $m$, what can we say about the eigenvalues $\lambda = \ell(\ell + m - 1)$ for which there exists a harmonic map arising from harmonic polynomial mappings (of degree $\ell$) of Euclidean space?

What I know

Trivially, for $m = 1$ every $\ell$ is possible. Also trivially, for any $m$ the case $\ell = 1$ is possible (corresponding to the identity map).

Less trivially, there are mappings generalizing the Hopf map that provides $\ell = 2$ solutions when $m = 3, 7, 15$.

Smith has a paper The Spherical Representations of Groups Transitive on $S^n$ that supposedly solves a portion of this problem. However, as far as I can tell the paper doesn't contain an explicit enumeration of the possible $(m,\ell)$ pairs arising from his study, instead relying on the reader to piece together the various results in Section 3 to get an answer. Unfortunately my Lie theory is not good enough to really understand what the conclusion is.

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    $\begingroup$ @LSpice thank you. I didn't know until just now that you can have hover hints for links in MarkDown. $\endgroup$ Commented Mar 6 at 14:46

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